General Relativity Fall 2019 Homework 2 Solutions

General Relativity Fall 2019 Homework 2 solutions

Exercise 1: Christoﬀel symbols

(i) In class we derived an expression for the Christoﬀel symbols in terms of the metric components and their derivatives, 1 Γα = gασ (g + g − g ) . (1) βγ 2 σβ,γ σγ,β βγ,σ

0 We also showed that, starting from a ICS {xµ }, and transforming to a general coordinate system {xµ} (note that here I chose to have primes for the ICS, and no primes for the general coordinates), we arrive at the geodesic equation with

0 ∂xα ∂2xα Γα = . (2) βγ ∂xα0 ∂xβ∂xγ

0 Show that this is indeed identical to the Christoﬀel symbol – don’t forget that {xµ } is deﬁned to be an ICS!

µ0 By deﬁnition, gµ0ν0 = ηµ0ν0 since {x } is an ICS. Changing coordinates, we have

0 0 ∂xµ ∂xν g = η 0 0 . (3) σβ ∂xσ ∂xβ µ ν Taking the derivative with respect to xγ we get

0 0 0 0 ! ∂2xµ ∂xν ∂xµ ∂2xν g = + η 0 0 . (4) σβ,γ ∂xσ∂xγ ∂xβ ∂xσ ∂xβ∂xγ µ ν

Thus we obtain

gσβ,γ + gσγ,β − gβγ,σ 0 0 0 0 0 0 0 0 0 0 0 0 ! ∂2xµ ∂xν ∂xµ ∂2xν ∂2xµ ∂xν ∂xµ ∂2xν ∂2xµ ∂xν ∂xµ ∂2xν = + + + − − η 0 0 . (5) ∂xσ∂xγ ∂xβ ∂xσ ∂xβ∂xγ ∂xσ∂xβ ∂xγ ∂xσ ∂xβ∂xγ ∂xβ∂xσ ∂xγ ∂xβ ∂xσ∂xγ µ ν

The third and ﬁfth terms cancel out. Since ηµ0ν0 is symmetric, the ﬁrst and last terms also cancel out. Moreover, the second and fourth terms are identical. So we are left with

0 0 ∂xµ ∂2xν g + g − g = 2 η 0 0 . (6) σβ,γ σγ,β βγ,σ ∂xσ ∂xβ∂xγ µ ν We also know how the inverse metric transforms: α σ ∂x ∂x 0 0 gασ = ηα σ , (7) ∂xα0 ∂xσ0

0 where we recall that {xµ } is an ICS. Thus we get

α σ µ0 2 ν0 1 ασ ∂x ∂x α0σ0 ∂x ∂ x g (g + g − g ) = η η 0 0 . (8) 2 σβ,γ σγ,β βγ,σ ∂xα0 ∂xσ0 ∂xσ ∂xβ∂xγ µ ν I highlighted the summed-over index σ. Upon summation, we get

α 2 ν0 α 2 ν0 α 2 ν0 α 2 α0 1 ασ ∂x α0σ0 µ0 ∂ x ∂x α0µ0 ∂ x ∂x ∂ x α0 ∂x ∂ x g (g + g − g ) = η δ 0 η 0 0 = η η 0 0 = δ 0 = , 2 σβ,γ σγ,β βγ,σ ∂xα0 σ ∂xβ∂xγ µ ν ∂xα0 ∂xβ∂xγ µ ν ∂xα0 ∂xβ∂xγ ν ∂xα0 ∂xβ∂xγ

0 which proves the equality. Note: it was key to use the fact that {xµ } is an ICS. 2

(ii) Using either expression, compute the transformation law of Christoﬀel symbols between general coordinate systems.

00 0 I will use the coordinate deﬁnition. Suppose now that {xµ } is an ICS, and that {xµ} and {xµ } are general coordinates. We use our very farvorite rule (the chain rule) to obtain

α0 2 α00 α0 α " γ α00 # α0 α " 2 γ α00 γ β 2 α00 # α0 ∂x ∂ x ∂x ∂x ∂ ∂x ∂x ∂x ∂x ∂ x ∂x ∂x ∂x ∂ x Γ 0 0 = = × = × + . β γ ∂xα00 ∂xβ0 ∂xγ0 ∂xα ∂xα00 ∂xβ0 ∂xγ0 ∂xγ ∂xα ∂xα00 ∂xγ0 ∂xβ0 ∂xγ ∂xγ0 ∂xβ0 ∂xγ ∂xβ

α 00 The second term can be rewritten in terms of Γβγ , and the ﬁrst term can be simpliﬁed when summing over α :

α0 2 α α0 β γ α0 ∂x ∂ x ∂x ∂x ∂x α Γ 0 0 = + Γ . (9) β γ ∂xα ∂xβ0 ∂xγ0 ∂xα ∂xβ0 ∂xγ0 βγ We will see later on that this is NOT the transformation rule for a tensor (it would be if the ﬁrst piece was not there), i.e. the Christoﬀel symbol is NOT a tensor.

(iii) Show that the following quantity transforms as a vector under general coordinate transformations: d2xµ dxρ dxσ aµ ≡ + Γµ . (10) dτ 2 ρσ dτ dτ These are the components of the 4-acceleration vector in a general coordinate system.

0 0 We already know that dxµ /dτ = (∂xµ /∂xµ) dxµ/dτ – this is just the chain rule. The next step is to compute how 0 d2xµ /dτ 2 transforms, (we basically already did this calculation in lecture 3):

0 0 ! 0 0 ! 0 0 d2xµ d ∂xµ dxµ ∂xµ d2xµ dxµ d ∂xµ ∂xµ d2xµ dxµ dxν ∂2xµ = = + = + . (11) dτ 2 dτ ∂xµ dτ ∂xµ dτ 2 dτ dτ ∂xµ ∂xµ dτ 2 dτ dτ ∂xµ∂xν

We see that this does not transform as a vector on its own. Combining with what we foud earlier, we now get

2 µ0 ρ0 σ0 µ0 2 µ µ ν 2 µ0 µ0 2 µ µ0 ρ σ ! ρ0 σ0 λ δ µ0 d x µ0 dx dx ∂x d x dx dx ∂ x ∂x ∂ x ∂x ∂x ∂x µ ∂x ∂x dx dx a = +Γ 0 0 = + + + Γ . dτ 2 ρ σ dτ dτ ∂xµ dτ 2 dτ dτ ∂xµ∂xν ∂xµ ∂xρ0 ∂xσ0 ∂xµ ∂xρ0 ∂xσ0 ρσ ∂xλ ∂xδ dτ dτ

µ This looks pretty awful, but can be simpliﬁed a lot. Combining the ﬁrst term and the term multiplying Γρσ just gives 0 ∂xµ µ ∂xµ a . So we wan to show that the other two terms actually cancel out. Upon renaming of dummy indices, we want to show that

0 0 0 0 ∂2xµ ∂xµ ∂2xλ ∂xρ ∂xσ + = 0. (12) ∂xµ∂xν ∂xλ ∂xρ0 ∂xσ0 ∂xµ ∂xν Using the chain rule, we rewrite

0 0 0 0 0 0 " 0 ! 0 # ∂xµ ∂2xλ ∂xρ ∂xσ ∂xµ ∂xσ ∂ ∂xλ ∂xµ ∂ ∂xσ ∂xλ ∂xλ ∂2xσ = = − . (13) ∂xλ ∂xρ0 ∂xσ0 ∂xµ ∂xν ∂xλ ∂xν ∂xµ ∂xσ0 ∂xλ ∂xµ ∂xν ∂xσ0 ∂xσ0 ∂xµ∂xν

Note that we had to be careful: while partial derivatives with respect to unprimed coordinates commute, mixed partial derivatives do not: ∂ ∂f ∂ ∂f 6= . (14) ∂xν0 ∂xµ ∂xµ ∂xν0

This is because the variables that are kept constant are diﬀerent in one set of coordinates and the other. Let us compute the diﬀerence explicitly: ∂ ∂f ∂ ∂f ∂xν ∂2f ∂ ∂xν ∂f ∂ ∂xν ∂f − = − = − × ∂xν0 ∂xµ ∂xµ ∂xν0 ∂xν0 ∂xν ∂xµ ∂xµ ∂xν0 ∂xν ∂xµ ∂xν0 ∂xν 0 ∂xµ ∂2xν ∂f = − . (15) ∂xµ ∂xν0 ∂xµ0 ∂xν 3

λ Now back to our calculation, the term in parentheses is just δν , which is a constant, and the last term simpliﬁes to

µ0 2 λ ρ0 σ0 2 σ0 2 µ0 ∂x ∂ x ∂x ∂x µ0 ∂ x ∂ x = −δ 0 = . (16) ∂xλ ∂xρ0 ∂xσ0 ∂xµ ∂xν σ ∂xµ∂xν ∂xµ∂xν So we have indeed proved Eq. (12), thus found that

µ0 0 ∂x aµ = aµ, (17) ∂xµ i.e. the 4-acceleration (including the Christoﬀel symbol part) transforms as a vector under a change of coordinates.

Exercise 2: geodesics in diﬀerent metrics

Consider the following two line elements: ds2 = −dt2 + dr2 + r2dθ2 + r2 sin2 θdϕ2, (18) ds2 = a(t)2 −dt2 + dx2 + dy2 + dz2 . (19) For each line element:

(i) Derive the geodesic equation by extremizing Z dxµ dxν I ≡ dτ g . (20) µν dτ dτ

Let us first consider the first line element: Z Z I = dτ −t˙2 +r ˙2 + r2θ˙2 + r2 sin2 θϕ˙ 2 ≡ dτ L(t, r, θ, ϕ, t,˙ r,˙ θ,˙ ϕ˙), (21) where an overdot means d/dτ. We use the Euler-Lagrange equations to find the 4 equations of motion: d (−2t˙) = 0, (22) dτ d (2r ˙) = 2r θ˙2 + sin2 θϕ˙ 2 , (23) dτ d (2r2θ˙) = 2r2 sin θ cos θϕ˙ 2, (24) dτ d (2r2 sin2 θϕ˙) = 0. (25) dτ These equations simplify to

t¨= 0, r¨ − r θ˙2 + sin2 θϕ˙ 2 = 0, r˙ θ¨ + 2 θ˙ − sin θ cos θϕ˙ 2 = 0, r r˙ϕ˙ cos θ ϕ¨ + 2 + 2 θ˙ϕ˙ = 0. r sin θ

Let’s do the same thing for the other metric. Denote the spatial coordinates {x1, x2, x3} ≡ {x, y, z}, so the metric is 2 2 2 i j ds = a(t) [−dt + δijdx dx ]. The Lagrangian is then

i i 2 2 i j L(t, x , t,˙ x˙ ) = a(t) −t˙ + δijx˙ x˙ (26) Euler-Lagrange equations then give d da (−2a(t)2t˙) = 2a(t) −t˙2 + δ x˙ ix˙ j , (27) dτ dt ij d (2a(t)2x˙ i) = 0. (28) dτ 4

This implies

2 i j t¨+ H(t)t˙ + H(t)δijx˙ x˙ = 0, x¨i + 2H(t)t˙x˙ i = 0, where 1 da H(t) ≡ . (29) a(t) dt

(ii) From this equation, and the general expression for the geodesic equation, infer the Christoﬀel symbols (i.e. do not use the explicit expression of the Christoﬀel symbols in this question).

We can just read off the Christoffel symbols from the geodesic equations. The only non-zero symbols are, for the first metric:

r r 2 Γθθ = −r, Γϕϕ = −r sin θ, 1 Γθ = , Γθ = − sin θ cos θ, rθ r ϕϕ 1 cos θ Γϕ = , Γϕ = . rϕ r θϕ sin θ

For the second metric, we ﬁnd the following only non-zero Christoﬀel symbols:

t Γtt = H(t), t Γij = H(t)δij, i i Γtj = H(t)δj.

(iii) Now compute the Christoﬀel symbols explicitly from their expression in terms of the metric components, and check that you got the right answer in (ii).

There are a priori 4 × 6 = 24 different Chirstoffel symbols (remember that they are symmetric in the lower two indices). So before starting to compute them all, let’s first try to eliminate as many as possible. Both metrics are diagonal, so we get 1 Γα = gαα (g + g − g ) [NOT summed over α] (30) βγ 2 βα,γ γα,β βγ,α Since the metrics are diagonal, it means that at least two indices must be the same for non-zero Christoffel symbols (otherwise all the terms in the parenthesis involve metric coefficients with unequal indices). Let us now look at each metric.

The coefficients of the first line element only explicitly depend on r and θ. This means that non-zero Christoffel symbols have at least one r index or one θ index (so there can be a derivative with repsect to either r or θ). With the result from above, it means that the only possibly non-zero coefficients are a priori

r r r r t θ ϕ r r r Γtt, Γrr, Γθθ, Γϕϕ, Γrt, Γrθ, Γrϕ, Γrt, Γrθ, Γrϕ (31) θ θ θ θ t r ϕ θ θ Γtt, Γrr, Γθθ, Γϕϕ, Γθt, Γθr, Γθϕ, Γθt, Γθϕ (32) These are still 19 coeﬃcients to compute explicitly. I will not write the calculations, as it merely amounts to applying the formula, but you can see that it is a lot more painful than extracting them from the geodesic equation!

For the second line element, the metric coefficients only depend on t. This means that non-zero Chirstoffel symbol must include at least one t. With the requirement that two indices are the same, we thus find that the non-zero t t i coefficients are Γtt, Γii, Γti [the i are NOT repeated here], which is indeed the ones we had found. Computing them gives the same answer. 5

Exercise 3: Uniformly accelerated motion

µ ν µ ν µ (i) Starting from gµν u u = −1, show that gµν u a = 0, where a is the 4-acceleration deﬁned in the ﬁrst exercise.

In a ICS,

dxµ dxν g uµuν = η , (33) µν µν dτ dτ thus, in an ICS,

d dxµ d2xν (g uµuν ) = 2η . (34) dτ µν µν dτ dτ 2 But, in an ICS, we also have aν = d2xν /dτ 2, hence

d (g uµuν ) = 2η uµaν = 2g uµaν . (35) dτ µν µν µν But the right-hand-side is now coordinate independent, as you can check explicitly. On the other hand, the left-hand- µ ν side is d/dτ(−1) = 0. Thus, regardless of the coordinates in which this expression is computed, gµν u a = 0.

(ii) For the remainder of this problem, suppose that {xµ} is a globally inertial coordinate system. Consider a particle moving along the 1-direction (i.e. the x-direction), starting at rest from the origin at t = 0. Suppose that in the particle’s instantaneous rest-frame, the components of its 4-acceleration are constant. Assuming the particle carries with it a cartesian coordinate system, whose axes are aligned with the spatial axes of the glocal ICS, write explicitly the components of the 4-acceleration in the particle’s instantaneous rest-frame. This is a long question, with a very short answer!

Assuming that the particle uses a ICS in its rest-frame, the metric in its reste-frame is the Minkowski metric. The µ0 µ0 ν0 components of the 4-velocity in its rest-frame are {u } = (1, 0, 0, 0). Imposing ηµ0ν0 a u = 0 in the particle’s rest- 0 frame, this implies a0 = 0. If we further assume that motion is along the 1-direction, we thus get {aµ } = (0, 1, 0, 0) in the particle’s instantaneous rest frame, where a is constant.

(iii) In the globally inertial frame of reference, in which the particle is initially at rest at the origin, describe the particle’s motion, i.e. compute its 3-velocity, position, and proper time as a function of coordinate time t. Discuss how these results match with the non-relativistic limit.

In the global ICS, the particle has 3-velocity ~v = (v, 0, 0), and 4-velocity p {uµ} = γ(1, v, 0, 0), γ ≡ 1/ 1 − v2. (36)

Doing a Lorentz transformation from the instantaneous rest-frame, to the global ICS, which has velocity −~v with respect to the particle’s rest-frame, we get the following components of the 4-acceleration in the global ICS:

0 a0 = γ(a0 + va) = γva, (37) 0 a1 = γ(a + va0 ) = γa, (38) a2 = a3 = 0. (39)

Now the 4-acceleration is just aµ = duµ/dτ, since we are in an ICS (otherwise it contains another piece, see ﬁrst exercise). Thus we have the following equations:

dγ d(γv) = γva, = γa. (40) dτ dτ Note that these two equations are redundant (hence self-consistent) because, using γ2v2 = γ2 − 1, we have

d(γv) 1 d(γv)2 1 dγ2 1 dγ = = = . (41) dτ 2γv dτ 2γv dτ v dτ 6

Differentiating the first equation with respect to τ again, we find

d2γ d(γv) = a = a2γ. (42) dτ 2 dτ Thus we ﬁnd

γ(τ) = γ(0) cosh(aτ) +γ ˙ (0) sinh(aτ). (43)

Now, sinceγ ˙ = aγv and the particle is initially at rest, we ﬁnd that γ(0) = 1, γ˙ (0) = 0, thus

γ(τ) = cosh(aτ). (44)

Now, remember that γ = dt/dτ is the ﬁrst component of the 4-velocity.

dt 1 1 = cosh(aτ) ⇒ t = sinh(aτ) ⇒ τ = sinh−1(at) . (45) dτ a a

Now we want to express the velocity as a function of coordinate time: p v = p1 − 1/γ2 = γ2 − 1/γ = tanh(aτ). (46)

Using hyperbolic trigonometry, we get

at v(t) = . (47) p1 + (at)2

Integrating this, we ﬁnd

1 x = p1 + (at)2 − 1 . (48) a

We see that the velocity asymptotes to 1 (in units of the speed of light), and becomes relativistic on a timescale 1 2 t ∼ 1/a. As long as t 1/a, we have τ ≈ t, v ≈ at and x ≈ 2 at , as we expect in the Newtonian limit.