Download Parallel Transport and Geodesics on a 2-Sphere and more Summaries Astronomy in PDF only on Docsity!
General Relativity: Homework 3 Solutions
1. Carroll Problem 3.
Verify these consequences of metric compatability (∇σ gμν = 0):
∇σ gμν^ = 0
∇λμνρσ = 0
Solution:
To verify the first claim, recall that gμν^ is the inverse of gμν so that we have:
gμν^ gμν = δμμ = n
where n is the dimension of the manifold. Now take the covariant derivative of both sides, noting that the covariant derivative of a constant is zero, ∇σ (gμν^ gμν ) = 0 gμν^ (∇σ gμν ) + (∇σ gμν^ )gμν = 0 ⇒ gμν (∇σ gμν^ ) = 0
And therefore ∇σ gμν^ = 0.
To verify the second claim note Carroll equations 2.69 and 2.70 which imply
μνρσ μνρσ^ = ˜μνρσ ˜μνρσ^ = −n!
where n is the dimension of the manifold we are working over, and we are assuming that the metric is Lorentzian (this assumption is uneccessary, it’s just why the negative sign is there, all we really need is that the right-hand side is a constant). Taking the covariant derivatives of both sides we obtain
∇λ(μνρσ )μνρσ^ + μνρσ ∇λ(μνρσ^ ) = 0
Since μνρσ^ is an honest to goodness tensor, we may relate it to μνρσ by factors of the gμν^. Specifically we have μνρσ^ = gμαgνβ^ gρδ^ gσγ^ αβδγ
By inserting this expression into the covariant derivative, using the Leibniz rule, and the fact that ∇σ gμν^ = 0, we arrive at the expression
∇λ(μνρσ )μνρσ^ + μνρσ gμαgνβ^ gρδ^ gσγ^ ∇λ(αβδγ ) = ∇λ(μνρσ )μνρσ^ + αβδγ^ ∇λ(αβδγ ) = 0
Replacing the dummy indices in the second expression we arrive at
2 ∇λ(μνρσ )μνρσ^ = 0
Multiplying both sides by μνρσ implies ∇λμνρσ = 0.
2. Carroll Problem 3.
Consider a 2-sphere with coordinates (θ, φ) and metric
ds^2 = dθ^2 + sin^2 θdφ^2
(a) Show that lines of constant longitude (φ = constant) are geodesics, and that the only line of constant latitude (θ = constant) that is a geodesic is the equator (θ = π/2).
(b) Take a vector with components V μ^ = (1, 0) and parallel-transport it once around a circle of constant latitude. What are the components of the resulting vector, as a function of θ?
Solution:
(a) To verify a curve is a geodesic we need to write the geodesic equation for this metric. First we need the connection coefficients, to which we appeal to Carroll equation (3.154), although it is a worthwhile exercise to derive these for yourself. The nonzero coefficients are:
Γθφφ = − sin θ cos θ
Γφθφ = Γφφθ = cot θ Now we may write out the geodesic equation(s):
d^2 xμ dλ^2
dxσ dλ
which give us the two equations: d^2 θ dλ^2
− sin θ cos θ
( (^) dφ dλ
d^2 φ dλ^2
dφ dλ
In the case that φ is a constant the first equation reads d
(^2) θ dλ^2 = 0, while the second equation is trivial. Thus the curve xμ^ = (λ, φ) is a geodesic. If instead we assume θ is a constant we get:
sin θ cos θ
( (^) dφ dλ
d^2 φ dλ^2
Since we can’t set dφdλ = 0 (because then the ’curve’ would just be a point), if the first equation is to be satisfied we must have θ = π/2 (note that θ = 0 is not a well-defined coordinate). Therefore xμ^ = (π/ 2 , λ) is a geodesic, and in fact it is the only geodesic (up to reparameterization) where θ is constant.
(b) In order to parallel transport a vector we must choose a path (with parameterization) to transport along. Any parameterization will work, however it is best to keep it simple, so choose xμ^ = (θ, λ). Now we must solve the parallel transport equation:
d dλ
V μ^ + Γμσρ
dxσ dλ
V ρ^ = 0
Note that dx
μ dλ = (0,^ 1) so that we get the two equations: dV θ dλ
− sin θ cos θ V φ^ = 0 (1)
(b) Solve for a geodesic corresponding to a circular orbit around the equator of the Earth (θ = π/2). What is dφ/dt?
(c) How much proper time elapses while a satellite at radius R 1 (skimming along the surface of the Earth, neglecting air resistance) completes one orbit? You can work to first order in Φ if you like. Plug in the actual numbers for the radius of the Earth and so on to get an answer in seconds. How does this number compare to the proper time elapsed on the clock stationary on the surface?
Solution:
(a) For a clock on the surface of the Earth, the world-line (parameterized by t) is xμ^ = (t, R 1 , θ, ωt), where ω is the angular velocity of the Earth (ω = 12 π rad hrs ). Now calculate the proper-time:
∆τ =
∫ (^) t
0
−gμν
dxμ dt
dxν dt
dt′^ =
∫ (^) t
0
(1 + 2Φ) − R 12 sin^2 θω^2 dt′
∆τ =
2 GM
R 1
− R^21 sin^2 θω^2 ∆t
Now let’s reinstate the factors of c. 2 GMR 1 has units of velocity squared, so we should divide that by c^2. The same goes for the second term under the square root. Also note that R 1 ω << c, so we can drop that term (we also have R^21 ω^2 << 2 GMR 1 , which justifies leaving in 2 GMR 1 ), leaving us with:
∆τ =
2 GM
c^2 R 1
∆t
For the clock on a tall building we replace R 1 with R 2 and see that the proper-time is decreased. Note that this effect is purely due to the difference in gravitational potential (the effect due to time dilation associated with the difference in speeds is negligible).
(b) To find a geodesic we must look at the geodesic equations. In fact, because we are looking for a specific type of geodesics, namely one with constant radius, r and θ = π/2 we do not need to look at all of the equations. First look at the geodesic equation for xμ^ = r which reads:
d^2 r dλ^2
dxρ dλ
dxσ dλ
Keeping in mind that we are restricting ourselves to paths with constant r and constant θ, this expres- sion simplifies to Γrtt
( (^) dt dλ
( (^) dφ dλ
dt dλ
Now calculate just the necessary connection coefficients
Γrtt =
grρ(∂tgtρ + ∂tgtρ − ∂ρgtt) = −
grr^ ∂r gtt
Γrtt =
∂r (1 −
2 GM
r
GM
r^2 (1 − 2Φ)
Γrφφ =
grρ(∂φgφρ + ∂φgφρ − ∂ρgφφ) = −
grr∂r gφφ
Γrφφ = −
∂r (r^2 sin^2 θ) = −
r sin^2 θ (1 − 2Φ)
Γrφt =
grρ(∂φgtρ + ∂tgφρ − ∂ρgφt) = 0
Now the geodesic equation reads:
GM r^2 (1 − 2Φ)
( (^) dt dλ
r sin^2 θ (1 − 2Φ)
( (^) dφ dλ
dφ dλ
GM
r^3
dt dλ
We still haven’t found the geodesic, to do so now consider the equation for xμ^ = t:
d^2 t dλ^2
dxρ dλ
dxσ dλ
Again, considering only paths such that r is constant and θ = π/2 this simplifies to:
d^2 t dλ^2
( (^) dt dλ
( (^) dφ dλ
dt dλ
dφ dλ
Now look at the equation for the necessary connection coefficients
Γtμν =
gtt(∂μgνt + ∂ν gμt − ∂tgμν )
and we can see that for μ, ν ∈ {t, φ} the connection coefficients will all be zero. The geodesic equation is now just d^2 t dλ^2
the simplest non-trivial solution to which is t = λ. Taking this as our parameterization, (3) then implies: dφ dt
GM
r^3 Putting it all together, the (up to reparameterization) geodesic corresponding to a circular orbit (of radius r) around the equator (θ = π/2) is:
xμ(t) = (t, r, π/ 2 ,
GM
r^3
t)
(c) For the satellite skimming the surface of the Earth (at θ = π/2) we have that the elapsed proper-time for a complete orbit is:
∆τ =
2 GM
c^2 R 1
R^21 ω^2 c^2
2 π ω
where ω =
GM R^31 and we have included the term related to the motion of the satellite because it is on the same order as the gravitational potential. Plugging in the expression for ω we have
∆τ = 2π
2 GM
c^2 R 1
R^21
c^2
GM
R^31
R^31
GM
= 2π
R^31
GM
3 R^21
c^2
∆τ = 2π
(6. 371 × 106 m)^3 (6. 674 × 10 −^11 N m 2 kg^2 )(5.^972 ×^10 (^24) kg) −^
3(6. 371 × 106 m)^2 (2. 998 × 108 ms )^2 = 5061s
Now to calculate the the elapsed proper-time for the clock stationary on the surface of the Earth, borrow the result from part (a), setting θ = π/2.
∆τ =
2 GM
c^2 R 1
2 π ω = 2π
2 GM
c^2 R 1
R^31
GM
= 2π
R^31
GM
2 R^21
c^2
Similar calculations show that all three of these vectors satisfy the Killing vector condition. All that’s left to do is raise their indices, but that’s easy for this metric:
Rxμ^ = (0, 0 , −z, y)
Ryμ^ = (0, z, 0 , −x) Rzμ^ = (0, −y, x, 0)
Finally, by looking at the calculations for the three rotation generators, let’s guess that the three vectors corresponding to Bxμ = (−x, t, 0 , 0) Byμ = (−y, 0 , t, 0) Bzμ = (−z, 0 , 0 , t) are also Killing vectors (corresponding to boosts along each axis). Again, verify by following the same argument as for the rotations. Let’s check the non-trivial case for Bxμ:
∇tBxx + ∇xBxt = ∂t(t) + ∂x(−x) = 1 + (−1) = 0
Now we need to again raise the indices, and in this case we need to be wary of the metric we are using:
Bxμ^ = (x, t, 0 , 0)
Byμ^ = (y, 0 , t, 0) Bzμ^ = (z, 0 , 0 , t)
In all we have 10 Killing vectors, which is in agreement with the fact that the Poincar´e group (isometries of the Minkowski metric) is 10 dimensional.
(b) In order to verify Killing’s equation we will need to calculate the connection coefficients, a quick observation and use of the hints will let us get away without calculating all of the coefficients. Since, as the hints indicate, the u-component of all of the Killing vectors is zero, by lowering the index, Kugμν = −Kv , we see that this implies Kv = 0 for all Killing vectors. Therefore, by looking at the covariant derivative of a 1-form ∇μKν = ∂μKν − Γρμν Kρ we conclude that we do not need to calculate the Γvμν coefficients (upper index equals v). We do however need to calculate the rest of them. We will proceed with the variational technique, that is consider variations of the integral (Carroll eqn. 3.49) I =
gμν
dxμ dτ
dxν dτ
dτ =
du dτ
dv dτ
( (^) dx dτ
( (^) dy dτ
)dτ
To get the coefficients with upper index u, we will actually need to consider variations of v (this is due to the cross terms in the metric). Under variations v 7 → v + δv the only term in the integrand that changes is dvdτ 7 → (^) dτdv + d( dτδv )and therefore the variation in I is
δI =
du dτ
d(δv) dτ
d^2 u dτ 2 δvdτ
where the second equality comes from integrating by parts. Requiring that the variation vanishes we are left with d^2 u dτ 2
which implies that all coefficients with upper index u vanish, Γuμν = 0. Now we want the coefficients with upper index x, for which we consider variations x 7 → x + δx. Under such variations the only change in the integrand is ( (^) dx dτ
( (^) dx dτ
d(δx) dτ
( (^) dx dτ
dx dτ
d(δx) dτ
Therefore the variation of the integral (up to first order in δx) is
δI =
a^2 (u)
dx dτ
d(δx) dτ dτ
Integrating by parts we obtain
δI = −
a^2 (u)
d^2 x dτ 2
da du
du dτ
dx dτ
dτ
Setting the variation equal to zero leads to the equation
a^2 (u)
d^2 x dτ 2
da du
du dτ
dx dτ
d^2 x dτ 2
a(u)
da du
du dτ
dx dτ
Now we may read off the connection coefficients
Γxux = Γxxu =
a(u)
da du
Similarly we obtain
Γyuy = Γyyu =
b(u)
db du
Now that we have all of the necessary connection coefficients we can now go about finding/guessing the Killing vectors. We readily obtain three of them by noting that the metric has no explicit dependence on the coordinates v, x, y, which implies that it is invariant under translations of those coordinates and therefore we have that the vectors (in (u, v, x, y) coordinates)
V μ^ = (0, 1 , 0 , 0)
Xμ^ = (0, 0 , 1 , 0) Y μ^ = (0, 0 , 0 , 1)
are Killing vectors. Let’s verify that all three of these vectors satisfy Killing’s equation. First let’s show it for Vμ. We need to be careful because lowering the index changes the components around, specifically Vμ = (− 1 , 0 , 0 , 0). Now we have
∇ν Vμ = ∂ν Vμ − ΓρνμVρ = 0 − Γuνμ(−1) = 0
where we’ve used that all of the components of Vμ are constants, hence all of the partial derivatives vanish, and that all Γuμν = 0.
Next we should verify that Xμ satisfies Killing’s equation (and by similar calculations, that Yμ does as well). This is a little trickier than in the previous case, because now lowering the index gives us Xμ = (0, 0 , a^2 (u), 0). ∇ν Xμ = ∂ν Xμ − ΓρνμXρ = ∂ν Xμ − a^2 (u)Γxνμ
The most interesting cases we need to check are ν = u, μ = x and vice versa. In the former we get
∇uXx = ∂ua^2 (u) − a^2 (u)
a(u)
da du
= a(u)
da du
Switching indices we have
∇xXu = ∂x(0) − a^2 (u)
a(u)
da du
= −a(u)
da du
In five-dimensional spacetime, xM^ = (xμ, y), the Randall-Sundrum metric is given by
ds^2 = gM N dxM^ dxN^ = e−^2 ky^ ημν dxμdxν^ + dy^2 where k is a constant and ημν =diag(− 1 , 1 , 1 , 1).
(a) Calculate the Christoffel connection coefficients. (Rather than direct calculation via Eq. (3.27), you will probably find it easier to use the variational technique!)
(b) Calculate the Riemann tensor, Ricci tensor, and Ricci Scalar.
(c) Is this a maximally symmetric space? Verify by using Eq. (3.191).
Solution: (a) Using the variational technique, we look for stationary points in the integral
I =
gμν
dxμ dτ
dxν dτ dτ =
∫ [
e−^2 ky
( (^) dx^0 dτ
( (^) dx^1 dτ
( (^) dx^2 dτ
( (^) dx^3 dτ
( (^) dy dτ
]
dτ
First let’s consider variations x^1 7 → x^1 + δx^1. Under such variations the only change in the integrand is
( (^) dx^1 dτ
( (^) dx^1 dτ
d(δx^1 ) dτ
( (^) dx^1 dτ
dx^1 dτ
d(δx^1 ) dτ
+ O
(δx^1 )^2
Therefore, up to first-order in δx^1 the variation of the integral is
δI =
∫ [
2 e−^2 ky^ dx^1 dτ
d(δx^1 ) dτ
]
dτ
Integrate by parts with u = e−^2 ky dx
1 dτ and^ dv^ =^
d(δx^1 ) dτ giving us
δI = e−^2 ky^ dx^1 dτ
δx^1 |boundary −
∫ [^
d^2 x^1 dτ 2
e−^2 ky^ − 2 ke−^2 ky^ dx^1 dτ
dy dτ
]
δx^1 dτ
Demanding that the variation of x^1 vanish at the boundary, and setting the variation of the integral equal to zero we get the following equation d^2 x^1 dτ 2 − 2 k
dx^1 dτ
dy dτ
which implies the non-zero Γ^1 ij are
Γ^114 = Γ^141 = −k
where we use the convention that y = x^4. Generalizing this calculation for other i’s we get
Γii 4 = Γi 4 i = −k, i = 0, 1 , 2 , 3
Note that the minus sign in front of
( (^) dx 0 dτ
does not affect the result of the calculation because there are no off-diagonal terms in the metric and we are setting the variation equal to zero. Now consider variations y 7 → y + δy. Every term in the integrand will change as a result of this variation. One should look familiar by now
( (^) dy dτ
( (^) dy dτ
d(δy) dτ
( (^) dy dτ
dy dτ
d(δy) dτ
+ O
(δy)^2
The other terms involve the exponential factor, whose variation is
e−^2 ky^7 → e−^2 k(y+δy)^ = e−^2 ky^ (1 − 2 k δy + O(δy^2 ))
Therefore the variation of the integral is (dropping terms O(δy^2 ))
δI =
∫ [
− 2 k e−^2 ky^ δy
( (^) dx^0 dτ
( (^) dx^1 dτ
( (^) dx^2 dτ
( (^) dx^3 dτ
dy dτ
d(δy) dτ
]
dτ
Integrating the last term by parts and dropping the boundary term we arrive at
δI =
∫ [
− k e−^2 ky
( (^) dx^0 dτ
( (^) dx^1 dτ
( (^) dx^2 dτ
( (^) dx^3 dτ
d^2 y dτ 2
]
δy dτ
δI = 0 ⇒ k e−^2 ky
( (^) dx^0 dτ
( (^) dx^1 dτ
( (^) dx^2 dτ
( (^) dx^3 dτ
d^2 y dτ 2
Now we read off the following connection coefficients
Γ^400 = −ke−^2 ky
Γ^4 ii = ke−^2 ky^ , i = 1, 2 , 3
All other connection coefficients equal 0.
(b) Since the Riemann tensor has so many terms, even if you include symmetries, it is often helpful to turn this calculation into a matrix calculation (matrices are the natural way of manipulating several equa- tions at once). For this particular metric it isn’t actually all that difficult to just do this term by term, since there are only so many non-zero connection coefficients, however for more complicated metrics this technique comes in quite handy. The definition of the Riemann tensor is:
Rρ σμν = ∂μΓρνσ − ∂ν Γρμσ + ΓρμλΓλνσ − ΓρνλΓλμσ
Define the matrices (Γμ)ρσ = Γρμσ (Rμν )ρσ = Rρ σμν
where ρ is the row index and σ is the column index. Rμν should not be confused with the Ricci tensor Rμν. What we have defined is merely a matrix that is useful for doing computations. With these definitions the definition of the Riemann tensor can be recast as the matrix equation
Rμν = ∂μΓν − ∂ν Γμ + ΓμΓν − Γν Γμ
The components of the Γ matrices can be written as
(Γ 0 )ρσ = −k δ 0 ρδ 4 σ − ke−^2 ky^ δ 4 ρδ 0 σ
(Γi)ρσ = −k δiρδ 4 σ + ke−^2 ky^ δ 4 ρδiσ ; i = 1, 2 , 3
(Γ 4 )ρσ = −k
∑^3
i=
δρiδσi
Also note that because of the antisymmetry of the Riemann tensor in the last two indices, it suffices to consider just the case where μ < ν. Let’s do some example calculations:
R 01 = ∂ 0 Γ 1 − ∂ 1 Γ 0 + Γ 0 Γ 1 − Γ 1 Γ 0
Because none of the connection coefficients depend on x^0 or x^1 the derivative terms drop out and we have
R 01 =
0 0 0 0 −k 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −ke−^2 ky^0 0 0
0 0 0 0 −k 0 0 0 0 0 0 0 0 0 0 0 ke−^2 ky^0 0
0 0 0 0 −k 0 0 0 0 0 0 0 0 0 0 0 ke−^2 ky^0 0
0 0 0 0 −k 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −ke−^2 ky^0 0 0
where
gμν^ =
−e^2 ky^0 0 0 0 e^2 ky^0 0 0 0 e^2 ky^0 0 0 0 e^2 ky^0 0 0 0 0 1
So we get for the Ricci scalar
R = −e^2 ky^ (4k^2 e−^2 ky^ ) + e^2 ky^ (− 4 k^2 e−^2 ky^ − 4 k^2 e−^2 ky^ − 4 k^2 e−^2 ky^ ) + 1(− 4 k^2 ) = − 20 k^2
(c) A metric defines a maximally symmetric space if it satisfies Carroll eq. 3.191, which for a five-dimensional manifold reads:
Rρσμν =
R
(gρμgσν − gρν gσμ)
Luckily the metric is diagonal and so we only need to worry about the terms Rμνμν and the symmetry of both expressions in the second pair of indices will take care of the rest. For i, j = 1, 2 , 3; i 6 = j we have
Rijij = gikRkjij = e−^2 ky^ (−k^2 e−^2 ky^ ) = −k^2 e−^4 ky
Compare this with R 20 (giigjj − gij gji) =
− 20 k^2 20 e−^2 ky^ e−^2 ky^ = −k^2 e−^4 ky
Now consider i = 1, 2 , 3
R 0 i 0 i = g 0 kRki 0 i = g 00 R^0 i 0 i = −e−^2 ky^ (−k^2 e−^2 ky^ ) = k^2 e−^4 ky
Compare with R 20
(giig 00 − gi 0 g 0 i) =
− 20 k^2 20
e−^2 ky^ (−e−^2 ky^ ) = k^2 e−^4 ky
Now we just take care of the terms involving 4’s, first considering i = 1, 2 , 3
Ri 4 i 4 = gikRk 4 i 4 = giiRi 4 i 4 = (e−^2 ky^ )(−k^2 ) = −k^2 e−^2 ky
Compare with R 20 (giig 44 − gi 4 g 4 i) =
− 20 k^2 20 e−^2 ky^ (1) = −k^2 e−^2 ky
...Just one last case to check
R 0404 = g 0 kRk 404 = g 00 R^0404 = −e−^2 ky^ (−k^2 ) = k^2 e−^2 ky
Compare with R 20
(g 00 g 44 − g 04 g 40 ) =
− 20 k^2 20
(−e−^2 ky^ )(1) = k^2 e−^2 ky
All cases check out, so this metric does indeed define a maximally symmetric space.
