Geneal Relativity Homewok 5. In the pesence of a cosmological constant, Einstein s Equation is (a) Calculate the gavitational potential point souce with = M 3 (). R µ Rg µ + g µ =GT µ. in the Newtonian limit outsie a spheically symmetic (b) A goo esciption of the gavitational foces in ou sola system is given by Newton s potential, = GM /, wheem is the mass of the sun. Use this fact to etemine an uppe limit fo the vacuum enegy ensity = /G, given that the aius of Pluto s obit is Pluto 6 0 m. Expess you answe in GeV 4. Solution: a) Fom Caoll 4., we have Poisson s equation, =4 G. Howeve, in the pesence of a cosmological constant, the Ricci scala an tenso now become R µ = R = GT + 4, G T µ Tg µ + g µ. If you followe the eivation of Poisson s Equation fom Einstein s Equation, you ll ecognize that R 00 =4 G( ) an that the only i eence hee is that = M 3 (). The spheical symmety of the istibution equies that the potential shoul only epen on, the istance fom the point souce, so we shoul look fo a solution of the fom (). Poisson s equation now eas =4 G(M 3 () ()). Integating both sies ove the volume of a sphee of aius 0, V =4 G (M 3 () ())V. Applying Stokes theoem to the left han sie, V = ( )V = ( ) S, whee the integal is now ove the suface of the sphee. Due to spheical symmety must point in the ˆ iection an be constant on the suface of the sphee, so the integal becomes V =4 0. Now, to aess the ight han sie, 4 G (M 3 () ())V =4 GM 3 G 03 3 Putting the two togethe we have: GM ( 0 )= 0 3 G 0 ˆ.
To solve fo, one must integate fom the ege of the univese (let s call it ) to. () = ()+ ˆ = ()+ GM 0 3 G 0 0 = GM 4 3 G Note that we ve cancelle the potential at infinity with the evaluation of the anti-eivative at the ege of the univese. b) To fin an uppe boun on the vacuum enegy ensity, we ll look at the value necessay to keep Pluto boun to the Sola System. We can wite this as the equiement that the negative gaient of the potential (i.e. the acceleation) be negative (so that the acceleation is towas the cente of the sola system). ( Pluto )= GM Pluto + 3 G Pluto < 0 Isolating the vacuum enegy ensity, < 3 M 3 Pluto = 3.989 0 30 kg GeV (6 0 m) 3.78 0 7 kg < 4.7 0 30 GeV 4 0 9 GeV 4.97 0 6 m GeV ) 3. Caoll poblem 5.3. Consie a paticle (not necessaily on a geoesic) that has fallen insie the event hoizon, <. Use the oinay Schwazschil cooinates (t,,, ). Show that the aial cooinate must ecease at a minimum ate given by Calculate the maximum lifetime fo a paticle along a tajectoy fom = to = 0. Expess this in secons fo a black hole with mass measue in sola masses. Show that this maximum pope time is achieve by falling feely with E! 0. Solution: Fo the Schwazschil metic, So s = = = t + + +, t. The best a paticle can o to escape once insie the Schwazschil aius is to tavel on a aial path ( / = 0) at the spee of light. Anothe way to think about it is that < 0 fo <, so / is minimal when = 0. = t. Now let s consie the maximum lifetime of a paticle moving along a tajectoy fom = to = 0, 0 0 p max = = q = p. max 0
Defining x =, max = 0 x / ( x) / x = B(3/, /), whee B(x, y) is the Beta function. Using the elationship between the Beta function an the Gamma function, max = ( ) ( 3 ) =( p p )( )= GM. () Reinstating the pope factos of c, fo M = n M, max = Gn M c 3 = n (6.674 0 Nm )(.989 0 30 kg) kg (.99 0 8 m =.56 0 5 n s. s )3 x Since E = K µ µ E! 0 by assumption. =( ) t, an since we ve taken t/! 0 to obtain the minimum /, 3. Caoll poblem 5.4. Consie Einstein s equations in vacuum, but with a cosmological constant, G µ + g µ = 0. a) Solve fo the most geneal spheically symmetic metic, in cooinates (t, ) that euce to the oinay Schwazschil cooinates when = 0. b) Wite own the equation of motion fo aial geoesics in tems of an e ective potential, as in (5.66). Sketch the e ective potential fo massive paticles. Solution: a) We still have full spheical symmety an a static metic, so the geneal fom of the metic will be the same as in the Schwazschil souce case, s = e () t + e () +. The calculation of the Chisto el symbols an the Riemann an Ricci tensos follows ientically to that fo the Schwazschil metic. What oes change is the fom of the tace-evese Einstein equation in a vacuum, which is now Taking the tace of both sies, R µ Rg µ + g µ =0. R R + 4 = 0 ) R = 4, so the tace-evese equation is just Fist consie the tt an components, R µ = g µ. We can combine them to get so In tems of the metic (so in tems of an R tt = e, R = e. e R tt + e R =0, e ( + ) R tt + R =0. ), we fin @ + @ =0. 3
So afte escaling t, =, just as we foun fo the Schwazschil metic. Now look at the component, R = 4 R. Using the elationship we foun fo the Ricci scala an the cosmological constant, this becomes, e [(@ @ ) ] + =. Using the elation between an we fin e [ @ ] = +, ) @ (e + 3 3 )=. The solution is e = 3, whee, by the equiement that when the cosmological constant vanishes we ecove the Schwazschil solution, must be the usual Schwazschil aius. The metic is theefoe s = b) The invaiants of geoesic motion ae 3 t + = g µ x µ x, E = g tt t, an L = g. We can wite own the analog of (5.63) as t g tt + g + =. 3 + Fixing = /, we have E + + L 3 + =0. So with Ẽ = E /, we have Ẽ = + V (), V () = 6 + 6 L GM L GML 3. See the attache Mathematica notebook fo a plot. fo 4
V[_, GM_, L_, Lamba_] := - 6 Lamba + - 6 Lamba L - GM + L GM L - ; 3 Plot[{V[,,, -], V[,,, 0], V[,,, ]}, {, 0, 5}, Ticks None]