Lots of Calculations in General Relativity Susan Larsen Tuesday, February 03, Introduction Special relativity... 7

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1 Susan Lasen Tuesday, Febuay 3, 5 Contents Intoduction... 6 Special elativity How to calculateδs: Example The fou velocity and the fou foce: uaka Metic and Vecto Tansfomations Flat Minkowski space: Othe ealizations of the flat space: Spheical pola coodinates Flat space with a singulaity Coodinate tansfomations Flat space in two dimensions The Penose Diagam fo Flat Space The line-element and metic of an ellipsoid: The signatue of a metic Thee-dimensional flat space in spheical coodinates and vecto tansfomation Static Weak Field Metic ates of Emission and eception Local inetial fames The metic of a Sphee at the Noth Pole Length, Aea, Volume and Fou-Volume fo Diagonal Metics Aea and Volume Elements of a Sphee Distance, Aea and Volume in the Cuved Space of a Constant Density Spheical Sta o a Homogenous Closed Univese Distance, Aea, Volume and fou-volume of a metic The dimensions of a peanut The dimensions of an egg Length and volume of the Schwazschild geomety Volume in the Womhole geomety... 4 Tenso Calculus Chistoffel symbols aba and aab in a diagonal metic... Page

2 Susan Lasen Tuesday, Febuay 3, Find the Chistoffel symbols of the -sphee with adius a Find the Chistoffel symbols of the Kahn-Penose metic (Colliding gavitational waves) Altenative solution: Show that cgab One-foms One-foms: why d The exteio deivative of a one-fom The geodesic equation Find the geodesic equations fo cylindical coodinates Use the geodesic equations to find the Chistoffel symbols fo the indle metic Geodesics Equations of the plane in pola coodinates Equations fo geodesics in a Womhole Geomety New - Geodesics and Chistoffel symbols of the Schwazschild metic with θ = π Solving the geodesic equation The tavel time though a womhole Geodesics in the Plane Using Pola Coodinates NEW - Geodesics Equations of the plane in Catesian coodinates New - Show that the geat cicle is a solution of the geodesic equation of a two-dimensional sphee NEW - Find all the time-like geodesic X(T) of the Flat Space metic in two dimensions ds = XdT + dx: Killing Vectos Show that if the Lie deivative of the metic tenso with espect to vecto X vanishes (LXgab ), the vecto X satisfies the Killing equation. - Altenative vesion Pove that b axb = acxc Constucting a Conseved Cuent with Killing Vectos Altenative vesion: Given a Killing vecto X the icci scala satisfiesxc c : The iemann tenso Independent elements in the iemann, icci and Weyl tenso Compute the components of the iemann tenso fo the unit -sphee Show that the icci scala = fo the unit -sphee Poof: if a space is confomally flat, i.e. gabx = fxηab the Weyl tenso vanishes The thee dimensional flat space in spheical pola coodinates Calculate the Chistoffel symbols of the thee dimensional flat space in spheical pola coodinates Page

3 Susan Lasen Tuesday, Febuay 3, The iemann tenso of the thee dimensional flat space in spheical pola coodinates A Lie deivative in the thee dimensional flat space in spheical pola coodinates The icci scala of the Penose Kahn metic A metic example : ds = ysinxdx + xtanydy The Chistoffel symbols of a metic example The icci scala of a metic example Calculate the Chistoffel symbols fo a metic example : ds = dψ + sinhψ dθ + sinhψsinθdφ A metic example 3: ds = u + νdu + u + νdν + uνdθ Calculate the Chistoffel symbols fo a metic example Calculate the iemann tenso of metic example Calculate the iemann tenso of metic example Altenative vesion Catan s Stuctue Equations icci otation coefficients fo the Tolman-Bondi- de Sitte metic (Spheical dust with a cosmological constant) The cuvatue two foms and the iemann tenso Find the icci scala using Catan s stuctue equations of the -sphee The thee dimensional flat space in spheical pola coodinates icci otation coefficients of the thee dimensional flat space in spheical pola coodinates Tansfomation of the icci otation coefficients bca into the Chistoffel symbols bca of the thee dimensional flat space in spheical pola coodinates icci otation coefficients of the indle metic The Einstein tenso fo the Tolman-Bondi- de Sitte metic Calculate the icci otation coefficients fo a metic example 3: ds = dψ + sinhψ dθ + sinhψsinθdφ The Einstein Field Equations The vacuum Einstein equations The vacuum Einstein equations with a cosmological constant Geneal emaks on the Einstein equations with a cosmological constant dimensions: Gavitational collapse of an inhomogeneous spheically symmetic dust cloud Find the components of the cuvatue tenso fo the metic in + dimensions using Catan s stuctue equations Page 3

4 Susan Lasen Tuesday, Febuay 3, Find the components of the cuvatue tenso fo the metic in + dimensions using Catan s stuctue equations altenative solution Find the components of the Einstein tenso in the coodinate basis fo the metic in + dimensions The Einstein equations of the metic in + dimensions Using the contacted Bianchi identities, pove that: bgab icci otation coefficients, icci scala and Einstein equations fo a geneal 4-dimensional metic: ds = dt + Lt, d + Bt, dφ + Mt, dz The Enegy-Momentum Tenso Pefect Fluids Altenative deivation The Gödel metic Null Tetads and the Petov Classification Constuct a null tetad fo the flat space Minkowski metic The Binkmann metic (Plane gavitational waves) The Schwazschild Solution The iemann and icci tenso of the geneal Schwazschild metic The iemann tenso of the Schwazschild metic Calculation of the scala abcdabcd in the Schwazschild metic Geodesics in the Schwazschild Spacetime The meaning of the integation constant: The choice of m Time Delay Use the geodesic equations to find the Chistoffel symbols fo the geneal Schwazschild metic. 9.8 The icci tenso fo the geneal time dependent Schwazschild metic The Schwazschild metic with nonzeo cosmological constant The icci otation coefficients and icci tenso fo the Schwazschild metic with nonzeo cosmological constant The geneal Schwazschild metic in vacuum with a cosmological constant: The icci scala The geneal Schwazschild metic in vacuum with a cosmological constant: Integation constants The geneal Schwazschild metic in vacuum with a cosmological constant: The spatial pat of the line element The effect of the cosmological constant ove the scale of the sola system The Petov type of the Schwazschild spacetime Page 4

5 Susan Lasen Tuesday, Febuay 3, 5 9. The deflection of a light ay in a Schwazschild metic with two diffeent masses The non-zeo Weyl scalas of the eissne-nodstöm spacetime... Black Holes The Path of a adially Infalling Paticle The Schwazschild metic in Kuskal Coodinates The Ke metic The Ke-Newman geomety The invese metic of the Ke Spinning Black Hole Cosmology Light tavelling in the Univese Spaces of Positive, Negative, and Zeo Cuvatue New The citical density The obetson-walke metic Find the components of the iemann tenso of the obetson-walke metic (Homogenous, isotopic and expanding univese) using Catan s stuctue equations The Einstein tenso and Fiedmann-equations fo the obetson Walke metic The Einstein tenso fo the obetson Walke metic Altenative vesion Manipulating the Fiedmann equations Paametes in an flat univese with positive cosmological constant: Stating with a = Ca + Λ3a use a change of vaiables u = Λ3Ca Gavitational Waves Gauge tansfomation - The Einstein Gauge Plane waves The iemann tenso of a plane wave The line element of a plane wave in the Einstein gauge The line element of a plane wave The osen line element Colliding gavity waves - coodinate tansfomation The delta δ(u) and heavy-side Θ(u) functions: pove that uδu Impulsive gavitational wave egion III Two inteacting waves The Naiai spacetime Page 5

6 Susan Lasen Tuesday, Febuay 3, 5.8 Collision of a gavitational wave with an electomagnetic wave The non-zeo spin coefficients 88.9 The Aichelbug-Sexl Solution The passing of a black hole Obsevations: The Futue Gavitational Wave detectos Bibliogafi Intoduction Woking with G means woking with diffeential equations at fou diffeent levels. It can be vey useful - wheneve one comes acoss a G calculation - to keep in mind, on which level you ae woking. The fou levels of diffeential equations ae:. The metic o line-element: Example: Gavitational ed shift : ds = g ab dx a dx b dτ = m dt Light emitted upwad in a gavitational field, fom an obseve located at some inne adius to an obseve positioned at some oute adius α = m m. Killing s equations ae consevation equations: b X a + a X b If you move along the diection of a Killing vecto, then the metic does not change. This leads to conseved quantities: A fee paticle moving in a diection whee the metic does not change will not fell any foces. If X is a Killing vecto,u = ( dt dτ, d dτ, dθ, dφ dτ dτ ) is the paticle fou velocity and p is the paticle fou impulse, then X u = g ab X a u b = const and X p = g ab X a p b = const along a geodesic. Tanslational symmety: Wheneve σ g μν fo some fixed σ (but fo all μ and ν) thee will be a symmety unde tanslation along x σ 3. Example: Killing vectos in the Schwazschild metic 4. The Killing vecto that coesponds to the independence of the metic of t is ξ = (,,,) and of φ is η = (,,,). The conseved enegy pe unit est mass: e = ξ u = g ab ξ a u b = g tt dt dτ = ( m ) dt. The conseved angula momentum pe unit est mass l = η u = g dτ abη a u b = g φφ dφ sin θ dφ dφ = fo θ = π dτ dτ 3. The Geodesic equation leads to equations of motion: K = g abx a x b dτ = (McMahon, p. 34) (McMahon, p. 68) 3 (Caoll, 4) 4 (McMahon, p. ) Page 6

7 Susan Lasen Tuesday, Febuay 3, 5 K x a = d ds ( K x a ) d x a ds + dx b dx c a bc ds ds Example: Planetay obits 5 Manipulating the geodesic equations of the Schwazschild metic leads to the following equation ( du dφ ) + u = k h + m u + mu3 h Which can be intepeted in tems of elliptic functions, u =, and h and k ae constants of integation. 4. The Einstein equations ae equations descibing the spacetime. G ab = ab g ab 8πGT ab = G ab + g ab Λ In case of a cosmological constant 6 : ab = ±g ab Λ a b = ±η a b Λ If n = 4, abcd has twenty independent component ten of which ae given by ab and the emaining ten by the Weyl tenso 7. Example: The Fiedmann equations A homogenous, isotopic and expanding univese descibed by the obetson-walke metic 8, in this case the Einstein equations becomes the Fiedmann equations: 8πρ 8πP = 3 a (k + a ) + Λ = a a + a (k + a ) + Λ Special elativity. 9How to calculate(δs) : Example In flat space calculate (Δs) fo the following pai of events: E = (,3,,4) and E = (4,,,) (Δs) = (Δt) (Δx) (Δy) (Δz) (.) = ( 4) (3 ) ( ( )) (4 ) = =. The fou velocity and the fou foce: u a K a The fou velocity u a = dxa dτ m du a dτ. Because u ak a = u a K a we can calculate u a K a = (dt, dx, dy, dz ), the fou impulse dτ dτ dτ dτ pa = m u a, the fou foce K a = dpa = dτ = (u ak a + u a K a ) = (m u a du a dτ + m u a du a dτ ) = m d(u a u a ) dτ 5 (A.S.Eddington, pp ) 6 (McMahon, p. 38) 7 (d'inveno, p. 87) 8 (McMahon, p. 6) 9 (McMahon, 6, p. 33), final exam. The answe to FE- is (c) (McMahon, 6, p. 34), final exam 4, and the answe to FE-4 is (a) Page 7

8 Susan Lasen Tuesday, Febuay 3, 5 u a u a is an invaiant and m d(u a u a ) = u dτ a K a 3 Metic and Vecto Tansfomations. 3. Flat Minkowski space: Flat Minkowski spacetime is the mathematical setting in which Einstein s special theoy of elativity is most conveniently fomulated. In Catesian coodinates with c = the line element is ds = dt + dx + dy + dz and the metic g μν = { } ds <, time-like, inside the light cone ds : null-vecto, on the light cone ds >, space-like, outside the light cone. 3. Othe ealizations of the flat space: 3.. Spheical pola coodinates The spheical pat of the metic can be tansfomed into spheical pola coodinates by x = sin θ cos φ y = sin θ sin φ (7.) z = cos θ dx = sin θ cos φ d + cos θ cos φ dθ sin θ sin φ dφ dy = sin θ sin φ d + cos θ sin φ dθ + sin θ cos φ dφ dz = cos θ d sin θ dθ (7.3) dx = (sin θ cos φ d + cos θ cos φ dθ sin θ sin φ dφ) = sin θ cos φ d + sin θ cos θ cos φ ddθ sin θ sin φ cos φ ddφ + cos θ cos φ dθ sin θ cos θ sin φ cos φ dθdφ + sin θ sin φ dφ dy = (sin θ sin φ d + cos θ sin φ dθ + sin θ cos φ dφ) = sin θ sin φ d + sin θ cos θ sin φ ddθ + sin θ sin φ cos φ ddφ + cos θ sin φ dθ + cos θ sin θ cos φ sin φ dθdφ + sin θ cos φ dφ dz = (cos θ d sin θ dθ) = cos θ d cos θ sin θ ddθ + sin θ dθ dx + dy + dz = sin θ cos φ d + sin θ cos θ cos φ ddθ sin θ sin φ cos φ ddφ + cos θ cos φ dθ sin θ cos θ sin φ cos φ dθdφ + sin θ sin φ dφ + sin θ sin φ d + sin θ cos θ sin φ ddθ + sin θ sin φ cos φ ddφ + cos θ sin φ dθ + cos θ sin θ cos φ sin φ dθdφ + sin θ cos φ dφ + cos θ d cos θ sin θ ddθ + sin θ dθ (McMahon, p. 86) (Hatle, 3, p. 35) Page 8

9 Susan Lasen Tuesday, Febuay 3, 5 = sin θ cos φ d + sin θ cos θ cos φ ddθ + cos θ cos φ dθ + sin θ sin φ dφ + sin θ sin φ d + sin θ cos θ sin φ ddθ + cos θ sin φ dθ + sin θ cos φ dφ + cos θ d cos θ sin θ ddθ + sin θ dθ = (sin θ cos φ + sin θ sin φ + cos θ)d + ( sin θ cos θ cos φ cos θ sin θ + sin θ cos θ sin φ)ddθ + ( cos θ cos φ + cos θ sin φ + sin θ)dθ + ( sin θ sin φ + sin θ cos φ)dφ = (sin θ (cos φ + sin φ) + cos θ)d + ( sin θ cos θ (cos φ + sin φ) cos θ sin θ)ddθ + (cos θ (cos φ + sin φ) + sin θ)dθ + sin θ (sin φ + cos φ)dφ = (sin θ + cos θ)d + ( sin θ cos θ cos θ sin θ)ddθ + (cos θ + sin θ)dθ + sin θ dφ = d + dθ + sin θ dφ The tansfomed line element is ds = dt + d + dθ + sin θ dφ (7.4) 3.. 3Flat space with a singulaity Look at the line element of the two-dimensional plane in pola coodinates (θ ) ds = d + dφ (7.6) and make the tansfomation, fo some constant a = a d = d ( a ) = a d d = a4 d 4 ds = a4 4 d + ( a ) dφ = a4 4 (d + dφ ) (7.7) This line element blows up at. Not because something physically inteesting happens hee, but simply because the coodinate tansfomation = a has mapped all the points at into. 4 We can show that that the distance between and a point with any finite value of is infinite, which coesponds to the distance between some finite value of and : ds = a4 4 (d + dφ ) = a d 4 ( + ( dφ d ) ) = a d dφ d 3 (Hatle, 3, p. 36) 4 (Hatle, 3, p. 63), poblem 7. Page 9

10 Susan Lasen Tuesday, Febuay 3, 5 = [ a ] Coodinate tansfomations The following line element coesponds to flat spacetime ds = dt + dxdt + dy + dz with the metic g ab = { } Find a coodinate tansfomation that puts the line element in the usual flat space fom ds = dt + dx + dy + dz a We want to find the matix, Λ b ab that tansfoms g a b into g ab g a ab = Λ b ab g a b we have a Λ b ab a = Λ b ab g a b ga b = g ab g a b = { } { } = { } t = t x x = t y = y z = z dt = dt dx dx = dt dy = dy dz = dz ds = dt + dxdt + dy + dz We check g ab = (dt dx ) + (dt )(dt dx ) + dy + dz = dt dx + dt dx + dt dt dx + dy + dz = dx + dt + dy + dz a = Λ b ab g a b = { } { } = { } 5 (Hatle, 3, p. 64), poblem 7. Page

11 Susan Lasen Tuesday, Febuay 3, Flat space in two dimensions Yet anothe ealization of flat space in two dimensions is the line element ds = X dt + dx (7.) This can be found fom the coodinate tansfomation t = X sinh(t) x = X cosh(t) dt = sinh(t) dx + X cosh(t) dt dx = cosh(t) dx + X sinh(t) dt ds = dt + dx = (sinh(t) dx + X cosh(t) dt) + (cosh(t) dx + X sinh(t) dt) = sinh (T) dx X cosh (T) dt X sinh(t) cosh(t)dx dt + cosh (T) dx + X sinh (T) dt + X cosh(t) sinh(t) dxdt = (cosh (T) sinh (T))dX X (cosh (T) sinh (T))dT = dx X dt The Penose Diagam fo Flat Space A Penose diagam is a method to map the infinite coodinates such as t, with the ange < t < +, and, with the ange < < +, into to coodinates with finite anges. Begin with the flat space line element in spheical pola coodinates ds = dt + d + dθ + sin θ dφ (7.4) and eplace t and by the coodinates u = t v = t + t = (u + v) = (u v) dt = (du + dv) d = (du dv) ds = dt + d + dθ + sin θ dφ = ( (du + dv)) + ( (du dv)) + ( (u v)) (dθ + sin θ dφ ) = dudv + 4 ((u v)) (dθ + sin θ dφ ) The (u, v) axes ae otated with espect to the (t, ) axes by 45. adial light ays tavel on lines of eithe constant u o constant v. emembe adial light ays has constant θ and φ and ds, which leaves us with dudv. Make a futhe tansfomation of u and v to new coodinates u and v. u = tan u v = tan v u = tan u v = tan v du = ( + tan u )du dv = ( + tan v )dv ds = dudv + 4 ((u v)) (dθ + sin θ dφ ) = ( + tan u )( + tan v )du dv + 4 ((tan u tan v )) (dθ + sin θ dφ ) Map these coodinates into a (t, ) diagam, whee u = tan u = t v = tan v = t + Because tan x lies between π and + π the anges fo (u, v ) and (t, ) ae finite. This is anothe example of how the infinite coodinates (t, ) is mapped into a finite egion. 6 (Hatle, 3, p. 43), example (Hatle, 3, p. 37), Box 7. Page

12 Susan Lasen Tuesday, Febuay 3, The line-element and metic of an ellipsoid: The line-element of an ellipsoid in Catesian coodinates ds = adx + bdy + cdz We use the paameteization x = cos φ sin θ y = sin φ sin θ z = cos θ With φ π in the xy-plane and θ π whee the zaxis coesponds to θ. dx = a(cos φ cos θ dθ sin φ sin θ dφ) dy = b(sin φ cos θ dθ + cos φ sin θ dφ) dz = c sin θ dθ dx = a (cos φ cos θ dθ sin φ sin θ dφ) = a [cos φ cos θ dθ + sin φ sin θ dφ cos φ cos θ sin φ sin θ dθdφ] dy = b[sin φ cos θ dθ + cos φ sin θ dφ + sin φ cos θ cos φ sin θ dθdφ] dz = c sin θ dθ Collecting the esults in tems of dθ, dφ and dθdφ we get the line element ds = [cos θ (a cos φ + b sin φ) + c sin θ]dθ + sin θ (a sin φ + b cos φ)dφ + (b a )(cos φ cos θ sin φ sin θ)dθdφ and the metic tenso g ab = { cos θ (a cos φ + b sin φ) + c sin θ (b a )(cos φ cos θ sin φ sin θ) (b a )(cos φ cos θ sin φ sin θ) sin θ (a sin φ + b cos φ) } As a funny obsevation you can now calculate the line-element and metic tenso of an idealized egg. Fo an idealized egg we can choose a = b = c ds = a [(cos θ + 4 sin θ)dθ + sin θ dφ ] g ab = a { cos θ + 4 sin θ sin θ } 3.4 9The signatue of a metic. Hee we ae going to investigate what happens to vaious quantities when a metic,g ab,changes signatue. a Chistoffel symbols bc = gad ( g db x c + g dc x b g bc x d ) No change a a a e a e a iemann tenso bcd = c bd d bc + bd ec bc ed No change c icci tenso ab = acb No change icci scala = g ab ab Changes sign Einstein tenso G ab = ab g ab No change Enegy tenso 8πGT ab = G ab No change Cosmological constant G ab = ab g ab + g ab Λ Changes sign 3.5 Thee-dimensional flat space in spheical coodinates and vecto tansfomation The line element ds = d + dθ + sin θ dφ 8 (Hatle, 3, p. 9), poblem (McMahon, 6, p. 36) (McMahon, 6, p. 46), Quiz -5, the answe to the quiz -5 is (c) Page

13 Susan Lasen Tuesday, Febuay 3, 5 The metic tenso: g ab = { sin θ} Given X a = (, sin θ, cos θ ) calculate X a. X a = g ab X a (.7) X = g X = ( )() = X θ = g θθ X θ = ( ) ( sin θ ) = sin θ X φ = g φφ X φ = ( sin θ) ( cos θ ) = tan θ 3.6 Static Weak Field Metic In this model the flat spacetime geomety of special elativity is modified to intoduce a slight cuvatue that will explain geometically the behavio of clocks. Futhe, the wold lines of extemal pope time in this modified geomety will epoduce the pedictions of Newtonian mechanics fo motion in a gavitational potential fo nonelativistic velocities. Φ(x i ) is a function of position satisfying the Newtonian field equation Φ(x ) = 4πGμ(x )and assumed to vanish at infinity. Fo example outside Eath Φ() = GM. This line element is pedicted by geneal elativity fo small cuvatues poduced by time-independent weak souces, and it is a good appoximation to the cuved spacetime geomety poduced by the Sun. ds = ( + Φ(xi ) c ) (cdt) + ( Φ(xi ) c ) (dx + dy + dz ) (6.) ates of Emission and eception We look at a system whee two light signals ae emitted in a system A, descibed by a wold line (ct, x A ), with a pope time sepaation Δτ A. We want to pedict: what is the pope time sepaation Δτ B in a system B, descibed by a wold line (ct, x B ) in a static weak field limit whee Φ c. This implies dx, Φ(xi ) = Φ(x i,,) = Φ i and dτ = ds c. Also notice that because the metic is independent of the coodinate t, Δt is the same in both systems. This leads to dτ i = ( + Φ i c ) (dt) Δτ A = ( + Φ A c ) Δt Δτ A ~ 4 ( + Φ A c ) Δt (6.) and Δτ B ~ ( + Φ B c ) Δt (6.) Eliminating Δt we get (Hatle, 3) (Hatle, 3) eq. (3.8) 3 (Hatle, 3, p. 7) 4 + x~ + x if x Page 3

14 Susan Lasen Tuesday, Febuay 3, 5 Δτ B = ( + Φ B c ) ( + Φ A c ) Δτ A ~ 5 ( + Φ B c ) ( Φ A c ) Δτ A Δτ B ~ ( + Φ B Φ A c ) Δτ A (6.3) which tells us the obseved fact, that when the eceive B is at a highe gavitational potential that the emitte A, the signals will be eceived moe slowly than they wee emitted and vice vesa Local inetial fames A local inetial fame is defined by the conditions g αβ (x P) = η αβ and g αβ x γ (7.3) x=x P This means that if you have a system descibed by a metic g αβ it can locally in a point P be tansfomed into the flat space metic η αβ. Futhemoe the fist deivatives of the tansfomed metic vanish. This is best illustated by an example The metic of a Sphee at the Noth Pole. The line element of the geomety of a sphee with adius a is ds = a (dθ + sin θ dφ ) (7.4) At the noth pole θ, and the metic doesn t look like the metic of a flat plane at all. Can we find a coodinate tansfomation so that ds = dx + y o g ij = { } Look at the coodinates x = aθ cos φ y = aθ sin φ (7.5) At the noth pole both x and y ae zeo. Next we calculate x + y = (aθ cos φ) + (aθ sin φ) = a θ (cos φ + sin φ) = a θ θ = a x + y (7.6) y x aθ sin φ = aθ cos φ = tan φ φ = tan ( y x ) (7.6) The diffeentials dθ = d ( a x + y ) = (xdx + ydy) a x + y 5 +x ~ x if x 6 (Hatle, 3, p. 4) 7 (Hatle, 3, p. 4) example 7. Page 4

15 Susan Lasen Tuesday, Febuay 3, 5 d(tan φ) = d ( y x ) = (xdx + ydy) a x + y ( + tan φ)dφ = x dy y x dx ( + ( y x ) ) dφ = x dy y x dx (x + y )dφ = xdy ydx dφ xdy ydx = x + y The line element ds = a (dθ + sin θ dφ ) The metic = a (( a = = (xdx + ydy)) x + y + sin xdy ydx θ ( x + y ) ) x + y (x dx + y dy + xydxdy) a sin θ + ( x + y ) (x dy + y dx xydxdy) x + y ((x + a sin θ x + y y ) dx + (y + a sin θ x + y x ) dy + xy ( a sin θ x + y ) dxdy) = ( x x + y + y (x + y ) a sin ( a x + y )) dx + ( y x + y + x (x + y ) a sin ( a x + y )) dy xy + ( x + y xy (x + y ) a sin ( a x + y )) dxdy g xx = x x + y + y (x + y ) a sin ( a x + y ) g yy = y x + y + x (x + y ) a sin ( a x + y ) g xy = g yx = xy x + y xy (x + y ) a sin ( a x + y ) If we evaluate these aound the noth pole whee x and y ae small 8 8 Use wxmaxima evaluate the Taylo polynomials to second ode: g_xx(x,y) := x^/(x^+y^)+a^*sin(/a*sqt(x^+y^))*sin(/a*sqt(x^+y^))*y^/(x^+y^)^; g_yy(x,y) := y^/(x^+y^)+a^*sin(/a*sqt(x^+y^))*sin(/a*sqt(x^+y^))*x^/(x^+y^)^; g_xy(x,y) :=x*y/(x^+y^)-a^*sin(/a*sqt(x^+y^))*sin(/a*sqt(x^+y^))*x*y/(x^+y^)^; taylo(g_xx(x,y),[x,y],[,],[,]); taylo(g_yy(x,y),[x,y],[,],[,]); taylo(g_xy(x,y),[x,y],[,],[,]); g_xx(x,y)=-y^/(3*a^)+... g_yy(x,y)= -x^/(3*a^) Page 5

16 Susan Lasen Tuesday, Febuay 3, 5 g xx = y 3a g yy = x 3a g xy = g yx = xy 3a At the noth pole i.e. (x, y) = (,) g αβ ((x, y) = (,)) = η αβ g αβ x γ (x,y)=(,) We have poved that the sphee is locally flat at the Noth Pole. Of couse, this is tue fo any set of coodinates on the sphee, because it is simply a matte of otation Length, Aea, Volume and Fou-Volume fo Diagonal Metics Fo a diagonal metic of the type ds = g dx dx + g dx dx + g dx dx + g 33 dx 3 dx 3 (7.7) you can define pope length elements in the vaious coodinates as dl = g dx dl = g dx dl 3 = g 33 dx 3 The aea element da = dl dl = g g dx dx (7.8) Notice, that it is not always the g and g that ae involved in the calculation of the aea. As we shall see below it can also be the g and g 33 The thee-volume element dv = dl dl dl 3 = g g g 33 dx dx dx 3 (7.9) The fou-volume element dv = g g g g 33 dx dx dx dx 3 (7.3) In the case of a non-diagonal metic the fou-volume element is dv = gd 4 x whee g is the deteminant of the matix g αβ Aea and Volume Elements of a Sphee The line element of flat space in pola coodinates ds = dt + d + dθ + sin θ dφ (7.3) The pope length elements dl = d dl = dθ dl 3 = sin θ dφ The aea element da = dl dl 3 = sin θ dθdφ (7.3) g_xy(x,y) =(y*x)/(3*a^) (Hatle, 3, p. 46) 3 (Hatle, 3, p. 47) example Page 6

17 Susan Lasen Tuesday, Febuay 3, 5 The aea A = da = sin θ dθdφ π = sin θ dθ = π [ cos θ] π = π ( ( )) = 4π The thee-volume element dv = dl dl dl 3 = dda = sin θ ddθdφ The thee-volume V π dφ = dv = sin θ ddθdφ = dda = 4π d = 4π [ 3 3 ] = 4π 3 3 Collection the esults A = 4π V = 4π 3 3 We ecognize the familia values Distance, Aea and Volume in the Cuved Space of a Constant Density Spheical Sta o a Homogenous Closed Univese The spheical line element is (a is a constant elated to the density of matte) ds = ( d + dθ + sin θ dφ a ) The pope length elements dl = ( d ) a dl = dθ dl 3 = sin θ dφ The cicumfeence aound equato whee = and θ = π C = dl 3 π = sin θ dφ = π The distance fom the cente ( ) to suface ( = ) along a line whee θ = const. and φ = const. 3 (Hatle, 3, p. 47) example Page 7

18 Susan Lasen Tuesday, Febuay 3, 5 S = dl = ( d ) a a = a d = 3 a [sin a ] = a sin a The aea of the two-suface at = A = da = dl dl 3 π = sin θ dθ π dφ (7.36) = 4π (7.37) The volume inside = is V = dv = dl dl dl 3 π π = ( d sin θ dθ dφ ) a = 4πa a d = 33 4πa [ a sin ( a ) a ] = 4πa 3 [ sin ( a ) a ( a ) ] = 34 4πa 3 ( sin ( a ) a ( a ) ) (7.38) Collecting the esults: The cicumfeence aound equato whee = and θ = π C = π The distance fom the cente ( ) to suface ( = ) along a line whee θ = const. and φ = const. S = a sin a The aea of the two-suface at = A = 4π 3 (Spiegel, 99) (4.37) 33 (Spiegel, 99) (4.39) 34 Fo /a : V = 4πa 3 ( sin ( a ) a ( a ) ) = 4πa 3 ( a + 4πa 3 ( 3 ( a )3 + ( a )3 ) = 4π 3 3 (Spiegel, 99) (.7) (.) 3 ( a )3 a ( ( a ) )) = Page 8

19 Susan Lasen Tuesday, Febuay 3, 5 The volume inside = is V = 4πa 3 ( sin ( a ) a ( a ) ) Distance, Aea, Volume and fou-volume of a metic ds = ( A ) dt + ( A ) d + dθ + sin θ dφ dl = ( A )dt dl = ( A )d dl = dθ dl 3 = sin θ dφ The pope distance along a adial line fom the cente to a coodinate adius = l = dl = ( A )d = [ 3 A3 ] = 3 A3 The aea of a sphee of coodinate adius = A = dl dl 3 π = sin θ dθ π dφ = 4π The thee-volume of a sphee of coodinate adius = V = dl dl dl 3 = ( A )d = 4π [ A5 ] = 4π ( A5 ) π sin θ dθ π dφ The fou-volume of a fou-dimensional tube bounded by a sphee of coodinate adius and two t = constant planes sepaated by a time T v = dl dl dl dl 3 T = dt ( A ) d π sin θ dθ = 4πT ( + A 4 A )d = 4πT [ A 7 7 A 5 5 ] = 4πT 3 (3 6A A 7 7 ) π dφ 35 (Hatle, 3, p. 66) poblem Page 9

20 Susan Lasen Tuesday, Febuay 3, The dimensions of a peanut The line-element of a peanut geomety is ds = a dθ + a f (θ)dφ f(θ) = sin θ ( 3 4 sin θ) dl = adθ dl = af(θ)dφ The distance fom pole to pole (φ ) d = dl = adθ = aπ The cicumfeence at a constant angle θ C = dl π π = af(θ)dφ π = af(θ) dφ = πaf(θ) At the cente o equato θ = π = πa sin θ ( 3 4 sin θ) C = πa sin ( π ) ( 3 4 sin ( π )) = πa ( 3 4 ) = π a The aea of a peanut A = dl dl π = a f(θ)dθ π π dφ = πa sin θ ( 3 4 sin θ) dθ π = πa ( 3 4 ( cos θ)) sin θ dθ = πa ( 3 4 ( x )) dx = π a ( + 3x )dx = π a [x + x 3 ] = π a ( ) = πa 36 (Hatle, 3, p. 66) poblem Page

21 Susan Lasen Tuesday, Febuay 3, The dimensions of an egg The line-element ds = a [(cos θ + 4 sin θ)dθ + sin θ dφ ] The cicumfeence of the egg at constant θ π C(θ) = a sin θ dφ = πa sin θ The distance fom pole to pole d pole to pole π = a cos θ + 4 sin θ dθ π = a + 3 sin θ dθ = 4,84a,77 πa The atio of the biggest cicle aound the axis to the pole-to-pole distance is C (θ = π ) πa = d pole to pole,77 πa =,3 The suface aea of an egg A π = a sin θ + 3 sin θ dθ = 3,4 π a π dφ Length and volume of the Schwazschild geomety The spatial pat of the Schwazschild line element is ds = ( m ) d dθ sin θ dφ (a) The adial distance between the sphee = M and the sphee = 3M l = dl 3M = M M d 3M = d M M = 39 [ ( M)] M 3M + M 3M M d ( M) 3M = 4 [ ( M) + M ln( + M)] M = M 3M + M ln( 3M + M) M ln( M) 37 (Hatle, 3, p. 9) poblem (Hatle, 3, p. 66) poblem px+q (ax+b)(px+q) dx = ax+b a 4 dx = (ax+b)(px+q) ap + aq bp a dx (ax+b)(px+q) (Spiegel, 99) (4.3) ln( a(px + q) + p(ax + b)) (Spiegel, 99) (4.) Page

22 Susan Lasen Tuesday, Febuay 3, 5 (b) = ( ln ( 3 )) M,5 M The spatial volume between the sphee = M and the sphee = 3M 3M V = dl dl dl 3 M 3M = M d M = 4π 7,5 M 3 = 6,43 4π 3 (M)3 π sin θ dθ π dφ Volume in the Womhole geomety The thee-dimensional volume on a t = constant slice of the womhole geomety bounded by two sphees of coodinate adius on each side of the thoat. The line-element ds = dt + d + (b + )(dθ + sin θ dφ ) (7.39) The volume V = dl dl dl 3 = (b + )d = 4π (b + )d π sin θ dθ = 4π [b ] π dφ = 4π ((b ) (( )b + 3 ( )3 )) = 4π 3 (3b + ) 4 Tenso Calculus 4. Chistoffel symbols aba and aab in a diagonal metic abc aba aab = ( g ab x c = ( g ab x a = ( g aa x b ) = ( g aa x b + g ac x b g bc x a ) (4.5) + g aa x b + g ab x a g ba x a ) g ab x a ) 4 (Hatle, 3)poblem (McMahon, 6, p. 34), final exam 6. The answe to FE-6 is (d) Page

23 Susan Lasen Tuesday, Febuay 3, 5 = ( g aa x b ) Find the Chistoffel symbols of the -sphee with adius a The line element: ds = a dθ + a sin θ dφ The metic tenso and its invese: g ab = { a a sin θ } gab = { a } a sin θ We have: abc = ( cg ab + b g ac a g bc ) (4.5) a bc = g ad dbc (4.6) θφφ = θ(a sin θ) = a θ sin θ cos θ φφ = g θθ θφφ = sin θ cos θ φθφ = φφθ = θ(a sin θ) = a φ sin θ cos θ θφ φ = φθ = g φφ φθφ = cot θ Find the Chistoffel symbols of the Kahn-Penose metic (Colliding gavitational waves) The line element: ds = dudv ( u) dx ( + u) dy The metic tenso: g ab = { ( u) } ( + u) and its invese: g ab = { ( u) ( + u) } abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) uxx = u( ( u) ) = ( u) v xx = g vu uxx = ( u) xux = xxu = u( ( u) ) = ( u) x x ux = xu = g xx xux = u uyy = u( ( + u) v ) = ( + u) yy = g vu uyy = ( + u) yuy = yyu = u( ( + u) y ) = ( + u) uy y = yu = g yy yuy = + u 43 (McMahon, 6, p. 74), example (McMahon, 6, p. 75), example 4-5, quiz 4-7, the answe to quiz 4-7 is (b) Page 3

24 Susan Lasen Tuesday, Febuay 3, Altenative solution: Show that c g ab p.69: c g d d ab = c g ab g db ac g ad cb If g db = g bd : d d = c g ab g bd ac g ad cb = c g ab bac acb since acb = abc (4.4): = c g ab ( bac + abc ) and bac + abc = c g ab (p. 73): = c g ab c g ab Q.E.D. 4.3 One-foms One-foms: why d Fo geneal foms, let α be a p-fom and β be a q-fom, we have α β = ( ) pq β α Fo one-foms this means α β = β α α α = α α This also holds fo df = f x a dxa because df is a one-fom as well. df a df b = f x a dxa f dxb xb = f x a x b dxa dx b = f x a x b dxb dx a df a df a = f x a x b dxa dx a and df a df a = f x a x b dxa dx a (ii) Now, because the patial deivatives commutate, this can only be tue if dx a dx a and d (4.5) The exteio deivative of a one-fom. The exteio deivative of a one-fom f a dx a : d(f a dx a ) = df a dx a (4.6) To use this equation it is impotant to notice, that the ight-hand side includes a summation of the patial deivatives times the diffeential in the usual way: df(x i ) = f(x x i )dx i. i Example σ = e f() dt dσ = d(e f() dt) = d(e f() ) dt = (ef() )d dt = f ()e f() d dt 45 (McMahon, 6, p. 78), example (McMahon, 6, p. 8) 47 (McMahon, 6, p. 8) equation (4.6) 48 You will find a moe extended use of equation (4.6) in e.g. the Binkmann metic, chapte 9, p.95 (McMahon, 6). (i) (ii) (i) Page 4

25 Susan Lasen Tuesday, Febuay 3, 5 ρ dρ = e g() cos θ sin φ d = d(e g() cos θ sin φ d) = d(e g() cos θ sin φ) d = (eg() cos θ sin φ)d d + θ (eg() cos θ sin φ)dθ d + φ (eg() cos θ sin φ)dφ d = θ (eg() cos θ sin φ)dθ d + φ (eg() cos θ sin φ)dφ d = e g() sin θ sin φ dθ d + e g() cos θ cos φ dφ d 4.4 The geodesic equation Find the geodesic equations fo cylindical coodinates The line element: ds = d + dφ + dz The metic tenso and its invese: g ab = { } g ab = { } abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) φφ = ( ) = φφ = g φφ = φφ = φφ = ( φ ) = φ φ = φ = g φφ φφ = The geodesics equation: d x a ds + dx b dx c a bc ds ds (4.33) x a = : x a = φ: x a = z: d φ ds + φ φ d ds + dx b dx c bc ds ds d (dφ ds ds ) d φ ds + φ dx b dx c bc ds ds d dφ ds ds + φ dφ d φ ds ds d φ ds + d dφ ds ds d z ds + dx b dx c z bc ds ds d z ds 49 (McMahon, 6, p. 83), example Page 5

26 Susan Lasen Tuesday, Febuay 3, 5 The Chistoffel symbols fom the geodesic equations We have K = g abx a x b = ( ) + (φ ) + (z ) (4.35) x a = : x a = φ: Now we need K x a K = d ds ( K x a ) (4.36) = d ds ( K ) φ = d ( ) = ds = φ K φ = d ds ( K ) φ = d ds ( φ ) = φ + φ x a = z: = φ + φ + φ K z = d ds ( K ) z = d (z ) = z ds Collecting the esults = φ = φ + φ + φ = z We can now find the Chistoffel symbols fom the geodesic equation : = φφ φ φ = φ φ = Use the geodesic equations to find the Chistoffel symbols fo the indle metic. The indle coodinate system o fame descibes a unifomly acceleating fame of efeence in Minkowski space. The line element: ds = ξ dτ dξ The metic tenso: g ab = { ξ } We have K = g abx a x b = ξ (τ ) ξ (4.35) 5 (McMahon, 6, p. 84), example 4- Page 6

27 Susan Lasen Tuesday, Febuay 3, 5 Now we need x a = ξ: K x a K ξ = d ds ( K x a ) (4.36) = d ds ( K ) ξ ξτ = d ds ( ξ ) = ξ = ξ + ξτ x a K = τ: = d τ ds ( K ) τ = d ds (ξ τ ) = ξξ τ + ξ τ = τ + ξ ξ τ + ξ τ ξ Collecting the esults = ξ + ξτ = ξξ τ + ξ τ We can now find the Chistoffel symbols fom the geodesic equation: ξ = ξ ττ τ ξτ = ξ τ τξ = ξ Geodesics Equations of the plane in pola coodinates We use the Eule-Lagange method. = d ds ( F x a ) F (.36) 5 x a F = g abx a x b (4.35) 53 The line element ds = d + dφ (8.) F = + φ x a = : F = φ F = = φ x a = φ: F φ F = φ φ 5 (Hatle, 3) Example 8-5 (McMahon, 6) 53 (McMahon, 6) Page 7

28 Susan Lasen Tuesday, Febuay 3, 5 = d ds ( φ ) = φ + φ φ = φ Collecting the esults = φ (8.6a) φ = φ (8.6b) 54 The non-zeo Chistoffel symbols ae φφ φ φ = = φ φ = (8.7) Equations fo geodesics in a Womhole Geomety We use the Eule-Lagange method. = d ds ( F x a ) F (.36) 56 x a F = g abx a x b (4.35) 57 The line element ds = dt + d + (b + )(dθ + sin θ dφ ) (8.) F = t + + (b + )θ + (b + ) sin θ φ x a = t: F t F = t t t x a = : F = (θ + sin θ φ ) F = = θ + sin θ φ x a = θ: F θ = (b + ) sin θ cos θ φ F = (b θ + )θ d ds ( F ) = θ + (b θ + )θ θ = sin θ cos θ φ (b + ) θ x a = φ: 54 (Hatle, 3) Example (Hatle, 3, p. 7), example (McMahon, 6) 57 (McMahon, 6) Page 8

29 Susan Lasen Tuesday, Febuay 3, 5 F φ F φ = (b + ) sin θ φ = d ds ((b + ) sin θ φ ) = sin θ φ + (b + ) sin θ cos θ θ φ + (b + ) sin θ φ φ = (b + ) φ cot θ θ φ Collecting the esults t (8.3a) = θ + sin θ φ (8.3b) θ = sin θ cos θ φ (b + ) θ (8.3c) φ = (b + ) φ cot θ θ φ (8.3d) 58 The non-zeo Chistoffel symbols ae θθ = φφ = sin θ θ φφ φ φ θ = sin θ cos θ θ = φ φ = (b + ) φ θφ θ = θ φ = φθ = (b + ) = cot θ (8.8) New - Geodesics and Chistoffel symbols of the Schwazschild metic with θ = π The metic ds = ( m ) dt + ( m ) d + dφ To find the geodesic we use the Eule-Lagange equation = d ds ( F x a ) F x a whee F = ( m ) t + ( m ) + φ x a = t: F t F = ( m t ) t d ds ( F ) = 4m t t ( m ) t = 4m t + ( m ) t m = t + ( m) t x a = : 58 (Hatle, 3) Example (Hatle, 3, p. 83) poblem Page 9

30 Susan Lasen Tuesday, Febuay 3, 5 F F d ds ( F = 6 m t m ( m ) + φ = ( m ) ) = ( m ) 4m ( m ) = ( m ) m ( m ) + m φ m = ( m) + ( m) m t 3 ( m)φ x a = φ: F φ F φ = φ d ds ( F ) φ = 4 φ + φ = 4 φ + φ = φ + φ Collecting the esults m = t + ( m) t m = ( m) + ( m) m t 3 ( m)φ = φ + φ We can now find the Chistoffel symbols: m m t t = t ( m) t = ( m) m = m( m) ( m) tt = 3 φ φ = φ φ = φφ = ( m) 4.5 Solving the geodesic equation The tavel time though a womhole Use the geodesic equations to calculate the pope tavel time of an astonaut tavelling though a womhole thoat along the coodinate adius fom = to =. The initial adial fou-velocity is u U, and because of spheically symmety u θ = u φ The fou-velocity is u a = (u t, u, u θ, u φ ) = ( dt dτ, d dτ, dθ dτ, dφ dτ ) 6 (( m t ) ) = m ( m ) = 6 (Hatle, 3) Example 8.5 m ( m) Page 3

31 Susan Lasen Tuesday, Febuay 3, 5 = 6 ( + U, U,, ) (8.) we will only look at u. We use the geodesic equation = θ + sin θ φ (8.3b) which we can ewite d = θ + sin θ φ dτ d dτ = d dτ (d dτ ) = du dτ (8.) which implies that u = U is a constant along the astonauts wold-line. So we can solve u = d dτ = U dτ = U d Δτ = U d = U [] = ( ( )) U = U (8.4) So the tavel time though the womhole Δτ = is vey much alike the usual time/speed calculation: U distance=time*velocity except with the velocity eplaced by the fou velocity NEW - Is the tajectoy time-like o space-like? The line-element dτ = dt d (b + )(dθ + sin θ dφ ) = dt ( ( d dt ) (b + ) (( dθ dt ) + sin θ ( dφ dt ) ) = dt ( ( d dτ dτ dt ) (b + )(() + sin θ () )) = dt ( (U + U ) ) = dt ( + U U + U ) = dt ( + U ) > i.e. the tajectoy is time-like. 6 u t is found fom the fact that the u a u a = is a conseved quantity: u a u a = u a η ij u a = (u t ) + (u ) + (u θ ) + (u φ ) = (u t ) + U = u t = + U Page 3

32 Susan Lasen Tuesday, Febuay 3, Geodesics in the Plane Using Pola Coodinates. We know that the geodesics in the plane ae staight lines. We can show this by using the line element, the Killing vecto ξ and the conseved quantity l = ξ u = const. The line element ds = d + dφ (8.) = ( d ds ) + ( dφ ds ) (8.33) The metic is independent of φ so we have a Killing vecto ξ = (,) And a conseved quantity l = ξ u = ξ a u a = g ab ξ b u a = g ξ u + g φφ ξ φ u φ = d ds + dφ ds = 64 dφ ds This we can inset into the line element and integate = ( d ds ) + ( dφ ds ) ( d ds ) = ( dφ ds ) = ( l ) = ( l ) d ds ) = ( (l ) (8.34) (8.35) We want to find φ as a function of dφ = dφ ds d ds d l = ( ( l ) ) (8.36) l = ( l ) dφ d = l ( l ) φ = 65 l l cos l + φ 63 (Hatle, 3) Example Notice this is equivalent with the geodesic equation we found befoe d ds ( φ ) 65 dx = x x n a n n a n cos an xn (Spiegel, 99) (4.334) Page 3

33 Susan Lasen Tuesday, Febuay 3, 5 = cos ( l ) + φ l = cos(φ φ ) (8.38) = cos φ cos φ + sin φ sin φ = x cos φ + y sin φ y = x cot φ + l sin φ = ax + b And the geodesic is a staight line as expected NEW - Geodesics Equations of the plane in Catesian coodinates. We use the Eule-Lagange method. = d ds ( F x a ) F (.36) 67 x a F = g abx a x b (4.35) 68 The line element ds = dx + dy F = x + y x a = x: F x F = x x x x a = y: F y F y = y y Collecting the esults x y The solution is obviously staight lines: x y x = k y = c x = k S + k y = c S + c y = K x + K New - Show that the geat cicle is a solution of the geodesic equation of a two-dimensional sphee. We use the Eule-Lagange method. 66 (Hatle, 3) Poblem (McMahon, 6) 68 (McMahon, 6) 69 (Hatle, 3) Poblem 8. Page 33

34 Susan Lasen Tuesday, Febuay 3, 5 The line element = d ds ( F x a ) F x a (.36) 7 F = g abx a x b (4.35) 7 ds = a dθ + a sin θ dφ F = a θ + a sin θ φ x a = θ: F θ = a sin θ cos θ φ F = a θ θ d ds ( F ) = a θ θ θ = sin θ cos θ φ x a = φ: F φ F φ = a sin θ φ d ds ( F ) φ = d ds (a sin θ φ ) = a sin θ cos θ θ φ + a sin θ φ φ = cot θ θ φ Collecting the esults θ = sin θ cos θ φ (I) φ = cot θ θ φ (II) The non-zeo Chistoffel symbols ae θ = sin θ cos θ φφ φ θφ φ = φθ = cot θ To show that the geat cicle is a solution to the geodesic equations we choose to wok in the plane θ = π, so the spheical pola coodinates ae x = a sin θ cos φ = a cos φ y = a sin θ sin φ = a sin φ z = a cos θ x = aφ sin φ y = aφ cos φ x + y = a φ The ight hand side we can find fom the second geodesic equation (II), which is educed to φ φ = constant. Next we use the law of consevation of the fou velocity in Catesian coodinates to show that the left hand side is a constant as well. dx u u a dx b = g ab ds ds = x + y = 7 (McMahon, 6) 7 (McMahon, 6) Page 34

35 Susan Lasen Tuesday, Febuay 3, 5 And we can conclude that the geat cicle is a solution NEW - Find all the time-like geodesic X(T) of the Flat Space metic in two dimensions ds = X dt + dx : To find X(T) we need a few integals which we can solve: a.the line element: ds = X dt + dx = ( dx ds ) X ( dt ds ) (I) b. Killing vectos: Because the metic is independent of T a Killing vecto is ξ = (ξ T, ξ X ) = (,) Accoding to (8.3) ξ u is a conseved quantity along a geodesic, whee u = (u T, u X ) = ( dt ds, dx ds ) ξ u = ξ a u a = g ab ξ b u a = g TT ξ T u T + g XX ξ X u X = X dt ds + dx ds = X dt ds X dt = constant = K (II) ds Inseting (II) into (I) dx ds Dividing (III) by (II) dx ds X dt ds dx dt = ( dx ds ) X ( K X ) = ( dx ds ) K X = + K X = + K X K = X K + K X = X K X + K K dt = X X + K dx T T dx = K X X + K = 73 ln ( K + X + K X ) ( K + X + K ) > if X > X (III) 7 (Hatle, 3, p. 84) poblem dx x x +a = a +a ln(a+ x ) (Spiegel, 99) (4.86) x Page 35

36 Susan Lasen Tuesday, Febuay 3, 5 Isolating X K + X + K X + ( K X ) + ( K X ) = exp( (T T )) = exp( (T T )) K X = (exp( (T T )) K X ) K X = exp( (T T )) + ( K X ) K X exp( (T T )) = exp( (T T )) exp( (T T )) = exp( (T T )) exp(t T ) = sinh(t T ) And we find the geodesics K X(T) = sinh(t T ) Ae these geodesics space-like o time-like: ds = X dt + dx = ( X + ( dx dt ) ) dt K K d ( = ( sinh(t T ) ) sinh(t T + ( ) ) ) dt dt ( ) K = ( sinh (T T ) + (K cosh(t T ) sinh (T T ) ) ) dt K = sinh (T T ) ( + coth (T T ))dt > because coth (T T ) > So these geodesics ae space-like, which we would expect because the metic is a flat-space metic, whee the geodesics in othe coodinates ae staight lines Is the wold-line X(T) = A cosh(t) time-like o space-like: ds = X dt + dx = ( X + ( dx dt ) ) dt = ( X d(a cosh T) + ( ) ) dt dt = A ( cosh T + sinh T)dT = A dt < i.e. the wold-line is time-like. 74 (Hatle, 3, p. 43) Example Page 36

37 Susan Lasen Tuesday, Febuay 3, Killing Vectos Show that if the Lie deivative of the metic tenso with espect to vecto X vanishes (L X g ab ), the vecto X satisfies the Killing equation. - Altenative vesion Expessions needed: Killings equation: b X a + a X b (8.) The Lie deivative of a (,) tenso: L V T ab = V c c T ab + T cb a V c + T ac b V c (4.8) c g ab (4.8) Now we can Calculate the Lie-deivative of g ab (4.8) and using (4.8) L X g ab = X c c g ab + g cb a X c + g ac b X c = a (g cb X c ) + b (g ac X c ) = a X b + b X a If L X g ab this implies that a X b + b X a, which is the Killing equation Pove that b a X b = ac X c We have c b X a a = bcdx d (8.6) Contacting with δ c a : δ c a ( c b X a ) = δ c a a ( bcdx d ) a b X a a = badx d a b X a = bd X d (8.7) Constucting a Conseved Cuent with Killing Vectos Altenative vesion: We wite J = a J a = T ab ( a X b ) and J = b J b = T ba ( b X a ) Now J = aj a + bj b = (Tab ( a X b ) + T ba ( b X a )) T ab is symmetic so T ab = T ba p.55 = Tab (( a X b ) + ( b X a )) Killings equation ( a X b ) + ( b X a ) (8.) Given a Killing vecto X the icci scala satisfiesx c c : We calculate the Lie deivative: L X ab = X c c ab + cb a X c + ac b X c (4.8) Multiplikation by g ab : g ab (L X ab ) = g ab (X c c ab ) + g ab ( cb a X c ) + g ab ( ac b X c ) Using c g ab, and L X g ab : L X g ab ab = X c c (g ab ab ) + g ab cb a X c + g ab ac b X c L X = X c c + g ab cb a X c + g ab ac b X c = X c c + a c a X c + c b X c = X c c + g bc ba a X c + g ac ba b X c = X c c + ba a (g bc X c ) + ba b (g ac X c ) = X c c + ba a X b + ba b X a = X c c + ba ( a X b + b X a ) 75 (McMahon, 6, p. 68), example 8-76 (McMahon, 6, p. 77) equation (8.7) 77 (McMahon, 6, p. 78) 78 (McMahon, 6, p. 79), quiz 8-3, the answe to Quiz 8-3 is (a) Page 37

38 Susan Lasen Tuesday, Febuay 3, 5 = X c c Now the Lie deivative of a scala is zeo L X, so X c c The iemann tenso e Pove: abcd = c abd d abc eac bd Fist we need c g d ab = c g ab ac e + ead bc (4.4) g db d cbg ad = c g ab bac abc (4.8) abcd e = g ae bcd e e = g ae ( c bd d bc + f e bd fc f e bc fd) (4.4) e e = g ae c bd g ae d bc + g ae f e bd fc g ae f e bc fd e = c (g ae bd) e e ( c g ae ) bd d (g ae bc) e + ( d g ae ) bc + f e bdg ae fc f e bcg ae fd e e = c abd ( eac + aec ) bd d abc + ( ead + aed ) bc + f bd afc f bc afd e e = c abd d abc ( eac + aec ) bd + ( ead + aed ) bc + e bd aec e bc aed e e = c abd d abc eac bd + ead bc (4.4) Pove: abcd = ( g ad + g bc g ac g bd x c x b x d x a x d x b x c x a) e e eac bd + ead bc (4.43) e e abcd = c abd d abc eac bd + ead bc (4.4) = c ( g ab x d + g ad x b g bd x a ) d ( g ab x c + g ac x b g bc x a ) e e eac bd + ead bc = ( g ab x c x d + g ad x c x b g bd x c x a) ( g ab x d x c + g ac x d x b g bc x d x a) e eac bd e + ead bc = ( g ad x c x b + g bc x d x a g ac x d x b g bd x c x a) e e eac bd + ead bc (4.43) Pove that (4.44): abcd = cdab = abdc = bacd and abcd + acdb + adbc abcd cdab = ( g ad x c x b + g bc x d x a g ac x d x b g bd x c x a) e e eac bd + ead bc (4.43) = ( g cb x a x d + g da x b x c g ca x b x d g db x a x c) e e eca db + ecb da = ( g cb x a x d + g da x b x c g ca x b x d g db x a x c) e f eca db + g ef cbg ef fda = ( g cb x a x d + g da x b x c g ca x b x d g db x a x c) e eca db + (g ef g ef f ) cb = 8 ( g cb x a x d + g da x b x c g ca x b x d g db x a x c) e f eca db + cb = ( g cb x a x d + g da x b x c g ca x b x d g db x a x c) e e eca db + cb eda = abcd Q.E.D. fda fda 79 (McMahon, 6, p. 86), equations (4.4), (4.43) and (4.44) 8 g ab g bc = δ a c if the metic is diagonal Page 38

39 Susan Lasen Tuesday, Febuay 3, 5 abdc bacd acdb adbc acdb + adbc = ( g ac x d x b + g bd x c x a g ad x c x b g bc x d x a) e e ead bc + eac bd = abcd Q.E.D. = ( g bd x c x a + g ac x d x b g bc x d x a g ad x c x b) e e ebc ad + ebd ac = ( g bd x c x a + g ac x d x b g bc x d x a g ad = ( g bd x c x a + g ac x d x b g bc x d x a g ad = ( g bd x c x a + g ac x d x b g bc x d x a g ad x c x b) g ef f bc x c x b) g efg ef x c x b) f bc g ef fad + g ef f bc fad + f bd fad + g ef g ef f bd fac g ef fac f bd fac = ( g bd x c x a + g ac x d x b g bc x d x a g ad x c x b) e e bc ead + bd eac = abcd Q.E.D. = ( g ab x d x c + g cd x b x a g ad x b x c g cb x d x a) e e ead cb + eab cd = ( g ac x b x d + g db x c x a g ab x c x d g dc x b x a) e e eab dc + eac db = ( g ab x d x c + g cd x b x a g ad x b x c g cb x d x a) e e ead cb + eab cd + ( g ac x b x d + g db x c x a g ab x c x d g dc x b x a) e e eab dc + eac db = ( g ac x b x d + g db x c x a g ad x b x c g cb x d x a) e e ead cb + eac db = abcd Q.E.D. Also notice e aaaa = c abd d abc eac bd e baaa = a baa a baa eba aa aaba aabb aabc abcc e + ead bc e + eba aa e = a aaa a aaa eaa aa e + eaa aa (4.4) = ( g ad x c x b + g bc x d x a g ac x d x b g bd x c x a) e e eac bd + ead bc (4.43) = ( g aa x b x a + g ab x a x a g ab x a x a g aa x b x a) e e eab aa + eaa ab f = g fe f abg fe faa + faa ab = g fe g fe f ab faa + faa ab = ( g ab x b x a + g ab x b x a g ab x b x a g ab x b x a) e e eab ab + eab ab = ( g ac x b x a + g ab x c x a g ab x c x a g ac x b x a) e e eab ac + eac ab = ( g ac x c x b + g bc x c x a g ac x c x b g bc x c x a) e e eac bc + eac bc f Page 39

40 Susan Lasen Tuesday, Febuay 3, Independent elements in the iemann, icci and Weyl tenso In n dimensions, thee ae N iemann = n (n )/ independent elements in the iemann tenso. In the icci tenso thee ae N icci = n(n + )/ independent elements, and in the Weyl tenso thee ae N Weyl independent elements if n = 4, if n < 4 thee ae none. Summaized: N iemann = n (n ) n(n + ) N icci = N Weyl n = 3 n = n = 4 33 abcd ab = c acb C abcd C 3 C 4 C 34 C 33 C 34 C 334 C C 44 C 434 C 334 C Compute the components of the iemann tenso fo the unit -sphee The numbe of independent elements in the iemann tenso in a metic of dimension n = is N = n (n ) θ = so we can choose to calculate φθφ 8 (McMahon, 6, p. 87) 8 The Weyl tenso possesses the same symmeties as the iemann tenso: C abcd = C abdc = C bacd = C cdab and C abcd + C adbc + C acdb. It possesses an additional symmety: C c acb. It follows that the Weyl tenso is tacefee, in othe wods, it vanishes fo any pai of contacted indices. One can think of the Weyl tenso as that pat of the cuvatue tenso fo which all contactions vanish (d'inveno, 99, p. 88) 83 Because: abcd + acdb + adbc 84 Because: C abcd + C acdb + C adbc 85 (McMahon, 6, p. 87), example 4- Page 4

41 Susan Lasen Tuesday, Febuay 3, 5 a a a The iemann tenso bcd = c bd d bc We choose a = c = θ, and b = d = φ e + bd a ec e a bc ed (4.4) θ φθφ Sum ove e: θ = θ φφ θ θ φ φθ = θ φφ + φφ θ θ = θ φφ + φφ θ = θ φφ e θ eθ θ θθ φ θ φθ φφ e + φφ e φθ θ eθ θ eφ θ θ φθ θφ e θ φθ eφ + φ θ φφ φθ = θ ( sin θ cos θ) cot θ ( sin θ cos θ) = cos θ + sin θ + cos θ = sin θ φ θ φθ φφ Show that the icci scala = fo the unit -sphee The line element: ds = dθ + sin θ dφ The metic tenso and its invese: g ab = { sin θ } gab = { } sin θ The icci scala: = g ab ab (4.47) Sum ove a: = g θb θb + g φb φb Sum ove b: = g θθ θθ + g θφ θφ + g φθ φθ + g φφ φφ g θφ = g φθ Sum ove c = g θθ θθ + g φφ φφ = g θθ c θcθ + g φφ c c φcφ ab = acb (4.46) φ + g φφ θ φθφ + g θθ θφθ + g φφ φ θ φ φφφ θθθ = φφφ + g φφ θ φθφ = g θθ g φφ φθφθ + g φφ g θθ θφθφ φθφθ = θφθφ (4.44) = g θθ θ θθθ = g θθ φ θφθ = g θθ g φφ θφθφ = g φφ φθφ = sin θ sin θ = θ θ φθφ = sin θ ex 4- emak that = g φφ θ φθφ is a geneal solution fo a -dimensional diagonal metic if we wite: = g (S) 86 (McMahon, 6, p. 89), example 4- Page 4

42 Susan Lasen Tuesday, Febuay 3, Poof: if a space is confomally flat, i.e. g ab (x) = f(x)η ab the Weyl tenso vanishes The Weyl scala C abcd = abcd + (g ad cb + g bc da g ac db g bd ca ) + 6 (g acg db g ad g cb ) (4.49) = abcd + (g ad e ceb + g bc e dea g ac e deb g bd e cea) + 6 (g acg db g ad g cb )g ef h ehf = abcd + (g adg ef fceb + g bc g ef fdea g ac g ef fdeb g bd g ef fcea ) + 6 (g acg db g ad g cb )g ef g hj jehf = abcd + f(x) (η ad η ef fceb + η bc η ef fdea η ac η ef fdeb η bd η ef fcea ) + 6 f(x)4 (η ac η db η ad η cb )η ef η hj jehf = abcd + f(x) (η ad eceb + η bc edea η ac edeb η bd ecea ) + 6 f(x)4 (η ac η db η ad η cb ) hehe = ( g ad x c x b + g bc x d x a g ac x d x b g bd x c x a) e e eac bd + ead bc + f(x) (η ad ( ( g eb x e x c + g ce x b x e g ee x b x c g cb x e x e) f fee cb + feb f ce) + η bc ( ( g ea x e x d + g de x a x e g ee x a x d g da x e x e) f fee da + fea f de) η ac ( ( g eb x e x d + g de x b x e g ee x b x d g db x e x e) f fee db + feb f de) η bd ( ( g ea x e x c + g ce x a x e g ee x a x c g ca x e x e) f fee ca + fea f ce)) + 6 f(x)4 (η ac η db η ad η cb ) ( ( g he x h x e + g eh x e x h g hh x e x e g ee x h x h) f fhh ee + fhe f eh) 87 (McMahon, 6, p. 9) Page 4

43 Susan Lasen Tuesday, Febuay 3, 5 = 88 ((η ad) c b + (η bc ) d a (η ac ) d b (η bd ) c a )f(x) e eac bd e = eac bd e + ead bc + f(x) (η ad ( ((η eb) e c + (η ce ) b e (η ee ) b c (η cb ) f e e )f(x) fee cb + feb f ce) + η bc ( ((η ea) e d + (η de ) a e (η ee ) a d (η da ) f e e )f(x) fee da η ac ( ((η eb) e d + (η de ) b e (η ee ) b d (η db ) f e e )f(x) fee db η bd ( ((η ea) e c + (η ce ) a e (η ee ) a c (η ca ) f e e )f(x) fee ca + fea f de) + feb f de) + fea f ce)) + 6 f(x)4 (η ac η db η ad η cb ) ( ((η he) h e + (η eh ) e h (η hh ) e e (η ee ) h h )f(x) f fhh ee e + ead bc + fhe f eh) + f(x) f (η ad ( fee cb f η ac ( fee db = g ef ( eac fbd + ead fbc ) + feb f f ce) + η bc ( fee da + fea f de) f f f + feb de ) ηbd ( fee ca + fea ce )) + 6 f(x)4 f (η ac η db η ad η cb )( fhh ee + fhe f eh) + f(x) (η ad g fh ( fee hcb + feb hce ) + η bc g fh ( fee hda + fea hde ) η ac g fh ( fee hdb + feb hde ) η bd g fh ( fee hca + fea hce )) + 6 f(x)4 g fj (η ac η db η ad η cb )( fhh jee + fhe jeh ) = f(x)η ef ( eac fbd + ead fbc ) + f(x)3 (η ad η fh ( fee hcb + feb hce ) + η bc η fh ( fee hda + fea hde ) η ac η fh ( fee hdb + feb hde ) η bd η fh ( fee hca + fea hce )) + 6 f(x)5 η fj (η ac η db η ad η cb )( fhh jee + fhe jeh ) 88 ((η ad ) c b + (η bc ) d a (η ac ) d b (η bd ) c a )f(x) = { + if a = b = c = d = x if a o b o c o d Page 43

44 Susan Lasen Tuesday, Febuay 3, 5 = f(x)η ef ( ( ( cg ea + a g ec e g ac )) ( ( dg fb + b g fd f g bd )) + ( ( dg ae + a g ed e g ad )) ( ( cg fb + b g fc f g bc ))) + f(x)3 (η ad η fh ( ( ( eg fe + e g fe f g ee )) ( ( bg hc + c g hb h g cb )) + ( ( bg fe + e g fb f g eb )) ( ( eg hc + c g he h g ce ))) + η bc η fh ( ( ( eg fe + e g fe f g ee )) ( ( ag hd + d g ha h g da )) + ( ( ag fe + e g fa f g ea )) ( ( eg hd + d g he h g de ))) η ac η fh ( ( ( eg fe + e g fe f g ee )) ( ( bg hd + d g hb h g db )) + ( ( bg fe + e g fb f g eb )) ( ( eg hd + d g he h g de ))) η bd η fh ( ( ( eg fe + e g fe f g ee )) ( ( ag hc + c g ha h g ca )) + ( ( ag fe + e g fa f g ea )) ( ( eg hc + c g he h g ce )))) + 6 f(x)5 η fj (η ac η db η ad η cb ) ( ( ( hg fh + h g fh f g hh )) ( ( eg je + e g je j g ee )) + ( ( eg fh + h g fe f g he )) ( ( hg je + e g jh j g eh ))) Page 44

45 Susan Lasen Tuesday, Febuay 3, 5 = f(x)η ef ( ( (η ea c + η ec a η ac e )) ( (η fb d + η fd b η bd f )) + ( (η ea d + η ed a η ad e )) ( (η fb c + η fc b η bc f ))) f(x) + f(x)3 (η ad η fh ( ( (η fe e + η fe e η ee f )) ( (η hc b + η hb c η cb h )) + ( (η fe b + η fb e η eb f )) ( (η hc e + η fe c η ce h ))) + η bc η fh ( ( (η fe e + η fe e η ee f )) ( (η hd a + η ha d η da h )) + ( (η fe a + η fa e η ea f )) ( (η hd e + η he d η de h ))) η ac η fh ( ( (η fe e + η fe e η ee f )) ( (η hd b + η hb d η db h )) + ( (η fe b + η fb e η eb f )) ( (η hd e + η he d η de h ))) η bd η fh ( ( (η fe e + η fe e η ee f )) ( (η hc a + η ha c η ca h )) + ( (η fe a + η fa e η ea f )) ( (η hc e + η he c η ce h )))) f(x) + 6 f(x)5 η fj (η ac η db η ad η cb ) ( ( (η fh h + η fh h η hh f )) ( (η je e + η je e η ee j )) + ( (η fh e + η fe h η he f )) ( (η je h + η jh e η eh j ))) f(x) Page 45

46 Susan Lasen Tuesday, Febuay 3, 5 = 89 f(x)η ef ( ( η ea c ) ( η fb d ) + ( η ea d ) ( η fb c )) f(x) + f(x)3 (η ad η fh ( ( η fe e ) ( η hc b ) + ( η fe b ) ( η hc e )) + η bc η fh ( ( η fe e ) ( η hd a ) + ( η fe a ) ( η hd e )) η ac η fh ( ( η fe e ) ( η hd b ) + ( η fe b ) ( η hd e )) η bd η fh ( ( η fe e ) ( η hc a ) + ( η fe a ) ( η hc e ))) f(x) + 6 f(x)5 η fj (η ac η db η ad η cb ) ( ( η fh h ) ( η je e ) + ( η fh e ) ( η je h )) f(x) = 4 f(x)ηef ( η ea η fb c d + η ea η fb d c )f(x) + 8 f(x)3 (η ad η fh ( η fe η hc e b + η fe η hc b e ) + η bc η fh ( η fe η hd e a + η fe η hd a e ) η ac η fh ( η fe η hd e b + η fe η hd b e ) η bd η fh ( η fe η hc e a + η fe η hc a e )) f(x) + 4 f(x)5 η fj (η ac η db η ad η cb )( η fh η je h e + η fh η je e h )f(x) 4. The thee dimensional flat space in spheical pola coodinates Calculate the Chistoffel symbols of the thee dimensional flat space in spheical pola coodinates The line element: ds = d + dθ + sin θ dφ The metic tenso and its invese: g ab = { sin θ} g ab = { sin θ} abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) θθ = ( ) = θθ = g θθ = 89 if e = c = a = x (η ec a η ac e )f(x) = { if e o c o a 9 (McMahon, 6, p. 9), quiz 4- and 4-3, the answe to quiz 4- is (a) sin θ cos θ, and the answe to quiz 4-3 is (c) cot θ Page 46

47 Susan Lasen Tuesday, Febuay 3, 5 θθ = θθ = ( θ ) = θ θ = θ = g θθ θθ = φφ = ( sin θ) = sin θ φφ = g φφ = sin θ φφ = φφ = ( sin θ) = sin φ θ φ φ = φ = g φφ φφ = θφφ = θ( sin θ) = θ sin θ cos θ φφ = g θθ θφφ = sin θ cos θ φθφ = φφθ = θ( sin θ) = φ sin θ cos θ θφ φ = φθ = g φφ φθφ = cot θ The iemann tenso of the thee dimensional flat space in spheical pola coodinates The numbe of independent elements in the iemann tenso in a metic of dimension n = 3 is N = n (n ) = 6 and we have to calculate: θθ ; φφ θ ; φθφ ; θφ θ φ ; θφ; φθ a a a The iemann tenso bcd = c bd d bc e + bd a ec e a bc ed (4.4) a, c =, b, d = θ: θθ = θθ e =, θ, φ: = θθ a, c =, b, d = φ: φφ = φφ e =, θ, φ: = φφ a, c = θ, b, d = φ: θ φθφ θ = θ φφ θ e =, θ, φ: = θ φφ a, c =, b = θ, d = φ: θφ = θφ e =, θ, φ: a, c = θ, b =, d = φ: θ θφ θ = θ φ e =, θ, φ: a, c = φ, b =, d = θ: φ = φ θ φθ θ θ θ θ φ φ θθ e + θθ e e θ eθ = ( ) ( ) e + φφ e e φ eφ φ φ φφ = sin θ ( ) ( sin θ) θ φ φθ θ + φφ θ e + φφ θ eθ φ θ φθ φφ e θ φθ eφ = cos θ + sin θ + ( sin θ) ( ) (cot θ)( sin θ cos θ) e e φ θ + θφ e θ eφ θ φ θ θ φ φ e θ + φ eθ e + θ φ eφ e =, θ, φ: = ( ) cot θ ( ) cot θ θ θθφ = θθφ = g θθ θφ e θ θ eφ e φ φ eθ A Lie deivative in the thee dimensional flat space in spheical pola coodinates Let w a = (, sin θ, sin θ cos φ) and v a = (, cos θ, sin φ) Calculate the Lie deivative u = L v w = v b b w w b b v (4.7) The covaiant deivative: b A a = Aa x b + bca (4.6) 9 (McMahon, 6, p. 9), quiz 4-4. We see that all the elements of the iemann tenso equals, and the answe to quiz 4-4 is (d) 9 (McMahon, 6, p. 9), quiz 4-6, the answe to quiz 4-6 is (a) Page 47

48 Susan Lasen Tuesday, Febuay 3, 5 u a = v b b w a w b b v a = v b ( wa x b + wa b = v x b + vb a bcw c va b w x b wb a bcv c wa va b = v wb xb x b + vb a bcw c w b a bcv c a = : u w v b = v wb xb x b + vb bcw c w b bcv c b =, θ, φ: = v θ θθw θ w θ θθv θ + v φ φφw φ w φ a bc w c ) w b ( va x b + a bcv c ) φφ v φ a = θ: u θ wθ vθ b = v wb xb x b + vb θ bcw c w b θ bcv c wθ vθ vθ b =, θ, φ: θ = v w wθ θ θ + v θ cw c w θ cv c + v θ θ θc w θ θ θcv c + v φ θ φcw c w φ θ φcv c c =, θ, φ: wθ vθ vθ θ = v w wθ θ θ + v w θ θ v + v φ θ w φ w φ θ θ φφ θ θ φv φ w θ w θ θ v θ + v θ w c θ θw wθ vθ vθ θ = v w wθ θ θ = ( cos θ)(cos θ) ( cos θ) (sin θ)( sin θ) = ( cos θ) a = φ: u φ wφ vφ b = v wb xb x b + vb φ bcw c w b φ bcv c wφ wφ vφ b =, θ, φ: θ = v + vφ wφ xθ xφ x φ + v φ cw c w φ cv c + v θ φ θc w θ φ θcv c + v φ φ φcw c w φ φ φcv c c =, θ, φ: wφ wφ vφ θ = v + vφ wφ xθ xφ x φ + v φ φw φ w φ φv φ + v θ φ θφ w θ φ θφv φ + v φ φ φw w φ φ φv + v φ φ φθw θ w φ φ φθv θ wφ wφ vφ θ = v + vφ wφ xθ xφ x φ = ( cos θ)(cos θ cos φ) + sin φ ( sin θ sin φ) sin θ cos φ (cos φ) = cos θ cos φ sin θ So we can conclude: u u = ( u θ ) = ( ( cos θ) ) u φ cos θ cos φ sin θ The icci scala of the Penose Kahn metic The icci scala: = g ab ab (4.47) The icci tenso ab c = acb (4.46) w c w φ Sum ove a = u, v, x, y: = g ub ub + g vb vb + g xb xb + g yb yb Sum ove b = u, v, x, y: = g uv uv + g vu vu + g xx xx + g yy yy = g uv c ucv + g vu c vcu + g xx c xcx + g yy c ycy Sum ove c = u, v, x, y: = g uv x uxv + g uv y uyv + g vu x vxu + g vu y vyu + g xx u xux + g xx v xvx 93 (McMahon, 6, p. 9), quiz 4-8.The answe to quiz 4-8 is (b) Page 48

49 Susan Lasen Tuesday, Febuay 3, 5 g uv = g vu : +g xx y xyx + g yy u yuy + g yy v yvy + g yy x yxy = g uv g xx xuxv + g uv g yy yuyv + g vu g xx xvxu + g vu g yy yvyu +g xx g uv vxux + g xx g vu uxvx + g xx g yy yxyx + g yy g uv vyuy +g yy g vu uyvy + g yy g xx xyxy abcd = cdab = 4g uv g xx = bacd = abdc : xuxv + 4g uv g yy yuyv + g xx g yy yxyx = 4g uv x uxv + 4g uv y uyv + g xx y xyx emak we can ewite this into to a geneal expession fo a non-diagonal metic of the type: g g g ab = g 33 { g 44 } We wite = 4g g g (S) x y y Now we need to calculate the thee elements in the iemann tenso: uxv; uyv; xyx x x x e x e x uxv = x uv v ux + uv ex ux ev y y y e y e y uyv = y uv v uy + uv ey uy ev y y y e y e y xyx = y xx x xy + xx ey xy ex So we can conclude A metic example : ds = y sin x dx + x tan y dy 4.. The Chistoffel symbols of a metic example The line element: ds = y sin x dx + x tan y dy The metic tenso and its invese: g ab = { y sin x x tan y } gab = { y sin x x tan y} abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) yxx = ( y yg xx ) = y sin x xx = g yy y sin x yxx = x tan y xyx = xxy = y sin x x x yx = xy = g xx xyx = y xxx = y x cos x xx = g xx xxx = cot x x xyy = x tan y yy = g xx x tan y xyy = y sin x y yxy = yyx = x tan y xy y = yx yyy = x ( + tan y y) yy = g yy yxy = x = g yy yyy = + tan y tan y 4.. The icci scala of a metic example Fom example 4- we know that fo a -dimensional diagonal metic: = g which x means we only have to calculate yxy 94 (McMahon, 6, p. 9), quiz 4-9, quiz 4-. The answe to quiz 4-9 is (c), and the answe to quiz 4- is (b) Page 49

50 Susan Lasen Tuesday, Febuay 3, 5 x x yxy = x yy e = x, y: x = x yy x y yx x y yx e + yy x + yy x ex x xx e yx x yx x ey x xy x tan y = x ( y sin x ) y ( x tan y ) + ( y tan y = y ( x cos x sin x sin x ) + ( y ) ( tan y + y x yy yx y sin x ) ( cot x) ( y ) y x yx yy x tan y cos x y sin x ) ( y ) + ( + tan y tan y ) ( y ) ( x tan y ) ( x y sin x ) + ( + tan y y tan y ) + ( y sin x ) xtan y cos x = ( y sin x ) ( x tan y cos x y sin x ) + ( + tan y y tan y ) = ( x cos x tan y + y sin x + y sin x tan y y sin ) x tan y = g yy x yxy = ( x tan y ) (x cos x tan y + y sin x + y sin x tan y y sin ) x tan y = ( x cos x tan y + y sin x + y sin x tan y x y sin x tan ) y Calculate the Chistoffel symbols fo a metic example : ds = dψ + sinh ψ dθ + sinh ψ sin θ dφ The line element: ds = dψ + sinh ψ dθ + sinh ψ sin θ dφ To find the Chistoffel symbols we calculate the geodesic fom the Eule-Lagange equation = d ds ( F x a ) F (.36) x a whee F = ψ + sinh ψ θ + sinh ψ sin θ φ x a = ψ: F ψ = cosh ψ sinh ψ θ + cosh ψ sinh ψ sin θ φ F = ψ ψ d ds ( F ) = ψ ψ = ψ cosh ψ sinh ψ θ cosh ψ sinh ψ sin θ φ = ψ cosh ψ sinh ψ θ cosh ψ sinh ψ sin θ φ x a = θ: F θ = cos θ sin θ sinh ψ φ F θ = sinh ψ θ d ds ( F ) θ = 4 cosh ψ sinh ψ ψ θ + sinh ψ θ = sinh ψ θ + 4 cosh ψ sinh ψ ψ θ + cos θ sin θ sinh ψ φ 95 (McMahon, 6, p. 35), final exam 9. The answe to FE-9 is (a) Page 5

51 Susan Lasen Tuesday, Febuay 3, 5 = θ + cosh ψ sinh ψ ψ θ + cos θ sin θ φ x a = φ: F φ F φ = sinh ψ sin θ φ d ds ( F ) φ = 4 cosh ψ sinh ψ ψ φ 4 cos θ sin θ θ φ + sinh ψ sin θ φ = 4 cosh ψ sinh ψ ψ φ 4 cos θ sin θ θ φ + sinh ψ sin θ φ coth ψ = φ + sin θ ψ φ cot θ sinh ψ θ φ Collecting the esults = ψ cosh ψ sinh ψ θ cosh ψ sinh ψ sin θ φ = θ + cosh ψ sinh ψ ψ θ + cos θ sin θ φ coth ψ = φ + sin θ ψ φ cot θ sinh ψ θ φ ψ θθ ψ φφ We can now find the non-zeo Chistoffel symbols: θ = cosh ψ sinh ψ φθ = cosh ψ sinh ψ sin θ θ φφ φ = cosh ψ sinh ψ ψφ φ = cos θ sin θ θφ coth ψ = sin θ cot θ = sinh ψ θ φ A metic example 3: ds = (u + ν )du + (u + ν )dν + u ν dθ 4.4. Calculate the Chistoffel symbols fo a metic example The line element: = (u + ν )du + (u + ν )dν + u ν dθ ds The metic tenso and its invese: (u + ν ) g ab = { (u + ν ) } g ab = u ν { (u + ν ) (u + ν ) u ν } abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) uuu = u u u(g uu ) = u uu = g uu uuu = (u + ν ) νuu = ν ν ν(g uu ) = ν uu = g νν νuu = (u + ν ) uνu = ν u ν(g uu ) = ν νu = g uu uνu = (u + ν ) 96 (McMahon, 6, p. 34), final exam 7 and 8. All the independent elements of the iemann tenso is zeo and the answe to FE-8 is (d) Page 5

52 Susan Lasen Tuesday, Febuay 3, 5 uνν = u u u(g νν ) = u νν = g uu uνν = (u + ν ) νuν = u ν u(g νν ) = u uν = g νν uνν = (u + ν ) ννν = ν ν ν(g νν ) = ν νν = g νν ννν = (u + ν ) uθθ = u(g θθ ) = uν u θθ θuθ = u(g θθ ) = uν θ uθ νθθ = ν(g θθ ) = u ν ν θθ = g uu uθθ = uν (u + ν ) = g θθ θuθ = u = g νν νθθ = u ν (u + ν ) θνθ = ν(g θθ ) = u θ ν νθ = g θθ θνθ = ν Collecting the esults we find the non-zeo Chistoffel symbols u u u ν uu = νν = uν = (u + ν ) ν ν ν u νν = uu = νu = (u + ν ) u θθ ν θθ θ uθ = uν (u + ν ) = u ν (u + ν ) = u θ νθ = ν 4.4. Calculate the iemann tenso of metic example In 3 dimensions the iemann tenso has six independent elements: ; 3 ; 3 ; 33 ; 33 ; 33 The iemann tenso abcd e e = c abd d abc eac bd + ead bc (4.4) uνuν e e = u uνν ν uνu euu νν + euν νu = u ( u) ν (ν) u u ν ν uuu νν + uuν νu νuu νν + νuν νu u = + (u + ν ) + ν (u + ν ) + ν (u + ν ) + u (u + ν ) uνuθ e e = u uνθ θ uνθ euu νθ + euθ νu uννθ e e = ν uνθ θ uνν euν νθ + euθ νν uθuθ e e = u uθθ θ uθu euu θθ + euθ θu = u ( uν u ν θ ) uuu θθ νuu θθ + θuθ θu = ν + u ν (u + ν ) + u ν (u + ν ) + ν uθνθ e e = ν uθθ θ uθν euν θθ + euθ θν = ν ( uν u ν θ ) uuν θθ νuν θθ + θuθ θν = uν + uν3 (u + ν ) + u3 ν (u + ν ) + uν Page 5

53 Susan Lasen Tuesday, Febuay 3, 5 ν = uν ( (u + ν ) + u (u + ν ) ) e e νθνθ = ν νθθ θ νθν eνν θθ + eνθ θν = ν ( u u ν θ ν) uνν θθ ννν θθ + θνθ θν = u u ν (u + ν ) + u ν (u + ν ) + u Calculate the iemann tenso of metic example Altenative vesion 97 The answe to Execise 8 in the Final Exam is given by abcd fo the following eason: Conside the global Minkowski spacetime d = du + dv + dθ, in some coodinates (u, v, θ ) [the coesponding iemannian cuvatue tenso identically vanishing, of couse], and conside the following coodinate tansfomation: u = uv cos θ v = uv sin θ θ = (u v ) The diffeentials ae elated as du = (udv + vdu) cos θ uv sin θ dθ dv = (udv + vdu) sin θ + uv cos θ dθ dθ = udu vdv fom which it eadily follows that du + dv + dθ = (udv + vdu) + (uv) dθ + (udu vdv) = (u + v )(du + dv ) + u v dθ Thus the line element ds = (u + v )(du + dv ) + u v dθ must coespond to an identically vanishing iemannian cuvatue tenso. 5 Catan s Stuctue Equations 5. 98icci otation coefficients fo the Tolman-Bondi- de Sitte metic (Spheical dust with a cosmological constant) The line element: ds = dt e ψ(t,) d (t, )dθ (t, ) sin θ dφ The Basis one foms = dt ω = e ψ(t,) d d = e ψ(t,) ω ω ω ω = (t, )dθ dθ = (t, ) ω = (t, ) sin θ dφ dφ = (t, ) sin θ ω η ij = { } 97 Kindly povided by M. John Fedsted: 98 (McMahon, 6, p. 6), example 5- Page 53

54 Susan Lasen Tuesday, Febuay 3, 5 a Catan s Fist Stuctue equation and the calculation of the icci otation coefficients b : dω a a = b ω b (5.9) a b a = b ω (5.) dω dω = d(e ψ(t,) d) = ψ e ψ(t,) dt d = ψ ω ω dω dω = d((t, )dθ) = dt dθ + d dθ = ω ω + eψ(t,) ω ω = d((t, ) sin θ dφ) = sin θ dt dφ + sin θ d dφ + (t, ) cos θ dθ dφ = ω ω + eψ(t,) ω ω cot θ + ω ω a b = Summaizing the cuvatue one foms in a matix: ψ ω ω ω ψ ω eψ(t,) ω ω eψ(t,) ω ω { eψ(t,) ω cot θ Whee a efes to column and b to ow. ω eψ(t,) ω cot θ ω } Now we can ead off the icci otation coefficients = ψ = ψ = = eψ(t,) = eψ(t,) = cot θ eψ(t,) = Which means the answe to quiz 5-6 is (b) = = = = eψ(t,) = cot θ a Tansfomation of the icci otation coefficients b into the Chistoffel symbols a bc a We have the tansfomation bc = (Λ ) d a d e f Λ e f bλ c (5.4) t = (Λ ) d d e f e f Λ tλ t = (Λ ) t d d e f Λ e f Λ = Λ t = (Λ ) t = ψ = ψ (Λ ) = ( ψ )(e ψ(t,) ) = ψ e ψ(t,) Page 54

55 Susan Lasen Tuesday, Febuay 3, 5 θ tθ t θθ θ θ = (Λ ) d = (Λ ) t d = (Λ ) d θθ = (Λ ) d φ tφ t φφ φ φ = (Λ ) d = (Λ ) t d = (Λ ) d φφ = (Λ ) d φ θφ θ φφ = (Λ ) d = (Λ ) d θ d e f d e f θ d e f d e f φ d e f d e f φ d e f d e f φ d e f θ d e f e f Λ tλ θ Λ e f θλ θ Λ e f Λ θ Λ e f θλ θ e f Λ tλ φ Λ e f φλ φ Λ e f Λ φ Λ e f φλ φ Λ e f θλ φ Λ e f φλ φ = Λ t = (Λ ) t (Λ θ ) = Λ = (Λ ) (Λ θ ) = Λ t = (Λ ) t (Λ φ ) = Λ = = = eψ(t,) e ψ(t,) = e ψ(t,) ( eψ(t,) ) = = sin θ = eψ(t,) e ψ(t,) = (Λ ) (Λ φ ) = e ψ(t,) ( = Λ θ = (Λ ) θ Howeve by this method we do not obtain = = = = e ψ(t,) = = sin θ = eψ(t,) ) sin θ = e ψ(t,) sin θ cot θ = = cot θ cot θ (Λ φ ) = ( ) sin θ = cos θ sin θ = ψ To check we calculate the Chistoffel symbols diectly fom the metic The line element: ds = dt e ψ(t,) d (t, )dθ (t, ) sin θ dφ The metic tenso g ab e ψ(t,) = { (t, ) } (t, ) sin θ e ψ(t,) and its invese: g ab (t, ) { (t, ) sin θ} abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) = ( e ψ(t,) ) = ψ e ψ(t,) = g = e ψ(t,) ψ e ψ(t,) = ψ t = t( e ψ(t,) ) = ψ e ψ(t,) t = g tt t = ( ψ e ψ(t,) ) = ψ e ψ(t,) t = t = t( e ψ(t,) ) = ψ e ψ(t,) Page 55

56 Susan Lasen Tuesday, Febuay 3, 5 t = t = g t = e ψ(t,) ψ e ψ(t,) = ψ tθθ t θθ θtθ θ tθ θθ θθ θθ θ θ tφφ t φφ φtφ φ tφ φφ φφ φφ φ φ θφφ = t( (t, )) = = g tt tθθ = = = θθt = t( (t, )) = θ = θt = g θθ θtθ = ( ) = = ( (t, )) = = g θθ = e ψ(t,) = θθ = ( (t, )) = θ = θ = g θθ θθ = ( ) = = t( (t, ) sin θ ) = sin θ = g tt tφφ = sin θ = sin θ = φφt = t( (t, ) sin θ ) = sin θ = φ φt = g φφ φtφ = ( sin θ ) ( sin θ) = ( (t, ) sin θ ) = sin θ = = g φφ = e ψ(t,) sin θ = φφ = ( (t, ) sin θ ) = sin θ = φ φ = g φφ φφ = ( sin θ ) ( sin θ) = θ( (t, ) sin θ ) = sin θ cos θ = θ φφ = g θθ θφφ = sin θ cos θ = sin θ cos θ φθφ φ θφ = φφθ = θ( (t, ) sin θ ) = sin θ cos θ φ = φθ = g φφ φθφ = sin θ ( sin θ cos θ) = cot θ 5. 99The cuvatue two foms and the iemann tenso To find the iemann tenso fom the cuvatue two foms, it can sometimes be moe convenient to use the following expession a b = a b d ω ω d a = b d ω ω d no summations 5.3 Find the icci scala using Catan s stuctue equations of the -sphee The line element: ds = dθ + sin θ dφ (5.8) 99 (McMahon, 6, p. 3) (McMahon, 6, p. 3), example 5- Page 56

57 Susan Lasen Tuesday, Febuay 3, 5 The Basis one foms ω ω = dθ dθ = ω = sin θ dφ dφ = sin θ ω η ij = { } Catan s Fist Stuctue equation and the calculation of the cuvatue one foms: dω a a = b dω dω ω b = d(sin θ dφ) = cos θ dθ dφ = cot θ ω ω (5.9) = cot θ ω Cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d = d(cot θ ω ) = d(cos θ dφ ) = sin θ dθ dφ = ω ω = = ω ω + Fom which we can identify the single independent element of the iemann tenso = And the icci scala = η = (S) 5.4 The thee dimensional flat space in spheical pola coodinates 5.4. icci otation coefficients of the thee dimensional flat space in spheical pola coodinates The line element: ds = d + dθ + sin θ dφ The Basis one foms ω = d d = ω η ij (McMahon, 6, p. ), quiz 5- and 5-, the answe to quiz 5- is (a) and to 5- is (c). Page 57

58 Susan Lasen Tuesday, Febuay 3, 5 ω ω = dθ dθ = ω = sin θ dφ dφ = sin θ ω = { } a Catan s Fist Stuctue equation and the calculation of the icci otation coefficients b : We have: dω a a = b ω b (5.9) a a = b ω (5.) b dω dω dω = d(dθ) = d dθ = ω ω = d( sin θ dφ) = sin θ d dφ + cos θ dθ dφ = ω ω cot θ + ω ω Summaizing the cuvatue one foms in a matix: ω ω a = b ω cot θ ω { ω cot θ ω } Whee a efes to column and b to ow. Now we can ead off the icci otation coefficients = = = cot θ = = cot θ = a 5.4. Tansfomation of the icci otation coefficients b into the Chistoffel symbols a bc of the thee dimensional flat space in spheical pola coodinates a We have the tansfomation bc = (Λ ) a d d e f Λ e f bλ c (5.4) θ θ θθ φ φ = (Λ ) d = (Λ ) d = (Λ ) d θ d e f d e f φ d e f Λ e f Λ θ Λ e f θλ θ e f Λ Λ φ = Λ = = = (Λ ) (Λ θ ) = ( ) = = Λ = = (McMahon, 6, p. ), quiz 5-3, the answe to quiz 5-3 is φφ = sin θ Page 58

59 Susan Lasen Tuesday, Febuay 3, 5 φφ φ θφ θ φφ = (Λ ) d = (Λ ) d = (Λ ) d d e f φ d e f θ d e f Λ e f φλ φ e f Λ θλ φ Λ e f φλ φ = (Λ ) (Λ φ ) = Λ θ = (Λ ) θ (Λ φ ) 5.5 3icci otation coefficients of the indle metic The line element: ds = u dv du = ( ) sin θ = sin θ cot θ = = cot θ = cot θ ( ) sin θ = sin θ cos θ The Basis one foms ω u = du du = ω u ω v = udv dv = ωv u η ij = { } a Catan s Fist Stuctue equation and the calculation of the icci otation coefficients b : We have dω a a = b ω b (5.9) a a = b ω (5.) b dω u dω v = d(udv) = du dv = ωu ω v v u u = u ωv v u v We calculate u v v = u = η u u u v v = η u u v u v = η u u η v v v u v = u 5.6 4The Einstein tenso fo the Tolman-Bondi- de Sitte metic The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d = d( ψ ω ) = d( ψ e ψ(t,) d) = [ ψ e ψ(t,) + (ψ ) e ψ(t,) ]dt d = [ ψ + (ψ ) ]ω ω = + = [ ψ + (ψ ) ]ω ω v u (McMahon, 6, p. ), quiz 5-4, the answe to quiz 5-4 is u v = v v 4 (McMahon, 6, p. ), quiz 5-7, the answe to quiz 5-7 is (d), almost! = u Page 59

60 Susan Lasen Tuesday, Febuay 3, 5 : d = d ( ω) = d( (t, )dθ) = dt dθ + ( ) d dθ = ω ω ( ) + eψ(t,) ω ω = = = eψ(t,) ω ( ψ ω ) = ω ω ( ) + ψ + e ψ(t,) ω ω d = d( ω ) = d( (t, ) sin θ dφ) = sin θ dt dφ + ( ) sin θ d dφ + cos θ dθ dφ = ω ω ( ) + eψ(t,) ω ω + cot θ ω ω = = eψ(t,) ω ( ψ ω ) cot θ + ω ω = ω ω ( ) + ψ + e ψ(t,) ω ω : d : d = d ( eψ(t,) ω ) = d( e ψ(t,) dθ) = + = [( ) e ψ(t,) + ψ e ψ(t,) ] dt dθ + [ e ψ(t,) + ψ e ψ(t,) ]d dθ = [( ) + ψ ] eψ(t,) ω ω + [ + ψ ] eψ(t,) ω ω = = = ω ( ψ ω ) = [( ) + ψ ] eψ(t,) ω ω + [( + ψ ) eψ(t,) = d ( eψ(t,) ω ) = d( e ψ(t,) sin θ dφ) = [( ) e ψ(t,) sin θ + ψ e ψ(t,) sin θ] dt dφ + ψ ] ω ω +[ e ψ(t,) sin θ + ψ e ψ(t,) sin θ]d dφ + e ψ(t,) cos θ dθ dφ = [( ) + ψ ] eψ(t,) ω ω + [ + ψ ] eψ(t,) ω ω + eψ(t,) cot θ ω ω = = + = ω ( ψ ω ) cot θ + ω eψ(t,) ω = [( ) + ψ ] eψ(t,) + ψ ] ω ω : d cot θ = d ( = + ω ω + [( + ψ ) eψ(t,) ω ) = d(cos θ dφ) = sin θ dθ dφ = + = ω ω + eψ(t,) ω ( + eψ(t,) ω ) ω ω = + Page 6

61 Susan Lasen Tuesday, Febuay 3, 5 = [ + ( ) ( ) eψ(t,) ] ω ω a = b Summaized in a matix: [ ψ + (ψ ) ]ω ω ω ω + ( ) + ψ e ψ ω ω S [( ) + ψ ] eψ ω ω + [( + ψ ) eψ + ψ ] ω ω ω ω + ( ) + ψ e ψ ω ω [( ) + ψ ] eψ ω ω + [( + ψ ) eψ + ψ ] ω ω S AS [ + ( ) ( ) eψ(t,) ] ω ω { S AS AS } Now we can find the independent elements of the iemann tenso in the non-coodinate basis: (A) = [ψ (ψ ) ] (B) (C) (D) = = [( ) + ψ ] eψ(t,) = [( + ψ ) eψ(t,) + ψ ] (B) (C) (D) = = [( ) + ψ ] eψ(t,) = [( + ψ ) eψ(t,) + ψ ] (E) = [ + ( ) ( ) eψ(t,) ] Whee A,B,C,D,E will be used late, to make the calculations easie The icci tenso: a b = a b = = = = + = A + B = = = [( ) + ψ ] eψ(t,) [( ) + ψ ] eψ(t,) = = = = + + = = (4.46) + + = [( ) + ψ ] eψ(t,) = = [ψ (ψ ) ] [( + ψ ) eψ(t,) = = = [( + ψ ) eψ(t,) + ψ ] + [ + ( ) + + ψ ] = A + D = + + = [ψ (ψ ) ] + = C ( ) eψ(t,) ] = B + D + E Page 6

62 Susan Lasen Tuesday, Febuay 3, 5 = = = [( + ψ ) eψ(t,) + ψ ] + [ + ( ) = + + ( ) eψ(t,) ] = B + D + E = a b = Summaized in a matix: [ψ (ψ ) ] { S [( ) + ψ ] eψ [ψ (ψ ) ] [( + ψ ) eψ + ψ ] Whee a efes to column and b to ow [( + ψ ) eψ + ψ ] + [ + ( ) ( ) eψ ] [( + ψ ) eψ + ψ ] + [ + ( ) ( ) eψ ]} The icci scala: = η a b a b (4.47) = η + η + η + η = = A + B ( A + D) ( B + D + E) = A + 4B 4D E = + 4 = [ψ (ψ ) ] [( + ψ ) eψ(t,) + ψ ] [ + ( ) ( ) eψ(t,) ] The Einstein tenso: G a b = a b η a b (4.48) G = η = = A + B (A + 4B 4D E) = D + E = = [( + ψ ) eψ(t,) + + ψ ] + [ + ( ) ( ) eψ(t,) ] G = [ ψ + ( ) ( + ψ + ( ) )e ψ(t,) ] = η = = G = η + = [( ) + ψ ] eψ(t,) G = η = + = A + D + (A + 4B 4D E) = B E = = [ + ( ) ( ) eψ(t,) ] = [( ) e ψ(t,) ( ) ] Page 6

63 Susan Lasen Tuesday, Febuay 3, 5 G = η = + = B + D + E + (A + 4B 4D E) = A + B D = + = [ψ (ψ ) ] = [ψ (ψ ) ] + [( + ψ )e ψ(t,) + ψ ] + [( + ψ ) eψ(t,) + ψ ] G = G = [ψ (ψ ) ] + [( + ψ )e ψ(t,) + ψ ] Summaized in a matix: G a b = [ ψ + ( ) ( + ψ + ( ) )e ψ ] { Whee a efes to column and b to ow S [( ) + ψ ] eψ [( ) e ψ ( ) ] [ψ (ψ ) ] + [( + ψ )e ψ + ψ ] [ψ (ψ ) ] + [( + ψ )e ψ + The Einstein tenso in the coodinate basis: The tansfomation: G ab = Λ aλ d b Gd (6.34) G tt G t d = Λ tλ t Gd = Λ d Λ t Gd = (Λ t) G = [ ψ + ( ) ( + ψ + ( ) )e ψ ] = Λ Λ tg = [ ( ) + ψ ] G = Λ d Λ G d = (Λ ) G = [( ) ( + + ( ) ) e ψ ] G θθ G φφ = Λ θ = Λ φ d Λ θg d Λ d φg d = (Λ θ ) G = ([ψ (ψ ) ] + [( + ψ )e ψ + ψ ]) = (Λ φ ) G = sin θ ([ψ (ψ ) ] + [( + ψ )e ψ + ψ ]) Summaized in a matix: G ab = [ ψ + ( ) ( + ψ + ( ) )e ψ ] [ ( ) { Whee a efes to column and b to ow S + ψ ] [( ) ( + + ( ) ) e ψ ] ([ψ (ψ ) ] + [( + ψ )e ψ + ψ ]) sin θ ([ψ (ψ ) ] + [( + ψ )e ψ + ψ ]) } 5.7 5Calculate the icci otation coefficients fo a metic example 3: ds = dψ + sinh ψ dθ + sinh ψ sin θ dφ The line element: ds = dψ + sinh ψ dθ + sinh ψ sin θ dφ 5 (McMahon, 6, p. 35), final exam Page 63

64 Susan Lasen Tuesday, Febuay 3, 5 ω ψ ω ω The Basis one foms = dψ dψ = ω ψ = sinh ψ dθ dθ = sinh ψ ω = sinh ψ sin θ dφ dφ = sinh ψ sin θ ω η ij = { } a Catan s Fist Stuctue equation and the calculation of the icci otation coefficients b : We have: dω a a = b ω b (5.9) a a = b ω (5.) b dω ψ dω dω = d(sinh ψ dθ) = cosh ψ dψ dθ = cosh ψ ω ψ ω sinh ψ = coth ψ ω ψ ω = d(sinh ψ sin θ dφ) = cosh ψ dψ dφ + cos θ dθ dφ = cosh ψ ω ψ ω + cos θ ω ω sinh ψ sin θ sinh ψ sinh ψ sin θ coth ψ = ωψ cot θ ω + ω ω sin θ sinh ψ a b = Summaizing the cuvatue one foms in a matix: coth ψ ω coth ψ ω sin θ coth ψ ω coth ψ ω cot θ ω { sin θ sinh ψ Whee a efes to column and b to ow. cot θ sinh ψ ω } ψ ψ Now we can find the non-zeo icci otation coefficients coth ψ = coth ψ ψ = coth ψ ψ = sin θ coth ψ cot θ = = sin θ sinh ψ cot θ = sinh ψ Page 64

65 Susan Lasen Tuesday, Febuay 3, 5 6 The Einstein Field Equations 6. 6The vacuum Einstein equations Pove that the Einstein field equations G ab = κt ab educes to the vacuum Einstein equations ab if we set T ab The Einstein tenso G ab = ab g ab (4.48) Now setting G ab = κt ab and calculating G ab = ab g ab ab = g ab Multiplying by g ab g ab ab = gab g ab using the definition = g ab ab (4.47) and that in 4 dimensions g ab g ab = 4 = 4 = Now this can only be tue if ab Q.E.D. 6. The vacuum Einstein equations with a cosmological constant Pove that the Einstein field equations G ab = κt ab educes to ab = g ab Λ and = 4Λ fo metics with positive signatue and ab = g ab Λ and = 4Λ fo metics with negative signatue in vacuum with a cosmological constant 7. The Einstein equation in vacuum with a cosmological constant and positive signatue is = ab g ab + g ab Λ (6.6) = g ab ab gab g ab + g ab g ab Λ = 4 + 4Λ = 4Λ Q.E.D. Next we ewite the Einstein equation = ab g ab + g ab Λ = ab g ab(4λ) + g ab Λ = ab g ab Λ ab = g ab Λ Q.E.D. In the non coodinate basis a b = η a b Λ 6 (McMahon, 6, p. 38) 7 An excellent qualitative explanation of the cosmological constant, you can find in (Geene, 4, s ) Page 65

66 Susan Lasen Tuesday, Febuay 3, 5 In the case of metics with negative signatue the Einstein equation in vacuum with a cosmological constant = ab g ab g ab Λ (6.6) and we can see that = 4Λ Q.E.D. ab = g ab Λ Q.E.D. In the non coodinate basis a b = η a b Λ 6.3 8Geneal emaks on the Einstein equations with a cosmological constant If we demand that the gavitational field equations ae () geneally covaiant () be of second diffeential ode in g ab (3) involve the enegy-momentum T ab linealy it can be shown that the only equation which meets these equiements is ab + μg ab Λg ab = κt ab whee μ, Λ, and κ ae constants. The demand that T ab satisfies the consevation equation b T ab leads to μ Poof: = if b T ab b ( ab + μg ab Λg ab ) b ab + μ b (g ab ) Λ b g ab 9 b ab + μ (( b )g ab + ( b g ab )) b ab + μ( b )g ab Next we use the Bianchi identity: a debc + b deca + c deab g db ( a debc + b deca + c deab ) a g db debc + b g db deca + c g db deab a b ebc + b b eca + c b eab a ec + b b eca c ea g ae ( a ec + b b eca c ea ) a g ae ec + b g ae b eca c g ae ea a a c + b b c c ( a a c c) g bc ( a a c c) ( a g bc a c ( c)g bc ) 8 (d'inveno, 99, p. 7) 9 b g ab g ae b eca = g ae b ab b e ac = ac = c Page 66

67 Susan Lasen Tuesday, Febuay 3, 5 ( a ab ag ab ) Now if we compae with we see that b ab + μ( b )g ab μ = dimensions: Gavitational collapse of an inhomogeneous spheically symmetic dust cloud Find the components of the cuvatue tenso fo the metic in + dimensions using Catan s stuctue equations The line element: ds = dt + e b(t,) d + (t, )dφ The Basis one foms = dt ω = e b(t,) d d = e b(t,) ω ω ω = (t, )dφ dφ = (t, ) ω η ij = { } Catan s Fist Stuctue equation and the calculation of the cuvatue two-foms dω a a = b ω b (5.9) a b a = b ω (5.) dω dω = d(e b(t,) d) = b e b(t,) dt d = b ω ω dω = d((t, )dφ) = dt dφ + d dφ = ω ω + e b(t,) ω ω a b Summaizing the cuvatue one foms in a matix: b ω ω = b ω { ω e b(t,) ω Whee a efes to column and b to ow. e b(t,) ω } (McMahon, 6, pp. 39-5), example 6-, example 6-3, example Page 67

68 Susan Lasen Tuesday, Febuay 3, 5 The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d : d : d = d(b ω ) = d(b e b(t,) d) = [b e b(t,) + (b ) e b(t,) ] dt d = [b + (b ) ] ω ω = + = [b + (b ) ] ω ω + + = d ( ω ) = d( (t, )dφ) = dt dφ + ( ) d dφ = ω ω ( ) e b(t,) ω ω = + = ω ω ( ( ) = b ) e b(t,) ω ω = d ( e b(t,) ω ) = d( e b(t,) dφ) = e b(t,) ω b ω = [( ) e b(t,) b e b(t,) ] dt dφ + [ e b(t,) + b e b(t,) ]d dφ = [( ) b ] e b(t,) = + = [( ) b ] e b(t,) ω ω [ + b ] e b(t,) ω ω + = = b ω ω ω ω ([ + b ] e b(t,) b ) ω ω a b Summaized in a matix: [b + (b ) ] ω ω ω ω ( ( ) b ) e b(t,) ω ω = [( ) b ] e b(t,) ω ω ([ + b ] e b(t,) b ) ω ω { } Now we can find the independent elements of the iemann tenso in the non-coodinate basis: (A) = [b + (b ) ] ω ω (B) (C) (D) = = [( ) + b ] e b(t,) = ([ + b ] e b(t,) b ) Page 68

69 Susan Lasen Tuesday, Febuay 3, 5 Whee A, B, C and D will be used late to make the calculations easie 6.4. Find the components of the cuvatue tenso fo the metic in + dimensions using Catan s stuctue equations altenative solution The line element: ds = dt + e b(t,) d + (t, )dφ Now we can compae with the Tolman-Bondi de Sitte line element, whee the pimes should not be mistaken fo the deivative d/d. ds = dt e ψ(t, ) d (t, )dθ (t, ) sin θ dφ And chose: dt = dt e ψ(t, ) d = e b(t,) d (t, )dθ (t, ) sin θ dφ = (t, )dφ Compaing the two metics we see: dφ = dφ, θ = π, (t, ) = (t, ), dt = dt Next we can use the fome calculations of the Tolman-Bondi de Sitte metic to find the iemann and Einstein tenso fo the + metic. But fist we need to find ψ = dψ(t, ) dt = e ψ(t, ) d dt (eψ(t, ) d d b(t,) ) = e d dt ψ = d ψ(t, ) dt = d dt ( b ) = b (t, ) ψ = dψ(t, ) d = e ψ(t, ) d d (eψ(t, ) ) = e ψ(t, ) d d d d (e b(t,) d d ) = e ψ(t, ) e b(t,) b (t, ) d(t, ) = dt = d(t, ) dt = d (t, ) dt = d(t, ) d = d (t, ) dt d = d (t, ) dt = (t, ) = (t, ) = d d(t, ) d = e ψ(t, ) e b(t,) (t, ) d = d, ) d (d(t dt ) = d d d d db(t, ) (e b(t,) ) = = b (t, ) d dt d ( (t, )) = e ψ(t, ) e b(t,) (t, ) The iemann tenso Tolman Bondi de Sitte + = [ψ (ψ ) ] (A) = [b + (b ) ] = = [( ) + ψ ] eψ(t,) Page 69

70 Susan Lasen Tuesday, Febuay 3, 5 = [( + ψ ) eψ(t,) = = [( ) + ψ ] eψ(t,) = [( + ψ ) eψ(t,) = [ + ( ) + ψ ] ( ) eψ(t,) ] (B) (C) + ψ ] (D) = Whee A, B, C and D will be used late to make the calculations easie = [( ) b ] e b(t,) = [( b ) e b(t,) b ] Find the components of the Einstein tenso in the coodinate basis fo the metic in + dimensions. The icci tenso: a b = a b = = = = = = = = = = [b + (b ) ] [( b ) e b(t,) = = B + D = + + = + = (4.46) + = [b + (b ) ] = A + B = [( ) b ] e b(t,) = C + b ] = A + D = + = [( b ) e b(t,) b ] a b = Summaized in a matix: [b + (b ) ] S [( ) b ] e b [b + (b ) ] [( b ) e b { Whee a efes to column and b to ow b ] [( b ) e b b ] } The icci scala: = η a b a b (4.47) Page 7

71 Susan Lasen Tuesday, Febuay 3, 5 = η + η + η = + + = (A + B) + ( A + D) + ( B + D) = A B + D = = [b + (b ) ] + + [( b ) e b(t,) b ] The Einstein tenso: G a b = a b η a b (4.48) G G = η = + = A + B + ( A B + D) = D = = [( b ) e b b ] = η = = G = η G G = [( ) b ] e b = η = = A + D ( A B + D) = B = = η = = B + D ( A B + D) = A = = = [b + (b ) ] Summaized in a matix: G a b = [( b ) e b S b ] [( ) b ] e b { [b + (b ) ] } Whee a efes to column and b to ow The Einstein tenso in the coodinate basis: The tansfomation: G ab = Λ aλ d b Gd (6.34) G tt G t G d = Λ tλ tg d = Λ d Λ tg d = Λ Λ d G d = (Λ t) G = [( b ) e b b ] = Λ Λ t G = [( ) b ] = (Λ ) G = eb Page 7

72 Susan Lasen Tuesday, Febuay 3, 5 G φφ = Λ φ Λ d φg d = (Λ φ ) G = [b + (b ) ] Summaized in a matix: G ab = [( b ) e b S b [( ] ) b ] { [b + (b ) ] } Whee a efes to column and b to ow eb The Einstein equations of the metic in + dimensions. Given the Einstein equation ( if c = G = ): G a b + Λη a b = κt a b (6.4) with Λ = λ you get G a b λ η a b = κt a b ρ and the stess-enegy tenso: T a b = κ { } You can find the Einstein equations [( b ) e b b ] { S [( ) b ] e b [b + (b ) ] } λ { } ρ = κ { } G : [( b ) e b b ] + λ = κρ p.5 G : [( ) b ] e b ( ) b p.5 G : λ + λ (6.4) G : [b + (b ) ] λ (6.4) 6.5 Using the contacted Bianchi identities, pove that: b G ab Expessions needed: Bianchi identity: = a debc + b deca + c deab (4.45) Poved page 78: = c g ab = c g ab g da g eb c g ab c g da g eb g ab (S3) (McMahon, 6, p. 5), quiz 6-, the answe to quiz 6- is (c) Page 7

73 Susan Lasen Tuesday, Febuay 3, 5 = c g de iemann tenso: abcd = abdc (4.44) c icci tenso: acb = ab (4.46) icci scala: = g ab a ab = a (4.47) g ae b eca = g ae b ab b e ac = ac = c (S4) The Einstein tenso: G ab = ab g ab (4.48) Konecke delta a g c = g ab a g bc = δ c (.5) a G a c = a ( a c g a c) = a a c ag a c use (.5) = a a c g a c a = a a c δ c a a = a a c c (S5) The poof: Multiply (4.45) by g db : = g db ( a debc + b deca + c deab ) (use (S3)) = a g db debc + b g db deca + c g db deab b = b b a ebc + b eca + c eab (use (4.44) and (4.46)) = b b a ec + b eca c eba (use (4.46)) = b a ec + b eca c ea Multiply by g ae = g ae b ( a ec + b eca c ea ) = a g ae ec + b g ae b eca c g ae ea = a c + b g ae b a eca c a (use (4.47)) = a c + b g ae b eca c (use (S4)) = a c + b c c = [ a a c c] (use (S5)) a = a G c Multiply by g bc = g bc a G a c = a g bc G a c = a G ab 6.6 3icci otation coefficients, icci scala and Einstein equations fo a geneal 4-dimensional metic: ds = dt + L (t, )d + B (t, )dφ + M (t, )dz The line element: ds = dt + L (t, )d + B (t, )dφ + M (t, )dz The Basis one foms ω = dt ω = L(t, )d d = L(t, ) ω ω = B(t, )dφ dφ = B(t, ) ω ω z = M(t, )dz dz = M(t, ) ωz η ij = { } 3 (McMahon, 6, pp. 5-53), quiz 6-5, 6-6, 6-7 and 6-8, the answe to quiz 6-5 is (a) and quiz 6-6 is (c), the answe to quiz 6-7 is (a), the answe to quiz 6-8 is (a) Page 73

74 Susan Lasen Tuesday, Febuay 3, 5 a Catan s Fist Stuctue equation and the calculation of the icci otation coefficients b : dω a a = b ω b (5.9) a b a = b ω (5.) dω dω dω dω z = d(l(t, )d) = L dt d = L ω ω L = d(b(t, )dφ) = B dt dφ + B d dφ = B ω ω B + ω ω B LB = d(m(t, )dz) = M dt dz + M d dz = M M ω ω z + M LM ω ω z a b Summaizing the cuvatue one foms in a matix: L B ω ω M ωz L B M = L ω L B B ω M ωz LB LM ω B ω B LB M { ωz M ωz M LM Whee a efes to column and b to ow } z z Now we can ead off the icci otation coefficients = L = L L L = B z z B = B B = M M z z = B LB = M LM = B LB z z = M M = M LM The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d = d ( L L ω) = d (L (t, )) d = L dt d + L d d = L ω ω L = = L L ω ω z z Page 74

75 Lots of Calculations in Geneal elativity Susan Lasen Tuesday, Febuay 3, 5 : d z z : d z z : d z z : d z z z z : d z = d ( B ω ) = d(b (t, )dφ) = B dt dφ + B B d dφ = B ω ω B + ω ω B BL = + + = B ω ω B + ω ω B + ω L B BL BL = B ω ω + ( B L B BL B BL ) ω ω + + z z L ω = = B ω L ω BL L = d ( M M ωz ) = d(m (t, )dz) = M dt dz + M d dz = M M ω ω z + M ML ω ω z z z z = z z = M ω ω z + M ω ω z + M ωz L ω M ML ML L = M M ωz ω + ( M L ML M ML ) ωz ω z z = = d ( B ω B ) = d ( BL L dφ) = B L B L L dt dφ + B L B L L = B L B L BL ω ω + B L B L BL 3 ω ω = + + = B L B L BL + z z ω ω + B L B L BL 3 ω ω + B L ω ω BL = B L B L BL ω ω + ( B L B L BL 3 + B L BL ) ω ω = d ( M ML ωz ) = d ( M L dz) = M L M L L = M L M L ML ω z ω + z = z + dt dz + M L M L ML 3 ω z ω z + = M L M L ML ω z ω M L M L + ML 3 = M L M L ML ω z ω M L M L + ( ML 3 z = z = z = z + = ( B M BM B M L BM ) ωz ω z + + z z z = M ML ωz L L ω d dφ = = B L ω ω BL M L M L L d dz z = ω z ω + M L ωz ω ML + M L ML ) ωz ω z + z + z z = M M ωz B B ω + M LM ωz ( B LB ω ) = M L ML ωz ω Page 75

76 Lots of Calculations in Geneal elativity Susan Lasen Tuesday, Febuay 3, 5 a b = Summaized in a matix: L ω ω B ω ω + ( B L L B BL B BL ) ω ω S ( B L BL B BL ) ω ω + ( B L B L BL 3 + B L BL ) ω ω M M ωz ω + ( M L ML M ML ) ωz ω ( M L ML M ML ) ωz ω M L M L + ( ML 3 + M L ML ) ωz ω S AS ( B M BM B M L BM ) ωz ω { S AS AS } Now we can find the independent elements of the iemann tenso in the non-coodinate basis: (A) = L L (B) = B z z (C) = M B M (D) = B L BL B z z (E) = M L BL ML M ML (F) = B L B L BL 3 + B L z M L M L z (G) = BL ML 3 z z (H) = B M BM B M L BM Whee A,B,C,D,E,F,G,H will be used late, to make the calculations easie + M L ML The icci tenso: a b z z z = a b = = L = B M M + + L B = A + B + C = = + + = B L BL B BL + M L ML M = D + E ML = = + + = z = = L = z = + z + + B L BL + z + z + z z + z z + z z + z z z + z (4.46) = = = + z + z + L + B L B L M L M L BL 3 + ML 3 + M L = A + F + G ML z = = z = = B B + B L B L BL 3 + B L BL + B M BM B M L = B + F + H BM z z z = z z = z z + z z + z + z z z z = z + z = M M L M L + M ML 3 + M L ML + B M BM B M L = C + G + H BM z + z + z + z z + z z + z Page 76

77 Lots of Calculations in Geneal elativity Susan Lasen Tuesday, Febuay 3, 5 a b = Summaized in a matix: { L L B B M M S L L + B L B L BL 3 B L BL B BL + M L ML M ML + B L BL M L M L + + M L ML 3 ML B B + B L B L + B L BL 3 BL + B M BM B M L BM Whee a efes to column and b to ow M M M L M L + + M L ML 3 ML + B M BM B M L BM} The icci scala: = η a b a b (4.47) = η + η + η + η = z z = (A + B + C) A + F + G B + F + H C + G + H = A B C + F + G + H = z z + z + z z + z = ( L L + B B + M M + B L B L BL 3 + B L M L M L + BL ML 3 + M L ML + B M BM B M L BM ) The Einstein tenso: G a b = a b η a b (4.48) G = η = + = A + B + C + ( A B C + F + G + H) = F + G + H z + z = B L B L = z + z BL 3 G = η = = B L BL B BL + M L ML M ML G = η G z = z η z G + B L M L M L + BL ML 3 + M L ML + B M BM B M L BM = η = = A + F + G ( A B C + F + G + H) = B + C + H = z + z G = η G z = z η z z + z = B B M M + B M BM B M L BM G = η = = B + F + H ( A B C + F + G + H) = A + C + G = z + z z + z = L L M M L M L M ML 3 + M L ML Page 77

78 Lots of Calculations in Geneal elativity Susan Lasen Tuesday, Febuay 3, 5 G z = z η z G z z = z z η z z = z z = C + G + H ( A B C + F + G + H) = A + B + F = + + = L L B B + B L B L BL 3 + B L BL Summaized in a matix: G a b = { B L B L + B L M L M L + + M L BL 3 BL ML 3 ML + B M BM B M L BM Whee a efes to column and b to ow S B L BL B BL + M L ML M ML B B M M + B M BM B M L BM L L M M M L M L + M L ML 3 ML L L B B + B L B L + B L BL 3 BL 7 The Enegy-Momentum Tenso 7. 4Pefect Fluids Altenative deivation The most geneal fom of the stess enegy tenso is T ab = Au a u b + Bg ab (7.8) In the local fame we know that ρ T a b P = { } (7.6) P P and u a = (,,,) Then we choose the metic with negative signatue η a b = { } This we can use to find the constants A and B T = Au u + Bη = A + B = ρ if i = j if i j T i j = Au i uj + Bη i j = B = { P B = P and A = ρ B = ρ + P Which leaves us with the most geneal fom of the stess enegy tenso fo a pefect fluid fo a metic with negative signatue T ab = (ρ + P)u a u b Pg ab (7.) If we instead choose the metic with positive signatue η a b = { } 4 (McMahon, 6, p. 6) Page 78

79 Susan Lasen Tuesday, Febuay 3, 5 = Au u + Bη = A B = ρ T i j = Au i uj + Bη i j = B = { P if i = j if i j B = P and A = ρ + B = ρ + P Which leaves us with the most geneal fom of the stess enegy tenso fo a pefect fluid fo a metic with negative signatue T ab = (ρ + P)u a u b + Pg ab (7.) T 7. 5The Gödel metic The Gödel metic is an exact solution of the Einstein field equations in which the stess-enegy tenso contains two tems, the fist epesenting the matte density of a homogeneous distibution of swiling dust paticles, and the second associated with a nonzeo cosmological constant. 6 The line element: ds = ω ((dt + e x dz) dx dy ex dz ) = ω (dt + e x dtdz dx dy + ex dz ) e x The metic tenso g ab = ω e { ex } e x and its invese g ab = ω { } e x e x The stess enegy tenso T ab = ρ ω { e x } e x e x The Einstein equation fo a metic with a negative signatue 8πGT ab = G ab g ab Λ 7 8πGg ab T ab = g ab G ab g ab g ab Λ 8πGρ = g ab ( ab g ab) 4Λ = g ab ab gab g ab 4Λ = 4 4Λ = 4Λ Λ = (8πGρ + ) 4 To find we wok in the non-coodinate basis The Basis one foms 5 (McMahon, 6, p. 36), final exam 4, the answe to Final Exam quiz 4 is (a) g ab T ab = ρ ω ω ( + e x e x + e x e x e x e x ) = ρ( + + ) = ρ Page 79

80 Susan Lasen Tuesday, Febuay 3, 5 ω = ω (dt + ex dz) dt = ωω ωω z ω x = dx ω dx = ωωx ω y = dy ω dy = ωωy ω z = ω ex dz dz = ωe x ω z η ij = { } Catan s Fist Stuctue equation: dω a a = b ω b (5.9) dω = d ω (dt + ex dz) = ex dx dz ω = ex ω ( ωωx ) (ωe x ω z ) = ωω x ω z = ωω z ω x + ωω x ω z = x ω x y ω y z ω z dω x dω y dω z = d ( ω ex dz) = ω ex dx dz = ω ex ( ωω x ) (ωe x ω z ) = ωω x ω z Summaizing the cuvatue one foms in a matix: ωω z ωω x a = ωω z ωω z b { ωω x ωω z } Whee a efes to column and b to ow. The cuvatue two foms and the iemann tenso: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d Page 8

81 Susan Lasen Tuesday, Febuay 3, 5 x x : d x x = y y : d y y = y + x x + y y + z z = d(ωω z ) = d (ω ω ex dz) = e x dx dz = ex ( ωω x ) ωe x ω z = ω ω x ω z x x x x = + x + y + z = ( ωω z ) ( ωω x ) = ω ω x ω z y + x x x y + y z z : d = d( ωω x ) = d ( ω ω dx) x z z z z = x : x d x x x x = x x x x : d y x y y x y y y y : d y y = y y y y z x y z + x x + x y + x x z x x x y z + y x + y y + y y y y z y x y y y + z z + z x + z y + z z z z z z x z y x = z z z z x : d x z x z x = d( ωω z ) = d ( ex dx dz) = ex ( ωω x ) (ωe x ω z ) = ω ω x ω z z z x z y = x + x x + y = ( ωω x ) (ωω z ) = ω ω x ω z = ω ω x ω z x z + z z x z = x Summaized in a matix: a ω = b { ω x ω z } ω ω x ω z Whee a efes to column and b to ow x Now we can see that the nonzeo elements of the iemann tenso in the non-coodinate basis ae z x z ω = The icci tenso, the nonzeo elements: Page 8

82 Susan Lasen Tuesday, Febuay 3, 5 a b x x z z = a b = x x = z z z = x z x x = z x z (4.46) x = z x z = ω = ω Summaized in a matix: ω a b = { } ω Whee a efes to column and b to ow The icci scala: = η a b a b (4.47) = η + η x x x x + η y y y y + η z z z z = x x z z = ω Now we can find Λ Λ = (8πGρ + ) 4 = 4 (8πGρ + ω ) If we use geometized units 8 i.e. 8πG = we get Λ = 4 (ρ + ω ) The fist tem ρ epesents the matte density of a homogeneous distibution of swiling dust paticles, 4 and the second tem Λ = ω is associated with a nonzeo cosmological constant. 8 Null Tetads and the Petov Classification 8. 9Constuct a null tetad fo the flat space Minkowski metic The line element: ds = dt d dθ sin θ dφ The metic tenso: g ab = { } sin θ and its invese: g ab = { sin θ} (McMahon, 6, p. 86), example 9-3 and Page 8

83 Susan Lasen Tuesday, Febuay 3, 5 The basis one foms ω ω ω ω = dt = d = dθ = sin θ dφ The null tetad Now we can use the basis one-foms to constuct a null tetad l n ( ) = ω dt + d m ( ω ) i ω = ( dt d dθ + i sin θ dφ ) (9.) m i ( ω ) dθ i sin θ dφ Witten in tems of the coodinate basis l a = (,,, ) n a = (,,, ) m a = (,,, i sin θ) m a = (,,, i sin θ) Next we use the metic to ise the indices l t l = g at l a = g tt l t = = = g a l a = g l = ( ) = l θ = l φ n t n = g at n a = g tt n t = = = g a n a = g n = ( ) ( ) = n θ = n φ m t = m m θ = g aθ m a = g θθ m θ = ( ) = m φ = g aφ m a = g φφ i sin θ m φ = ( sin ) θ = i sin θ Collecting the esults l a = (,,, ) la = (,,, ) n a = (,,, ) na = (,,, ) m a = (,,, i sin θ) ma = (,,, i sin θ ) Page 83

84 Susan Lasen Tuesday, Febuay 3, 5 m a = (,,, i sin θ) ma = (,,, i sin θ ) 8. The Binkmann metic (Plane gavitational waves) The line element: ds = H(u, x, y)du + dudv dx dy H The metic tenso: g ab = { } and its invese: g ab = { H } The basis one foms Finding the basis one foms is not so obvious, we wite: ds = H(u, x, y)du + dudv dx dy = (ω u ) (ω v ) (ω x ) (ω y ) du[h(u, x, y)du + dv] dx dy = (ω u + ω v )(ω u ω v ) (ω x ) (ω y ) ω u + ω v = du ω u ω v = Hdu + dv ω x = dx = dy ω y ω u = (H + )du + dv du = ωu + ω v ω v = ( H)du dv dv = ( H)ωu ( + H)ωv ω x = dx dx = ω x ω y = dy dy = ω y η ij = { } The othonomal null tetad Now we can use the basis one-foms to constuct a othonomal null tetad l n ( ) = ω u ω u + ω v du m ( ) ( ω v i ω y ) x = ( ω u ω v ω x + iω y ) = ( Hdu + dv ) (9.) dx + idy m i ω ω x iω dx idy Witten in tems of the coodinate basis l a = (,,, ) n a = (H,,, ) m a = (,,, i) m a = (,,, i) (McMahon, 6, p. 95), example 9-5. The answe to quiz 9-3 is (a) and to quiz 9-4 is (d) Page 84

85 Susan Lasen Tuesday, Febuay 3, 5 Next we use the metic to ise the indices l u = g au l a = g vu l v = l v = g av l a = g uv l u + g vv l v = ( ) + ( H) = l x = l y n u n v = g au n a = g vu n v = ( ) = = g av n a = g uv n u + g vv n v = ( H) + ( H) ( ) = H n x = n y m u = m v m x = g ax m a = g xx m x = ( ) = m y = g ay m a = g yy m y = ( ) i = i Collecting the esults l a = (,,, ) la = (,,, ) n a = (H,,, ) na = (, H,, ) m a = (,,, i) ma = (,,, i) m a = (,,, i) m a = (,,, i) The non-zeo Chistoffel symbols abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) uuu = H u uux = uxu = H x xuu = H x H uuy = uyu = y yuu = H y v uu = g vu uuu = H u v v ux = xu = g vu uux = H x x uu = g xx xuu = H x v v uy = yu = g vu uuy = H y y uu = g yy yuu = H y The spin coefficients calculated fom the null tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) (9.5) Page 85

86 Susan Lasen Tuesday, Febuay 3, 5 ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m al b = v n a m al v = v n x m xl v v n y m ylv c = ( v n x xvn c )m xl v c ( v n y yvn c )m yl v ν = b n a m an b = u n a m an u v n a m anv = u n x m xn u u n y m yn u v n x m xn v v n y m ynv c = ( u n x xun c )m xn u c ( u n y yun c )m yn u c ( v n x xv v = xun v m xn u v + yun v m ynu v = ( xum x v + yum y)n v n u = ( H n c )m xn v c ( v n y yvn c )m ynv x ( ) + H y i ) = H H ( + i x y ) λ = b n a m am b = x n a m am x y y n a m am = x n x m xm x x n y m ym x y n x m xm y y y n y m ym c = ( x n x xxn c )m xm x c ( x n y xyn c )m ym x c y ( y n x yxn c )m xm c y ( y n y yyn c )m ym μ = b n a m am b κ = b l a m a l b = v l a m a l v = v l x m x l v + v l y m y l v c = ( v l x vxl c )m x l v c + ( v l y vyl c )m y l v τ = b l a m a n b = u l a m a n u + v l a m a n v = u l x m x n u + v l x m x n v + u l y m y n u + v l y m y n v c = ( u l x uxl c )m x n u c + ( v l x vxl c )m x n v c + ( u l y uyl c )m y n u c + ( v l y vyl c ) v l y m y n v ρ = b l a m a m b = x l a m a m x + y l a m a m y = x l x m x m x + y l x m x m y + x l y m y m x + y l y m y m y = ( x l x c xxl c )m x m x c + ( y l x yxl c )m x m y c + ( x l y xyl c )m y m x c + ( y l y yyl c )m y m y σ = b l a m a m b ε = ( bl a n a l b b m a m al b ) = ( vl a n a l v v m a m al v ) = ( vl u n u l v v m x m xl v ) + ( vl v n v l v v m y m yl v ) = (( c vl u vul c )n u l v c ( v m x vxm c )m xl v ) + (( c vl v vvl c )n v l v c ( v m y vym c )m yl v ) γ = ( bl a n a n b b m a m an b ) = ( ul a n a n u u m a m an u ) + ( vl a n a n v v m a m an v ) = ( ul u n u n u u m x m xn u ) + ( vl u n u n v v m x m xn v ) + ( ul v n v n u u m y m yn u ) + ( vl v n v n v v m y m yn v ) = (( c ul u uul c )n u n u c ( u m x uxm c )m xn u ) Page 86

87 Susan Lasen Tuesday, Febuay 3, 5 α + (( c vl u vul c )n u n v c ( v m x vxm c )m xn v ) + (( c ul v uvl c )n v n u c ( u m y uym c )m yn u ) + (( c vl v vvl c )n v n v c ( v m y vym c )m yn v ) = ( bl a n a m b b m a m am b) = ( xl a n a m x x m a m am x) + ( yl a n a m y y m a m am y) = ( xl u n u m x x m x m xm x) + ( yl u n u m y y m x m xm y) + ( xl v n v m x x m y m ym x) + ( yl v n v m y y m y m ym y) = (( c xl u xu + (( c yl u yu + (( c xl v xv + (( c yl v yv l c )n u m x c ( x m x xxm c )m xm x) l c )n u m y c ( y m x yxm c )m xm y) l c )n v m x c ( x m y xym c )m ym x) β = ( bl a n a m b b m a m am b ) l c )n v m y c ( y m y yym c )m ym y) Collecting the esults π κ ε H ν + i ) τ γ = H ( x y λ ρ α μ σ β The Weyl Scalas and Petov classification Ψ = Dσ δκ σ(ρ + ρ ) σ(3ε ε ) + κ(π π + α + 3β) (3.) Ψ = Dβ δε σ(α + π) β(ρ ε ) + κ(μ + γ) + ε(α π ) (3.3) Ψ = δ τ Δρ ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Λ (3.4) Ψ 3 = δ γ Δα + ν(ρ + ε) λ(τ + β) + α(γ μ ) + γ(β τ ) (3.5) Ψ 4 = δ ν Δλ + λ(μ + μ ) λ(3γ γ ) + ν(3α + β + π τ ) (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) We see that Ψ = Ψ = Ψ = Ψ 3 and Ψ 4 = δ ν = m a a ν = m a a ( H H ( + i x y )) = [m x x ( H H + i x y ) + m y y ( H H + i x y )] Page 87

88 Susan Lasen Tuesday, Febuay 3, 5 = [( ) ( H x + i H x y ) + (i ) ( H x y + i H y )] = H [ x H y i H x y ] Ψ 4 : This is a Petov type N, which means thee is a single pincipal null diection of multiplicity 4. This coesponds to tansvese gavity waves. The icci tenso Φ = δν Δμ μ λλ μ(γ + γ ) + ν π ν(τ 3β α ) = δν = m a a ν (9.4) = m a a ( H H ( + i x y )) = m x x ( H H ( + i x y )) + my y ( H H ( + i x y )) = ( ) ( ) ( H x + i H x y ) + ( i ) ( ) ( H x y + i H y ) = [ H x + H y ] Φ = abn a n b = ubn u n b vbn v n b (9.) = uun u n u vun v n u uvn u n v vvn v n v = uu uv ( H) vv ( H) ( H) = uu + H uv H 4 vv = [ H x + H y ] uu = H [ x + H y ] uv = vv The Weyl tenso calculated fom the null tetad found in example 9-5. This calculation show that the spin coefficients and the Weyl scalas depend on the chosen null tetad, and the icci tenso does not (of couse). The null tetad (9.6), (9.7), (9.8) l a = (,,, ) l a = (,,, ) n a = ( H,,, ) na = (, H,, ) m a = (,,, i) ma = (,,, i) m a = (,,, i) m a = (,,, i) Accoding to the Catan calculation futhe below the sign is wong Page 88

89 Susan Lasen Tuesday, Febuay 3, 5 ν = b n a m an b = u n a m an u v n a m anv = u n x m xn u u n y m yn u v n x m xn v v n y m ynv c = ( u n x xun c )m xn u c ( u n y yun c )m yn u c ( v n x xvn c )m xnv c ( v n y yvn c )m ynv v = xun v m xn u v + yun v m yn u v = ( xum x v + yum y)n v n u = ( H x ( ) + H y ( i )) = ( H H + i x y ) (9.3) Ψ 4 = δ ν = m a a ν = m a a ( ( H H + i x y )) = ( ) [m x x ( H H + i x y ) + m y y ( H H + i x y )] = ( ) [( ) H ( x + i H x y ) + ( i ) ( H x y + i H y )] = 4 [ H x H y + i H x y ] (9.3) Φ = δν = m a a ν = m a a ( ( H x = m x x ( ( H H + i x y )) + my y ( ( H H + i x y )) H + i )) (9.4) y = ( ) [( ) ( H x + i H x y ) + (i ) ( H x y + i H y )] = 4 [ H x + H y ] (9.3) Φ = abn a n b (9.) n a = (, H,, ) = uun u n u vun v n u uvn u n v vvn v n v = uu vu ( H) vv ( H) = uu + H vu H 8 vv = H 4 [ x + H y ] uu = H [ x + H y ] uv = vv Finding the icci tenso of the Binkmann metic using Catan s stuctue equation Catan s Fist Stuctue equation and the calculation of the cuvatue one-foms Accoding to the Catan calculation futhe below the sign is wong Page 89

90 Susan Lasen Tuesday, Febuay 3, 5 dω a a = b ω b (5.9) a b a = b ω (5.) dω u dω v dω x dω y = d ( (H(u, x, y) + )du + dv) = ( H x = ( H x ωx (ω u + ω v ) + H y ωy (ω u + ω v )) = d ( ( H)du dv) = ( H x H dx du + dy du) y H dx du + dy du) y = ( H x ωx (ω u + ω v ) + H y ωy (ω u + ω v )) a b The cuvatue one-foms summaized in a matix: H (ωu + ω v )(A) x H (ωu + ω v )(A) x = H (ωu + ω v )(A) H (ωu + ω v )( A) x x H (ωu + ω v )(B) H (ωu + ω v )( B) { y y Whee a efes to column and b to ow and A and B will be used late, to make the calculations easie H (ωu + ω v )(B) y H (ωu + ω v )(B) y } The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) Fist we will calculate A A = H (ωu + ω v ) H (ωu + ω v ) x x = ( H x ) (ω u ω u + ω u ω v + ω v ω u + ω v ω v ) = ( H x ) (ω u ω v + ω v ω u ) = ( H x ) (ω u ω v ω u ω v ) B B = H (ωu + ω v ) H (ωu + ω v ) y y da = d ( H (ωu + ω v )) = d ( H(u, x, y) du) x x Page 9

91 Susan Lasen Tuesday, Febuay 3, 5 = H(u, x, y) x dx du + H(u, x, y) dy du x y = H(u, x, y) x ω x (ω u + ω v ) + H(u, x, y) ω y (ω u + ω v ) x y db = d ( H (ωu + ω v )) = d ( H(u, x, y) du) = H y y x y dx du + H dy du y = H ωx (ω u + ω v ) + H ωy x y y (ω u + ω v ) Now we ae eady to calculate the cuvatue two-foms u u u : d u u u u u v u x u y u x u = u u + v u + x u + y u = x u = A A + B B u u u u = d u + u v v u : d u v v u v v v x v y v x u = u u + v u + x u + y u = x u = ( A) (A) + ( B) (B) v u x x u : d u = da = H ωx x (ω u + ω v ) + H ωy (ω u + ω v ) x y x x u x v x x x y u = u u + v u + x u + y u x u = H ωx x (ω u + ω v ) + H ωy (ω u + ω v ) x y y y u : d u = db = H ωx (ω u + ω v ) + H ωy x y y (ω u + ω v ) y y u y v y x y y u = u u + v u + x u + y u y u = H ωx (ω u + ω v ) + H ωy x y y (ω u + ω v ) v v v d v v v u v v v x v y v x v = u v + v v + x v + y v = x v = (A) (A) (B) (B) v v x x v : d v x v x v y y v : d v y = u v y v x x : d x x x = da = H ωx x (ω u + ω v ) + H ωy (ω u + ω v ) x y x u x v x x x y = u v + v v + x v + y v = H ωx x (ω u + ω v ) + H ωy (ω u + ω v ) x y = db = H ωx (ω u + ω v ) + H ωy x y y (ω u + ω v ) y u y v y x y y v + v v + x v + y v = H ωx (ω u + ω v ) + H ωy x y y (ω u + ω v ) x x = u x x u + v x x v x x + x x x x + y y x u y + y u v + y v + y y u y v = (A) (A) (A) (A) Page 9

92 Susan Lasen Tuesday, Febuay 3, 5 y y x : d x y y x = u y x y y y u x + v v x y + x x x y + y y x = (B) (A) (B) (A) a b Summaized in a matix: H ωx x (ω u + ω v ) + H ωy (ω u + ω v ) H ωx (ω u + ω v ) + H ωy x y x y y (ω u + ω v ) = H ωx x (ω u + ω v ) + H ωy (ω u + ω v ) H ωx (ω u + ω v ) + H ωy x y x y y (ω u + ω v ) S AS { S AS } Now we can wite down the independent elements of the iemann tenso in the noncoodinate basis: x u x u (C) = H y x u y u (E) = H y x u x v (C) = H y x u y v (E) = H y x u y u (D) = H y v y v (E) = H x y y x u y v (D) = H x y x v x v (C) = H x x v y u (D) = H x y x v y v (D) = H x y Whee C,D and E will be used late, to make the calculations easie The icci tenso: a b u u v u = a b = u u = v u x u y u v v = v v x v y v u = u u u u = v u u u = v u v v + u v u v + v v u v + v v v x + u x u x + v x u x + v x v y + u y u y + v y u y + v y v (4.46) = C + E = H x + H y = C + E = H x + H y = C + E = H x + H y Page 9

93 Susan Lasen Tuesday, Febuay 3, 5 x x y x y y = x x = y x = y y u = x u x u = y u x u = y u y v + x v x v + y v x v + y v y x + x x x x + y x x x + y x y y + x y x y + y y x y + y y y = C + C = D + D = E + E Summaized in a matix: H x + H H y x + H y a b = H x + H H y x + H y { } Whee a efes to column and b to ow The icci tenso in the coodinate basis: The tansfomation: ab = Λ aλ d b d uu uv = Λ uλ d u d = Λ u uλ d u u d + Λ v uλ d u v d = Λ u uλ u u u u + Λ u uλ v u u v + Λ v u u v u + Λ u Λ u v Λ v u v v = ( (H + )) (C + E) + (H + ) ( H)(C + E) + ( H) (H + )(C + E) + ( ( H)) (C + E) = (C + E) = H x + H y = Λ uλ d v d = Λ u uλ d v u d + Λ v uλ d v v d = Λ u uλ u v u u + Λ u uλ v v u v + Λ v u u v u + Λ u Λ v v Λ v v v v = (H + ) (C + E) + (H + )( )(C + E) + ( H) (C + E) + ( H)( )(C + E) vv = Λ vλ d v d = Λ u vλ d v u d + Λ v vλ d v v d = Λ u vλ u v u u + Λ u vλ v v u v + Λ v vλ u v v u + Λ v vλ v v v v = (C + E) + ( )(C + E) + ( )(C + E) + ( )( )(C + E) xx = yy = xy = g ab ab uu Summaized in a matix: Page 93

94 Susan Lasen Tuesday, Febuay 3, 5 H x + H y ab { } Whee a efes to column and b to ow G uu = H x + H y 9 The Schwazschild Solution 9. 3The iemann and icci tenso of the geneal Schwazschild metic The line element: ds = e ν() dt e λ() d dθ sin θ dφ The Basis one foms ω = e ν() dt dt = e ν() ω ω = e λ() d d = e λ() ω ω ω = dθ dθ = ω = sin θ dφ dφ = sin θ ω η ij = { } a Catan s Fist Stuctue equation and the calculation of the icci otation coefficients b : dω a a = b ω b (5.9) a b a = b ω (5.) dω = d(e ν() dt) = ν e ν() d dt = ν e λ() ω ω dω = d(e λ() d) dω dω = d(dθ) = d dθ = e λ() ω ω = d( sin θ dφ) = sin θ d dφ + cos θ dθ dφ = e λ() ω ω cot θ + ω ω Summaizing the cuvatue one foms in a matix: 3 (McMahon, 6, p. 4) Page 94

95 Susan Lasen Tuesday, Febuay 3, 5 a b = ν e λ() ω ν e λ() ω e λ() ω { e λ() ω cot θ ω Whee a efes to column and b to ow. e λ() ω e λ() ω cot θ ω } Now we can ead off the icci otation coefficients = ν e λ() = ν e λ() = e λ() = e λ() = e λ() cot θ = = e λ() = cot θ The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d : d = d(ν e λ() ω ) = d(ν e ν() λ() dt) = (ν + ν (ν λ ))e ν() λ() d dt = (ν + ν (ν λ ))e λ() ω ω = + + = (ν + ν (ν λ ))e λ() ω ω = : d = e λ() ω ν e λ() ω = ν e λ() ω ω = ν e λ() ω ω = + = ν e λ() ω ω = ν e λ() ω ω + + = = : d = d ( e λ() ω ) = d(e λ() dθ) = λ e λ() d dθ = λ e λ() ω ω = = λ e λ() ω ω Page 95

96 Susan Lasen Tuesday, Febuay 3, 5 : d = d ( e λ() ω ) = d(e λ() sin θ dφ) = λ e λ() sin θ d dφ + e λ() cos θ dθ dφ = λ e λ() ω ω + e λ() cot θ ω ω = + cot θ = ω e λ() ω : d = λ e λ() ω ω + + = cot θ = d ( ω ) = d(cos θ dφ) = sin θ dθ dφ = ω ω = = = e λ() ω ω = ( e λ() ) ω ω a b = Summaized in a matix: (ν + ν (ν λ ))e λ() ω ω ν S S AS e λ() ω ν ω e λ() ω ω λ e λ() ω λ ω e λ() ω ω ( e λ() ) ω ω { S AS AS } Now we can find the independent elements of the iemann tenso in the non-coodinate basis: = (ν + ν (ν λ ))e λ() = ν e λ() = λ e λ() = ν e λ() = λ e λ() = ( e λ() ) The icci tenso: a b = a b = = + + = (ν + ν (ν λ ) + ν ) e λ() = = = (ν + ν (ν λ ) λ ) e λ() + (4.46) = = Page 96

97 Susan Lasen Tuesday, Febuay 3, 5 = = = ν e λ() + λ e λ() + ( e λ() ) = = = ν e λ() + λ e λ() + ( e λ() ) = = a b = Summaized in a matix: { (ν + ν (ν λ ) + ν ) e λ (ν + ν (ν λ ) λ ) e λ ν e λ + λ e λ + ( e λ ) ν e λ + λ e λ + ( e λ ) } Whee a efes to column and b to ow 9. 4The iemann tenso of the Schwazschild metic Solving the vacuum equations we find ν = λ and e ν = m, eλ = ( m ) The line element: ds = ( m ) dt ( m ) d dθ sin θ dφ Now we can find the independent elements of the iemann tenso in the non-coodinate basis: Fist we calculate ν = d e ν d (eν ) = d e ν d ( m ) = e ν ( m ) = m m ν = d d ( d e ν d (eν )) = ν e ν d d (eν ) + d e ν d (eν ) = ν ν + d e ν d (m ) = ν ν m 3 e ν = (ν + ν (ν λ ))e λ() = (ν + ν ν )e ν() = ( ν ν m 3 e ν + ν ν ) e ν() = m 3 = ν e λ() = ν = λ e λ() = m 3 = ν e λ() = m 3 = λ e λ() = m 3 eν() = e ν ( m ) eν() = m 3 4 (McMahon, 6, p. 5) Page 97

98 Susan Lasen Tuesday, Febuay 3, 5 = ( e λ() ) = ( ( ( m eν() ) )) = = m 3 Collecting the esults = m 3 = m 3 = m 3 = m 3 = m 3 = m Calculation of the scala abcd abcd in the Schwazschild metic abcd abcd = = ( m 3 ) (m m m ) + ( ) ( ) + (m 3 ) (m m m ) + ( ) ( ) + ( m 3) ( m 3) + (m 3) (m 3) + (m 3) (m 3) + ( m 3) ( m 3) + ( m 3) (m 3) + ( m 3) ( m 3) + (m 3) (m 3) + ( m 3) ( m 3) + ( m 3) ( m 3) + (m 3) (m 3) + ( m 3) ( m 3) + (m 3) (m 3) + ( m 3) (m 3) + ( m 3) ( m 3) + (m 3) (m 3) + ( m 3) ( m 3) + ( m m ) ( 3 3 ) + (m 3 ) (m m m ) + ( ) ( ) + (m 3 ) (m 3 ) = 48m (.35) Geodesics in the Schwazschild Spacetime To find the geodesic we use the Eule-Lagange equation = d ds ( F x a ) F (.36) x a whee F = ( m ) t ( m ) θ sin θ φ x a = t: F t F t = ( m ) t 5 (McMahon, 6, p. 6), equation (.35) 6 (McMahon, 6, p. 6) Page 98

99 Susan Lasen Tuesday, Febuay 3, 5 d ds ( F ) t = 4m t + ( m ) t = 4m t + ( m ) t m = t + ( m) t (.37) x a = : F = m m + ( m) θ sin θ φ F = ( m ) d ds ( F ) = ( m ) + ( m ) m = ( m ) 4m + ( m) = ( m ) m + ( m) m + θ + sin θ φ m = ( m) m( m) + 3 ( m)θ ( m) sin θ φ (.38) x a = θ: F θ = cos θ sin θ φ F = θ θ d ds ( F ) = 4 θ θ θ = 4 θ θ + cos θ sin θ φ = θ + θ cos θ sin θ φ (.39) x a = φ: F φ F φ = sin θ φ d ds ( F ) φ = 4 sin θ φ 4 cos θ sin θ θ φ sin θ φ = 4 sin θ φ 4 cos θ sin θ θ φ sin θ φ = φ + φ + cot θ θ φ (.4) 9.5 7The meaning of the integation constant: The choice of m We can use the geodesic equations to justify the choice of m by investigating the geodesic equations in the classical limit i.e. m, dt d and v = c, whee v is the velocity and τ is the pope time. dτ dτ We want to investigate the case of a adially infalling paticle i.e. dθ and dφ. We also want to wok in SI-units so we have to substitute m by Gm. Also emembe that t = dt = dt d, = = d = c ds cdτ c ds cdτ v and = d = d = a c ds c dτ c, whee a is the paticle acceleation. We use equation (.38): 7 (McMahon, 6, p. 8) Page 99

100 Susan Lasen Tuesday, Febuay 3, 5 m = ( m) m( m) + t 3 ( m)θ ( m) sin θ φ Now befoe we cay on with the physics we also have to be sue that each tem in this equation has the same dimension. It tuns out that they don t and theefoe the thid tem has to be multiplied by c, in which case each tem gets the dimension 8. length m = ( m) m( m) + 3 c t ( m)θ ( m) sin θ φ m = ( m) m( m) + 3 c t dθ, dφ = m + m c t m = a c m v = a Gm (v c ) a = Gm c + m = c (a m v + mc ) t = c, = v c, = a c + Gm Multiplying with M on both sides we get pecisely the Newtonian gavitational law F = Ma = GMm Gm m c v c 9.6 9Time Delay To descibe the time delay of a light ay outside a massive body, like the Sun we can use the Schwazschild metic. We can choose to wok in the plane with θ = π, and togethe with ds (light ays ae placed on the cone) we can ewite the Schwazschild line element: = ( m ) dt ( m ) d dφ Now we would like to descibe the time delay of the light ay as a function of the distance, fom the massive body, and theefoe we have to get id of the φ s. It s not that difficult. If we use pola coodinates the least distance,, the light ay passes the massive body is: = sin φ φ = sin dφ d ( dφ d ) = d d (sin ) = 3 = 4 ( ) ( ) Now we can begin ewiting the Schwazschild line element and solve the diffeential equation = ( m ) dt ( m ) d dφ 8 This actually oiginates fom the line element of the Schwazschild metic itself, because in ode to get the same dimension of each tem, the fist tem has to be multiplied by c : ds = ( m ) c dt ( m ) d dθ sin θ dφ 9 (McMahon, 6, p. 9) 3 d dx sin u = (3.) (Spiegel, 99) du u dx Page

101 Susan Lasen Tuesday, Febuay 3, 5 ( m ) ( dt d ) ( dt d ) dt = = + ( m ) (dφ d ) = + ( m ) 4 ( ) = + ( m ) ( ) ( ( ) = ) + ( m ) ( ) ( ) = ( m = 3 ) ( ( ) ) ( m ) d m 3 ( m ) ( ) 3 3 d ( ( + m ) ) ( m 3 ) d ( ( + m ) m 3 ) d m 3 ( ( + m ) ) ( m 3 ) ( ) To get the total time delay between points (, ) we have to integate fom to and fom to t ( + m m dt = 3 ) d ( + m m t ( + 3 ) d ) ( ) = + = + ( d + ) ( d + ) d + m d + m t t = 33 m ( d ) m ( d ) d m d m m 3 ( ) d m 3 ( ) d d + 34 m [ln ( + ) ln ( + )] d 3 ε + ε fo ε (.8) (Spiegel, 99) 3 ε ε fo ε (.) (Spiegel, 99) 33 xdx = x a x a (4.) (Spiegel, 99) 34 dx = ln(x + x a x a ) (4.) (Spiegel, 99) Page

102 Susan Lasen Tuesday, Febuay 3, 5 35 m [ ] + +m [ln ( + ) ln ( + )] m [ ] = + +m [ ln ( ( + )( + ) ) ] We can use this fomula to calculate the time delay fom e.g. Venus to Eath. The fist tem is the odinay flat space distance, and the delay is chaacteized by the emaining tems. We have = e, = v. emembe to get the ight unit we must multiply by G c 3. t delay = mg c 3 [ ln (( v + v )( e + e ) ) v v e ] e Use the geodesic equations to find the Chistoffel symbols fo the geneal Schwazschild metic. To find the geodesic we use = d ds ( K x a ) K (4.36) x a whee K = g abx a x b = eν() t eλ() θ sin θ φ (4.35) x a = t: K t K = e t t d ds ( K dν ν() ) = e t d t + e ν() t dν ν() = e d t + e ν() t = t + dν d t x a = : K K d ds ( K dν ν() = e t dλ λ() e d d θ sin θ φ = e λ() ) = e = e λ() dλ d e λ() dλ λ() d e λ() dν ν() e t + θ + sin θ φ d 35 dx x a = x x a a x (4.4) (Spiegel, 99) 36 (McMahon, 6, p. 3) quiz -. Answe: - is (a) Page

103 Susan Lasen Tuesday, Febuay 3, 5 = + dλ d + e x a = θ: K θ = cos θ sin θ φ K θ = θ d ds ( K ) θ = θ θ = θ θ + cos θ sin θ φ = θ + θ cos θ sin θ φ (ν() λ()) dν d t e λ() θ sin θ e λ() φ x a = φ: K φ K φ = sin θ φ d ds ( K ) φ = sin θ φ cos θ sin θ θ φ sin θ φ = sin θ φ cos θ sin θ θ φ sin θ φ = φ + φ + cot θ θ φ Collecting the esults = t + dν d t = + dλ d dν (ν() λ()) + e t e λ() θ sin θ e λ() φ d = θ + θ cos θ sin θ φ = φ + φ + cot θ θ φ Now we can find the Chistoffel symbols fom the equation d x a ds + a dxb dx c (4.33) bc ds ds t t t = t = dν d = dλ d tt = e d = e λ() = sin θ e λ() θθ φφ (ν() λ()) dν θ θ θ φφ θ = θ φ φ = φ = cos θ sin θ θφ φ = φ = φ = φθ = cot θ Page 3

104 Susan Lasen Tuesday, Febuay 3, The icci tenso fo the geneal time dependent Schwazschild metic. The line element: ds = e ν(t,) dt e λ(t,) d dθ sin θ dφ The Basis one foms ω = e ν(t,) dt dt = e ν(t,) ω ω = e λ(t,) d d = e λ(t,) ω ω ω = dθ dθ = ω = sin θ dφ dφ = sin θ ω η ij = { } a Catan s Fist Stuctue equation and the calculation of the Catan stuctue coefficients b : dω a a = b ω b (5.9) a b a = b ω (5.) dω dω dω dω = d(e ν(t,) dt) = ν e ν(t,) d dt = ν e λ(t,) ω ω = d(e λ(t,) d) = λ e λ(t,) dt d = λ e ν(t,) ω ω = d(dθ) = d dθ = e λ(,t) ω ω = d( sin θ dφ) = sin θ d dφ + cos θ dθ dφ = e λ(,t) ω ω cot θ + ω ω In this case we have to be paticula caeful in eading off the cuvatue one foms. The cuvatue onefoms ae antisymmetic in the sense that: a b = b a (5.). This means that = η = η = η η =. But in the fome calculation we found that = ν e λ(t,) ω and = λ e ν(t,) ω, which means that we in ode to fulfill the antisymmetic popeties need to equie that = = λ e ν(t,) ω + ν e λ(t,) ω, because = ν e λ(t,) ω + (something that makes antisymmetic), and = λ e ν(t,) ω + (something that makes antisymmetic). a b Summaizing the cuvatue one foms in a matix: λ e ν(t,) ω + ν e λ(t,) ω λ e ν(t,) ω + ν e λ(t,) ω = e λ(t,) ω { e λ(t,) ω Whee a efes to column and b to ow. e λ(t,) ω e λ(t,) ω cot θ ω cot θ ω } 37 (McMahon, 6, p. 3), quiz -. Answe to quiz -: t = (dλ dt ) Page 4

105 Susan Lasen Tuesday, Febuay 3, 5 The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d = d(λ e ν(t,) ω + ν e λ(t,) ω ) = d(λ e λ(t,) ν(t,) d + ν e ν(t,) λ(t,) dt) = (λ + λ (λ ν )e λ(t,) ν(t,) dt d + (ν + ν (ν λ ))e ν(t,) λ(t,) d dt = [ (λ + λ (λ ν )) e ν(t,) + (ν + ν (ν λ ))e λ(t,) ] ω ω = : d = : d = : d = [ (λ + λ (λ ν )) e ν(t,) + (ν + ν (ν λ ))e λ(t,) ] ω ω = e λ(t,) ω (λ e ν(t,) ω + ν e λ(t,) ω ) = λ e λ(t,) ν(t,) ω ω + ν e λ(t,) ω ω = e λ(t,) ω (λ e ν(t,) ω + ν e λ(t,) ω ) = λ e λ(t,) ν(t,) ω ω + ν e λ(t,) ω ω = = = d ( e λ(t,) ω ) = d(e λ(t,) dθ) = λ e λ(t,) dt dθ λ e λ(t,) d dθ = λ e ν(t,) λ(t,) ω ω + λ e λ(t,) ω ω = = λ e ν(t,) λ(t,) ω ω + λ e λ(t,) ω ω : d = d ( e λ(t,) ω ) = d(e λ(t,) sin θ dφ) = λ e λ(t,) sin θ dt dφ λ e λ(t,) sin θ d dφ + e λ(t,) cos θ dθ dφ = λ e ν(t,) λ(t,) ω ω + λ e λ(t,) ω ω + e λ(t,) cot θ ω ω = = cot θ = ω e λ() ω = λ e ν(t,) λ(t,) ω ω + λ e λ(t,) ω ω cot θ : d = d ( ω ) = d(cos θ dφ) = sin θ dθ dφ = ω ω = = Page 5

106 Susan Lasen Tuesday, Febuay 3, 5 = e λ(t,) ω ω = ( e λ(t,) ) ω ω a = b Summaized in a matix: λ e λ(t,) ν(t,) ω ω + ν e λ(t,) ω ω λ [ (λ + λ (λ ν )) e ν(t,) + (ν + ν (ν λ ))e λ(t,) ] ω ω e λ(t,) ν(t,) ω ω + ν e λ(t,) ω ω λ S e ν(t,) λ(t,) ω ω + λ λ e λ(t,) ω ω e ν(t,) λ(t,) ω ω + λ e λ(t,) ω ω ( e λ(t,) ) S AS ω ω { S AS AS Now we can find the independent elements of the iemann tenso in the coodinate basis: = (λ + λ (λ ν )) e ν(t,) + (ν + ν (ν λ ))e λ(t,) = ν e λ(t,) = λ e λ(t,) ν(t,) = λ e λ(t,) = ν e λ(t,) = λ e λ(t,) ν(t,) = λ e λ(t,) = ( e λ(t,) ) The icci tenso: a b = a b = = (4.46) = + + = (λ + λ (λ ν )) e ν(t,) + (ν + ν (ν λ ) + ν ) e λ(t,) = = = = = = + + = λ = (λ + λ (λ ν )) e ν(t,) (ν + ν (ν λ ) λ ) e λ(t,) = = = ( ν + λ ) e λ(t,) + ( e λ(t,) ) = = + + = ( ν + λ ) e λ(t,) + ( e λ(t,) ) + = = + + e λ(t,) ν(t,) Summaized in a matix: Page 6

107 Susan Lasen Tuesday, Febuay 3, 5 a b = { (λ + λ (λ ν )) e ν + (ν + ν (ν λ ) + ν λ e λ ν ) e λ Whee a efes to column and b to ow λ e λ ν (λ + λ (λ ν )) e ν (ν + ν (ν λ ) λ ) e λ ( ν + λ ) e λ + ( e λ ) ( ν + λ ) e λ + ( e λ The icci tenso in the coodinate basis: The tansfomation: ab = Λ aλ d b d tt t θθ d = Λ tλ t d = (Λ t) = e ν(t,) ( (λ + λ (λ ν )) e ν(t,) + (ν + ν (ν λ ) + ν ) e λ(t,) ) = (λ + λ (λ ν )) + (ν + ν (ν λ ) + ν ) eν(t,) λ(t,) = Λ Λ t d d = Λ = Λ Λ d d = (Λ ) Λ t = e λ(t,) e ν(t,) λ e λ(t,) ν(t,) = λ = e λ(t,) ((λ + λ (λ ν )) e ν(t,) (ν + ν (ν λ ) λ ) e λ(t,) ) = (λ + λ (λ ν )) e ν(t,)+λ(t,) (ν + ν (ν λ ) λ ) = Λ θλ d θ d = (Λ θ ) = (( ν + λ ) e λ(t,) + ( e λ(t,) ) ) = (( ν + λ ) )e λ(t,) + φφ = Λ φ Λ d φ d = (Λ φ ) = sin θ (( ν + λ ) e λ(t,) + ( e λ(t,) ) ) = ((( ν + λ ) )e λ(t,) + ) sin θ ab = Summaized in a matix: { (λ + λ (λ ν )) + (ν + ν (ν λ ) + ν ) eν λ Whee a efes to column and b to ow λ λ (λ + λ (λ ν )) e ν+λ (ν + ν (ν λ ) λ ) (( ν + λ ) )e λ + ((( ν + λ ) )e λ + ) sin θ } Page 7

108 Susan Lasen Tuesday, Febuay 3, The Schwazschild metic with nonzeo cosmological constant The icci otation coefficients and icci tenso fo the Schwazschild metic with nonzeo cosmological constant. The line element: ds = f()dt + f() d + dθ + sin θ dφ Whee f() = m 3 Λ Now we can compae with the line element of the Schwazschild metic with zeo cosmological constant, whee the pimes should not be mistaken fo the deivative d/d. ds = e ν( ) dt e λ( ) d dθ sin θ dφ And choose: e ν( ) dt = f()dt e λ( ) d = f() d dθ = dθ sin θ dφ = sin θ dφ Compaing the two metics we see: φ = φ, θ = θ, =, e ν( ) = f(), ν = λ, t = t Next we can use the fome calculations of the Schwazschild metic with zeo cosmological constant to find the icci otation coefficients and the icci tenso fo the Schwazschild metic with non-zeo cosmological constant. But fist we need to calculate dν( ) d = e ν( ) d d (eν( ) ) = e ν( ) = e ν( ) df() d d ν( ) d = d d [ df() f() df() d d f() d = d d (m 4m Λ) = 3 d dt ( f() d dt ) = e ν( ) d d ( f()) f() = df() f() d d ] = d d [ f() df() d ] = f () (df() d ) + d f() f() d = d d ( m 3 Λ ) = m 3 Λ = (f() + Λ ) 3 3 Λ = (df() d + Λ) The icci otation coefficients = = dν( ) d e λ( ) = df() f() d f() = df() = f() d = m 3 Λ3 m 3 Λ3 = 3m Λ 3 9 8m 3Λ 3 m 3 Λ m 3 Λ 38 (McMahon, 6, pp. 3-3), quiz -3 and -4. The answe to quiz -3 is (b) and the answe to quiz -4 is (a) Page 8

109 Susan Lasen Tuesday, Febuay 3, 5 = = = = e λ( ) = f() = m 3 Λ = cot θ cot θ = = The icci tenso = = [ d ν d + ( dν d ) ( dν dλ d ) ( d ) + dν d ] e λ( ) (.4) = [ d ν d + ( dν d ) + dν d ] eν( ) = [ f () (df() d ) d f() = [ f() d + d f() f() d + ( df() f() d ) + f() df() d ] f() = [ f() ( (df() d + Λ)) + = [ d ν d + ( dν d ) f() df() ] f() = Λ d + df() f() d ] f() ( dν dλ d ) ( d ) dλ d ] e λ( ) = = Λ (.5) = [ dν d + dλ d ] e λ( ) + e λ( ) = dν d e λ( ) + e λ( ) = df() f() d f() + f() = = ( (f() + Λ )) + f() = Λ df() d + f() = ((f() + Λ )) + f() (.6) Altenatively we could use the fomula calculated ealie on page 38: a b = η a b Λ valid in vacuum systems with a cosmological constant and positive signatue, fom which we immediately can see that = = = = Λ The geneal Schwazschild metic in vacuum with a cosmological constant: The icci scala The metic ds = e ν() dt e λ() d (dθ + sin θ dφ ) 39 (McMahon, 6, p. 77), quiz -. And the answe to quiz - is (c) Page 9

110 Susan Lasen Tuesday, Febuay 3, 5 In this case we can wite the Einstein equation in the local fame (non-coodinate basis) we name the cosmological constant 4 : = a b η a b + η a b (6.6) = η a b a b ηa b η a b + η a b η a b = = The geneal Schwazschild metic in vacuum with a cosmological constant: Integation constants We know that λ() = ln k ν() λ () = ν () and e λ() = k eν() We also need to find (e ν() ) = e ν() + ν ()e ν() ν () = (eν() ) e ν() As in quiz - we use the Einstein equation in the non-coodinate basis, but this time fo the coodinate = η + η (6.6) = + = + 4 = + = Ealie (p.) we calculated the icci tenso: = ν e λ() + λ e λ() + ( e λ() ) enaming = ν k eν() + eν() k = ( (eν() ) e ν() g() = e ν() ) eν() k = ( g () g() ) g() k + g() k 3 k + k = g () + eν() k 4 If you compae this to quiz -3 and -4 page 3-3 you can conclude that = Λ. The eason is that the metic in the two cases changes signatue, which implies that in the fist case = 4Λ, and in the second = 4 = 4Λ. You might also check the poofs on page (McMahon, 6, p. 77), quiz - and -3. The answes to quiz - is: e ν() = A + B + 3 k 3 and quiz -3 is B = k. Page

111 Susan Lasen Tuesday, Febuay 3, 5 We guess the solution (polynomials with exponents highe than 3 cannot contibute): g() = e ν() = A + B + C + D 3 g () = B + C + 3D k + k = B + C + 3D Now compaing the coefficients we find B = k C and we can conclude that D = 3 k e ν() = A + k + 3 k 3 e ν() = A + k + 3 k and the line element ds = e ν() dt e λ() d dθ sin θ dφ becomes ds = ( A + k + 3 k ) dt ( A + k + 3 k ) d dθ sin θ dφ If k = and this should be identical to the odinay Schwazschild vacuum metic, which means that A has to be equal to: A = m The geneal Schwazschild metic in vacuum with a cosmological constant: The spatial pat of the line element. The line element ds = e ν() dt e λ() d dθ sin θ dφ can in Gaussian nomal coodinates be witten as ds = dt a (t)dσ In this case we want to find the spatial pat of the line element dσ = g ij dx i dx j and to do that we will use the method outlined on page 6, whee the metic is found fom the iccitenso: ij = Kg jl g ij = K ij g = K = K (Λ ) = K (e λ() ) ( ) = K eλ() In the fome quiz we found that e λ() = k eν() 4 (McMahon, 6, p. 77), quiz -4 Page

112 Susan Lasen Tuesday, Febuay 3, 5 = k ( m + k + 3 k 3 ) = m k + 3 g = K m k + 3 g θθ = K θθ = K (Λ θ ) = K () ( ) = K g φφ = K φφ = K (Λ φ ) = K ( sin θ) ( ) = K sin θ dσ = d K m k + K dθ K sin θ dφ 3 whee we can omit the common facto = and finally get if we choose k = K d dσ = m + + dθ + sin θ dφ The effect of the cosmological constant ove the scale of the sola system To check the effect of the cosmological constant ove the scale of the sola system we will look at the line element of a adially moving (i.e. dφ, dθ ) light ay (ds ) in the sola system with a sun mass m in a univese with a cosmological constant Λ popotional to the size of the Univese. The line element: ds = 44 ( + 3 Λ ) dt d d dt = (±) + 3 Λ dt = d + 3 Λ = 3 Λ d 3 Λ Λ 43 (McMahon, 6, p. 77), Quiz -5, answe (c) 44 Notice: If we set Λ we get the usual Minkowski line element. Page

113 Susan Lasen Tuesday, Febuay 3, 5 = 45 [ 3 Λ Λ 3 tan ( 3 Λ )] = 3 Λ tan (( ) 3 Λ ) if ( ) Λ So the effect of the cosmological constant on a small scale is negligible The Petov type of the Schwazschild spacetime m The metic tenso: g ab = ( m ) { ( m ) sin θ} and its invese: g ab = ( m ) { sin θ} The basis one foms ω = m dt dt = m ω ω ω ω = m d d = dθ dθ = ω = m ω = sin θ dφ dφ = sin θ ω η ij = { } The othonomal null tetad = a +x a tan x (Spiegel, 99) eq.( 4.6) a 46 (McMahon, 6, p. 3), quiz dx Page 3

114 Susan Lasen Tuesday, Febuay 3, 5 Now we can use the basis one-foms to constuct a othonomal null tetad (9.) m dt + d l n ( m ) = ω ω + ω m ( ω ) i ω = ω ω ω = + iω i ( ω ) ( ω iω ) Witten in tems of the coodinate basis l a = ( m, m ( m m dt,, ) na = ( m, d m dθ + i sin θ dφ dθ i sin θ dφ ) m m a = (,,, i sin θ) m a = (,,, i sin θ) Next we use the metic to ise the indices,, ) l t = g tt l t = ( m ) m = m l = g l = ( m ) m l θ = l φ n t = g tt n t = ( m ) m = m = m n = g n = ( m ) m ) = m ( n θ = n φ m t = m m θ = g θθ m θ = ( ) = m φ = g φφ m φ = sin θ i sin θ = i sin θ Collecting the esults l a = ( m, m,, ) l a = ( m, m,, ) Page 4

115 Susan Lasen Tuesday, Febuay 3, 5 n a = ( m, m,, ) n a = ( m, m,, ) m a = (,,, i sin θ) ma = (,,, i sin θ ) m a = (,,, i sin θ) m a = (,,, i sin θ ) The spin coefficients calculated fom the null tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m alb = t n a m al t n a m al = t n θ m θl t n θ m θl t n φ m φl t n φ m φl c = ( t n θ tθn c )m θl t c ( n θ θn c )m θl c ( t n φ tφn c )m φlt c ( n φ φn c )m φl ν = b n a m anb λ b = b n a m am = θ n a m am θ φ φ n a m am = θ n θ m θm θ φ n θ m θm φ θ n φ m φm θ φ φ n φ m φm c = ( θ n θ θθn c )m θm θ c ( φ n θ φθn c )m θm φ c θ ( θ n φ θφn c )m φm c φ ( φ n φ φφn c )m φm = θθn m θm θ φ + φφn m φm = ( m)n ( ( )) ( m) sin θ n ( (i sin θ )) μ κ = b n a m amb = b l a m a l b = t l a m a l t + l a m a l = t l θ m θ l t + l θ m θ l + t l φ m φ l t + l φ m φ l c = ( t l θ tθl c )m θ l t c + ( l θ θ l c )m θ l c + ( t l φ tφ l c )m φ l t c + ( l φ φl c )m φ l Page 5

116 Susan Lasen Tuesday, Febuay 3, 5 τ ρ = b l a m a n b = b l a m a m b = θ l a m a m θ + φ l a m a m φ = θ l θ m θ m θ + φ l θ m θ m φ + θ l φ m φ m θ + φ l φ m φ m φ c = ( θ l θ θθl c )m θ m θ c + ( φ l θ φθl c )m θ m φ c + ( θ l φ θφl c )m φ m θ c + ( φ l φ φφl c )m φ m φ = θθl m θ m θ φφl m φ m φ = ( m ) ( ( )) m + ( m ) sin θ m ( ( i sin θ )) ( i sin θ ) σ = ( m ) = b l a m a m b = θθl m θ m θ φφl m φ m φ ε = ( bl a n a l b b m a m al b ) = ( tl a n a l t t m a m al t ) + ( l a n a l m a m al ) = ( tl t n t l t t m θ m θl t ) + ( l t n t l m θ m θl ) + ( tl n l t t m φ m φl t ) = (( tl t c tt = (( tt + ( l n l m φ m φl ) l c )n t l t c ( t m θ tθ + (( tl c t + (( c l l )n t l t ) + (( t l t t )m θl t ) + (( l t c tl c )n t l c ( m θ θ)m θl ) l c )n l t c ( t m φ tφ)m φl t ) l c )n l c ( m φ φ)m φl ) l t )n t l ( m θ θ θm θ )m θl ) + (( t tl t )n l t ) + (( l l )n l ( m φ φ φm φ ) m φl ) = 4 (( ttl ) m ) + 4 ( ( t l t t l t ) ( m θ θ θm θ ) m ) + 4 (( t tl t )) + 4 ( ( l l ) ( m ) + ( m φ φ φm φ ) m i sin θ ) Page 6

117 Susan Lasen Tuesday, Febuay 3, 5 = 5 m ( m ) ( ( m m ) ) + 5 ( ( ( m ) m m m ) ( () ) m ) + 5 ( m m m ) + 5 ( ( m + m m ( ) m ) ( m ) + ( (i sin θ) i sin θ) m i sin θ ) = 5 m ( m + m 5 ) ( m ( m m m + 5 ) ( m ) ) + 5 = m 3 m ( ( γ = ( bl a n a n b b m a m an b ) m ( m ) 3 + m = ( tl a n a n t t m a m an t ) + ( l a n a n m a m an ) ( m 3 ( m ) ) ) ) = ( tl t n t n t ) + ( l t n t n ) + ( tl n n t ) + ( l n n ) = ( tl t c tt = ( tt l c )n t n t + ( c l t t l )n t n t + ( t l t t l c )n t n + ( c tl t l t )n t n + ( t t l c )n n t + ( c l l c )n n l t )n n t + ( l l )n n = tt = 5 m = 3 m l (n t ) + l t n t n t tl t n t n + l (n ) l (n ) m m m m + 5 m m 3 m 5 m + 5 m m Page 7

118 Susan Lasen Tuesday, Febuay 3, 5 α = ( bl a n a m b b m a m am b) = ( θl a n a m θ θ m a m am θ) + ( φl a n a m φ φ m a m am φ) = ( θl t n t m θ θ m θ m θm θ) + ( φl t n t m φ φ m θ m θm φ) + ( θl n m θ θ m φ m φm θ) = (( c θl t θt + ( φl n m φ φ m φ m φm φ) l c )n t m θ ( θ m θ + (( c φl t φt + (( c θl θ + (( c φl φ c θθ m c )m θm θ) l c )n t m φ ( φ m θ l c )n m θ ( θ m φ l c )n m φ ( φ m φ c φθ c θφ c φφ m c )m θm φ) m c )m φm θ) m c )m φm φ) = φ φθm φ m θm φ + ( ( θm φ φ θφm φ ) m φm θ) + θ φ φφm θ m φm = 5 cot θ + ( ( θ(i sin θ) cot θ (i sin θ))m φm θ) + 5 cos θ sin θ sin θ = 3 cot θ β = ( bl a n a m b b m a m am b ) = φ φθm φ m θm φ + ( ( θm φ θφ = 5 φ m φ ) m φm θ ) + θ φφ cot θ + ( ( θ(i sin θ) cot θ (i sin θ))m φm θ ) 5 = 3 cot θ m θ m φmφ cot θ Collecting the esults π κ ε = ν τ γ = 3 m 3 m m m λ ρ = ( m ) α = 3 cot θ μ σ β = 3 cot θ The Weyl Scalas and Petov classification Ψ = Dσ δκ σ(ρ + ρ ) σ(3ε ε ) + κ(π π + α + 3β) (3.) Page 8

119 Susan Lasen Tuesday, Febuay 3, 5 Ψ = Dβ δε σ(α + π) β(ρ ε ) + κ(μ + γ) + ε(α π ) (3.3) Ψ = δ τ Δρ ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Λ (3.4) Ψ 3 = δ γ Δα + ν(ρ + ε) λ(τ + β) + α(γ μ ) + γ(β τ ) (3.5) Ψ 4 = δ ν Δλ + λ(μ + μ ) λ(3γ γ ) + ν(3α + β + π τ ) (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) Ψ Ψ = Dβ δε βρ = l a a β m a a ε βρ = l a a β m a a ε βρ = ( m ) ( 3 cot θ) ( 3 cot θ) ( ( m ) ) = 4 ( m ) cot θ [ ( ) + ] Ψ = Δρ + ργ = n a a ρ + ργ = n a a ρ + ργ = ( m ) ( ( m ) ) + ( ( m ) ) ( 3 m m ) Ψ 3 = ( m ) m 3 ( m = m 3 = δ γ Δα = m a a γ n a a α = ( m ) ( 3 cot θ) ( m ) m 3 ) = 47 4 cot θ m Ψ 4 Ψ : This is a Petov type D, which means thee ae two pincipal null diections. The Petov type D is associated with the gavitational field of a sta o a black hole. The two pincipal null diections coespond to ingoing and outgoing conguence of light ays. 47 This should be. Howeve, the esult is epoduced in FE- concening the eissne-nodstöm metic. Page 9

120 Susan Lasen Tuesday, Febuay 3, The deflection of a light ay in a Schwazschild metic with two diffeent masses The metic ds = ( m ) dt ( m ) d dθ sin θ dφ Copying the method pp 4-9 = ( m ) t ( m e = 49 ( m ) t l = 4 φ = ( m ) ) φ (.5 ) e ( m ) l (.53 ) = φ = l = lu if u = ( ) = = e ( m ) ( m ) l u ( m ) l u = e ( m u) ( m u) l u ( m u)l u Diffeentiating with espect to = m u ( m u) l u m u ( m u) ( m u) l u u u ( m u) ( m u) l + m u l u u u( m u)l = u l m ( m u) ( ( m u) u m ( m u) u u ( m u) ( m u) + m u u( m u)) = m ( m u)u m ( m u)u ( m u)( m u)u + m ( m u) u ( m u)( m u) u = (m m )u ( m u)( m u)u + m ( m u) u ( m u)( m u) u = (m m )u ( m u)( m u)u + (3m u )( m u) u In the limit whee mu = m = (m m )u u + u Fom which we can conclude, that the deflection is somehow dependent on (m m ) 9. 5 The non-zeo Weyl scalas of the eissne-nodstöm spacetime The eissne-nodstöm spacetime is a static solution to the Einstein-Maxwell field equations 5, which coesponds to the gavitational field of a chaged non-otating, spheically symmetic of mass m 5. The metic: ds = ( m + e ) dt ( m + e ) d dθ sin θ dφ 48 (McMahon, 6, p. 36) final exam 3 49 Killing vectos p. 5 (McMahon, 6, p. 35), final exam, If e the esults fom Quiz -5 ae epoduced Page

121 Susan Lasen Tuesday, Febuay 3, 5 The metic tenso: g ab = { ( m + e ) ( m + e ) ( m + e ) sin θ} and its invese: g ab = ( m + e ) { sin θ} The Chistoffel symbols To find the geodesic we use the Eule-Lagange equation = d ds ( F x a ) F (.36) x a whee F = ( m + e ) t ( m + e ) θ sin θ φ x a = t: F t F t = ( m + e ) t d ds ( F m e ) = 4 ( t 3 ) t + ( m + e ) t m e = 4 ( 3 ) t + ( m + e ) t = t + ( m + e ) x a = : F F ( = ( m e 3 ) t + ( m m e ) t 3 m e = ( 3 ) t + ( m + e = ( m + e ) + e ) ( m e 3 ) θ sin θ φ ) d ds ( F ) = ( m + e m e ) + ( + ) m e ( 3 ) θ sin θ φ ( m e 3 ) Page

122 Susan Lasen Tuesday, Febuay 3, 5 = ( m + e m e ) + 4 ( + ) = ( m + e m e ) + 4 ( + ) ( ( m m e ( 3 ) m e ) m e ( ) t + e m e ) ( 3 ) + θ + sin θ φ = ( m + e m e m e ) + ( + ) ( 3 + θ + sin θ φ = ( m + e m e ) ( 3 3 ) + ( m + e ( m + e ) θ ( m x a = θ: F θ = cos θ sin θ φ F = θ θ d ds ( F ) = 4 θ θ θ = 4 θ θ + cos θ sin θ φ 3 ) m e ( ) t e ) (m ) t 3 + e ) sin θ φ = θ + θ cos θ sin θ φ (.39) x a = φ: F φ F φ = sin θ φ d ds ( F ) φ = 4 sin θ φ 4 cos θ sin θ θ φ sin θ φ = 4 sin θ φ 4 cos θ sin θ θ φ sin θ φ = φ + φ + cot θ θ φ (.4) 3 Collecting the esults = t + ( m + e ) ( m e ) t 3 = ( m + e m e ) ( 3 ) + ( m + e e ) (m 3 ) t ( m + e ) θ ( m = θ + θ cos θ sin θ φ = φ + φ + cot θ θ φ + e ) sin θ φ Page

123 Susan Lasen Tuesday, Febuay 3, 5 We can now find the Chistoffel symbols: t t = ( m + e m e ) ( 3 ) = ( m + e m e ) ( 3 ) = ( m + e e ) (m 3 ) tt θθ φφ θ θ θ φφ φ φ φ θφ = ( m + e ) = ( m + e ) sin θ = = cos θ sin θ = = cot θ The basis one foms ω = ( m + e ) dt dt = ( m + e ) ω ω = ( m + e ) d d = ( m + e ) ω ω = dθ dθ = ω = sin θ dφ dφ = sin θ ω ω η ij = { } The othonomal null tetad Now we can use the basis one-foms to constuct a othonomal null tetad (9.) Page 3

124 Susan Lasen Tuesday, Febuay 3, 5 l n ( ) m m = ( ) i i = ω ω ω ( ω ) ( Witten in tems of the coodinate basis l a = ( ( m + e ) n a = ( ( m + e ) ω + ω = ω ω ω + iω ( ω iω ) ( m + e ) ( m + e ), ( m m, ( m a = (,,, i sin θ) m a = (,,, i sin θ) Next we use the metic to ise the indices l t = g tt l t = ( m + e ) ( m l = g l = ( m + e ) ( m l θ = l φ n t = g tt n t = ( m + e ) ( m dt + ( m dt ( m n = g n = ( m + e ) ( ( m + e ) d + e ) d dθ + i sin θ dφ dθ i sin θ dφ ) + e ),, ) + e ),, ) n θ = n φ m t = m m θ = g θθ m θ = ( ) = m φ = g φφ m φ = sin θ i sin θ = i sin θ + e ) = ( m + e ) = ( m + e ) = ( m + e ) + e ) + e ) + e ) ) = ( m + e ) Collecting the esults l a = ( ( m + e ) n a = ( ( m + e ) m, ( m, ( + e ),, ) l a = ( ( m + e ), + e ),, n a = ( ( m + e ), m ( + e ),, m ( + e ),, ) Page 4

125 Susan Lasen Tuesday, Febuay 3, 5 m a = (,,, i sin θ) ma = (,,, i sin θ ) m a = (,,, i sin θ) m a = (,,, i sin θ ) The spin coefficients calculated fom the null tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m alb = t n a m al t n a m al = t n θ m θl t n θ m θl t n φ m φl t n φ m φl c = ( t n θ tθn c )m θl t c ( n θ θn c )m θl c ( t n φ tφn c )m φlt c ( n φ φn c )m φl ν = b n a m an b λ b = b n a m am = θ n a m am θ φ φ n a m am = θ n θ m θm θ φ n θ m θm φ θ n φ m φm θ φ φ n φ m φm c = ( θ n θ θθn c )m θm θ c ( φ n θ φθn c )m θm φ c θ ( θ n φ θφn c )m φm c φ ( φ n φ φφn c )m φm = θθn m θm θ φ + φφn m φm = ( m + e ) n ( ( )) ( m + e ) sin θ n ( (i sin θ )) μ κ τ ρ = b n a m amb = b l a m a l b = t l a m a l t + l a m a l = t l θ m θ l t + l θ m θ l + t l φ m φ l t + l φ m φ l c = ( t l θ tθl c )m θ l t c + ( l θ θl c )m θ l c + ( t l φ tφ = b l a m a n b = b l a m a m b = θ l a m a m θ + φ l a m a m φ = θ l θ m θ m θ + φ l θ m θ m φ + θ l φ m φ m θ + φ l φ m φ m φ l c )m φ l t c + ( l φ φl c )m φ l Page 5

126 Susan Lasen Tuesday, Febuay 3, 5 c = ( θ l θ θθ = θθ l c )m θ m θ c + ( φ l θ φθ c + ( φ l φ φφl c )m φ m φ l m θ m θ φφl m φ m φ = ( m + e ) ( m l c )m θ m φ c + ( θ l φ θφl c )m φ m θ + e ) ( ( )) σ + ( m + e ) sin θ ( m = ( m + e ) = b l a m a m b = θθl m θ m θ φφl m φ m φ ε = ( bl a n a l b b m a m al b ) + e ) ( ( i sin θ )) ( i sin θ ) = ( tl a n a l t t m a m al t ) + ( l a n a l m a m al ) = ( tl t n t l t t m θ m θl t ) + ( l t n t l m θ m θl ) + ( tl n l t t m φ m φl t ) = (( tl t c tt = (( tt + ( l n l m φ m φl ) l c )n t l t c ( t m θ tθ + (( tl c t + (( c l l )n t l t ) + (( t l t t )m θl t ) + (( l t c tl c )n t l c ( m θ θ)m θl ) l c )n l t c ( t m φ tφ)m φl t ) l c )n l c ( m φ φ)m φl ) l t )n t l ( m θ θ θm θ )m θl ) + (( t tl t )n l t ) + (( l l )n l ( m φ φ φm φ ) m φl ) = (( ttl )n t l t ) + (( ( m + e ) t tl t ) n t l ( () θ θm θ ) m θl ) + (( t tl t )n l t ) + (( ( m + e ) l ) n l ( (i sin θ) φ φm φ ) m φl ) Page 6

127 Susan Lasen Tuesday, Febuay 3, 5 = (( ttl )n t l t ) + (( (m e 3 ) ( m + e ) t tl t ) n t l ( θ θm θ ) m θl ) + (( t tl t )n l t ) + (( (m e 3 ) ( ) ( m + e 3 ) l ) n l ( (i sin θ) φ φm φ ) m φl ) = (( ttl )n t l t ) + (( e (m 3 ) ( m + e ) ( m + e m e ) ( 3 ) ( m + e ) ) n t l ( ) m θl ) + (( t tl t )n l t ) + (( e (m 3 ) ( m + e 3 ) + ( m + e m e ) ( 3 ) ( m + e ) ) n l ( (i sin θ) i sin θ) m φl ) = (( ttl )n t l t ) + (( t tl t )n l t ) = ( tt l n t t + tl t n )l t Page 7

128 Susan Lasen Tuesday, Febuay 3, 5 = (( m + e e ) (m 3 ) ( m + e ) ( m + e ) + ( m + e m e ) ( 3 ) ( m + e ) ( m + e ) ) l t = e (m 3 ) ( m + e ) = m e 3 ( 3 ) ( m + e ) γ = ( bl a n a n b b m a m an b ) = ( tt l n t t + tl t n )n t α = m e 3 ( 3 ) ( m + e ) = ( bl a n a m b b m a m am b) = ( θl a n a m θ θ m a m am θ) + ( φl a n a m φ φ m a m am φ) = ( θl t n t m θ θ m θ m θm θ) + ( φl t n t m φ φ m θ m θm φ) + ( θl n m θ θ m φ m φm θ) = (( c θl t θt + ( φl n m φ φ m φ m φm φ) l c )n t m θ ( θ m θ + (( c φl t φt + (( c θl θ + (( c φl φ = ( φ φθ mφ m θm φ ( θ m φ θφ c θθ m c )m θm θ) l c )n t m φ ( φ m θ l c )n m θ ( θ m φ l c )n m φ ( φ m φ c φθ c θφ c φφ φ mφ ) m φm θ θ + φφ m c )m θm φ) m c )m φm θ) m c )m φm φ) m θ m φm φ) = (cot θ i sin θ ( ) ( θ(i sin θ) cot θ θ i sin θ) m cos θ sin θ i φ ) m sin θ = φ i cos θ m = i cos θ i sin θ = 3 cot θ β = ( bl a n a m b b m a m am b ) Page 8

129 Susan Lasen Tuesday, Febuay 3, 5 = i cos θ mφ = i cos θ ( i sin θ ) = 3 cot θ Collecting the esults π κ ε = m e 3 ( 3 ) ( m + e ) ν τ γ = m e 3 ( 3 ) ( m + e ) λ ρ = ( m + e ) α = 3 cot θ μ σ β = 3 cot θ The Weyl Scalas and Petov classification Ψ = Dσ δκ σ(ρ + ρ ) σ(3ε ε ) + κ(π π + α + 3β) (3.) Ψ = Dβ δε σ(α + π) β(ρ ε ) + κ(μ + γ) + ε(α π ) (3.3) Ψ = δ τ Δρ ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Λ (3.4) Ψ 3 = δ γ Δα + ν(ρ + ε) λ(τ + β) + α(γ μ ) + γ(β τ ) (3.5) Ψ 4 = δ ν Δλ + λ(μ + μ ) λ(3γ γ ) + ν(3α + β + π τ ) (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) Ψ Ψ = Dβ δε β(ρ ε ) + εα = l a a β m a a ε βρ + βε + εα = l t t β + l β m θ θ ε m φ φ ε βρ = l β βρ = ( m + e ) ( Ψ 3 = ( m + e ) ( = Δρ + ρ(γ + γ ) = n t t ρ n ρ + ργ = n ρ + ργ 3 cot θ) ( 3 cot θ) ( 3 cot θ) ( m cot θ) + e ) Page 9

130 Susan Lasen Tuesday, Febuay 3, 5 Ψ 3 = n ( ( m + e ) ) + ( ( m + e ) ) ( m e 3 ( 3 ) ( m + e ) ) = ( m + e ) (( m ( e (m 3 )) = (( m + e ) ( ) + (m e 3 ) + e ) ( ) + (m e 3 ) ( m ) ( = (( + m e 3) + (m e e 3) + (m 3 )) = m 3 + 3e 4 = 4m + 3e 4 = δ γ Δα + +αγ + γβ = m a a γ n a a α = m θ θ γ + m φ φ γ n t t α n α = ( ( m + e ) ) ( cot θ) = 4 cot θ ( m + e ) Ψ 4 3 (m e 3 )) + e ) ) Black Holes. 53 The Path of a adially Infalling Paticle 54 Paticle obits in the Schwazschild space time ae descibed by = ( m ) (dt dτ ) ( m ) ( d dτ ) ( dθ dτ ) sin θ ( dφ dτ ) (.44) whee τ is the local paticle time (pope time) and t can be descibed as a distant obseves (ou) time. Fo paths along adial lines we can set dθ = dφ, and eaange the Schwazschild line element: = ( m ) (dt dτ ) ( m ) ( d dτ ) 53 (McMahon, 6, p. 38) 54 A moe thoough eview of the physical intepetation of the equations can be found hee: Page 3

131 Susan Lasen Tuesday, Febuay 3, 5 m = ( m ) ( dt dτ ) ( d dτ ) On page we investigated the Killing vecto ξ = (,,,) and found that ( m ) dt is a conseved quantity and, since ( m ) dt dτ fo, ( m ) dt dτ dτ must have the oveall value : = ( m ) (dt dτ ) = ( m ) and fom the Schwazschild line element above we can conclude m = ( d dτ ) d dτ = ± m ( dt dτ ) Notice that fo : dt =, and the paticle pope time, τ, and the time of the distant obseve, t, ae dτ equal. Notice also that fo : d, which means the velocity of the paticle is zeo in this limit. Also dτ notice that d has to be negative because ddeceases as the paticle moves inwads. Next we eaange dτ the two equations into a diffeential equation: dt = d m ( m ) (i) We can solve this by integation fom (fa out) to (in the vicinity of m), and find the t() which descibes the paticles path fom ou distant point of view, o moe popula: what happens to the poo astonaut as he appoaches the black holes event hoizon fom viewed fom ou distant position. t t = m ( m d ) = m = m = 55 m 3/ ( m) d (x + m)3/ dx x + m)3/ ([(x ] + m 3 = + m)3/ ([(x + m (x + m)] m 3 (x + m) dx) x + (m) dx ) x x + m x = m + m)3/ = 56 [(x + 4m (x + m) + 4m x + m m ln m 3 m x + m + m ] = [3/ + 4m + (m) 3/ m ln m 3 + m ] = 3 m [3/ m + 6m ] + [m ln + m ] = 3 m ( 3/ 3/ + 6m 6m ) + m ln m + m if m: We set ε = m, ε = m = x + m + m m 55 (ax+b)n/ x dx = (ax+b)n/ n 56 dx = ax+b b ln x ax+b b ax+b+ b + b (ax+b)(n )/ dx (4.) (Spiegel, 99) x if b > (4.87) (Spiegel, 99) Page 3

132 Susan Lasen Tuesday, Febuay 3, 5 t t = 3 m ( m 6m ) + m ln m + m + m m m ln m + m + m m m + m = m ln m + m m ln ε ε + ε + ε m ln ε ε + ε + ε = m ln ε ε ε ε m ln ε ε = m ln m m = m ln m m m = 57 (m ) exp ( m (t t )) Anothe way to come aound this is to inset ε = in the diffeential equation (i) and solve it. m dt d = m ( m ) = m ( m ) ε ( + ε ) ε ( ε) = ε ε ( ε) ε dt d ε = m fo ε 57 (d'inveno, 99, p. 9) Page 3

133 Susan Lasen Tuesday, Febuay 3, 5 t t = d m mx = m x dx mx mx = m[ln x] mx = m [ln ( m )] = m (ln ( m = m ln ( m ) ( m ) = m ln ( m) ( m) m = ( m)e t t m ) ln ( m )) x = /m. 58 The Schwazschild metic in Kuskal Coodinates. The Kuskal coodinates > m whee We calculate u v u v u t v t u = e 4m t cosh m 4m = e 4m t sinh m 4m (.5) = e m ( ) (.8) m = e4m t sinh 4m m 4m = = e4m t cosh 4m m 4m = v 4m u 4m = e4m t cosh + e 4m 4m m 4m m t cosh m 4m = v = e4m t sinh + e t 4m 4m m 4m m sinh m 4m = Now we can use the chain ule du = u u dt + t d dv Witten as a matix u { du dv } = { t v t 58 (McMahon, 6, p. 4) = v v dt + t d u } { dt v d } u 4m ( m v 4m ( m ) ) Page 33

134 Susan Lasen Tuesday, Febuay 3, 5 With the invese u { dt d } = 59 { t v t = u } v { du dv } = u v t u v t ( v 4m ) ( m ) ( u 4m ) ( m = 4m v u v u { u ( m ) ) { v ( m dt = 4m v (vdu udv) u dt = 6m (v u ) (v du + u dv uvdudv) { v v t v 4m ( m u 4m )} {du dv } d = 4m v u ( m ) ( udu + vdv) d = 6m (v u ) ( m ) (u du + v dv uvdudv) Next we find ( m ) dt = ( m ) 6m (v u ) (v du + u dv uvdudv ) ( m ) d Inseting into the Schwazschild metic = ( m ) 6m (v u ) (u du + v dv uvdudv) u } { du u dv } t ds = ( m ) dt ( m ) d (dθ + sin θ dφ) ) u 4m ( = ( m ) 6m (v u ) (v du + u dv (u du + v dv )) (dθ + sin θ dφ) = ( m ) 6m (u v ) (dv d ) (dθ + sin θ dφ) = 6m ( m ) m { du dv } v 4m } ) e m ( m ) (du dv ) (dθ + sin θ dφ) = 3m3 e m(dv du ) (dθ + sin θ dφ) (.7) The Kuskal coodinates < m u = e 4m m sinh t 4m (.6) 59 a b { c d } = { d b ad bc c a } Page 34

135 Susan Lasen Tuesday, Febuay 3, 5 whee We calculate v u v u t v t u v As befoe we find u { dt d } = { t v t = e 4m m cosh t 4m = e m ( ) (.8) m = e4m 4m = e4m 4m = e4m 4m = e4m 4m = u } v m m m m cosh t 4m = sinh t 4m = v 4m u 4m t sinh e 4m 4m m t cosh e 4m 4m m { du dv } = u v t u v t ( v 4m ) ( u m ) ( 4m ) ( m ) { m m { v v t v 4m ( m = 4m v u v u { u ( m ) v (m )} {du dv } dt = 4m v (vdu udv) u dt = 6m (v u ) (v du + u dv uvdudv) d = 4m v u (m d = 6m Next we find ( m ) dt = ( m ) 6m ( m ) d ) ( udu + vdv) sinh t 4m = cosh t 4m = u } { du u dv } t u 4m (v u ) (m ) (u du + v dv uvdudv) (v u ) (v du + u dv uvdudv ) = ( m ) 6m (v u ) (u du + v dv uvdudv) Inseting into the Schwazschild metic we find as befoe ds = ( m ) dt ( m ) d (dθ + sin θ dφ) u 4m ( ) m v 4m ( ) m u ) 4m ( ) m v 4m } { du dv } Page 35

136 Susan Lasen Tuesday, Febuay 3, 5.3 The Ke metic = 3m3 e m(dv du ) (dθ + sin θ dφ) (.7).3. 6 The Ke-Newman geomety A moe geneal metic is the Ke-Newman geomety, coesponding to a simultaneously otating and electically chaged black hole of mass m, chage Q and angula momentum S. ds = Δ Σ (dt a sin θ dφ) sin θ (( + a )dφ adt) Σ Σ Δ d Σdθ (33.) = Δ Σ [dt + (a sin θ) dφ a sin θ dtdφ] sin θ [a dt + ( + a ) dφ a( + a )dtdφ] Σ Σ Δ d Σdθ = Σ [Δ a sin θ]dt + Σ [ Δa sin θ + sin θ a( + a )]dtdφ Σ Δ d Σdθ + Σ [Δ(a sin θ) sin θ ( + a ) ]dφ = Σ [Δ a sin θ]dt + a sin θ [ Δ + ( + a )]dtdφ Σ Σ Δ d Σdθ + sin θ [Δa sin θ ( + a ) ]dφ Σ = Σ [(Σ m + a sin θ + Q ) a sin θ]dt = ( m Σ = ( m Σ = ( m Σ = ( m Σ + a sin θ [ (Σ m + a sin θ) + (Σ + a sin θ)]dtdφ Σ Σ Δ d Σdθ + sin θ [(Σ m + a sin θ)a sin θ (Σ + a sin θ) ]dφ Σ Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ + sin θ [Σa sin θ ma sin θ + (a sin θ) Σ (Σ + (a sin θ) + Σa sin θ)]dφ Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ + sin θ [ Σa sin θ ma sin θ Σ ]dφ Σ Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ sin θ [a sin θ + ma sin θ + Σ] dφ Σ Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ ( + a + a m sin θ ) sin θ dφ + Q + Q + Q + Q Δ = m + a + Q Σ = + a cos θ Σ 6 (C.W.Misne, 973) chapte 33 Page 36

137 Susan Lasen Tuesday, Febuay 3, 5 = + a a sin θ = Δ Q + m a sin θ a = S m We look at thee special cases:.3.. Q In the case of Q we see immediately that the Ke-Newman geomety educes to the Ke geomety descibing a non-chaged otating black hole. = ( m ds Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ ( + a + a m sin θ ) sin θ dφ Σ Δ = m + a Σ = + a cos θ = + a a sin θ = Δ + m a sin θ a = S m.3.. S In the case of S we see immediately that the Ke-Newman geomety educes to the eissne- Nodstøm geomety descibing a chaged non-otating black hole. = ( m ds Σ + Q Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ ( + a + a m sin θ ) sin θ dφ Σ = ( m Σ = ( m + Q ) dt Δ = m + Q Σ = + Q Σ ) dt Σ Δ d Σdθ sin θ dφ m + Q d dθ sin θ dφ.3..3 Q and S In the case of Q and S we see immediately that the Ke-Newman geomety educes to the Schwazschild geomety descibing a non-chaged non-otating black hole. = ( m ds Σ ) dt + 4am sin θ dtdφ Σ Σ Δ d Σdθ ( + a + a m sin θ ) sin θ dφ Σ = ( m ) dt m d dθ sin θ dφ Δ = m Σ = Page 37

138 Susan Lasen Tuesday, Febuay 3, The invese metic of the Ke Spinning Black Hole The Ke metic of a spinning black hole with mass m and angula momentum S. whee ds = ( m Σ ) dt + 4am sin θ Σ Δ = m + a Σ = + a cos θ = + a a sin θ = Δ + m a sin θ a the metic tenso = S m dtdφ Σ Δ d Σdθ ( + a + a m sin θ ) sin θ dφ (.9) Σ ( m Σ ) am sin θ Σ g ab = am sin θ { Σ with the invese g tt Σ Δ g tφ Σ ( + a + a m sin θ ) sin θ Σ } Δ g ab = Σ Σ { g φt g φφ } ΣΔ (( + a ) Δa sin θ) am ΣΔ Δ = Σ Σ am { (Δ a sin θ) ΣΔ ΣΔ sin θ } whee we can calculate g tt, g tφ, g φφ fom the invese 6 g ab = g tt g φφ (g φt ) { g φφ g tφ g φt g } tt Fist we calculate the common facto g tt g φφ (g φt ) = ( ( m Σ ) ( + a + a m sin θ ) sin θ ( am sin θ ) Σ Σ = ( ( m Σ = ( ( + a + a m sin θ Σ ) ( + a + a m sin θ ) ( am Σ Σ ) sin θ) ) + m Σ ) sin θ ( + a + a m sin θ ) ( am Σ Σ ) sin θ) sin θ 6 (McMahon, 6, p. 46) 6 a b { c d } = { d b ad bc c a } Page 38

139 Susan Lasen Tuesday, Febuay 3, 5 = ( ( + a + a m sin θ ) + m Σ Σ ( + a )) sin θ = ( ( + a + m Σ a sin θ) + m Σ ( + a )) sin θ = ( ( + a + m Σ = ( ( + a ) + m Σ ( + a cos θ)) = ( ( + a ) + m Σ = ( ( + a ) + m) sin θ = Δ sin θ Now we can calculate the invese metic g φφ g tt = g tt g φφ (g φt ) g tφ g φφ = a sin θ) + m Σ ( + a cos θ + a sin θ)) Σ) sin θ sin θ sin θ = g φφ Δ sin θ ( + a + a m sin θ Σ ) sin θ = Δ sin θ = Δ ( + a + a m sin θ ) Σ = ΣΔ (Σ( + a ) + a m sin θ) = ΣΔ (( + a cos θ)( + a ) + ( + a Δ)a sin θ) = ΣΔ (( + a cos θ)( + a ) + ( + a )a ( cos θ) Δa sin θ) = ΣΔ (( + a ) Δa sin θ) (.3) g tφ = g tt g φφ (g φt ) = g tφ Δ sin θ = Δ sin θ = am ΣΔ am sin θ Σ g tt g tt g φφ (g φt ) = g tt Δ sin θ m = Δ sin ( θ Σ ) = ΣΔ sin (Σ m) θ (.3) Page 39

140 Susan Lasen Tuesday, Febuay 3, 5 = ΣΔ sin θ (Δ + m a sin θ m) = (Δ a sin θ) ΣΔ sin θ (.3) Cosmology. 63 Light tavelling in the Univese Light tavelling in the univese can be descibed by the line element ds = dt a (t)dx, whee dx = dx + dx + dx 3 and the x i ae commoving coodinates. Light tavel along null geodesic i.e. ds. We can now wite t t dt a(t) fo the total commoving distance light emitted at time t can tavel by time t. If we multiply this by the value of the scale facto a(t ) at time t, then we will have calculated the physical distance that the light has taveled in this time inteval. This algoithm can be widely used to calculate how fa light can tavel in any given time inteval, evealing whethe to points in space, fo example ae in causal contact. As you can see, fo acceleated expansion, even fo abitaily lage t the integal is bounded, showing that the light will neve each abitaily distant commoving locations. Thus, in a univese with acceleated expansion, thee ae locations with which we can neve communicate.. 64 Spaces of Positive, Negative, and Zeo Cuvatue Accoding to (.5) the spatial pat of a homogenous, isotopic metic is dσ = d k + dθ + sin θ dφ (.5) ewiting it in a moe geneal fom dσ = dχ + (χ)dθ + (χ) sin θ dφ (.6) we see that dχ = d k d dχ = ± k and in ode to identify the metic fo the diffeent k-values we solve the latte diffeential equation. k : dχ = ± d χ = ± = ±χ (χ) = χ and the metic dσ = dχ + χ dθ + χ sin θ dφ k > : d dχ = ± k d χ = ± k = ± k dx x = 65 ± k sin x x = k 63 (Geene, s. 56) note 64 (McMahon, 6, p. 6) 65 dx = a x sin x (4.37) (Spiegel, 99) a Page 4

141 Susan Lasen Tuesday, Febuay 3, 5 = ± k sin k = k sin(± kχ) = ± k sin( kχ) (χ) = k sin ( kχ) if k = we get (χ) = sin (χ) and the metic dσ = dχ + sin (χ) dθ + sin (χ) sin θ dφ k < : d dχ = ± + K d = ± χ + ( K) = ± K dx + x = 66 ± K ln (x + x + ) = 67 ± K sinh x = ± k sinh ( k) = k sinh(± kχ) = ± k sinh( kχ) (χ) = k sinh ( kχ) if k = we get (χ) = sinh (χ) and the metic dσ = dχ + sinh (χ) dθ + sinh (χ) sin θ dφ K = k x = K = k Note: We have omitted the constants of integation because of symmety easons. The metics have to fulfill the equiement of homogeneity and isotopy.3 New The citical density 68 The citical density ρ c = 3 H 8π G can be found fom some simple Newtonian consideations 69. (.8) dx 66 = ln(x + x a x a ) (4.) (Spiegel, 99) 67 sinh x = ln(x + x + ) (8.55) (Spiegel, 99) 68 (McMahon, 6, p. 67) 69 (Weinbeg, 979, s. 57) Page 4

142 Susan Lasen Tuesday, Febuay 3, 5 Conside a galaxy with mass m on the suface of a sphee with adius, density ρ and an expansion ate coesponding to the Hubble expansion H. The velocity of the galaxy is given by the Hubble law v = H. The potential enegy of the galaxy is E pot = 4π 3 3 ρ mg = 4π ρmg 3 The kinetic enegy of the galaxy is E kin = mv = mh The total enegy E tot = E kin + E pot = mh 4π ρmg 3 = m ( H 4πρG 3 ) Setting the total enegy to zeo defines the citical density = m ( H 4πρ cg ) 3 ρ c = 3H 8πG.4 7 The obetson-walke metic.4. 7 Find the components of the iemann tenso of the obetson-walke metic (Homogenous, isotopic and expanding univese) using Catan s stuctue equations The metic: ds = dt + a (t) k d + a (t) dθ + a (t) sin θ dφ The Basis one foms ω = dt ω = a(t) k d d = ω ω k ω a(t) = a(t)dθ dθ = a(t) ω = a(t) sin θ dφ dφ = a(t) sin θ ω η ij = { } Catan s Fist Stuctue equation and the calculation of the cuvatue one foms: 7 (McMahon, 6, p. 6), example 7-7 (McMahon, 6, p. 6), example Page 4

143 Susan Lasen Tuesday, Febuay 3, 5 dω a a = b dω ω b dω = d ( a(t) k d) = dω dω a k dt d = a a ω ω = d(a(t)dθ) = a dt dθ + ad dθ = a ω ω k + ω ω a a = d(a(t) sin θ dφ) = a sin θ dt dφ + a sin θ d dφ + a cos θ dθ dφ = a sin θ ω ω k + a sin θ ω ω + a cos θ a sin θ a a sin θ ω a sin θ = a ω ω k + ω ω cot θ + ω ω a a a a ω a = b The cuvatue one-foms summeized in a matix a ω a ω a ω a a a a ω k ω k a a a a ω k ω cot θ ω a a a a ω k { ω cot θ ω a a a Whee a efes to the column and b the ow ω } Cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) : d : d = d ( a a ω) = d ( a a a = ω k = = a a ω ω + a a d) = d ( k k d) = a dt d k k ω = a ω ω a a + + = d ( a a ω) a = d ( adθ) = d(a dθ) = a dt dθ + a d dθ a = a ω ω k + a ω ω a = ω ω k + a a a a a a ω ω Page 43

144 Susan Lasen Tuesday, Febuay 3, 5 = : d : d = a k = a ω ω = a ω ω a = d( a ω a a ) = d ( a sin θ dφ) = d(a sin θ dφ) a = a sin θ dt dφ + a sin θ d dφ + a cos θ dθ dφ = a ω ω k + a a a ω ω cot θ + a ω ω a = = + a k a ω ω + = a a ω ω + + φ a cot θ ω ω a = k = d ( ω ) k = d ( adθ) = d ( k a a dθ) = k k d dθ = ω k a ω = = = ( a a ) ω ω = (a + k) ω ω a k : d = d ( ω ) = d ( k a sin θ dφ) = k k sin θ d dφ + k cos θ dθ dφ = k ω a ω k cot θ + a ω ω = = = a ω a ω k cot θ a ω ω = (a + k) ω ω : d a + + cot θ = d ( ω cot θ ) = d ( a sin θ dφ) = d(cos θ dφ) = sin θ dθ dφ a a = ω ω a = = + = ( a a ) ω k ω ω ω (a) = (a + k) a ω ω Summaized in a matix Page 44

145 Susan Lasen Tuesday, Febuay 3, 5 a b = a ω ω a ω ω a ω ω a a a (a + k) S a ω ω (a + k) a ω ω (a + k) S AS ω ω { S AS AS } Now we can ead off the elements in the iemann tenso in the non-coodinate basis = a a = a a = a a a = (a + k) a = (a + k) a = a + k a.4. The Einstein tenso and Fiedmann-equations fo the obetson Walke metic The icci scala: = η a b a b = = = = = a a + a a + a a + (a + k) a (a + k) a + (a + k) a = ( a a + (a + k) a ) + + The Einstein tenso: G a b = a b η a b (4.48) G G = η = = = η = + ( = 3 ( (a + k) a ) + G = G = G = G = G = G = G = G = G = G = G + + ) G = η = = = ( = a a a a a + k a = ( a a + a + k a ) + + ) Page 45

146 Susan Lasen Tuesday, Febuay 3, 5 G G = η = = = = η = = = ( = a a a a a + k a = ( a a + a + k a ) ( = a a a a a + k a = ( a a + a + k a ) ) + ) Summaized in a matix: G a b = { 3 ( (a + k) a ) ( a a + a + k a ) ( a a + a + k a ) ( a Whee a efes to column and b to ow a + a + k a ) } The Fiedmann equations: Given the Einstein equation ( if c = G = ): G a b Λη a b = 8πT a b (7.4) ρ and the stess-enegy tenso: T P a b = 8π { } (7.6) P P You can find the Fiedmann- equations { 3 ( (a + k) a ) ( a a + a + k a ) ( a a + a + k a ) ( a a + a + k a ) } Λ { ρ P } = 8π { } P P Page 46

147 Susan Lasen Tuesday, Febuay 3, 5 3 a (k + a ) + Λ = 8πρ (7.7) a a + a (k + a ) + Λ = 8πP (8.8).4.3 The Einstein tenso fo the obetson Walke metic Altenative vesion. The line element: ds = dt + a (t) k d + a (t) dθ + a (t) sin θ dφ Now we can compae with the Tolman-Bondi de Sitte line element, whee the pimes should not be mistaken fo the deivative d/d. ds = dt e ψ(t, ) d (t, )dθ (t, ) sin θ dφ And chose: dt = dt e ψ(t, ) d = a(t) k d (t, )dθ = a(t)dθ (t, ) sin θ dφ = a(t) sin θ dφ Compaing the two metics we see: dφ = dφ, dθ = dθ, θ = θ, (t, ) = a(t), dt = dt Next we can use the fome calculations of the Tolman-Bondi de Sitte metic to find the Einstein tenso fo the obetson-walke metic. But fist we need to find ψ = dψ(t, ) dt = e ψ(t, ) d dt (eψ(t, ) ) = a(t) d d k k d ( dt a(t) ψ = d a (t) dt ( a(t) ) = d dt a (t) (a (t)a(t) a (t)a (t)) ( ) = a(t) a(t) = ( a (t) a(t) ) ψ = dψ(t, ) d = e ψ(t, ) d d (eψ(t, ) ) = e ψ(t, ) d d d d k = a(t) k e ψ(t, ) = d(t, ) dt = da (t) dt = d(t, ) d = da(t) dt = da (t) dt = d d d d = a (t) = a (t) = d d ( k e ψ(t, ) ) = d d d = d d d (a (t)) = d d d d d ) = a (t) a(t) a (t) a(t) k d ( a(t) d ) k (a(t)) = e ψ(t, ) a(t) = k a(t) e ψ(t, ) a(t) d ( k a (t) k (a (t)) = e ψ(t, ) a(t) k d d ) The Einstein tenso: Page 47

148 Susan Lasen Tuesday, Febuay 3, 5 Tolman Bondi de Sitte G = [ ψ + ( ) ( + ψ + ( ) )e ψ(t,) ] obetson-walke G = 3 a (t) + k a(t) G = [( ) + ψ ] eψ(t,) G G = [( ) e ψ(t,) ( ) a (t) ] G = ( a(t) + a (t) + k a(t) ) G = [ψ (ψ ) ] + a (t) [( + ψ )e ψ(t,) + ψ ] G = ( a(t) + a (t) + k a(t) ) G = [ψ (ψ ) ] + a (t) [( + ψ )e ψ(t,) + ψ ] G = ( a(t) + a (t) + k a(t) ).5 7 Manipulating the Fiedmann equations. Show that the two Fiedman equations 3 a (k + a ) + Λ = 8πρ (7.7) a a + a (k + a ) + Λ = 8πP (8.8) can be manipulated into: d dt (ρa3 ) + P d dt (a3 ) ewiting (7.7): 8πρ = 3 a (k + a ) + Λ 8πρa 3 = 3a(k + a ) + Λa 3 8π d dt (ρa3 ) = d dt (3a(k + a ) + Λa 3 ) = 3a (k + a ) + 6aa a + 3Λa a ewiting (7.8): 8πP = a a + a (k + a ) + Λ 8πP d dt (a3 ) = ( a a + a (k + a ) + Λ) d dt (a3 ) = ( a a + a (k + a ) + Λ) 3a a = 6aa a + 3a (k + a ) + 3Λa a 8πP d dt (a3 ) = (6aa a + 3a (k + a ) + 3Λa a ) Now adding 8π d dt (ρa3 ) + 8πP d dt (a3 ) = 3a (k + a ) + 6aa a + 3Λa a (6aa a + 3a (k + a ) + 3Λa a ) Q.E.D Paametes in an flat univese with positive cosmological constant: Stating with a = C a + Λ 3 a use a change of vaiables u = Λ 3C a3 We have 7 (McMahon, 6, p. 65), quiz (McMahon, 6, p. 78), quiz -6. The answe to quiz -6 is (a) Page 48

149 Susan Lasen Tuesday, Febuay 3, 5 a ( a a ) and = C a + Λ 3 a = C a 3 + Λ 3 u = Λ 3C a3 u = Λ C = 3 u a a a a u u = 3 a a eaanging we get ( u 3 u ) = Λ 3u + Λ 3 u = 3 Λ 3 u + Λ 3 u = 3Λ u u + du dt = 3Λ u u + t t = 74 ln( u + + u) 3Λ = ln( u + + u) 3Λ = ln(u + + u u + ) 3Λ = 3Λ ln (u + + u + u) = 3Λ ln (u + + (u + ) ) = 3Λ ln (u + + (u + ) ) = 3Λ (ln + ln (u + + (u + ) )) = 75 3Λ (ln + cosh (u + )) u = cosh( 3Λ(t t ) ln ) a 3 = 3C Λ [cosh( 3Λ(t t ) ln ) ] Leaving out the constants of integation 3Λt ln we get a 3 = 3C [cosh( 3Λt) ] Λ 74 dx = (ax+b)(px+q) ap ln( a(px + q) + p(ax + b)) (4.8) (Spiegel, 99) 75 cosh x = ln(x + x ) (8.56) (Spiegel, 99) Page 49

150 Susan Lasen Tuesday, Febuay 3, 5 Gavitational Waves. 76 Gauge tansfomation - The Einstein Gauge a equiing that bcd, ab and ae unchanged unde a gauge-tansfomation of fist ode in ε, show that this is fulfilled by the coodinate tansfomations x a x a = x a + εφ a (3.) h ab h ab = h ab φ a,b φ b,a a ψ b,a ψ a a b,a = ψ b,a φ b whee φ a is a function of position and φ a,b. We have a bcd = εηae ( h bd x c x f + h df x c x b + h bc x d x f h cf x d xb) (3.4) ab c c h = ε ( h a x b x c + h b x a x c Wh ab x a xb) (3.5) = ε ( h cd x c Wh) xd (3.6) ψ ab = h ab η abh (3.8) a The Einstein gauge tansfomation is a coodinate tansfomation that leaves bcd, ab and unchanged. The coodinate tansfomation that will do this is x a x a = x a + εφ a (3.) In ode to show this you only have to convince youself that the line element is unchanged. Checking ds = g ab dx a dx b = (η ab + εh ab )dx a dx b ds b = g a b dxa dx = (η ab + εh ab )dx a dxb η ab = η ab = (η ab + εh ab )d(x a + εφ a )d(x b + εφ b ) = (η ab + εh ab ) ( xa x c dxc + ε φa x c dxc ) ( xb x d dxd + ε φb x d dxd ) = (η ab + εh ab )(δ a c dx c + εφ a,cdx c )(δ b d dx d + εφ b,ddx d ) = (η ab + εh ab )(δ a c δ b d dx c dx d + εφ a,cδ b d dx c dx d + εφ b,dδ a c dx c dx d ) + ε = (η ab + εh ab )(dx a dx b + εφ a,cdx c dx b + εφ b,ddx a dx d ) = (η ab + εh ab )(dx a dx b + εη ae φ e,c dx c dx b + εη bf φ f,d dx a dx d ) = (η ab + εh ab )dx a dx b + η ab (εη ae φ e,c dx c dx b + εη bf φ f,d dx a dx d ) + ε = (η ab + εh ab )dx a dx b + (εδ e b φ e,c dx c dx b + εδ f a φ f,d dx a dx d ) = (η ab + εh ab )dx a dx b + (εφ b,c dx c dx b + εφ a,d dx a dx d ) enaming the dummy vaiables = (η ab + εh ab )dx a dx b + (εφ b,a dx a dx b + εφ a,b dx a dx b ) = (η ab + εh ab + εφ b,a + εφ a,b )dx a dx b = (η ab + εh ab )dx a dx b if h ab = h ab φ b,a φ a,b Q.E.D 76 (McMahon, 6, p. 86) Page 5

151 Susan Lasen Tuesday, Febuay 3, 5 a Next we ae going to investigate the tansfomation of the deivative of the tace evese ψ b,a ψ a a b,a = ψ b,a φ b ψ a = η ac ψ cb,a b,a = η ac (h cb,a η cbh,a ) = η ac (h cb,a η d cbh d,a) = η ac (h cb,a η cbη ed h ed,a) = η ac (h cb,a φ b,ca φ c,ba ) ηac η cb η ed (h ed,a φ d,ea φ e,da ) = η ac (h cb,a η cbη ed h ed,a ) η ac (φ b,ca + φ c,ba η cbη ed (φ d,ea + φ e,da )) = η ac (h cb,a η d cbh d,a) η ac φ b,ca η ac (φ c,ba η cbη ed (φ d,ea + φ e,da )) = η ac (h cb,a η cbh,a ) φ b (η ac φ c,ba ηac η cb η ed (φ d,ea + φ e,da )) = η ac ψ cb,a φ b (η ac φ c,ba δ b a η ed (φ d,ea + φ e,da )) a = ψ b,a a = ψ b,a a φ b (φ,ba a φ b (φ,ba enaming the dummy vaiables δ b a e (φ,ea (φ e,eb d + φ,da )) d + φ,db )) ψ a b,a a = ψ b,a a φ b (φ,ba (φ a,ab a + φ,ab )) a = ψ b,a φ b Q.E.D. P.88: The choice of ψ ψ a b,a = η ac ψ cb,a. Plane waves a b,a leads to 77 = η ac (h cb,a η cbh,a) = η ac h cb,a δ b a h,a = h ab,a h,b.. 78 The iemann tenso of a plane wave Hee we want to show that the iemann tenso only depends on h xx, h xy, h yx and h yy. Fo symmety easons it is only necessay to show that the iemann tenso does not depend on h tt and h tx. The iemann tenso 77 Howeve I don t know how to show that the iemann-tenso keeps the same fom if we make this choice 78 (McMahon, 6, pp. 88,3) Page 5

152 Susan Lasen Tuesday, Febuay 3, 5 a bcd = εηaf ( h df x c x b h bd x c x f + Fo plane waves we have h ab h ab x We also need h ab z and h ab t h ab = h ab (t z) = h ab y (t z) h ab = z (t z) = h ab (t z) (t z) h ab = t (t z) = h ab (t z) = h ab z t h ab z = h ab t h tt h tx h ty h tz h xt h xx h xy h xz h ab = h yt h yx h yy h yz h ( zt h zx h zy h zz ) h bc x d x f h cf x d x h tt h tx h ty (h tt + h zz ) b) (3.4) = h tx h xx h xy h tx h ty h xy h xx h ty (3.6) The Minkowski η ab = ( The dependence on h tt d = f = t ( a = t): ( (h tt + h zz ) h tx h ty h zz ) t bct = εηtt ( h tt x c x b h bt x c t + h bc t b = t: t tct = ε ( h tt x c t h tt x c t + h tc t b = x: t xct = ε ( h tt x c x h xt x c t + h xc t c = t: t xtt = ε ( h xt t + h xt t ) c = x: t xxt = ε ( h xt x t + h xx t ) = ε h xx t c = y: ) h ct t x b) h ct t ) h ct t x ) = ε ( h xt x c t + h xc t ) Page 5

153 Susan Lasen Tuesday, Febuay 3, 5 t xyt = ε ( h xt y t + h xy t ) = ε h xy t c = z: t xzt = ε ( h xt z t + h xz t ) = ε h xt ( t b = y: t yct = ε ( h tt x c y h yt x c t + h yc t c = t: t ytt = ε ( h yt t + h yt t ) c = x: t yxt = ε ( h yt x t + h yx t ) = ε h yx t c = y: t yyt = ε ( h yt y t + h yy t ) = ε h yy t c = z: t yzt = ε ( h yt z t + h yz t ) = ε h yt ( t b = z: b = d = t: t zct = ε ( h tt x c z h zt x c t + h zc t c = t: t ztt = ε h tt ( t z h zt t c = x: + h zt t t zxt = ε h tt ( x z h zt x t + h zx t c = y: t zyt = ε h tt ( y z h zt y t + h zy t c = z: h tx t ) h ct t y ) = ε ( h yt x c t + h yc t ) h ct t z ) h tt t z ) t zzt = ε h tt ( z h zt z t + h zz t h zt t z ) = ε ( a tct = εηaf ( h tf a = x( f = x): h yt t ) h xt t z ) = ε ( h tx t + h xt t ) h yt t z ) = ε ( h ty t + h yt t ) ( h tt t + (h tt + h zz )) ( + h zz t t + (h tt + h zz )) t x c t h tt x c x f + h tc t x f h cf t ) ) Page 53

154 Susan Lasen Tuesday, Febuay 3, 5 x tct = εηxx ( h tx x c t h tt x c x + h tc t x h cx t ) = ε h tx ( x c t h cx t ) c = x: x txt = ε h tx ( x t h xx t ) = ε h xx t c = y: x tyt = ε h tx ( y t h yx t ) = ε h yx t c = z: x tzt = ε h tx ( z t h zx t ) = ε ( h tx t a = y( f = y): y tct c = x: y txt c = y: y tyt c = z: + h tx t ) = εηyy ( h ty x c t h tt x c y + h tc t y h cy t ) = ε ( h ty x c t h cy t ) = ε h ty ( x t h xy t ) = ε h xy t = ε h ty ( y t h yy t ) = ε h yy t y tzt = ε h ty ( z t h zy t ) = ε ( h ty t a = z( f = z): z tct c = x: = εηzz ( h tz x c t h tt x c z + h tc t z h cz t ) + h ty t ) = ε ( h tz x c t h tt x c z + h tc t z h cz t ) z txt = ε h tz ( x t h tt x z + h tx t z h xz t ) c = y: z tyt = ε h tz ( y t h tt y z + h ty t z h yz t ) c = z: z tzt = ε h tz ( z t h tt z + h tz t z h zz t ) The dependence on h tx d = t, f = x ( a = x): x bct = εηxx ( h tx x c x b h bt x c x + h bc t x h cx t x b) = ε ( h tx x c x b h cx t x b) b = x: x xct b = y: x b = z: yct Page 54

155 Susan Lasen Tuesday, Febuay 3, 5 x zct = ε ( h tx x c z h cx t z ) b = t, d = x: a tcx = εηaf ( h xf x c t h tx x c x f + h tc x x f h cf x t ) = εηaf ( h xf x c t h tx x c x f) a = t( f = t): t tcx = εηtt ( h xt x c t h tx x c t ) a = x( f = x): x tcx = εηxx ( h xx x c t h tx x c x ) = ε h xx x c t a = y( f = y): y tcx = εηyy ( h xy x c t h tx x c y ) = ε h xy x c t a = z( f = z): z tcx = εηzz ( h xz x c t h tx x c z ) = εηzz ( h tx x c t + h tx x c t ) b = t, c = x: a txd = εηaf ( h df x t h td x x f + a = t( f = t): t txd = εηtt ( h tx x d t h xt x d t ) a = x( f = x): h tx x txd = εηxx ( h tx x d x h xx x d t ) = ε h xx x d t a = y( f = y): y txd = εηyy ( h tx x d y h xy x d t ) = ε h xy x d t a = z( f = z): x d x f h xf x d t ) = εηaf ( h tx x d x f h xf x d t ) z txd = εηzz ( h tx x d z h xz x d t ) = ε ( h tx x d t + h tx x d t ) c = t, f = x( a = x): x btd = εηxx ( h dx t x b h bd t x + h bt x d x h tx h tx x d x b) = ε h dx ( t x b x d x b) The nonzeo calculated elements of the iemann tenso, fom which we can conclude that the iemann tenso only depends on h xx, h xy, h yx and h yy. t xxt = ε h xx t t xyt = ε h xy t t yxt = ε h yx t x txt = ε h xx t x txz = ε h xx t x tyt = ε h yx t y txt y txz y tyt = ε h xy t = ε h xy t = ε h yy t Page 55

156 Susan Lasen Tuesday, Febuay 3, 5 t yyt = ε h yy t x tzx = ε h xx t x zxt = ε h xx t x zyt = ε h yx t y tzx = ε h xy t.. 79 The line element of a plane wave in the Einstein gauge The petubation h tt h tx h ty (h tt + h zz ) h ab = h tx h xx h xy h tx h ty h xy h xx h ty (3.6) ( (h tt + h zz ) h tx h ty h zz ) the petubation in the Einstein gauge h xx h xy h ab = ( h xy h ) (3.7) xx with the tansfomation h ab = h ab φ b,a φ a,b whee we can assume that φ a = φ a (t z) How to pove (3.7) 8 : ) h xx and h xy ae unchanged by the tansfomation h xx = h xx φ x,x φ x,x = h xx φ x x = h xx h xy = h xy φ y,x φ x,y = h xy φ y x φ x y = h xy ) Choosing the emaining elements h ab leaves h xx and h xy unchanged h tt = h tt φ t,t φ t,t = h tt φ t t h tt = φ t t h tx = h tx φ x,t φ t,x = h tx φ x t φ t x = h tx φ x t h tx = φ x t h ty = h ty φ y,t φ t,y = h ty φ y t φ y x = h ty φ y t h ty = φ y t h tz = h tz φ z,t φ t,z = h tz φ z t φ t z h zt = h zt φ t,z φ z,t = h zt φ t z φ z t h tz = h zt = φ z t + φ t z 79 (McMahon, 6, pp. 9,) 8 (d'inveno, 99, pp ) Page 56

157 Susan Lasen Tuesday, Febuay 3, 5 h zz = h zz φ z,z φ z,z = h zz φ z z h zz = φ z z..3 8 The line element of a plane wave With h ab h xx h xy = ( ) h xy h xx (3.7) we find the line element ds = g ab dx a dx b = (η ab + εh ab )dx a dx b = (η tt + εh tt )dt + (η tx + εh tx )dtdx + (η ty + εh ty )dtdy + (η tz + εh tz )dtdz + (η xt + εh xt )dxdt + (η xx + εh xx )dx + (η xy + εh xy )dxdy + (η xz + εh xz )dxdz + (η yt + εh yt )dydt + (η yx + εh yx )dydx + (η yy + εh yy )dy + (η yz + εh yz )dydz + (η zt + εh zt )dzdt + (η zx + εh zx )dzdx + (η zy + εh zy )dzdy + (η zz + εh zz )dy = dt ( εh xx )dx + εh xy dxdy + εh yx dydx ( + εh xx )dy dz = dt ( εh xx )dx ( + εh xx )dy dz + εh xy dxdy h xy : ds = dt ( εh xx )dx ( + εh xx )dy dz (3.8) h xx : ds = dt dx dy dz + εh xy dxdy (3.9) Consideing the following tansfomation dx dx dy = dy dx + dy = dx = (dx + dy ) dy = (dx dy ) dx = dx + dy + dx dy dy = dx + dy dx dy dxdy = (dx dy ) dx + dy = dx + dy we can ewite the line element ds = dt dx dy dz + εh xy dxdy = dt dx dy dz εh xy (dx dy ) = dt ( + εh xy )dx ( εh xy )dy dz (3.)..4 8 The osen line element The line element: ds = dudv a (U)dx b (U)dy 8 (McMahon, 6, p. 9) 8 (McMahon, 6, p. 98) Page 57

158 Susan Lasen Tuesday, Febuay 3, 5 The metic tenso: g ab = { a (U) b (U)} The basis one foms Finding the basis one foms is not so obvious, we wite: ds = dudv a (U)dx b (U)dy = (ω ) (ω ) (ω ) (ω 3 ) = (ω + ω )(ω ω ) (ω ) (ω 3 ) du = ω + ω dv = ω ω ω = (du + dv) du = ω + ω ω ω ω 3 = (du dv) dv = ω ω = a(u)dx dx = a(u) ω = b(u)dy dy = b(u) ω3 η ij = { } Catan s Fist Stuctue equation and the calculation of the cuvatue one-foms dω a a = b ω b (5.9) dω = d ( (du + dv)) dω = d ( (du dv)) dω dω 3 = d(a(u)dx) = da da du dx = (ω + ω ) ω = du du a(u) a = d(b(u)dy) = db db du dy = (ω + ω ) ω3 = du du b(u) b da ω (ω + ω ) du db ω3 (ω + ω ) du The cuvatue one-foms summaized in a matix: Page 58

159 Susan Lasen Tuesday, Febuay 3, 5 da a du ω (A) db b du ω3 (B) da a = a du ω (A) db b du ω3 (B) b da a du ω (A) da a du ω ( A) db { b du ω3 (B) db b du ω3 ( B) } Whee a efes to column and b to ow and A and B will be used late, to make the calculations easie The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) Fist we will calculate da = d ( da a du ω ) = d ( da du dx) = d a du du dx = d a (ω du + ω ) ω a = d a (ω a du ω + ω ω ) db = d ( db b du ω3 ) = d ( db du dy) = d b du du dy = d b (ω du + ω ) ω3 b = d b (ω b du ω 3 + ω ω 3 ) Now we ae eady to calculate the cuvatue two-foms = d + = = d + = = d + = d d a = (ω a du ω + ω ω ) 3 = d + 3 = d b (ω b du ω 3 + ω ω 3 ) = d + = = d + d a 3 = d = d = (ω a du ω + ω ω ) 3 = d = d = d b (ω b du ω 3 + ω ω 3 ) = d + = = d = = db ω3 da ω + db ω3 ( da b du a du b du a du ω ) Page 59

160 Susan Lasen Tuesday, Febuay 3, = d = Summaized in a matix: d a (ω a du ω + ω ω ) d b (ω b du ω 3 + ω ω 3 ) a = d a b (ω a du ω + ω ω ) d b (ω b du ω 3 + ω ω 3 ) S AS { S AS } Whee a efes to column and b to ow Now we can wite down the independent elements of the iemann tenso in the noncoodinate basis: = d a a du 3 3 = d b b du = d a a du 3 3 = d b b du = d a a du 3 3 = d b b du The icci tenso: a b = (4.46) a b 3 = = = ( d a a du + d b b du ) 3 = = = ( d a a du + d b b du ) 3 = = = 3 = = = = ( d a a du + d b b du ) 3 = = = ( d a a du + d b b du ) 3 = = = 3 = = = = = = = = d a a du d a a du = 3 3 = 3 = = 3 = Page 6

161 Susan Lasen Tuesday, Febuay 3, = 3 3 = = d b b du d b b du Summaized in a matix: ( d a a du + d b b du ) ( d a a du + d b b du ) a b = ( d a a du + d b b du ) ( d a a du + d b b du ) { } Whee a efes to column and b to ow.3 83 Colliding gavity waves - coodinate tansfomation The metic of a plane gavitational wave ds = δ(u)(x Y )du + dud dx dy (3.4) can be witten in tems of the null coodinates u and ν by using the following coodinate tansfomation 84 u = u X Y du d dx dy = ν Θ(u)( u)x + Θ(u)( + u)y = ( uθ(u))x = ( + uθ(u))y = du = dν + ν u du + dx + x y dy = dν ( (Θ (u)( u) Θ(u))x (Θ (u)( + u) + Θ(u))y ) du Θ(u)( u)xdx + Θ(u)( + u)ydy = 85 dν ( (δ(u) Θ(u))x (δ(u) + Θ(u))y ) du Θ(u)( u)xdx + Θ(u)( + u)ydy = dν (δ(u)(x y ) Θ(u)(x + y ))du Θ(u)( u)xdx + Θ(u)( + u)ydy = ( Θ(u) uθ (u))xdu + ( uθ(u))dx = Θ(u)xdu + ( uθ(u))dx = (Θ(u) + uθ (u))ydu + ( + uθ(u))dy = Θ(u)ydu + ( + uθ(u))dy X Y = ( uθ(u)) x ( + uθ(u)) y δ(u)(x Y ) = ( + u Θ (u))(x y ) uθ(u)(x + y ) = δ(u)( + u Θ (u))(x y ) δ(u)uθ(u)(x + y ) = δ(u)(x y ) dx = ( Θ(u)xdu + ( uθ(u))dx) 83 (McMahon, 6, p. 34) uθ (u) = uδ(u) Page 6

162 Susan Lasen Tuesday, Febuay 3, 5 dx + dy = Θ (u)x du + ( uθ(u)) dx Θ(u)( uθ(u))xdudx dy = (Θ(u)ydu + ( + uθ(u))dy) = Θ (u)y du +( + uθ(u)) dy + Θ(u)( + uθ(u))ydudy = Θ (u)x du + ( uθ(u)) dx Θ(u)( uθ(u))xdudx + Θ (u)y du + ( + uθ(u)) dy + Θ(u)( + uθ(u))ydudy = Θ (u)(x + y )du + ( uθ(u)) dx Θ(u)( uθ(u))xdudx + ( + uθ(u)) dy + Θ(u)( + uθ(u))ydudy ds = δ(u)(x Y )du + dud dx dy = δ(u)(x y )du + du (dν (δ(u)(x y ) Θ(u)(x + y ))du Θ(u)( u)xdx + Θ(u)( + u)ydy) (Θ (u)(x + y )du + ( uθ(u)) dx Θ(u)( uθ(u))xdudx + ( + uθ(u)) dy + Θ(u)( + uθ(u))ydudy) = dudν + du( Θ(u)( u)xdx + Θ(u)( + u)ydy) (( uθ(u)) dx Θ(u)( uθ(u))xdudx + ( + uθ(u)) dy + Θ(u)( + uθ(u))ydudy) = 86 dudν ( uθ(u)) dx ( + uθ(u)) dy (3.43).4 87 The delta δ(u) and heavy-side Θ(u) functions: pove that uδ(u) Definitions + if u δ(u) = { if u ; δ(u)du = if u Θ(u) = { if u > ; dθ(u) = 88 Θ u (u) = Θ(u) = δ(u)du du δ(u); We calculate f(u)δ(u)du = f(u)θ (u)du = [f(u)θ(u)] f (u)θ(u)du if f(u) = u we find f(u)δ(u)du = f( ) f (u)du = f( ) (f( ) f()) = f() = uδ(u)du Next we assume that uδ(u). Multiplying both sides with a test function f(u) and integating we get f(u)uδ(u)du = f(u) du 86 Θ (u) = Θ(u) 87 (McMahon, 6, p. 34) 88 (3.44) Page 6

163 Susan Lasen Tuesday, Febuay 3, 5 f() which is consistent with ou initial assumption and we can theefoe conclude that uδ(u) Next we calculate f(u)δ (u)du = [f(u)δ(u)] f (u)δ(u)du if f(u) = u we find f(u)δ (u)du = [f(u)δ(u)] f (u)θ (u)du = [f(u)δ(u)] (f ( ) f (u)du) = (f ( ) (f ( ) f ())) = 89 f () = uδ (u)du ([f (u)θ(u)] f (u)θ(u)du) = Next we assume that δ(u) = uδ (u). Multiplying both sides with a test function f(u) and integating we get f(u)δ(u)du = f(u)uδ (u)du f() = (f(u)u) (u ) = (f (u) u + f(u))(u ) = f() which is consistent with ou initial assumption and we can theefoe conclude that δ(u) = uδ (u).5 9 Impulsive gavitational wave egion III The line element: ds = dudν [ νθ(ν)] dx [ + νθ(ν)] dy The Chistoffel symbols To find the Chistoffel symbols we calculate the geodesic fom the Eule-Lagange equation = d ds ( F x a ) F (.36) x a whee F = u ν [ νθ(ν)] x [ + νθ(ν)] y x a = u: F u F = ν u d ds ( F ) = ν u 89 The geneal fomula is f(u)δ (n) (u)du = ( ) n f (n) () 9 (McMahon, 6, p. 35), example 3- Page 63

164 Susan Lasen Tuesday, Febuay 3, 5 = ν x a = ν: F = Θ(ν)[ νθ(ν)]x Θ(ν)[ + νθ(ν)]y ν F = u ν d ds ( F ) = u ν = u Θ(ν)[ νθ(ν)]x + Θ(ν)[ + νθ(ν)]y x a = x: F x F x d ds ( F x = [ νθ(ν)] x dθ(ν) ) = ( ν Θ(ν) νν dν ) [ νθ(ν)]x [ νθ(ν)] x = 9 4(Θ(ν) + νδ(ν))[ νθ(ν)]x ν [ νθ(ν)] x = 9 [ νθ(ν)] x Θ(ν)[ νθ(ν)]x ν = x Θ(ν) [ νθ(ν)] x ν x a = y: F y F y = [ + νθ(ν)] y d ds ( F dθ(ν) ) = (ν Θ(ν) + νν y dν ) [ + νθ(ν)]y [ + νθ(ν)] y = 4(Θ(ν) + νδ(ν))[ + νθ(ν)]y ν [ + νθ(ν)] y = [ + νθ(ν)] y + Θ(ν)[ + Θ(ν)]y ν = y + Θ(ν) [ + νθ(ν)] y ν Collecting the esults = ν = u Θ(ν)[ νθ(ν)]x + Θ(ν)[ + νθ(ν)]y = x Θ(ν) [ νθ(ν)] x ν = y + Θ(ν) [ + νθ(ν)] y ν We can now find the Chistoffel symbols: u xx = Θ(ν)[ νθ(ν)] x νx Θ(ν) = [ νθ(ν)] u yy = Θ(ν)[ + νθ(ν)] y yν = Θ(ν) [ + νθ(ν)] 9 dθ(ν) dν = δ(ν) 9 νδ(ν) Page 64

165 Susan Lasen Tuesday, Febuay 3, 5 The Petov type The line element ds = dudν [ νθ(ν)] dx [ + νθ(ν)] dy The metic tenso: g ab = { [ νθ(ν)] } [ + νθ(ν)] and its invese: g ab = { [ νθ(ν)] [ + νθ(ν)] } The basis one foms Finding the basis one foms is not so obvious, we wite: ds = dudν [ νθ(ν)] dx [ + νθ(ν)] dy = (ω u ) (ω ν ) (ω x ) (ω y ) = (ω u + ω ν )(ω u ω ν ) (ω x ) (ω y ) du = (ω u + ω ν ) dν = (ω u ω ν ) ω u = (du + dν) du = (ωu + ω ν ) ω ν = (du dν) dν = (ωu ω ν ) ω x = ( νθ(ν))dx dx = ωx νθ(ν) ω y = ( + νθ(ν))dy dy = + νθ(ν) ωy η ij = { } The othonomal null tetad Now we can use the basis one-foms to constuct a othonomal null tetad (9.) Page 65

166 Susan Lasen Tuesday, Febuay 3, 5 l n ( ) m m = ( i i ω u ω u + ω ν ) ( ω ν ω x y ) = ( ω u ω ν ω x + iω y ) ω ω x iω du = dν ( νθ(ν))dx + i( + νθ(ν))dy (( νθ(ν))dx i( + νθ(ν))dy ) Witten in tems of the coodinate basis l a = (,,, ) n a = (,,, ) m a = (,, ( νθ(ν)), i( + νθ(ν)))m a = (,, ( νθ(ν)), i( + νθ(ν))) Next we use the metic to ise the indices l u = g au l a = g νu l ν = l ν = g aν l a = g uν l u = = l x = l y n u = g au n a = g νu n ν = = n ν = g aν n a = g uν n u = n x = n y m ν = m u m x = g xx m x = [ νθ(ν)] ( νθ(ν)) = ( νθ(ν)) m y = g yy m y = [ + νθ(ν)] i( + νθ(ν)) = i ( + νθ(ν)) Collecting the esults l a = (,,, ) l a = (,,, ) n a = (,,, ) n a = (,,, ) m a = (,, ( νθ(ν)), i( + νθ(ν))) ma = (,, ( νθ(ν)), i ( + νθ(ν)) ) m a = (,, ( νθ(ν)), i( + νθ(ν))) m a = (,, ( νθ(ν)), i ( + νθ(ν)) ) The spin coefficients calculated fom the othonomal tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients Page 66

167 Susan Lasen Tuesday, Febuay 3, 5 π = b n a m al b = ν n a m al ν = ν n x m xl ν ν n y m yl ν ν = b n a m an b = u n a m an u = u n x m xn u u n y m yn u λ = b n a m am b = x n a m am x y y n a m am = x n x m xm x y n x m xm y x n y m ym x y y n y m ym c = ( x n x xxn c )m xm x c ( y n x yxn c )m xm y c x ( x n y xyn c )m ym c y ( y n y yyn c )m ym = u xxn u m xm x + u y yyn u m ym μ = b n a m am b = u xxn u m xm x + u yyn u m ym y κ = b l a m a l b = ν l a m a l ν = ν l x m x l ν + ν l y m y l ν τ = b l a m a n b = u l a m a n u = u l x m x n u + u l y m u n ν ρ = b l a m a m b = x l a m a m x + y l a m a m y = x l x m x m x + y l x m x m y + x l y m y m x + y l y m y m y c = ( x l x xxl c )m x m x c + ( y l x yxl c )m x m y c + ( x l y xyl c )m y m x c + ( y l y yyl c )m y m y = ( u xxl u m x m x + u yyl u m y m y) σ = ( Θ(ν)[ νθ(ν)] ( ( νθ(ν)) ) ( ( νθ(ν)) ) + Θ(ν)[ + νθ(ν)] ( i ( + νθ(ν)) ) (i ( + νθ(ν)) )) = Θ(ν) ( ( νθ(ν)) ( + νθ(ν)) ) = νθ(ν) 93 ( + νθ(ν))( νθ(ν)) = b l a m a m b = Θ(ν) ( ( νθ(ν)) + ( + νθ(ν)) ) Θ(ν) = ( + νθ(ν))( νθ(ν)) ε = ( bl a n a l b b m a m al b ) = ( νl a n a l ν ν m a m al ν ) = ( νl u n u l ν ν m x m xl ν ν m y m yl ν ) = (( νm x c νx m c )m x c + ( ν m y νym c )m y) = (( ν( νθ(ν)) x νxm x )m xn ν + ( ν (i( + νθ(ν))) y νym y ) m yn ν ) 93 Θ (ν) = Θ(ν) Page 67

168 Susan Lasen Tuesday, Febuay 3, 5 = (( Θ(ν) νδ(ν) ( Θ(ν) ) ( νθ(ν))) m xnν [ νθ(ν)] γ = ( bl a n a n b b m a m an b ) α Θ(ν) + (i(θ(ν) + νδ(ν)) [ + νθ(ν)] i( + νθ(ν))) m yn ν ) = ( ul a n a n u u m a m an u ) = ( ul u n u n u u m x m xn u u m y m yn u ) = ( bl a n a m b b m a m am b) = ( xl a n a m x x m a m am x) + ( yl a n a m y y m a m am y) = ( xl u n u m x x m x m xm x x m y m ym x + y l u n u m y y m x m xm y y m y m ym y) β = ( bl a n a m b b m a m am b ) Collecting the esults π κ ε ν τ γ λ ρ νθ(ν) = ( + νθ(ν))( νθ(ν)) α μ σ Θ(ν) = ( + νθ(ν))( νθ(ν)) β The Weyl Scalas and Petov classification Ψ = Dσ δκ σ(ρ + ρ ) σ(3ε ε ) + κ(π π + α + 3β) (3.) Ψ = Dβ δε σ(α + π) β(ρ ε ) + κ(μ + γ) + ε(α π ) (3.3) Ψ = δ τ Δρ ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Λ (3.4) Ψ 3 = δ γ Δα + ν(ρ + ε) λ(τ + β) + α(γ μ ) + γ(β τ ) (3.5) Ψ 4 = δ ν Δλ + λ(μ + μ ) λ(3γ γ ) + ν(3α + β + π τ ) (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) Ψ = Dσ δκ σ(ρ + ρ ) σ(3ε ε ) + κ(π π + α + 3β) = Dσ σρ = l a Θ(ν) a ( ( + νθ(ν))( νθ(ν)) ) ( Θ(ν) ( + νθ(ν))( νθ(ν)) ) νθ(ν) ( + νθ(ν))( νθ(ν)) Page 68

169 Susan Lasen Tuesday, Febuay 3, 5 = l ν Θ(ν) ν ( ( νθ(ν))( + νθ(ν)) ) νθ(ν) ( ν Θ(ν)) Θ(ν) = ν ( ν Θ(ν) ) νθ(ν) ( ν Θ(ν)) = δ(ν)( ν Θ(ν)) Θ(ν)( νθ(ν) ν δ(ν)) νθ(ν) ( ν Θ(ν)) ( ν Θ(ν)) = δ(ν) Ψ Ψ = δ τ Δρ ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Λ = Δρ = n a a ρ = n u u ρ Ψ 3 Ψ 4 94Ψ : This is a Petov type N, which means thee is a single pincipal null diection of multiplicity 4. This coesponds to tansvese gavity waves in egion III Two inteacting waves The line element: ds = dudν cos aν dx cosh aν dy The Chistoffel symbols To find the Chistoffel symbols we calculate the geodesic fom the Eule-Lagange equation = d ds ( F x a ) F (.36) x a whee F = u ν cos aν x cosh aν y x a = ν: F = a cos aν sin aν x a cosh aν sinh aν y ν F = u ν d ds ( F ) = u ν = u a cos aν sin aν x + a cosh aν sinh aν y x a = u: F u F = ν u d ds ( F ) = ν u = ν x a = x: F x (McMahon, 6, p. 33), example 3- Page 69

170 Susan Lasen Tuesday, Febuay 3, 5 F x = cos aν x d ds ( F ) x = 4a cos aν sin aν ν x cos aν x = a cos aν sin aν ν x cos aν x = x a tan aν ν x x a = y: F y F y = cosh aν y d ds ( F ) y = 4a cosh aν sinh aν ν y cos aν y = a cosh aν sinh aν ν y cosh aν y = y + a tanh aν ν y Collecting the esults = ν = u a cos aν sin aν x + a cosh aν sinh aν y = x a tan aν ν x = y + a tanh aν ν y We can now find the Chistoffel symbols: u u = a cos aν sin aν yy = a cosh aν sinh aν x x = a tan aν νx = a tan aν y y = a tanh aν νy = a tanh aν xx xν yν The Petov type The line element ds = dudν cos aν dx cosh aν dy The metic tenso: g ab = { cos aν } cosh aν and its invese: g ab = cos aν { cosh aν} The basis one foms Finding the basis one foms is not so obvious, we wite: ds = dudν cos aν dx cosh aν dy Page 7

171 Susan Lasen Tuesday, Febuay 3, 5 = (ω u ) (ω ν ) (ω x ) (ω y ) = (ω u + ω ν )(ω u ω ν ) (ω x ) (ω y ) du = (ω u + ω ν ) dν = (ω u ω ν ) ω x = cos aν dx = cosh aν dy ω y ω u = (du + dν) du = (ωu + ω ν ) ω ν = (du dν) dν = (ωu ω ν ) ω x = cos aν dx dx = cos aν ωx ω y = cosh aν dy dy = cosh aν ωy η ij = { } The othonomal null tetad Now we can use the basis one-foms to constuct a othonomal null tetad (9.) l n ( ) = ω u ω u + ω ν du m ( ) ( ω ν i ω x y ) = ( ω u ω ν ω x + iω y ) = dν cos aν dx + i cosh aν dy m i ω ω x iω ( cos aν dx i cosh aν dy) Witten in tems of the coodinate basis l a = (,,, ) n a = (,,, ) m a = (,, cos aν, i cosh aν) m a = (,, cos aν, i cosh aν) Next we use the metic to ise the indices l u = g au l a = g νu l ν = l ν = g aν l a = g uν l u = = l x = l y n u = g au n a = g νu n ν = = n ν = g aν n a = g uν n u = n x = n y m ν = m u m x = g xx m x = cos aν cos aν = cos aν m y = g yy m y = cosh aν i cosh aν = i cosh aν Collecting the esults l a = (,,, ) l a = (,,, ) n a = (,,, ) n a = (,,, ) m a = (,, cos aν, i cosh aν) ma = (,, cos aν, i cosh aν ) Page 7

172 Susan Lasen Tuesday, Febuay 3, 5 m a = (,, cos aν, i cosh aν) m a = (,, cos aν, i cosh aν ) The spin coefficients calculated fom the othonomal tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m al b = ν n a m al ν = ν n x m xl ν ν n y m yl ν ν = b n a m an b = u n a m an u = u n x m xn u u n y m yn u λ b = b n a m am = x n a m am x y y n a m am = x n x m xm x y n x m xm y x n y m ym x y y n y m ym μ = b n a m am b κ = b l a m a l b = ν l a m a l ν = ν l x m x l ν + ν l y m y l ν τ = b l a m a n b = u l a m a n u = u l x m x n u + u l y m y n u ρ = b l a m a m b = x l a m a m x + y l a m a m y = x l x m x m x + y l x m x m y + x l y m y m x + y l y m y m y c = ( x l x xxl c )m x m x c + ( y l x yxl c )m x m y c + ( x l y xyl c )m y m x c + ( y l y yyl c )m y m y = ( u xxl u m x m x + u yyl u m y m y) = ( a cos aν sin aν ( cos aν ) = a (tan aν tanh aν) + a cosh aν sinh aν ( i cosh aν ) (i cosh aν )) σ = b l a m a m b = ( u xx l u m x m x + u yyl u m y m y ) = a (tan aν + tanh aν) ε = ( bl a n a l b b m a m al b ) = ( νl a n a l ν ν m a m al ν ) = ( νl u n u l ν ν m x m xl ν ν m y m yl ν ) = (( νm x c νxm c )m xl ν c + ( ν m y νym c )m yl ν ) = (( ν cos aν x νxm x ) m xl ν + ( ν i cosh aν y νym y ) m yl ν ) Page 7

173 Susan Lasen Tuesday, Febuay 3, 5 = (( a sin aν a tan aν) cos aν) m x + (ia sinh aν a tanh aν i y cosh aν) m γ = ( bl a n a n b b m a m an b ) α = ( ul a n a n u u m a m an u ) = ( ul u n u n u u m x m xn u u m y m yn u ) = ( bl a n a m b b m a m am b) = ( xl a n a m x x m a m am x) + ( yl a n a m y y m a m am y) β = ( bl a n a m b b m a m am b ) Collecting the esults π κ ε ν τ γ λ ρ = a (tan aν tanh aν) α μ σ = a (tan aν + tanh aν) β The Weyl Scalas and Petov classification Ψ = Dσ δκ σ(ρ + ρ ) σ(3ε ε ) + κ(π π + α + 3β) (3.) Ψ = Dβ δε σ(α + π) β(ρ ε ) + κ(μ + γ) + ε(α π ) (3.3) Ψ = δ τ Δρ ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Λ (3.4) Ψ 3 = δ γ Δα + ν(ρ + ε) λ(τ + β) + α(γ μ ) + γ(β τ ) (3.5) Ψ 4 = δ ν Δλ + λ(μ + μ ) λ(3γ γ ) + ν(3α + β + π τ ) (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) Ψ = Dσ σ(ρ + ρ ) = Dσ σρ = l a a σ σρ = l ν ν σ σρ = a Ψ = l ν ν ( a (tan aν + tanh aν)) (a (tan aν + tanh aν)) (a (tan aν tanh aν)) = ( a ( + tan aν + tanh aν)) ( a (tan aν tanh aν)) Page 73

174 Susan Lasen Tuesday, Febuay 3, 5 Ψ = Δρ = n a a ρ = n u u ρ Ψ 3 Ψ 4 Ψ : This is a Petov type N, which means thee is a single pincipal null diection (n a ) of multiplicity The Naiai spacetime The line element: ds = Λν du + dudv (dx + dy ) = + Λ (x + y ) The metic tenso and its invese: g ab = { Λν } g ab = { Λν } The Chistoffel symbols abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) uuν = ν ν(g uu ) = Λν uν = g νu uuν = Λν νuu = ν ν(g uu ) = Λν uu = g νν νuu = Λ ν 3 xxx xxy yxx yyy yyx = x(g xx ) = Λx 3 xx = y(g xx ) = Λy 3 xy = y(g xx ) = Λy 3 y xx = y(g yy ) = Λy 3 y yy = x(g yy ) = Λx 3 y yx = x(g yy ) = Λx u uu = g uν νuu = Λν x = g xx xxx = Λx x = g xx xxy = Λy = g yy yxx = Λy = g yy yyy = Λy = g yy yyx = Λx 3 x yy = g xx xyy = Λx Collecting the esults we find the non-zeo Chistoffel symbols ν u = uu = Λν ν = Λ ν 3 xyy uν uu x y x xx = yx = yy = Λx x y y xy = xx = yy = Λy 96 (McMahon, 6, p. 38), example Page 74

175 Susan Lasen Tuesday, Febuay 3, 5 The basis one foms Finding the basis one foms is not so obvious, we wite: ds = Λν du + dudv (dx + dy ) = (ω u ) (ω v ) (ω x ) (ω y ) du[ Λν du + dv] dx dy = (ω u + ω v )(ω u ω v ) (ω x ) (ω y ) ω u + ω v = du ω u ω v = Λν du + dv ω x = dx ω y = dy ω u = ( Λν + )du + dv du = ω u + ω v ω v = (Λν + )du dv dv = ( + Λν )ω u ( Λν )ω v ω x = dx dx = ωx ω y = dy dy = ωy η ij = { } The othonomal null tetad Now we can use the basis one-foms to constuct a othonomal null tetad l n ( m ) = ω u ω u + ω v m ( ) ( ω v i ω y ) x = ( ω u ω v ω x + iω y ) = i ω ω x iω ( Witten in tems of the coodinate basis du Λν du + dv dx + i dy l a = (,,, ) n a = ( Λν,,, ) m a = (,,, i ) m a = (,, Next we use the metic to ise the indices l u = g au l a = g vu l v = l v = g av l a = g uv l u + g vv l v = ( ) + Λν = l x = l y n u = g au n a = g vu n v = ( ) = dx i dy ), i ) (9.) Page 75

176 Susan Lasen Tuesday, Febuay 3, 5 n v = g av n a = g uv n u + g vv n v = ( ( Λν )) + Λν ( ) = Λν n x = n y m u = m v m x = g ax m a = g xx m x = ( ) = m y = g ay m a = g yy m y = ( ) i = i Collecting the esults l a = (,,, ) la = (,,, ) n a = ( Λν,,, ) n a = (, Λν,, ) m a = (,, m a = (,,, i ) ma = (,,, i), i ) m a = (,,, i) The spin coefficients calculated fom the othonomal tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m alb = ν n a m alν = ν n x m xl ν ν n y m ylν = ( ν n x c νxn c )m xl ν ( ν n y c νyn c )m ylν ν = b n a m an b = u n a m an u = u n x m xn u u n y m yn u λ b = b n a m am = x n a m am x y y n a m am = x n x m xm x y n x m xm y x n y m ym x y y n y m ym μ = b n a m am b κ = b l a m a l b = ν l a m a l ν = ν l x m x l ν + ν l y m y l ν τ = b l a m a n b = u l a m a n u = u l x m x n u + u l y m y n u ρ = b l a m a m b Page 76

177 Susan Lasen Tuesday, Febuay 3, 5 = x l a m a m x + y l a m a m y = x l x m x m x + y l x m x m y + x l y m y m x + y l y m y m y c = ( x l x xxl c )m x m x c + ( y l x yxl c )m x m y c + ( x l y xyl c )m y m x c + ( y l y yyl c )m y m y σ = b l a m a m b ε = ( bl a n a l b b m a m al b ) = ( νl a n a l ν ν m a m al ν ) = ( νl u n u l ν ν m x m xl ν ν m y m yl ν ) γ = ( bl a n a n b b m a m an b ) = ( ul a n a n u u m a m an u ) = ( ul u n u n u u m x m xn u u m y m yn u ) = ( c ul u uul c )n u n u = u uul u n u n u = Λν ( ) = Λν α = ( bl a n a m b b m a m am b) = ( xl a n a m x x m a m am x) + ( yl a n a m y y m a m am y) = ( xm x m xm x + y m x m xm y + x m y m ym x + y m y m ym y) = ([ c xm x xxm c ]m xm x + [ y m x yx c + [ y m y yym c ]m ym y) = ([ x ( ) x y xxm x xxm y ] m xm x + [ y ( ) = + [ x ( i ) x y xym x xy + [ y ( i ) x y yym x yy c m c ]m xm y c + [ x m y xy x m y ] m ym m y ] m ym y) x y yxm x yx Λx Λx ([( ( ) ) ( ) (Λy ) (i x )] m xm + [( Λy Λy ( ) ) ( Λx ) ( ) (i y )] m xm Λx Λy + [( i ( ) ) ( Λx ) ( ) (i x )] m ym Λy + [( i (Λx ) ) ( Λy ) ( ) (i )] m ym y) x m c ]m ym y m y ] m xm Page 77

178 Susan Lasen Tuesday, Febuay 3, 5 = ([ (Λy ) (i )] m xm x + [ ( Λx ) (i )] m xm y + [ ( Λy ) ( x )] m ym + [ ( Λx ) ( )] m ym y) = (( i Λy ) ( ) Λx + (i ) ( ) (i ) + ( Λy ) (i ) ( ) + ( Λx ) (i ) ) = Λy (( i ) + ( Λx Λy ) (i ) + ( Λx )) = Λ ( x + iy) β = ( bl a n a m b b m a m am b ) = Λy (( ) ( i ) m xm x + ( Λx ) ( i ) m xm y + ( Λy ) ( ) m ym x ( Λx ) ( ) m ym y ) = Λy (( ) ( i ) ( ) ( Λx ) ( ) (i i ) ( )) = iλy ( Λx iλy Λx ) = Λ (x + iy) + ( Λx ) ( i ) ( i ) ( ) + (Λy ) ( ) (i ) ( ) Collecting the esults π κ ε ν τ γ = Λν λ ρ α = Λ (x iy) μ σ β = Λ (x + iy) Newman-Penose identities Dγ Δε = α(τ + π ) + β(τ + π) γ(ε + ε ) + τπ νκ + Ψ Λ NP + Φ (3.58) δα δ β = ρμ σλ + αα + ββ αβ + γ(ρ ρ ) + ε(μ μ ) Ψ + Λ NP + Φ (3.59) Δρ δ τ = ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Ψ Λ NP (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) Page 78

179 Susan Lasen Tuesday, Febuay 3, 5 educes to Dγ = Ψ Λ NP + Φ δα δ β = αα + ββ αβ Ψ + Λ NP + Φ = Ψ Λ NP These we can solve Ψ Λ NP + Φ Ψ Λ NP Φ Collecting the esults = Dγ = l a a ( Λν) = lν ν ( Λν) = ( Λ) = Λ = αα + ββ αβ δα + δ β = ( Λ (x iy)) ( Λ Λ Λ (x + iy)) + ( (x + iy)) ( (x iy)) ( Λ Λ (x iy)) ( (x + iy)) m a a ( Λ (x iy)) + m a a ( Λ (x + iy)) = Λ 8 (x + y ) + Λ 8 (x + y ) + Λ 8 (x + y ) m x x ( Λ (x iy)) m y y ( Λ (x iy)) + m x x ( Λ (x + iy)) + m y y ( Λ (x + iy)) = Λ (x + y ) + ( Λ ) + i (i Λ ) + ( ) ( Λ ) + (i ) (i Λ ) = Λ (x + y ) Λ 4 Λ 4 Λ 4 Λ 4 = Λ (x + y ) Λ = Λ (x + y ) ( + Λ (x + y )) Λ = Λ Λ = Ψ Λ NP + Φ Λ = Ψ Λ NP Φ = Ψ Λ NP Ψ = Λ (3.65) Λ NP = 4 Λ (3.64) Φ = 4 Λ (3.65) Checking Φ and ab = g ab Λ Page 79

180 Susan Lasen Tuesday, Febuay 3, 5 Φ = 4 ab(l a n b + m a m b) (9.) = 4 g abλ(l a n b + m a m b) ab = g ab Λ = 4 Λ(g abl a n b + g ab m a m b) = 4 Λ(g uul u n u + g uν l u n ν + g νu l ν n u + g xx m x m x + g yy m y m y) = 4 Λ(g νul ν n u + g xx m x m x + g yy m y m y) = 4 Λ (lν n u (mx m x + m y m y)) = 4 Λ ( (( ) + ( i ) i )) = Λ( ) 4 And we can conclude that ab g ab Λ Instead we will look at a genealized Naiai spacetime The line element: ds = AΛν du + Bdudv C (dx + dy ) = D + EΛ(x + y ) Finding the basis one foms is not so obvious, we wite: ds = AΛν du + Bdudv C (dx + dy ) = (ω u ) (ω v ) (ω x ) (ω y ) du[ AΛν du + Bdv] C dx C dy = (ω u + ω v )(ω u ω v ) (ω x ) (ω y ) ω u + ω v = du ω u ω v = AΛν du + Bdv ω x = C dx ω y = C dy The basis one-foms ω u = ( AΛν + )du + Bdv du = ωu + ω v ω v = (AΛν + )du Bdv dv = B ( + AΛν )ω u B ( AΛν )ω v ω x ω y = C dx dx = C ωx = C dy dy = C ωy Page 8

181 Susan Lasen Tuesday, Febuay 3, 5 η ij = { } Catan s Fist Stuctue equation and the calculation of the cuvatue one-foms dω a a = b ω b (5.9) dω u = d ( ( AΛν + )du + Bdv) = AΛνdν du = AΛν ( B ( + AΛν )ω u B ( AΛν )ω v ) (ω u + ω v ) = A B Λνωu ω v dω v dω x dω y = d ( (AΛν + )du Bdv) = AΛνdν du = AΛν ( B ( + AΛν )ω u B ( AΛν )ω v ) (ω u + ω v ) = A B Λνων ω u = d ( C = d ( C C dx) = d ( D + EΛ(x + y ) C dy) = d ( D + EΛ(x + y ) dx) = yλe C dy) = xλe C dy dx = yλe ωx ω y C dx dy = xλe ωy ω x C a b The cuvatue one-foms summaized in a matix: A Λν(ωu + ω ν ) B A Λν(ωu + ω ν ) B = ΛE (xωy yω x ) C { Whee a efes to column and b to ow ΛE C (xωy yω x ) } The cuvatue two foms: a b a = d b a + b = a b d ω ω d (5.7), (5.8) Page 8

182 Susan Lasen Tuesday, Febuay 3, 5 Fist we see that u u = d ν ν a b = d ( A B Λν(ωu + ω ν )) = d ( A B Λνdu) fo all combinations = A B Λdν du ν u x y y x = A B Λ ( B ( + AΛν )ω u B ( AΛν )ω v ) (ω u + ω v ) = 4A B Λωu ω v ν = d u x = d y u = d ν = 4A B Λων ω u = d ( ΛE C (xωy yω x )) = d ( ΛE (xdy ydx)) = ( ΛE 4Λ E x ) dx dy + ( ΛE + 4Λ E y ) dy dx = ( ΛE C 4Λ E x ) ω x ω y + ( ΛE C C + 4Λ E y ) ω y ω x C = ( 4ΛE C 4Λ E (x + y )) ω x ω y C = 4ΛE C ( ΛE(x + y ))ω x ω y = 4ΛE C (D + EΛ(x + y ) ΛE(x + y ))ω x ω y = 4ΛED ωx ω y C y = d x x = d y = 4ΛED ωy ω x C a b = Summaized in a matix: 4A Λων B ω u 4A Λωu B ω v 4ΛED ωy ω x C 4ΛED { ωx ω y C } Now we can wite down the independent elements of the iemann tenso in the noncoodinate basis: Page 8

183 Susan Lasen Tuesday, Febuay 3, 5 ν u ν u u ν u v = 4A B Λ = 4A B Λ x y x y y x y x = 4ΛED C = 4ΛED C The icci tenso: a b u u = a b = u u v u = v u x u y u v v = v v x v y v x x y x y y = x x = y x = y y u = u u u u = v u u u = v u v u = x u x u = y u x u = y u y v + u v u v + v v u v + v v v v + x v x v + y v x v + y v y x + u x u x + v x u x + v x v x + x x x x + y x x x + y x y (4.46) y + u y u y + v y u y + v y v y + x y x y + y y x y + y y y v = u v u u = v u v y = x y x x = y x y = 4A B Λ = 4A B Λ = 4ΛED C = 4ΛED C Summaized in a matix: 4A Λ B 4A a b = Λ Λ B Λ 4ΛED = { } = η Λ a b Λ C Λ 4ΛED { C } Whee a efes to column and b to ow Compaed with 97 a b = η a b Λ we can see the that we can choose the coefficients ae A = ; B = ; C =, D = and E = 4, which coesponds to a Naiai line element consistent with ab = g ab Λ ds = Λν du + dudv (dx + dy ) = + Λ 4 (x + y ) So let s copy the Chistoffel, spin coefficient and Newman-Penose identity calculations with this new = + Λ 4 (x + y ). The null tetad is unchanged. The Chistoffel symbols 97 Page 38 Page 83

184 Susan Lasen Tuesday, Febuay 3, 5 abc = ( a cg ab + b g ac a g bc ) (4.5) bc = g ad dbc (4.6) uuν = ν ν(g uu ) = Λν uν = g νu uuν = Λν νuu = ν ν(g uu ) = Λν uu = g νν νuu = Λ ν 3 xxx xxy yxx yyy yyx = x(g xx ) = Λx 3 xx = y(g xx ) = Λy 3 xy = y(g xx ) = Λy 3 y xx = y(g yy ) = Λy 3 y yy = x(g yy ) = Λx 3 y yx = x(g yy ) = Λx u uu = g uν νuu = Λν x = g xx xxx = Λx x = g xx xxy = Λy = g yy yxx = Λy = g yy yyy = Λy = g yy yyx = Λx 3 x yy = g xx xyy = Λx Collecting the esults we find the non-zeo Chistoffel symbols ν u = uu = Λν ν = Λ ν 3 xyy uν uu x y x xx = yx = yy = Λx x y y xy = xx = yy = Λy The spin coefficients calculated fom the othonomal tetad π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m alb = ν n a m alν = ν n x m xl ν ν n y m ylν = ( ν n x c νxn c )m xl ν ( ν n y c νyn c )m ylν ν = b n a m an b = u n a m an u = u n x m xn u u n y m yn u λ b = b n a m am = x n a m am x y y n a m am = x n x m xm x y n x m xm y x n y m ym x y y n y m ym Page 84

185 Susan Lasen Tuesday, Febuay 3, 5 μ = b n a m am b κ = b l a m a l b = ν l a m a l ν = ν l x m x l ν + ν l y m y l ν τ = b l a m a n b = u l a m a n u = u l x m x n u + u l y m y n u ρ = b l a m a m b = x l a m a m x + y l a m a m y = x l x m x m x + y l x m x m y + x l y m y m x + y l y m y m y c = ( x l x xxl c )m x m x c + ( y l x yxl c )m x m y c + ( x l y xy c + ( y l y yyl c )m y m y σ = b l a m a m b ε = ( bl a n a l b b m a m al b ) = ( νl a n a l ν ν m a m al ν ) = ( νl u n u l ν ν m x m xl ν ν m y m yl ν ) γ = ( bl a n a n b b m a m an b ) = ( ul a n a n u u m a m an u ) = ( ul u n u n u u m x m xn u u m y m yn u ) = ( c ul u uul c )n u n u = u uul u n u n u = Λν ( ) = Λν l c )m y m x α = ( bl a n a m b b m a m am b) = ( xl a n a m x x m a m am x) + ( yl a n a m y y m a m am y) = ( xm x m xm x + y m x m xm y + x m y m ym x + y m y m ym y) = (( c xm x xxm c )m xm x + ( y m x yx c + ( y m y yym c )m ym y) c m c )m xm y c + ( x m y xy x m c )m ym = (( x ( ) x y xxm x xxm y ) m xm x + ( y ( ) x y y yxm x yxm y ) m xm + ( x ( i ) x y x xym x xym y ) m ym + ( y ( i ) x y yym x yym y ) m ym y) Page 85

186 Susan Lasen Tuesday, Febuay 3, 5 = Λx ([( + (Λx ) ) ( ) (Λy ) (i x )] m xm + [( Λy + (Λy ) ) ( ) (Λx ) (i y )] m xm Λx + [( i + (Λy ) ) ( ) + (Λx ) (i x )] m ym Λy + [( i (Λx ) ) ( ) + (Λy ) (i )] m ym y) = Λy ([( ) (i )] m xm x + [( Λx ) (i )] m xm y + [( Λy ) ( x )] m ym + [ ( Λx ) ( )] m ym y) = (( i Λy ) ( ) ( Λx ) (i ) ) = Λy ( (i 4 ) + ( Λx Λy ) (i 4 4 ) + ( Λx 4 )) = Λ (x iy) 4 β = ( bl a n a m b b m a m am b ) Λx + (i ) ( ) (i ) + ( Λy ) (i ) ( ) = Λy ([( ) (i )] m xm x + [( Λx ) (i )] m xm y + [( Λy ) ( )] m ymx + [ ( Λx ) ( )] m ym y ) = Λy (( ) ( i ) ( ) ( Λx ) ( ) (i i ) ( )) = iλy ( 4 Λx 4 iλy 4 Λx 4 ) = Λ (x + iy) 4 + ( Λx ) ( i ) ( i ) ( ) + (Λy ) ( ) (i ) ( ) Collecting the esults π κ ε ν τ γ = Λν λ ρ α = Λ (x iy) 4 μ σ β = Λ (x + iy) 4 Page 86

187 Susan Lasen Tuesday, Febuay 3, 5 Newman-Penose identities Dγ Δε = α(τ + π ) + β(τ + π) γ(ε + ε ) + τπ νκ + Ψ Λ NP + Φ (3.58) δα δ β = ρμ σλ + αα + ββ αβ + γ(ρ ρ ) + ε(μ μ ) Ψ + Λ NP + Φ (3.59) Δρ δ τ = ρμ σλ + τ(β α τ ) + ρ(γ + γ ) + κν Ψ Λ NP (3.6) Whee D = l a a Δ = n a a δ = m a a δ = m a a (9.3) educes to Dγ = Ψ Λ NP + Φ δα δ β = αα + ββ αβ Ψ + Λ NP + Φ = Ψ Λ NP These we can solve Ψ Λ NP + Φ = Dγ = l a a ( Λν) = lν ν ( Λν) = ( Λ) = Λ Ψ Λ NP Φ = αα + ββ αβ δα + δ β = ( Λ (x iy)) ( Λ Λ Λ (x + iy)) + ( (x + iy)) ( (x iy)) ( Λ Λ (x iy)) ( 4 4 (x + iy)) ma a ( Λ 4 (x iy)) + m a a ( Λ (x + iy)) 4 = Λ 3 (x + y ) + Λ 3 (x + y ) + Λ 3 (x + y ) m x x ( Λ (x iy)) 4 m y y ( Λ 4 (x iy)) + m x x ( Λ 4 (x + iy)) + m y y ( Λ (x + iy)) 4 = Λ 8 (x + y ) + ( Λ 4 ) + i (i Λ 4 ) + ( ) ( Λ 4 ) + (i ) (i Λ 4 ) = Λ 8 (x + y ) Λ 8 Λ 8 Λ 8 Λ 8 = Λ 8 (x + y ) Λ = Λ 8 (x + y ) ( + Λ 4 (x + y )) Λ = Λ Collecting the esults Λ = Ψ Λ NP + Φ Λ = Ψ Λ NP Φ = Ψ Λ NP Ψ = 3 Λ (3.65) Λ NP = 6 Λ (3.64) Φ (3.65) Page 87

188 Susan Lasen Tuesday, Febuay 3, 5 Which is consistent with the fome calculation of Φ, and we can conclude that if ab = g ab Λ so should ightfully be = + Λ 4 (x + y ).8 98 Collision of a gavitational wave with an electomagnetic wave The nonzeo spin coefficients The line element in egion ν : ds = dudν cos aν (dx + dy ) The metic tenso: g ab = { cos aν } cos aν and its invese: g ab = { cos aν cos aν} The Chistoffel symbols: To find the Chistoffel symbols we calculate the geodesic fom the Eule-Lagange equation = d ds ( F x a ) F (.36) x a whee F = u ν cos aν (x + y ) x a = u: F u F = ν u d ds ( F ) = ν u = ν x a = ν: F = a cos aν sin aν (x + y ) ν F = u ν d ds ( F ) = u ν = u a cos aν sin aν (x + y ) x a = x: F x F x = cos aν x d ds ( F ) x = 4a cos aν sin aν ν x cos aν x = a cos aν sin aν ν x cos aν x = x a tan aν ν x 98 (McMahon, 6, p. 3), quiz 3-. The answe to quiz 3- is (a) Page 88

189 Susan Lasen Tuesday, Febuay 3, 5 x a = y: F y F y = cos aν y d ds ( F ) y = 4a cos aν sin aν ν y cos aν y = a cos aν sin aν ν y cos aν y = y a tan aν ν y Collecting the esults = ν = u a cos aν sin aν (x + y ) = x a tan aν ν x = y a tan aν ν y We can now find the Chistoffel symbols: u u xx = a cos aν sin aν yy = a cos aν sin aν x x xν = a tan aν νx = a tan aν y y yν = a tan aν νy = a tan aν The basis one foms: ds = dudν cos aν (dx + dy ) = (ω u ) (ω ν ) (ω x ) (ω y ) = (ω u + ω ν )(ω u ω ν ) (ω x ) (ω y ) du = (ω u + ω ν ) dν = (ω u ω ν ) ω x = cos aν dx = cos aν dy ω y ω u = (du + dν) du = (ωu + ω ν ) ω ν = (du dν) dν = (ωu ω ν ) ω x = cos aν dx dx = cos aν ωx ω y = cos aν dy dy = cos aν ωy η ij = { } The othonomal null tetad: Now we can use the basis one-foms to constuct a othonomal null tetad Page 89

190 Susan Lasen Tuesday, Febuay 3, 5 l n ( ) = ω u ω u + ω ν du m ( ) ( ω ν i ω y ) x = ( ω u ω ν ω x + iω y ) = dν cos aν dx + i cos aν dy m i ω ω x iω ( cos aν dx i cos aν dy) Witten in tems of the coodinate basis l a = (,,, ) n a = (,,, ) m a = (,, cos aν, i cos aν) m a = (,, cos aν, i cos aν) Next we use the metic to ise the indices l u = g au l a = g νu l ν = l ν = g aν l a = g uν l u = = l x = l y n u = g au n a = g νu n ν = = n ν = g aν n a = g uν n u = n x = n y m ν = m u m x = g xx m x = cos aν cos aν = cos aν m y = g yy m y = cos aν i cos aν = i cos aν (9.) Collecting the esults: l a = (,,, ) l a = (,,, ) n a = (,,, ) n a = (,,, ) m a = (,, cos aν, i cos aν) ma = (,, cos aν, i cos aν ) m a = (,, cos aν, i cos aν) m a = (,, cos aν, i cos aν ) The spin coefficients calculated fom the othonomal tetad: π = b n a m al b κ = b l a m a l b ε = ( bl a n a l b b m a m al b ) ν = b n a m an b τ = b l a m a n b γ = ( bl a n a n b b m a m an b ) λ = b n a m am b ρ = b l a m a m b α = ( bl a n a m b b m a m am b) (9.5) μ = b n a m am b σ = b l a m a m b β = ( bl a n a m b b m a m am b ) Calculating the spin-coefficients π = b n a m al b = ν n a m al ν = ν n x m xl ν ν n y m yl ν ν = b n a m an b = u n a m an u = u n x m xn u u n y m yn u λ b = b n a m am = x n a m am x y y n a m am = x n x m xm x y n x m xm y x n y m ym x y y n y m ym μ = b n a m am b Page 9

191 Susan Lasen Tuesday, Febuay 3, 5 κ = b l a m a l b = ν l a m a l ν = ν l x m x l ν + ν l y m y l ν τ = b l a m a n b = u l a m a n u = u l x m x n u + u l y m y n u ρ = b l a m a m b = x l a m a m x + y l a m a m y = x l x m x m x + y l x m x m y + x l y m y m x + y l y m y m y σ c = ( x l x xx = ( u xx l c )m x m x c + ( y l x yx c + ( y l y yyl c )m y m y l u m x m x + u yyl u m y m y) = ( a cos aν sin aν ( cos aν ) = a tan aν = b l a m a m b = ( u xxl u m x m x + u yyl u m y m y ) = ( a cos aν sin aν ( cos aν ) ε = ( bl a n a l b b m a m al b ) = ( νl a n a l ν ν m a m al ν ) = ( νl u n u l ν ν m x m xl ν ν m y m yl ν ) l c )m x m y c + ( x l y xy l c )m y m x a cos aν sin aν ( i cos aν ) (i cos aν )) a cos aν sin aν ( i cos aν ) ) = (( νm x c νxm c )m xl ν c + ( ν m y νym c )m yl ν ) = (( ν cos aν x νxm x ) m xl ν + ( ν i cosh aν y νym y ) m yl ν ) = (( a sin aν ( a tan aν) cos aν) m x + ( ia sin aν + a tan aν i cos aν) m y) γ = ( bl a n a n b b m a m an b ) α = ( ul a n a n u u m a m an u ) = ( ul u n u n u u m x m xn u u m y m yn u ) = ( bl a n a m b b m a m am b) = ( xl a n a m x x m a m am x) + ( yl a n a m y y m a m am y) β = ( bl a n a m b b m a m am b ) Collecting the esults π κ ε Page 9

192 Susan Lasen Tuesday, Febuay 3, 5 ν τ γ λ ρ = a tan aν α μ σ β This means that e(ρ) and thee is expansion (o pue focusing=divegence) The Aichelbug-Sexl Solution The passing of a black hole The line element ds = 4μ log(x + y ) du + dud dx dy Compaing with the Binkmann metic ds = H(u, x, y)du + dudv dx dy We see that we can copy the esults fom the Binkmann calculations p.95 if H(u, x, y) = 4μ log(x + y ) The only non-zeo spin-coefficient is: ν = ( H H + i x y ) (9.3) = ( (4μ log(x + y )) x = 4μ ( x x + y + i y x + y ) x = μ ( x + y + i y x + y ) + i (4μ log(x + y )) ) y. Obsevations: The Futue Gavitational Wave detectos. GWs ae faint defomations of the space-time geomety, popagating at the speed of light and geneated by catastophic events in the Univese, in which stong gavitational fields and sudden acceleation of asymmetic distibution of lage masses ae involved. GWs have a quadupola natue and have two polaizations, h + and h x, whee h is the so called space-time stain h = L L L = δl L, the elative dimensional8distotion of an extended mass distibution. The effect of these two polaizations on a cicula mass distibution is shown in the figue. GWs ae ceated by acceleating masses, but because gavity is the weakest of the fou fundamental foces, GWs ae extemely small. Fo this eason, only extemely massive and compact objects having intense and asymmetic gavitational fields, like neuton sta and black hole binay systems, ae expected to be able to geneate detectable GW emission. The diect detection of GWs is still missing and it is quite easy to undestand why. Fo example, the expected amplitude on Eath of the GW emitted by a coalescing binay system of neuton sta located in the Vigo cluste is of the ode of h~. This means that a detecto having a dimension of a mete expeiences an oscillating defomation of m, an astonishingly small quantity. In the 96s, the fist GW detectos wee based on a (multi)-ton esonant ba, that should esonate when excited by the passage of a GW. These detectos evolved, opeating at cyogenic tempeatue to minimize the distubance of the themal Bownian vibation and being ead by 99 (McMahon, 6, p. 3), quiz 3-. The answe to quiz 3- is (b) This is an extact of the aticle: Opening a New Window on the Univese The Futue Gavitational Wave detectos (Michele Puntuo - 3). Page 9