University of California at San Diego { Department of Physics { Prof. John McGreevy General Relativity (225A) Fall 2013 Assignment 4 { Solutions Posted October 16, 2013 Due Monday, October 28, 2013 1. Non-relativistic limit of a perfect fluid The stress-energy tensor for a perfect fluid in Minkowski space is T µν = (( + p) uµuν + pηµν) : µν Consider the continuity equation @µT = 0 in the nonrelativistic limit, p (recall that includes the rest mass!). Show that it implies the conservation of mass, and Euler's equation: ~ ~ ρ @t~v + ~v · r ~v = −rP: (See section 4.2 of Wald for more on this. Note that he uses ρ for and sets c = 1. ) The continuity equation says µν µ ν µ ν µ ν µν 0 = @µT = (@µ( + p)) u u + ( + p) ((@µu ) u + u @µu ) + (@µp) η : (1) The vector field u satisfies ν ν ν −1 = u uν =) 0 = @µ(uνu ) = 2 (@µu ) uν (in flat space). Projecting the conservation equation onto uν gives µ ν µ 0 = −u @µ ( + p) − ( + p)@νu + 0 + @µpu µ µ = −u @µ − @µu ( + p) (2) To find the content of the conservation law ? to uν add uν times (2) ν ν µ 0 = u (−( + p)@νu + 0 + @µpu ) to both sides of (1) to get: µ ν µ ν µν 0 = +u u @µp + ( + p) u @µu + @µpη ν ν µν µ ν = ( + p) u @µu + @µp (η + u u ) : (3) µ µ µ ~ Now take the nonrelativistic limit: p; u = (1;~v) in which case @µv ! −0+r·~v. Eqn (2) becomes µ µ p 1 @ 0 = u @ + @ u ( + p) ' +@ 0 + ~v · r~ + r~ · v 1 + ' + + r~ · (~v) µ µ x c @t 1 which is energy conservation. Eqn (3) becomes µ ν µν µ ν ~ ν ~ ν µν 0 = ( + p) u @µu +@µp (η + u u ) ' @0 + ~v · r u + (@0 + ~v · r)p u +@µpη Now look at the time and space components: 1 1 ν = 0 : 0 = (0) + @ p + ~v · r~ p − @ p = ~v · r~ p c t c t 1 1 ν 6= 0 : 0 = @ ~v + ~v · r~ ~v + ~v @ p + ~v · r~ p + r~ p c t c t In the NR limit v c, so the important terms are: 1 ν 6= 0 : 0 = @ ~v + ~v · r~ ~v + r~ p c t which is the Euler equation. 2. Stress tensors for fields in Minkowski space (a) Given a (translation-invariant) lagrangian density L(φ, @µφ) for a scalar field φ, define the energy-momentum tensor as µ @L µ Tν = − @νφ + δν L: @ (@µφ) µ Show that the equation of motion for φ implies the conservation law @µTν . The EoM for φ is δS @L @L 0 = = (x) − @µ (x) δφ(x) @φ @@µφ (see the lecture notes section 3.1), so µ @L @L @µTν = − @µ − @µ@νφ + @νL @ (@µφ) @ (@µφ) |{z} @L @L | {z } =@ν φ @φ +@ν (@µφ) @L @(@µφ) =− @φ @L @L @L @L = − @νφ − @µ@νφ + @νφ + @µ@νφ = 0: (4) @φ @ (@µφ) @φ @ (@µφ) (b) Show that the energy-momentum tensor for the Maxwell field 1 1 T µν = F µρF ν − ηµνF 2 EM 4πc ρ 4 µ is traceless, that is (TEM )µ = 0. 0 1 µ 1 µρ 1 2 Tµ = @F Fµρ − 4F A = 0: 4πc | {z } 4 =F 2 2 ν An apology: the notation Fρ in the pset statement is ambiguous: when raising and lowering indices on an object with multiply (non-symmetrized) indices, you ν have to keep track of which one was raised or lowered. I should have written F ρ (which is harder to TeX). (c) Show that the energy-momentum tensor for the Maxwell field 1 1 T µν = F µρF ν − ηµνF 2 EM 4πc ρ 4 in the presence of an electric current jµ obeys µν ρ ν @µTEM = −j F ρ: Explain this result in words. I am setting c = 1. Recall that Maxwell's equations with a source are µνρσ ρ 0 = @νFρσ;@ Fµρ = 4πjµ: We will need both. 1 2 @ T µν = (@ F µρ) F ν + F µρ@ F ν − ηµν @ F F ηαγηβδ µ EM 4π µ ρ µ ρ 4 µ αβ γδ 1 1 = − (@µF ) F νρ + F µρ @ F ν − (@νF ) F µρ 4π ρµ µ ρ 2 µρ 1 1 = −j F νρ + F µρηνσ @ F − @ F (5) ρ 4π µ σρ 2 σ µρ µρ 1 µρ But by antisymmetry of F : F (@µFσρ) = 2 F (@µFσρ − @ρFσµ), so µν νρ 1 µρ νσ ρ ν @µTEM = −jρF + F η (@µFσρ + @ρFµσ + @σFρµ) = −j F ρ : 8π | {z } ναβγ /µσρν @αFβγ =0 The EM field can do work on the currents and vice versa; therefore the energy and momentum in the EM field can turn into energy of the charges and vice versa, and is therefore not conserved. The RHS of the ν = 0 component is minus the work done on the particles by the fields; the RHS of the ν = i component is minus the force (change in momentum per unit time) in the i direction exerted on the particles by the Maxwell field. µ (d) Optional: show that tracelessness of Tµ implies conservation of the dilatation µ ν µ current D = x Tν . Convince yourself that the associated conserved charge R 0 space D is the generator of scale transformations. µ µ ν µ µ @µD = Tµ + x @µTν = Tµ : Z Z 0 0 i 0 i i @ D = tT0 + x Ti = tH + x P = −i t@t + x i space space @x α β α β which actions on functions of t; x by : (t@t + x@x) t x = (α + β)t x and expo- nentiates to eat@t f(t) = f(eat) (it shifts log t by a). 3 3. Polyakov form of the worldline action (a) Consider the following action for a particle trajectory xµ(t): Z dxµ dxν S [x] = −m dt g (x) : ? dt dt µν (Here gµν is some given metric. You may set gµν = ηµν if you like.) Convince yourself that the parameter t is meaningful, that is: reparametrizing t changes S?. Reparemetrizing by t~ = t~(t) changes the measure by one factor of J ≡ @tt~, but multipliesx _ 2 by J −2. Now consider instead the following action Z 1 dxµ dxν S[x; e] = − ds g (x) − m2e(s) : e(s) ds ds µν The dynamical variables are xµ(s) (positions of a particle) and e(s); e is called an einbein1: 2 2 2 ds1d = e (s)ds : 2 (b) Show that S[x; e] is reparametrization invariant if we demand that ds1d is an invariant line element. p Notice that the second term is just R g our usual way of making a coordinate- @ invariant integration measure. In the first term the @s s each transform like e, so 1 2 e x_ transforms like e. (c) Derive the equations of motion for e and xµ. Compare with other reparametrization- invariant actions for a particle. δS p 0 = = −e−2x_ 2 + m2 =) e = m x_ 2 δe(s) Z p S[x; e] = −m ds x_ 2 is the reparam-invariant action we discussed before, so obviously the EoM agree. Even the values of the action agree when we set e equal to the solution of its EoM. (d) Take the limit m ! 0 to find the equations of motion for a massless particle. In this limit, e is a Lagrange multiplier which enforces that the tangent vector is lightlike: δS 0 = m=0 / x_ 2 : δe(s) The EoM for x is still the geodesic equation. Now we can't choose s to be the proper time, but we have to demand that the lengths of the tangent vectors are independent of s { they are all zero. So any parametrization satisfying the lightlike-constraint is affine. 1that's German for `the square root of the metric in one dimension' 4 4. Show that S2 using stereographic projections (aka Poincar´emaps)for the coordinate charts (see the figure) 2 2 2 2 xN : S − fnorth poleg ! IR xS : S − fsouth poleg ! IR is a differentiable manifold of dimension two. More precisely: 3 (a) Write the Poincar´emaps explicitly in terms of the embedding in IR (f(x1; x2; x3) 2 3 P 2 2 IR j i xi = 1g ! IR ). If we place the center of the unit sphere at the origin, the plane in the picture is fx3 = −1g. The line LN in the picture passes through i(N) = (0; 0; 1) (the embedding of the north pole) and i(p) = (x; y; z), the embedding of the point of interest p. The N N intersection of LN with z = −1 determines xN (p) = (x1 ; x2 ; −1). We can parametrize ~ ~ ~ it as LN (t) so LN (0) = i(N) = (0; 0; 1) and LN (1) = (x; y; z), which gives ~ N N N LN (t) ≡ (x1 (t); x2 (t); x3 (t)) = (tx; ty; 1 − t + tz): N 2 The value of t = tN where −1 = x3 (tN ) = 1 − (1 − z)tN tN is then tN = 1−z . Notice that this is a nice function as long as z 6= 1, and that's why we must exclude the north pole (where (x; y; z) = (0; 0; 1)) from this patch. The coordinates of the stereographic projection from N are then 2 x (p) = (xN (t ); xN (t )) = (t x; t y) = (x; y): N 1 N 2 N N N 1 − z For the south pole, the plane is at x3 = +1 and the line in question is LS(t) = (tx; ty; (1 − t)(−1) + tz) −2 so the value of t where x3(tS) = +1 is tS = 1+z and −2 x (p) = (xN (t ); xN (t )) = (x; y): s 1 S 2 S 1 + z −1 2 2 (b) Show that the transition function xS ◦ xN : IRN ! IRS is differentiable on the overlap of the two coordinate patches (everything but the poles).) 5 The overlap UN \US where both coord systems are defined is everywhere but the north N N x1 z+1 x2 and south poles.