Solutions Ph 236a Week 4

Solutions Ph 236a Week 4 Page 1 of 13 Solutions Ph 236a Week 4 Kevin Barkett, Jonas Lippuner, and Mark Scheel October 30, 2015 Contents Problem 1................................... 2 Part a................................... 2 Part b................................... 2 Problem 2................................... 3 Part a................................... 3 Part b................................... 4 Problem 3................................... 5 Part a................................... 5 Partb................................... 6 Partc.................................... 7 Problem 4................................... 9 Part a................................... 9 Part b................................... 9 Problem 5................................... 10 Part a................................... 10 Part b................................... 11 Problem 6................................... 13

Solutions Ph 236a Week 4 Page 2 of 13 Part a Problem 1 We are given that the first law of thermodynamics for a relativistic fluid can be written as dρv = P dv + T ds 1.1 which uses the fact that the total number of baryons, N, is conserved. That means we can write some of the quantities in terms of the number of baryons and the density of baryons. Let n be the baryon number density. Then we can write the volume and entropy as V = N/n S = sn 1.2 where s is the entropy per baryon. Then we can rewrite the first law as d ρ N N = P d + T dns 1.3 n n but since N is constant, we can divide it out. ρ 1 d = P + T ds n n dρ n ρdn dn n 2 =P d n 2 + T ds dρ =P + ρ dn n + nt ds 1.4 which is the desired result. Part b We are given that for a perfect fluid that T µν = P + ρu µ u ν + P g µν 1.5 and that the conservation of baryons is nu α,α = 0. In the fluid s rest frame, write out the 0 component of the equation of motion. T 0ν ;ν =P,ν g 0ν + P g 0ν ;ν + P + ρ,ν u ν u 0 + + P + ρ[u ν ;νu 0 + u 0 ;νu ν ] = 0 1.6 Now we know that g µν ;ν = 0 in general, and also that in the rest frame u 0 = 1 and u 0 ;ν = u α u α ;ν = 1 2 u αu α ;ν = 0 1.7

Solutions Ph 236a Week 4 Page 3 of 13 so that the equation of motion reduces to 0 = d dt P + d dt P + ρ + P + ρuν ;ν 1.8 Now let the baryon number density in the fluid s rest frame be n. number-flux vector of baryons is conserved, nu α ;α = 0 = n,ν u ν + nu ν ;ν = dn dt + nuν ;ν u ν ;ν = 1 n dn dt Now the This equation can be used to eliminate u ν ;ν in the equation of motion to give dρ dt = P + ρ dn n dt 1.9 1.10 If we compare this to the first law of thermodynamics from Part a, it must be that ds dt = 0 which is the condition for isentropy. Therefore the perfect fluid is isentropic. Part a Problem 2 For each case, we need to find the parameter value which evaluates the curve to the point P 0 = 0, 1, 0 and then compute the derivative of the function and evaluate at that point. x α λ = λ, 1, λ : First, set 0, 1, 0 = λ, 1, λ and we can see that here λ = 0. Now fx α λ = x 2 y 2 + z 2 = λ 2 1 + λ 2 = 2λ 2 1 2.1 Now take the derivative with respect to λ and evaluate at λ = 0 df dλ = 4λ λ=0 = 0 2.2 λ=0 x α ξ = sin ξ, cos ξ, ξ : Follow the same procedure as above to see that sin ξ, cos ξ, ξ = 0, 1, 0 ξ = 0 fx ξ ξ = x 2 y 2 + z 2 = sin ξ 2 cos ξ 2 + ξ 2 = 1 2 cos ξ 2 + ξ 2 df dξ = 4 sin ξ cos ξ + 2ξ λ=0 = 0 2.3 ξ=0

Solutions Ph 236a Week 4 Page 4 of 13 x ρ ξ = sinh ρ, cosh ρ, ρ + ρ 3 : sinh ρ, cosh ρ, ρ + ρ 3 = 0, 1, 0 ρ = 0 fx ρ ρ = x 2 y 2 + z 2 = sinh ρ 2 cosh ρ 2 + ρ + ρ 3 2 =ρ 6 + 2ρ 4 + ρ 2 1 df dρ = 6ρ 5 + 8ρ 3 + 2ρ 0 = 0 2.4 ρ=0 Part b Computing the components of the vectors d/dλ, d/dξ, and d/dρ is the same thing as asking for the directional derivatives along each of the curves evaluated at the same point as in part a. The directional derivative at a point P 0 is defined as d = dλ P 0 dx α dλ P 0 d dx α 2.5 So in the case we are considering with Catesian coordinates and implied Minkowski metric d dx d = dλ P 0 dλ dx + dy d dλ dy + dz d 2.6 dλ dz P 0 Now apply this to each of the curves given in the problem d d = dλ λ=0 dx + 0 + d dz λ=0 d = cos ξ d dξ ξ=0 dx sin ξ d dy + d dz ξ=0 d = cosh ρ d dρ ρ=0 dx sinh ρ d dy + 1 + 3ρ2 d dz ρ=0 = d dx + d dz = d dx + d dz = d dx + d dz 2.7 2.8 2.9

Solutions Ph 236a Week 4 Page 5 of 13 Part a Problem 3 Let f : M R where M is the manifold. dropping the vector arrows for convienence, Let s expand all of the terms [U, [V, W ]] =UV W f UW V f V W Uf + W V Uf [V, [W, U]] =V W Uf V UW f W UV f + UW V f [W, [U, V ]] =W UV f W V Uf UV W f + V UW f 3.1 Sum up the above 3 equations and find that all of the terms cancel. Thus we have [U, [V, W ]] + [V, [W, U]] + [W, [U, V ]] = 0

Solutions Ph 236a Week 4 Page 6 of 13 Partb

Solutions Ph 236a Week 4 Page 7 of 13 Partc First, let s do this the easy way. If e α are basis vectors, so [e α, e β ] = cαβ e, 3.2 [e λ, [e α, e β ]] = [e λ, c αβ e ] = cαβ,λ e + cαβ [e λ, e ] = cαβ,λ e + c = cαβ,λ + c αβ c ρ λ e ρ ρ αβ c λρ e, 3.3 where cαβ,λ or equivalently λcαβ means e λ acting on cαβ, i.e. the directional derivative along e λ of the scalar cαβ. Then, the Jacobi identity says that the sum of 3.3 for cyclic combinations of λ, α, β is zero. You get cαβ,λ + c βλ,α + c λα,β + c ρ αβ c λρ + c ρ βλ c αρ + c ρ λα c βρ = 0. 3.4 It is instructive to derive this the hard way, using full vector components. Define the vectors as follows, U = u α e α V = v β e β V = w σ e σ 3.5 where each of the e α are basis vectors. vectors can be written as Then the commutator between two [V, W ] = [v β e β, w σ e σ ] = v β w σ [e β, e σ ] + v β e σ β w σ w σ e β σ v β = v β w σ cβσ e + v σ σ w w σ σ v e, = v β w σ cβσ + vσ σ w w σ σ v e, = Q e, 3.6 where in the last line I defined Q as a shortcut. The commutator between three vectors is [U, [V, W ]] = [u α e α, Q e ] = u β Q σ cβσ σ Q Q σ σ u e, 3.7 where I have repeated the steps in Eq. 3.6. Now let s evaluate the derivative of Q, which we need in the above equation: λ Q = cβα wα λ v β + v β λ w α + v β w α λ c βα + λ v β β w λ w β β v + v β λ β w w β λ β v 3.8

Solutions Ph 236a Week 4 Page 8 of 13 Then ω, [U, [V, W ]] = v β w α cβα + vα α w w α α v u δ c δλ λu + u λ cβα wα λ v β + v β λ w α + v β w α λ c βα + λ v β β w λ w β β v + v β λ β w w β λ β v. 3.9 To evaluate the Jacobi identity, we sum the previous equation for cyclic combinations of U, V, and W, and set the result to zero. This is straightforward, but a bit tedious. When we do this, we get 0 = u α v β w cαβ,λ + c βλ,α + c λα,β + c ρ αβ c λρ + c ρ βλ c ρ αρ + c u β v α λ w + v β w α λ u + w β u α λ v cβα + u λ v β λ β w + v λ w β λ β u + w λ u β λ β v u λ w β λ β v v λ u β λ β w w λ v β λ β u + 12 terms with c βα times 1st derivs of u, v, or w λ λα c βρ + 12 terms with no cβα and two 1st derivs of u, v, or w. 3.10 The terms on the last two lines on Eq. 3.10, which I didn t write out explicitly, contain either one or two first derivatives of u, v, or w. These terms cancel out Just write them out, there are 24 of them. Each term appears twice, but with βα opposite sign, taking into account the fact that c is antisymmetric on the first two indices. However, we need a bit more work to simplify the remaining lines. Let s first look at the two terms that have a second derivative of w: the first term on the 3rd line of Eq. 3.10, and the second term on the 4th line of Eq. 3.10. These terms can be written u λ v β λ β w v λ u β λ β w = u λ v β λ β w β λ w = u λ v β [eλ,e β ]w = u λ v β c ρ λβ eρw ρ = u λ v β cλβ ρw. 3.11 Note that this expression is the same as minus the first term on line 2 of Eq. 3.10, so it cancels out that term. Likewise, the two terms on line 3 and 4 of Eq. 3.10 that have second derivatives of u cancel out another term on line 2 of Eq. 3.10, and the terms that have second derivatives of v cancel out the remaining term on line 2. So in the end, only the first line of Eq. 3.10 survives, and this is line is the same as Eq. 3.4 since it must hold for arbitrary u, v, and w.

Solutions Ph 236a Week 4 Page 9 of 13 Part a Expanding the derivative yields Problem 4 g αβ, = g µα g νβ g µν, = g µα, g νβ g µν + g µα g νβ, g µν + g µα g νβ g µν, = g µα, δ β µ + g νβ, δ α ν + g µα g νβ g µν, = g αβ, + g αβ, + g µα g νβ g µν, g αβ, = g µα g νβ g µν,, 4.1 where we have used the fact that g νβ g µν = g β µ = δ β µ, which was shown on the second homework. Thus we have shown the desired result. Part b Suppose we have a matrix C = Cɛ, for small ɛ, we can Taylor-expand Cɛ around ɛ = 0. We find Cɛ C0 + ɛ dc dɛ + Oɛ 2 = A + ɛb + Oɛ 2, 4.2 ɛ=0 where we have defined A = C0 and B = dc dɛ ɛ=0. To first order in ɛ, we find detcɛ = deta + ɛb = deta1 + ɛa 1 B = deta det1 + ɛa 1 B = deta1 + ɛ tra 1 B = deta + ɛ detatra 1 B. 4.3 And so ddetcɛ detcɛ detc0 dɛ = lim ɛ 0 ɛ=0 ɛ 1 dc = detc0tr [C0] dɛ, 4.4 ɛ=0 since A = C0 and B = dc dɛ ɛ=0. Because we can always shift the argument of the function, we can write the above as Thus for the matrix C = g µν we have det C = det C trc 1 C. 4.5 g,α = g trg µν g νλ,α = gg µν g νµ,α = gg µν g µν,α, 4.6 because C 1 = g µν, α C = g λ,α, matrix multiplication is contraction over inner indices i.e. C 1 α C = g µν g νλ,α, taking the trace is contracting the indices of a matrix i.e. tra = A µ µ, and finally we used symmetry of g µν in the last equality.

Solutions Ph 236a Week 4 Page 10 of 13 Part a Problem 5 On Homework 2 we found that the 1-forms dr = sin θ cos φ dx + sin θ sin φ dy + cos θ dz, dθ = cos θ cos φ r are dual to the basis vectors dx + cos θ sin φ r dy sin θ r dφ = sin φ r sin θ dx + cos φ r sin θ dy 5.1 dz, e r = r = sin θ cos φ e x + sin θ sin φ e y + cos θ e z e θ = θ = r cos θ cos φ e x + r cos θ sin φ e y r sin θ e z, e φ = φ = r sin θ sin φ e x + r sin θ cos φ e y. 5.2 Denote e r, e θ, e φ by e i, and denote eˆr, eˆθ, e by eî. We have eî = L i î e i, 5.3 where 1 0 0 L i î = 0 1/r 0. 5.4 0 0 1/r sin θ Denote dr, dθ, dφ by dx i, and define ωî = Lîi dx i, 5.5 where Lîi = 1 0 0 0 r 0. 5.6 0 0 r sin θ We find that ωî, eĵ = Lîi dx i, L j e ĵ j = LîiL j dx i, e ĵ j = LîiL j ĵ δi j = LîiL i ĵ 1 0 0 = 0 1 0 = δîĵ, 5.7 0 0 1

Solutions Ph 236a Week 4 Page 11 of 13 where we used the fact that dx i are dual to e j. We have thus shown that the basis 1-forms ωîdefined in 5.5 are dual to the orthonormal basis vectors eĵ. Evaluating 5.5 yields ωˆr = dr = sin θ cos φ dx + sin θ sin φ dy + cos θ dz, ω ˆθ = r dθ = cos θ cos φ dx + cos θ sin φ dy sin θ dz, ω = r sin θ dφ = sin φ dx + cos φ dy. 5.8 Part b Recall that c ˆk îĵ where is defined as [ eî, eĵ] = c ˆk îĵ eˆk, 5.9 [ eî, eĵ] = eî eĵ eĵ eî = eî eĵ eĵ eî. 5.10 Note that [ eî, eĵ] is antisymmetric and so c ˆk îĵ for î = ĵ and c ˆk îĵ We get 0 = cˆrˆr ˆr = c ˆθ ˆrˆr which means that We also find = c ˆk ĵî. Thus is antisymmetric. Hence c ˆk îĵ = 0 = cˆrˆr = cˆθ ˆθ ˆr = c ˆθ ˆθ ˆθ = c ˆr = c = c ˆθ = c. 5.11 [ eˆr, eˆθ] = eˆr eˆθ eˆθ eˆr = r = 1 r 2 θ + 1 r c c c ˆθ ˆθ 1 r 2 rθ 1 r ˆr ˆr = c ˆr ˆθ ˆθˆr ˆr ˆr ˆθ [ eˆr, e ] = eˆr e e eˆr = r = 1 r 2 sin θ ˆθ θ = 0, ˆθ = c = 1 ˆθ r, = c φ + 1 r sin θ ˆθˆr ˆθˆr 1 r sin θ 1 r θ r 2 θr = 1 r eˆθ, 5.12 = 0. 5.13 1 φ r sin θ 2 rφ 1 r sin θ φ r 2 φr = 1 r e, 5.14

Solutions Ph 236a Week 4 Page 12 of 13 which gives Finally, we obtain [ eˆθ, e ] = eˆθ e e eˆθ = 1 r and so = cos θ r 2 sin 2 θ To show that cij k of basis. We find ˆr c ˆr = c ˆr = 0, ˆr ˆθ c ˆr = c ˆθ = 0, ˆr c ˆr = c = 1 ˆr r. 5.15 1 θ r sin θ φ + 1 r 2 sin θ ˆr 1 φ r sin θ 2 θφ 1 r 2 sin θ cˆθ = c ˆr ˆθ = 0, ˆθ cˆθ = c ˆθ ˆθ = 0, cˆθ = c ˆθ 1 φ r θ 2 φθ = cos θ r sin θ e, 5.16 = cos θ r sin θ. 5.17 is not a tensor, consider its transformation under a change c k ī j e k = [ e ī, e j] = [ ī, k] = ī j j ī and thus = L i ī il j j j L j j jl i ī i = L i ī L j j,i j + L j j i j L L j j iī,j i + L iī j i = L i ī Lj j i j j i + L i ī Lk j,i Lj j Lk ī,j k = L i ī Lj j c ij k k + L i ī Lk j,i Lj j Lk ī,j L k k k = L i ī Lj j L k k cij k e k + L i ī Lk j,i Lj j Lk ī,j c k ī j L k k e k, 5.18 = L i ī Lj j L k k cij k + L i ī Lk j,i Lj j Lk ī,j L k k, 5.19 which means that the commutation coefficient cij k tensor because of the extra second term. does not transform like a Alternatively, one can use the following simpler argument. In some bases e.g. e x, e y, e z it is obvious that all commutation coefficients vanish, i.e. cij k = 0. If cij k were a tensor, then we would have c k ī j = L i ī Lj j L k k c k ij = 0, 5.20 in all other coordinate systems. But we have seen that this is not true in the orthonormal spherical coordinate system, thus cij k is not a tensor.

Solutions Ph 236a Week 4 Page 13 of 13 Problem 6 Let L α ᾱ be the transformation matrix from the unbarred into the barred frame, i.e. eᾱ = L α ᾱ e α. Then we find Γ ᾱ β e = e β eᾱ = L β β e β Lα ᾱ e α = L β β eβ L α ᾱ e α, 6.1 where we have used linearity of the covariant derivative. Using the chain rule, we get Γ ᾱ β e = L β eβ L β α ᾱ e α + L α ᾱ eβ e α, 6.2 where we have treated L α ᾱ like a scalar field, because it is not a tensor in the sense that it transforms like a tensor under a coordinate change. Therefore, eβ L α ᾱ = eβ L α ᾱ = L α ᾱ,β, 6.3 and so Γ ᾱ β e = L β β L α ᾱ,β e α + L α ᾱγ αβ e = L β βl α ᾱ,βl α e + L β βl α ᾱγ L αβ e = L β βl α ᾱl Γ αβ + Lβ βl αl α ᾱ,β e, 6.4 where we have just rearranged terms and expanded unbarred vectors in terms of barred ones. We have thus found Γ ᾱ β = Lβ βl α ᾱl Γ αβ + Lβ βl αl α ᾱ,β. 6.5 If the second term was not there, Γ αβ would indeed transform as a tensor, but because the second term is not 0 in general, Γ αβ does not transform like a tensor.