General Relativity (225A) Fall 2013 Assignment 4 – Solutions

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 · ∇ ~v = −∇P.

(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+∇·~v. Eqn (2) becomes

µ µ p 1 ∂ 0 = u ∂ + ∂ u ( + p) ' +∂ 0 + ~v · ∇~ + ∇~ · v 1 + ' + + ∇~ · (~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 · ∇ u + (∂0 + ~v · ∇)p u +∂µpη

Now look at the time and space components: 1 1 ν = 0 : 0 = (0) + ∂ p + ~v · ∇~ p − ∂ p = ~v · ∇~ p c t c t 1 1 ν 6= 0 : 0 = ∂ ~v + ~v · ∇~ ~v + ~v ∂ p + ~v · ∇~ p + ∇~ p c t c t In the NR limit v c, so the important terms are: 1 ν 6= 0 : 0 = ∂ ~v + ~v · ∇~ ~v + ∇~ 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.   µ 1 µρ 1 2 Tµ = F Fµρ − 4F  = 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. √ 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 √ 0 = = −e−2x˙ 2 + m2 =⇒ e = m x˙ 2 δe(s) Z √ 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 − {north pole} → IR xS : S − {south pole} → IR

is a diﬀerentiable manifold of dimension two. More precisely: 3 (a) Write the Poincar´emaps explicitly in terms of the embedding in IR ({(x1, x2, x3) ∈ 3 P 2 2 IR | i xi = 1} → IR ).

If we place the center of the unit sphere at the origin, the plane in the picture is {x3 = −1}. 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. Notice that S = = S . The transition functions are nice and x1 z−1 x2 diagonal N S S x1 z + 1 1 0 x1 S S x1 N = S ≡ M(x1 , x2 ) S x2 z − 1 0 1 x2 x2 S – notice that we must think of z here as a function of the coordinates x1,2. Explicitly it is ﬁndable by 2 2 2 2 2 x + y 1 − z r2 ≡ xS + xS = 4 = 4 1 2 (1 + z)2 (1 + z)2 −2r2 ± p4r4 − 4(r2 − 4)(r2 + 4) r2 ∓ 4 =⇒ z = = − . 2(r2 + 4) r2 + 4 The bottom root is just z = −1 which is not what we want – −1 < z < 1 on the overlap of patches. So the top root it is: 2 2 4 − r2 4 − xS − xS z = = 1 2 . 4 + r2 S 2 S 2 4 + (x1 ) + (x2 ) And z+1 = −4 = −4 and therefore the transition map is z−1 r2 S 2 S 2 (x1 ) +(x2 ) xN −4 xS 1 = 1 xN S 2 S 2 xS 2 (x1 ) + (x2 ) 2 The inverse map is obtainable noticing that xN 2 + xN 2 = 16 , so it has 1 2 S 2 S 2 (x1 ) +(x2 ) exactly the same form, but with N ↔ S. The transition function −1 φ = xN ◦ xS : xS(US ∩ UN ) → xN (US ∩ UN ) N 2 N 2 is then perfectly smooth away from the points where 0 = x1 + x2 or 0 = S2 S2 x1 + x2 which are just S and N.

5. Verify explicitly that if ωµ is a one-form (cotangent vector), then ∂µων −∂νωµ transforms as a rank-2 covariant tensor. µ a µ ˜ Under the coordinate change x → x˜ (x), ωµ(x) → ωã(˜x) = Ja ωµ(x), and ∂µ → ∂a = µ µ ∂xµ ˜ µ Ja ∂µ with Ja ≡ ∂xã = ∂ax . ˜ ˜ (dω)µν = ∂µων − ∂νωµ → ∂aω˜b − ∂bωã ˜ ˜ µ = ∂a ∂bx ωµ − (a ↔ b) ˜ ˜ µ ˜ ˜ µ ˜ µ ˜ ˜ µ ˜ = ∂a∂bx − ∂b∂ax ωµ + ∂bx ∂aωµ − ∂ax ∂bωµ | {z } =0 chain rule ˜ µ ˜ ν ˜ µ ˜ ν = ∂bx ∂ax ∂νωµ − ∂ax ∂bx ∂νωµ relabel dummy indices ˜ ν ˜ µ ˜ µ ˜ ν = ∂bx ∂ax ∂µων − ∂ax ∂bx ∂νωµ factorize ˜ µ ˜ ν = ∂bx ∂ax (∂µων − ∂νωµ) (6)

6 6. Lie brackets. The commutator or Lie bracket [u, v] of two vector ﬁelds u, v on M is deﬁned as follows, by its action on any function on M:

[u, v](f) = u(v(f)) − v(u(f)) .

(a) Show that its components in a coordinate basis are given by

µ ν µ ν µ [u, v] = u ∂νv − v ∂νu .

µ ν µ ν [u, v](f) = u ∂µ(v ∂ν(f)) − v ∂µ(u ∂ν(f)) product rule µ ν µ ν µ ν µ ν = u (∂µv )∂νf − v (∂µu )∂νf + u v ∂µ∂νf − v u ∂µ∂νf rename dummy indices µ ν µ ν µ ν ν µ = (u (∂µv )∂ν − v (∂µu )∂ν) f + u v ∂µ∂νf − v u ∂µ∂νf µ ν µ ν = (u (∂µv )∂ν − v (∂µu )∂ν) f (7)

So the components of the bracket are as given above. (b) Using the fact that uµ, vµ transform as contravariant vectors, show explicitly that [u, v]µ also transforms this way. This is very similar to the problem about the transformation of dω above. I will do this one in a more geometric way, by studying the vector itself, rather than the components :

µ ν µ ν a ˜ b ˜ u (∂µv )∂ν − v (∂µu )∂ν → u˜ (∂av˜ )∂b − (u ↔ v) ∂x˜a (J a ≡ )= J auµ((J −1)σ∂ J bvν)(J −1)ρ∂ − (u ↔ v) µ ∂xµ µ a σ ν b ρ a −1 σ σ σ ρ (Jµ(J )a = δµ)= u (∂σv )∂ρ − (u ↔ v) rename dummy indices µ ν µ ν = u (∂µv )∂ν − v (∂µu )∂ν. (8)

(c) (Optional extra bit) Convince yourself from the general deﬁnition of Lie derivative given in lecture that Luv = [u, v]. See the footnote in section 5.3.1 of the lecture notes.