Solutions for the FINAL EXAM

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1 Ludwig Maximilian University of Munich LMU General Relativity TC1 Prof. Dr. V. Mukhanov WS 014/15 Instructors: Dr. Ted Erler Dr. Paul Hunt Dr. Alex Vikman Solutions for the FINAL EXAM I. ENERGY MOMENTUM TENSOR 7 POINTS Energy momentum tensor EMT of plasma with a heat transfer is given by T µν = 4 3 εu µu ν 1 3 εg µν σq µ u ν σq ν u µ where the local rest frame is defined by u µ which is as usual u µ u µ = 1 the heat transfer vector is specified through q µ u µ = 0 q µ q µ = 1 and σ = const. 1. 1P Find the trace of this EMT.. P Find the four-momentum density of the plasma in the local rest frame u µ. 3. P Find the energy density of plasma in a boosted frame where r µ r µ = 1 r µ u µ = r µ q µ = 0. V µ = u µ + vr µ 1 v 4. P Let us study an arbitrary covariantly conserved and symmetric EMT: T µν. Now suppose that there is a Killing vector field ξ µ in the spacetime. Prove that J α = T α β ξβ is a covariantly conserved current. Under which conditions T αβ ξ α ξ β is an energy density measured by an observer? Solution I This problem is similar to problem 5.3 see also 5. and to problem 6. and the last point uses P Find the trace of this EMT. TrT = T = T µ µ = g µν T µν = 4 3 εu µu µ 1 3 εgµ µ σq µ u µ σq µ u µ now we use the standard normalization of the four-velocity u µ u µ = 1 orthogonality q µ u µ = 0 and g µ µ = g µν g νµ = δ µ µ = 4 so that T µ µ = 4 3 ε 4 3 ε = 0. Thus this T µν is trace-free exactly in the same way as the EMT for an ideal gas of photons σ = 0 and as the EMT for electromagnetic field.. P Find the four-momentum density of the plasma in the local rest frame u µ. The EMT has information on all flows of energy and momentum of the system in all reference frames. Per definition the four-momentum density of a system with respect to a particular observer u µ is defined as a one-form P u V on four-vectors V : P u V = T V u 1

2 note that we use symmetric EMT or in terms of slots and machines P u... = T... u. Thus writing this expression in terms of abstract index notation or components we have P µ = T µν u ν = 4 3 εu µ 1 3 εg µνu ν σq µ σq ν u µ u ν = εu µ σq µ where we have again used u µ u µ = 1 and orthogonality q µ u µ = P Find the energy density of plasma in a boosted frame V µ = u µ + vr µ 1 v where r µ r µ = 1 r µ u µ = r µ q µ = 0. Following the discussion from the previous point we have thus ε V = P V V = T V V = T µν V µ V ν ε V = 4 3 ε u µv µ 1 3 ε σ V µ q µ V ν u ν = 4 ε 3 1 v 1 3 ε = ε 3 + v 3 1 v where in the last equality we have used that r µ u µ = r µ q µ = 0 so that V µ q µ = P Let us study an arbitrary covariantly conserved and symmetric EMT: T µν. Now suppose that there is a Killing vector field ξ µ in the spacetime. Prove that J α = T α β ξβ is a covariantly conserved current. Under which conditions T αβ ξ α ξ β is an energy density measured by an observer? µ J µ = α T α β ξ β + T α β α ξ β = T α β α ξ β = T αβ α ξ β = 1 T αβ α ξ β + β ξ α = 0 where we have used that the EMT is conserved α T α β = 0 symmetric T αβ = T βα and finally that ξ µ is a Killing vector: α ξ β + β ξ α = 0. The useful property of this conserved current is that contrary to the EMT it provides us conserved charges. Indeed so that for Q = V d3 x gj t we have ˆ d dt Q = d 3 x t gj t ˆ = V µ J µ = 1 g µ gj µ V d 3 x i gj i ˆ = dσ i gj i = 0 V where we have used the Gauss theorem and assumed that on the boundary V i.e. at spatial infinity the current is vanishing. Thus Q is conserved. T αβ ξ α ξ β is an energy density measured by an observer if ξ µ is a four-velocity: i.e. it is future-directed ξ t > 0 time-like vector ξ µ ξ µ > 0 which is normalized ξ µ ξ µ = 1. If this property is satisfied at least at infinity then Q can be interpreted as total conserved energy. II. THE SCHWARZSCHILD SPACETIME AND VECTOR FIELDS 13 POINTS Consider the metric of the Schwarzschild spacetime written in the synchronous frame the Lemaître coordinates ds = dτ F dr F 4 rg sin θdφ + dθ where c = 1 r g is the Schwarzschild radius and with the coordinate τ < R. [ ] 1/3 3 R τ F R τ =. r g

3 1. 1P Consider two observers moving on the trajectories R 1 τ = R τ = R = const θ 1 τ = 0 θ τ = π φ 1 τ = φ τ = φ = const. Find the spatial distance l θ τ between these two observers along the path φ = const R = const τ = const.. P Consider two other observers moving on the trajectories R 1 τ = R 1 = const R τ = R = const R 1 < R and θ 1 τ = θ τ = θ = const φ 1 τ = φ τ = φ = const. Find the physical spatial distance l R τ between these two observers on the path φ = const θ = const τ = const. 3. 1P What are the accelerations of these observers? 4. P Find the position of the horizon in these coordinates. To do this recall that the horizon is given by a null light-like hypersurface: f τ R θ φ = 0 and g µν µ f ν f = 0 and then assume that in fact f τ R θ φ = f R τ. 5. 3P Consider following four vector fields Find the scalar products of all of them. V τ = τ 1 V R = F R V θ = F r g 1 θ V φ = F r g sin θ 1 φ. 6. P Can these vector fields 1 build a coordinate basis? For an extra support of your answer calculate [V τ V R ]. 7. P Consider new vector fields given by V X = V R + vv τ V T = V τ + vv R 1 v 1 v with v < 1. Find all scalar products in the new system of vector fields {V T V X V θ V φ }. Solution II This problem is related to the homework exercises v. Useful reading: Landau & Lifschitz and P Consider two observers moving on the trajectories R 1 τ = R τ = R = const θ 1 τ = 0 θ τ = π φ 1 τ = φ τ = φ = const. Find the spatial distance l θ τ between these two observers along the path φ = const R = const τ = const. For the tangent vector V along some space-like path we have the that the square of its absolute value is dl V = g ik dx i V dx k V to calculate the distance along a path between two points P 1 and P parametrized by a parameter λ we need to take V = dλ λ and calculate ˆ P ˆ λ ˆ λ l = dl V = dλ g ik dx i λ dx k dx λ = dλ g i dx k ik P 1 λ 1 λ 1 dλ dλ. Along the path from this point we can take θ as our parameter so that so that and therefore Note that this distance is decreasing with time. dl θ τ = F 4 r gdθ dl θ τ = F r g dθ [ ] /3 3 R τ l θ τ = πr g F = πr g. r g 3

4 . P Consider two other observers moving on the trajectories R 1 τ = R 1 = const R τ = R = const R 1 < R and θ 1 τ = θ τ = θ = const φ 1 τ = φ τ = φ = const. Find the physical spatial distance l R τ between these two observers on the path φ = const θ = const τ = const. Along this path we can take R as our parameter so that and consequently l R τ = ˆ R R 1 dr 1 F = ˆ R R 1 dl R τ = F dr [ ] 1/3 3 R τ dr = ˆ X r g 3 r g dx X 1/3 = r g X /3 X /3 1 X 1 [ 3 /3 R τ l R τ = r g r g ] /3 3 R1 τ. r g 3. 1P What are the accelerations of these observers? Following 7.1 we know that the observers who do not change spatial coordinates in a synchronous frame are moving on geodesics. Thus the acceleration is null. It is easy to prove this. Indeed in that case u µ = δ τ µ and for the acceleration we have so that for a synchronous coordinates the acceleration for a comoving observer vanishes: a µ = u λ λ u µ = u λ λ u µ + Γ µ λν uν u λ ds = dτ γ ik dx i dx k a µ = Γ µ ττ = 1 gµν g µττ + g µττ g ττµ = P Find the position of the horizon in these coordinates. To do this recall that the horizon is given by a null light-like hypersurface: f τ R θ φ = 0 and g µν µ f ν f = 0 and then assume that in fact f τ R θ φ = f R τ. This is a natural assumption because of the spherical symmetry and because the metric only depends on the combination R τ. Let s denote R τ = v so that f v. In this case Thus the horizon should be located at or at g µν µ f ν f = g RR + g ττ f = 1 F f. F = ±1 R τ = ± 3 r g. For a given R which corresponds to a freely falling observer first one crosses +r g /3 which is therefore the true horizon. 5. 3P Consider following four vector fields V τ = τ V R = F R V θ = F r g 1 θ V φ = F r g sin θ 1 φ. Find the scalar products of all of them. This vector fields build the so-called tetrad or vierbein. For their scalar products we have g V τ V τ = 1 4

5 g V R V R = g µν dx µ V R dx ν V R = F g µν dx µ R dx ν R = F g RR = 1 the other way is to write these vector fields in components so that V µ τ = dx µ V τ = δ µ τ 3 V µ R = dxµ V R = F dx µ R = F δ µ R V µ θ = F r g 1 δ µ θ V µ φ = F r g sin θ 1 δ µ φ. g V φ V φ = g µν V µ φ V ν φ = F r g sin θ gφφ = 1. In this way one finds that all these vectors are orthogonal because the metric does not have any cross terms so that e.g. and g V R V τ = g µν dx µ V R dx ν V τ = F g µν δ µ R δν τ = 0 g V φ V φ = g V θ V θ = g V R V R = g V τ V τ = P Can these vector fields build a coordinate basis? For an extra support of your answer calculate [V τ V R ]. Of course these vector fields cannot build a coordinate basis. Indeed in this basis the metric g = η AB ω A ω B the Einstein summation convention is assumed has components η AB = diag where the 1-forms ω A are defined as dual to V A : ω A V B = δb A. If this were a coordinate basis it would imply that the metric is everywhere Minkowski and the spacetime is Minkowski with zero curvature tensor. However we started from the Schwarzschild spacetime where the Riemann tensor is nonvanishing: R αβµν 0. Thus this cannot be a coordinate basis. On the other hand if this were a coordinate basis then there should exist such coordinates - 4 scalar functions τ R φ θ - that V τ = τ V R = R V θ = θ V φ = φ. The partial derivatives commute - thus all commutators should vanish. As suggested in the problem let us calculate [V τ V R ] = τ F R F R τ = τ F R = 1 [ ] /3 3 R τ r 1 g R = 1 r g r 1 g F 3 V R 0. Therefore τ R do not exist and this basis is not a coordinate basis. 7. P Consider new vector fields given by V X = V R + vv τ V T = V τ + vv R 1 v 1 v with v < 1. Find all scalar products in the new system of vector fields {V T V X V θ V φ }. This set of vector fields is again a tetrad. The transformation above is again an active Lorentz transformation. {V T V X V θ V φ } build another orthonormal system: g V X V T = g V R + vv τ V τ + vv R 1 v = g V R V τ + vg V τ V τ + v g V R V τ + vg V R V R 1 v g V X V T = vg V τ V τ + vg V R V R 1 v = 0 g V T V T = g V τ + vv R V τ + vv R 1 v = g V τ V τ + v g V R V R 1 v = 1 g V X V X = g V R + vv τ V R + vv τ 1 v = v g V τ V τ + g V R V R 1 v = 1. 5

6 III. GEODESICS 8 POINTS Consider the spacetime where the radial coordinate r 0 and H = const. ds = 1 H r dt 1 H r 1 dr r sin θdφ + dθ Consider a free photon with conserved angular momentum L and conserved energy E. Assume that the orbit is at θ = π/. i 3P Find the turning points - the points where the derivative of the radius r with respect to the affine parameter vanishes: dr/dλ = 0. ii 3P Using an analogy with classical mechanics find an effective potential for the motion and prove that the photon cannot move with r = const in the spacetime above.. P Suppose a static observer at r = R emits light which is then received at r = 0 by another static observer. Calculate the redshift of the signal in the form of the ratio of the frequencies ω. You can use the standard textbook formulas or you can think about the redshift in the following way below. The photon has four momentum k µ = dx µ /dλ where λ is the affine parameter along the geodesic. On geodesics k µ is parallel transported. The frequency measured an observer is ω = u µ k µ where u µ is the four velocity of the observer. Now compare the frequencies observed by the static observers do not forget about the conserved energy of the photon. Solution III This problem is related to homework problems 13. but nothing from this problem is used in the solution 8.3 a including the derivation done in the lecture and mostly 9.1. This spacetime is a patch of the de Sitter spacetime written in static coordinates from 13.. The wave vector / four-momentum of the photon is k µ = dxµ dλ where λ is the affine parameter along the light-like geodesic. In this parametrization the geodesic equation takes the form k α α k µ = 0. This spacetime 4 is spherically symmetric and static. Thus there are Killing vectors corresponding to rotation along φ : ξ φ = φ = ξ µ φ µ = δ µ φ µ and time translation: ξ t = t. The conserved energy is E = k µ ξ µ t = k t and the conserved angular momentum is L = k µ ξ µ φ = k φ the minus in this definition appears because of our signature convention. The four-momentum of the photon is light-like vector therefore 0 = g µν k µ k ν = g tt k t dr + grr + g φφ k φ dλ where we have used that it θ = π/. Further k µ = g µν k ν so that k t = g tt k t = E/g tt while k φ = g φφ k φ = L/g φφ. Therefore dr g rr + E + L = 0. dλ g tt g φφ Similarly to the Schwarzschild spacetime for 4 g rr g tt = 1 therefore dr + L = E dλ g rr g φφ which is exactly the same equation as in the Schwarzschild spacetime. This looks like the total conserved energy E / for a 1d motion in the potential L V L r = 1 g rr r g φφ r A turning point r 0 occurs when V r 0 = E /. A circular geodesic is possible only if r c = const or in other words at such positions of equilibrium where the effective force V is vanishing: V r c = 0. 6

7 For 4 and θ = π/ we have dr + 1 H r L dλ r = E so that the effective potential is For a turning point we have V L r = 1 1 H r L r. 1 H r 0 L r 0 = E L = E + HL r 0 thus r 0 = L E + HL. For a possible circular orbit we have: V = 1 [ d 1 H r L ] dr r = r d dr [ 1 H r L ] r d dr [ 1 H r L thus V 0 for r > 0. Therefore there is no circular orbit. For the redshift of the signal between static observers we have ω = u µ k µ = along the light geodesic E is conserved therefore ξ µ t r g µν ξ µ t ξν t ] = L r 4 k µ = E gtt E = ω x 1 g tt x 1 = ω x g tt x so that we get the formula which one can also find in the standard textbooks and in the lecture notes g tt x 1 ω x = ω x 1 g tt x. For the spacetime 4 we get if r = 0 ω r = ω R 1 H R 1 H r ω 0 = ω R 1 H R < ω R thus the frequency is lower red-shift because of the work of gravity on the photon. 7

8 IV. COSMOLOGY 6 POINTS Suppose a spatially-flat Friedmann universe is completely dominated by cold dark matter dust. Thus it is almost our universe where we live in. 1. P Find the scale factor a t as a function of cosmological time assuming that the current Hubble parameter is H 0 the scale factor now is unity and the beginning of time t = 0 corresponds to the Big Bang a 0 = 0.. 3P Suppose a supernova explosion happens at the radial comoving coordinate r at the moment t of cosmological time. Find the cosmological time t r when the observers at the radial comoving coordinate r = 0 will see the explosion. Hint: do not forget about conformal time. 3. 1P Calculate the redshift as a ratio of the frequencies for the light from the explosion. Solution IV This problem is related to 13.1 from the homework problems. 1. If the universe is filled by dust or cold dark matter we have from the energy conservation / continuity equation thus which implies that ε = 3H ε + p = 3ȧ a ε dε = 3εd ln a ε = ε 0 a 3 where for our normalization ε 0 means the energy density of dust now when a = 1. The first Friedmann equation gives G N = 1 ȧ 8π a = 3 ε 0a 3 = H 0 a 3/ thus this first order ODE gives which yields a 1/ da = H 0 dt a 3/ = 3 H 0 t t 0 with t 0 being one constant of integration. In our choice of the beginning of time t 0 = 0. Thus the scale factor evolves as /3 3 a t = H 0t. 5. Here we need to consider the geodesic equation for light either directly or using the conformal time as suggested in the problem. The metric is ds = dt a t δ ik dx i dx k = dt a t dr + r sin θdφ + dθ. 6 Let s consider a radial geodesic fro which k µ = k t δ µ t + k r δ µ r. Let s recall that g µν k µ k ν = k t a k r = 0 8

9 so that k t = ±a t k r. 7 The photon is future directed so that k t = dt dλ > 0 but the photon propagates from larger r to r = 0 so that Therefore k r = dr dλ < 0. Thus we have or in other terms canceling dλ on both sides Note that the conformal time is defined as so that In our case k t = a t k r. dt dr = a t dλ dλ ˆ t dt t 1 a t = r r 1. ˆ tr t dt = a t dη η = r. /3 3 dt H 0t = r. so that ˆ tr t dt t /3 = /3 3 H 0 r t 1/3 r t 1/3 = r /3 3 3 H 0. so that This can be written as t r = r 3 /3 3 H 0 + t 1/3 3 = t 1 + r 3t t r = t 1 + a t 3 r. 3t /3 3 3 H 0t. 9

10 3. Now let us calculate the redshift. The simplest physical way to get the result is to say that the wavelength of the photon get s stretched λ a hence thus the ratio of frequencies is ω r ω = a t a t r = t ω ω 1 = λ 1 λ = a 1 a 8 t r /3 = 1 + a t r < 1. 3t Thus the photon looses energy because of the stretching of the space. This is the solution expected. We could also proceed more formally and compare ω = u µ k µ = k t = k t in synchronous coordinate frame for both observers which are comoving in this synchronous reference frame with u µ = δ µ t. The geodesic equation for the photon four-momentum is We need to solve it only for k t = ω because of the equation 7. We have dω dλ + Γ t tt + 1 a Γt rr a Γt rt ωk t = 0. For the Christoffel symbols we get: dk µ dλ + Γµ αβ kα k β = 0 9 Γ t tt = 1 gtµ g µtt + g µtt g ttµ = 0 Γ t rr = 1 gtµ g µrr + g µrr g rrµ = 1 g rrt = aȧ and Γ t rt = 1 gtµ g µtr + g µrt = 1 g ttr + g trt = 0. The geodesics equation after substitution of Christoffel symbols takes the form further we can use that and obtain which implies dω dλ + ȧ a ωkt = 0. k t = dt dλ dω dλ + 1 da a dλ ω = 0 d ln ω = d ln a and consequently along the geodesic ωa = const which is exactly the relation used in 8. 10

11 V. SPACECRAFT AND ROTATING STAR 8 POINTS Consider an approximate metric far outside of a rotating star ds = dt + 4J sin θ dtdφ r 1 dr r sin θdφ + dθ where M and J are constants and c = G N = 1. Suppose there is a spacecraft at r = r and some angle θ moving with the the four-velocity where Ω = const. u t + Ω φ 1. P Find components u µ of the four-velocity of the spacecraft.. 1P Find the duration of the proper time of for the pilots τ corresponding to a short time interval t. 3. 5P Find components a µ of the acceleration experienced by the pilots of the spacecraft. Evaluate the expression at Ω = 4J/r 3. Solution V This is the metric far outside of any rotating body. In particular this is the asymptotic behavior for the Kerr metric from the problem sheet 1. This problem is related to what we have learned in this problem sheet. 1. The four-velocity has to be normalized thus u µ = γ δ µ t + Ωδ µ φ where we need to find γ. 1 = g µν u µ u ν = γ + 4J sin θ Ω r Ω sin θ = γ r 4J + r 3 Ω Ωr sin θ so that Thus u t = γ and u φ = γω.. Now we have γ = 1/ 4J + r 3 Ω Ωr sin θ. u t = dt dτ thus for sufficiently short infinitesimal time-intervals we get τ = t u t = 1/ 4J + r 3 Ω Ωr sin θ t. 3. The acceleration is a µ = u λ λ u µ = u λ λ u µ + Γ µ αβ uα u β. u µ does not depend on t or φ therefore u λ λ u µ = u t t u µ + u φ φ u µ = 0 and the acceleration is a µ = Γ µ αβ uα u β = Γ µ tt u t + Γ µ φφ u φ + Γ µ tφ ut u φ = γ Γ µ tt + Ω Γ µ φφ + Γµ tφ Ω. 11

12 Thus the main task is to calculate Christoffel symbols: Γ µ tt = 1 gµν g νtt + g νtt g ttν = 1 g ttνg µν = 1 g ttrg µr = 1 g ttrg rr δ r µ where we have used that the metric is stationary and nothing depends on time. Γ µ φφ = 1 gµν g νφφ + g νφφ g φφν = 1 g φφνg µν = 1 g φφrg µr 1 g φφθg µθ where we have used that the metric does not depend on φ. Further Γ µ φφ = 1 g φφrg µr 1 g φφθg µθ = 1 g φφrg rr δ r µ 1 g φφθg θθ δ µ θ. And finally Γ µ tφ = 1 gµν g νtφ + g νφt g tφν = 1 g tφνg µν = 1 g tφθg θθ δ µ θ 1 g tφrg rr δ r µ. Now let s take derivatives of the metric. Γ µ tt = 1 g ttrg rr δ µ r = 1 d 1 M dr r δ r µ = M r δµ r Γ µ φφ = 1 g φφrg rr δ µ r 1 g φφθg θθ δ µ θ = r sin θδ µ r sin θ δ µ θ Γ µ tφ = 1 g tφθr δ µ θ + 1 g tφr δ µ r = θ J sin θ r r δ µ θ + r J sin θ r δ µ r Γ µ sin θ tφ = J r 3 δ µ θ sin θ r δ r µ. a µ = γ M r δµ r Ω r sin θδ µ r + sin θ δ µ sin θ θ + ΩJ r 3 δ µ θ sin θ r δ r µ a µ = γ [ M r Ωr sin θ Ω + J r 3 δ r µ + Ω ] 4J sin θδµ θ r 3 Ω. For Ω = 4J/r 3 we have a θ = 0 and γ = 1/ 4J + r 3 Ω Ω r sin θ = 1/ or in other terms a r = M r 4J r sin θ 6J r 3 = M r 4J r 5 sin θ a r = M r 1 4J r 3 M sin θ. 1