General Relativity: Homework 3 Solutions 1. Carroll Problem 3.1 Verify these consequences of metric compatability (∇ σ g μν = 0): ∇ σ g μν =0 ∇ λ μνρσ =0 Solution: To verify the first claim, recall that g μν is the inverse of g μν so that we have: g μν g μν = δ μ μ = 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 (g μν g μν )=0 g μν (∇ σ g μν )+(∇ σ g μν )g μν =0 ⇒ g μν (∇ σ g μν )=0 And therefore ∇ σ g μν = 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 g μν . Specifically we have μνρσ = g μα g νβ g ρδ g σγ αβδγ By inserting this expression into the covariant derivative, using the Leibniz rule, and the fact that ∇ σ g μν = 0, we arrive at the expression ∇ λ (μνρσ )μνρσ + μνρσ g μα g νβ g ρδ g σγ ∇ 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 ∇ λ μνρσ = 0. 1
Transcript
General Relativity: Homework 3 Solutions
1. Carroll Problem 3.1
Verify these consequences of metric compatability (∇σgµν = 0):
∇σgµν = 0
∇λεµνρσ = 0
Solution:
To verify the first claim, recall that gµν is the inverse of gµν so that we have:
gµνgµν = δµµ = n
where n is the dimension of the manifold. Now take the covariant derivative of both sides, noting that thecovariant derivative of a constant is zero,
∇σ(gµνgµν) = 0
gµν(∇σgµν) + (∇σgµν)gµν = 0
⇒ gµν(∇σgµν) = 0
And therefore ∇σgµν = 0.
To verify the second claim note Carroll equations 2.69 and 2.70 which imply
εµνρσεµνρσ = εµνρσ ε
µνρσ = −n!
where n is the dimension of the manifold we are working over, and we are assuming that the metric isLorentzian (this assumption is uneccessary, it’s just why the negative sign is there, all we really need is thatthe right-hand side is a constant). Taking the covariant derivatives of both sides we obtain
∇λ(εµνρσ)εµνρσ + εµνρσ∇λ(εµνρσ) = 0
Since εµνρσ is an honest to goodness tensor, we may relate it to εµνρσ by factors of the gµν . Specifically wehave
εµνρσ = gµαgνβgρδgσγεαβδγ
By inserting this expression into the covariant derivative, using the Leibniz rule, and the fact that∇σgµν = 0,we arrive at the expression
Replacing the dummy indices in the second expression we arrive at
2∇λ(εµνρσ)εµνρσ = 0
Multiplying both sides by εµνρσ implies ∇λεµνρσ = 0.
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2. Carroll Problem 3.5
Consider a 2-sphere with coordinates (θ, φ) and metric
ds2 = dθ2 + sin2 θdφ2
(a) Show that lines of constant longitude (φ = constant) are geodesics, and that the only line of constantlatitude (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 constantlatitude. 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 needthe connection coefficients, to which we appeal to Carroll equation (3.154), although it is a worthwhileexercise 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):
d2xµ
dλ2+ Γµρσ
dxρ
dλ
dxσ
dλ= 0
which give us the two equations:d2θ
dλ2− sin θ cos θ
(dφdλ
)2= 0
d2φ
dλ2+ 2 cot θ
dθ
dλ
dφ
dλ= 0
In the case that φ is a constant the first equation reads d2θdλ2 = 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(dφdλ
)2= 0
d2φ
dλ2= 0
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). Thereforexµ = (π/2, λ) is a geodesic, and in fact it is the only geodesic (up to reparameterization) where θ isconstant.
(b) In order to parallel transport a vector we must choose a path (with parameterization) to transportalong. Any parameterization will work, however it is best to keep it simple, so choose xµ = (θ, λ). Nowwe must solve the parallel transport equation:
d
dλV µ + Γµσρ
dxσ
dλV ρ = 0
Note that dxµ
dλ = (0, 1) so that we get the two equations:
dV θ
dλ− sin θ cos θ V φ = 0 (1)
2
dV φ
dλ+ cot θ V θ = 0 (2)
To solve this system differentiate (2) with respect to λ (keeping in mind that θ is a constant), solve fordV θ
dλ , plug it into (1) and do a little algebra to obtain:
d2V φ
dλ2+ cos2 θ V φ = 0
We can do the same procedure (differentiating (1) and plugging into (2)) to obtain a differentialequation for V θ:
d2V θ
dλ2+ cos2 θ V θ = 0
These equations are just those for a simple harmonic oscillator (with frequency cos θ), so the form ofthe solutions are:
V θ(λ) = A cos(λ cos θ) +B sin(λ cos θ)
V φ(λ) = C cos(λ cos θ) +D sin(λ cos θ)
Now apply the boundary condition V (0) = (1, 0) which gives us that A = 1 and C = 0. To pin downthe last two constants, appeal to one of the original equations, say (1):
dV θ
dλ= sin θ cos θ V φ
cos θ(− sin(λ cos θ) +B cos(λ cos θ)) = sin θ cos θ(D sin(λ cos θ))
− sin(λ cos θ) +B cos(λ cos θ) = sin θD sin(λ cos θ)
Comparing the coefficients of sin(λ cos θ) and cos(λ cos θ) on both sides we see that we must have B = 0and D = −1
sin θ . Thus we have the solution to the parallel transport equation:
V µ(λ) =
(cos(λ cos θ),− sin(λ cos θ)
sin θ
)Plugging in λ = 2π we see that after parallel transporting V µ around a circle of constant latitude θthe resulting vector is:
V µ(2π) =
(cos(2π cos θ),− sin(2π cos θ)
sin θ
)As a check we can see that the vector is unchanged (i.e. V µ(2π) = (1, 0)) if we parallel transportaround the equator (θ = π/2), as expected.
3. Carroll Problem 3.6
A good approximation to the metric outside the surface of the Earth is provided by
may be thought of as the familiar Newtonian gravitational potential. Here G is Newton’s constant and Mis the mass of the Earth. For this problem Φ may be assumed to be small.
(a) Imagine a clock on the surface of the Earth at a distance R1 from the Earth’s center, and another clockon a tall building at distance R2 from the Earth’s center. Calculate the time elapsed on each clock asa function of the coordinate time t. Which clock moves faster?
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(b) Solve for a geodesic corresponding to a circular orbit around the equator of the Earth (θ = π/2). Whatis dφ/dt?
(c) How much proper time elapses while a satellite at radius R1 (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 inthe actual numbers for the radius of the Earth and so on to get an answer in seconds. How does thisnumber 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, R1, θ, ωt), whereω is the angular velocity of the Earth (ω = π rad
12 hrs ). Now calculate the proper-time:
∆τ =
∫ t
0
√−gµν
dxµ
dt
dxν
dtdt′ =
∫ t
0
√(1 + 2Φ)−R2
1 sin2 θω2dt′
∆τ =
√1− 2GM
R1−R2
1 sin2 θω2∆t
Now let’s reinstate the factors of c. 2GMR1
has units of velocity squared, so we should divide that by c2.The same goes for the second term under the square root. Also note that R1ω << c, so we can dropthat term (we also have R2
1ω2 << 2GM
R1, which justifies leaving in 2GM
R1), leaving us with:
∆τ =
√1− 2GM
c2R1∆t
For the clock on a tall building we replace R1 with R2 and see that the proper-time is decreased. Notethat this effect is purely due to the difference in gravitational potential (the effect due to time dilationassociated 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 specifictype of geodesics, namely one with constant radius, r and θ = π/2 we do not need to look at all of theequations. First look at the geodesic equation for xµ = r which reads:
d2r
dλ2+ Γrρσ
dxρ
dλ
dxσ
dλ= 0
Keeping in mind that we are restricting ourselves to paths with constant r and constant θ, this expres-sion simplifies to
Γrtt( dtdλ
)2+ Γrφφ
(dφdλ
)2+ 2Γrφt
dφ
dλ
dt
dλ= 0
Now calculate just the necessary connection coefficients
Γrtt =1
2grρ(∂tgtρ + ∂tgtρ − ∂ρgtt) = −1
2grr∂rgtt
Γrtt =1
2
(1
(1− 2Φ)
)∂r(1−
2GM
r) =
GM
r2(1− 2Φ)
Γrφφ =1
2grρ(∂φgφρ + ∂φgφρ − ∂ρgφφ) = −1
2grr∂rgφφ
Γrφφ = −1
2
(1
(1− 2Φ)
)∂r(r
2 sin2 θ) = − r sin2 θ
(1− 2Φ)
Γrφt =1
2grρ(∂φgtρ + ∂tgφρ − ∂ρgφt) = 0
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Now the geodesic equation reads:
GM
r2(1− 2Φ)
( dtdλ
)2 − r sin2 θ
(1− 2Φ)
(dφdλ
)2= 0
⇒ dφ
dλ=
√GM
r3dt
dλ(3)
We still haven’t found the geodesic, to do so now consider the equation for xµ = t:
d2t
dλ2+ Γtρσ
dxρ
dλ
dxσ
dλ= 0
Again, considering only paths such that r is constant and θ = π/2 this simplifies to:
d2t
dλ2+ Γttt
( dtdλ
)2+ Γtφφ
(dφdλ
)2+ Γttφ
dt
dλ
dφ
dλ= 0
Now look at the equation for the necessary connection coefficients
Γtµν =1
2gtt(∂µgνt + ∂νgµt − ∂tgµν)
and we can see that for µ, ν ∈ {t, φ} the connection coefficients will all be zero. The geodesic equationis now just
d2t
dλ2= 0
the simplest non-trivial solution to which is t = λ. Taking this as our parameterization, (3) thenimplies:
dφ
dt=
√GM
r3
Putting it all together, the (up to reparameterization) geodesic corresponding to a circular orbit (ofradius r) around the equator (θ = π/2) is:
xµ(t) = (t, r, π/2,
√GM
r3t)
(c) For the satellite skimming the surface of the Earth (at θ = π/2) we have that the elapsed proper-timefor a complete orbit is:
∆τ =
√1− 2GM
c2R1− R2
1ω2
c22π
ω
where ω =√
GMR3
1and 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
√1− 2GM
c2R1− R2
1
c2GM
R31
√R3
1
GM= 2π
√R3
1
GM− 3R2
1
c2
∆τ = 2π
√(6.371× 106m)3
(6.674× 10−11Nm2
kg2 )(5.972× 1024kg)− 3(6.371× 106m)2
(2.998× 108ms )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 =
√1− 2GM
c2R1
2π
ω= 2π
√1− 2GM
c2R1
√R3
1
GM= 2π
√R3
1
GM− 2R2
1
c2
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∆τ = 2π
√(6.371× 106m)3
(6.674× 10−11Nm2
kg2 )(5.972× 1024kg)− 2(6.371× 106m)2
(2.998× 108ms )2= 5061s
For a clock stationary on the surface the elapsed proper-time (during the orbit of the satellite) isexactly the same as that of the satellite! This is because both are the same distance from the center ofthe Earth, and therefore they both have the same gravitational potential (and the effect due to timedilation clearly has a negligible effect).
To fully convince ourselves that this effect is due purely to gravity, consider a stationary clock at thetop of Mount Everest (≈ 9000m high). Repeating the previous calculation we obtain ∆τ = 5072s.Gravity slows the passage of proper-time! A true means for slowing the effects of aging.
4. Carroll Problem 3.13
Find explicit expressions for a complete set of Killing vector fields for the following spaces:
(b) A spacetime with coordinates (u, v, x, y) and metric
ds2 = −(dudv + dvdu) + a2(u)dx2 + b2(u)dy2
where a and b are unspecified functions of u. This represents a gravitational wave spacetime. (Hints,which you need not show: there are five Killing vectors in all, and all of them have a vanishing ucomponent Ku.)
Be careful, in all of these cases, about the distinction between upper and lower indices.
Solution:
All answers are vectors with upper indices.
(a) First note that all of the metric components are independent of t, x, y, z and therefore the four trans-lations of those coordinates correspond to symmetries of the metric. These symmetries are generatedby the four Killing vectors:
Tµ = (1, 0, 0, 0)
Xµ = (0, 1, 0, 0)
Y µ = (0, 0, 1, 0)
Zµ = (0, 0, 0, 1)
To see that each of these satisfy ∇(µKν) = 0 first note that all of the connection coefficients vanish forthe Minkowski metric, and therefore ∇(µKν) = ∂(µKν). Since all of the components of these vectors areconstant, all of the partial derivatives will vanish as well and the Killing vector condition is satisfied.
Next consider the three generators of purely spatial rotations (see Carroll pages 138-9 for a moredetailed discussion of these). They are (with lowered indices):
Rxµ = (0, 0,−z, y)
Ryµ = (0, z, 0,−x)
Rzµ = (0,−y, x, 0)
Let’s explicitly verify that Rxµ satisfies the Killing vector condition. First check that ∇µRxµ = 0, butthis is clear since the µth component of Rxµ is independent of the µth coordinate (e.g. the y componentis independent of y). Next condsider terms where µ 6= ν, in fact the only non-trivial combination is:
∇yRxz +∇zRxy = ∂y(y) + ∂z(−z) = 1 + (−1) = 0
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Similar calculations show that all three of these vectors satisfy the Killing vector condition. All that’sleft 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 threevectors 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 sameargument 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 Poincare group (isometriesof the Minkowski metric) is 10 dimensional.
(b) In order to verify Killing’s equation we will need to calculate the connection coefficients, a quickobservation 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 thecovariant derivative of a 1-form ∇µKν = ∂µKν − ΓρµνKρ we conclude that we do not need to calculatethe 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 =1
2
∫gµν
dxµ
dτ
dxν
dτdτ =
1
2
∫(−2
du
dτ
dv
dτ+ a2(u)
(dxdτ
)2+ b2(u)
(dydτ
)2)dτ
To get the coefficients with upper index u, we will actually need to consider variations of v (this is dueto the cross terms in the metric). Under variations v 7→ v + δv the only term in the integrand that
changes is dvdτ 7→
dvdτ + d(δv)
dτ and therefore the variation in I is
δI =1
2
∫−2
du
dτ
d(δv)
dτ=
∫d2u
dτ2δvdτ
where the second equality comes from integrating by parts. Requiring that the variation vanishes weare left with
d2u
dτ2= 0
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. Undersuch variations the only change in the integrand is(dx
dτ
)2 7→ (dxdτ
+d(δx)
dτ
)2=(dxdτ
)2+ 2
dx
dτ
d(δx)
dτ+O(δx2)
7
Therefore the variation of the integral (up to first order in δx) is
δI =
∫a2(u)
dx
dτ
d(δx)
dτdτ
Integrating by parts we obtain
δI = −∫ (
a2(u)d2x
dτ2+ 2a(u)
da
du
du
dτ
dx
dτ
)dτ
Setting the variation equal to zero leads to the equation
a2(u)d2x
dτ2+ 2a(u)
da
du
du
dτ
dx
dτ= 0
⇒ d2x
dτ2+
2
a(u)
da
du
du
dτ
dx
dτ= 0
Now we may read off the connection coefficients
Γxux = Γxxu =1
a(u)
da
du
Similarly we obtain
Γyuy = Γyyu =1
b(u)
db
du
Now that we have all of the necessary connection coefficients we can now go about finding/guessing theKilling vectors. We readily obtain three of them by noting that the metric has no explicit dependenceon the coordinates v, x, y, which implies that it is invariant under translations of those coordinates andtherefore 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’sshow 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
∇νVµ = ∂νVµ − ΓρνµVρ = 0− Γuνµ(−1) = 0
where we’ve used that all of the components of Vµ are constants, hence all of the partial derivativesvanish, and that all Γuµν = 0.
Next we should verify that Xµ satisfies Killing’s equation (and by similar calculations, that Yµ doesas well). This is a little trickier than in the previous case, because now lowering the index gives usXµ = (0, 0, a2(u), 0).
∇νXµ = ∂νXµ − ΓρνµXρ = ∂νXµ − a2(u)Γxνµ
The most interesting cases we need to check are ν = u, µ = x and vice versa. In the former we get
∇uXx = ∂ua2(u)− a2(u)
( 1
a(u)
da
du
)= a(u)
da
du
Switching indices we have
∇xXu = ∂x(0)− a2(u)( 1
a(u)
da
du
)= −a(u)
da
du
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⇒ ∇uXx +∇xXu = a(u)da
du− a(u)
da
du= 0
Unfortunately those are the only obvious symmetries of the metric, and as the hint suggests, we stillhave two more Killing vectors to find. To find the last two vectors let’s explore a bit with the covariantderivative. Look at
∇uKu = ∂uKu + ΓρuuKρ
Since all of the relevant connection coefficients with lower labels uu are zero, we see that in orderfor Kµ to satisfy Killing’s equation we must have ∂uKu = 0, that is, Ku must be independent of u.Following the same logic with ∇xKx we see that Kx must be independent of x. Now let’s look at theimplications of the cross terms.
In order for Kµ to satisfy Killing’s equation we must have
∂uKx −2
a(u)
da
duKx = −∂xKu
Because Ku is independent of u, and Kx is independent of x, the right hand side must equal a constant(i.e. Ku = λx for some constant λ, which we can set to -1 without loss of generality). Therefore wehave the linear first-order differential equation
∂uKx −2
a(u)
da
duKx = 1
Following the procedure for solving such equations (i.e. looking it up in your favorite differentialequations book) we solve for the integrating factor µ(u)
µ(u) = exp
{∫− 2
a(u)
da
dudu
}= exp{−2 ln |a(u)|+ C} = Da−2(u)
for some constant D. Now that we have the integrating factor the solution is given by
Kx = µ(u)−1(∫
µ(u)du)
=1
Da2(u)
∫Da−2(u)du = a2(u)
∫1
a2(u)du
Note that for a problem with boundary conditions there should be some constants hanging aroundfrom the improper integrals, however since we do not have any such conditions we can just seek thesimplest such solution. Thus with lower indices our solution is
Kµ =
(− x, 0, a2(u)
∫du
a2(u), 0
)You should check for yourself that this indeed solves the equation. Raising the indices we get theKilling vector
Kµ =
(0, x,
∫du
a2(u), 0
)Following the same procedure we obtain the final Killing vector
Kµ =
(0, y, 0,
∫du
b2(u)
)
9
5.
In five-dimensional spacetime, xM = (xµ, y), the Randall-Sundrum metric is given by
ds2 = gMNdxMdxN = e−2kyηµνdx
µdxν + dy2
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), youwill 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 =1
2
∫gµν
dxµ
dτ
dxν
dτdτ =
1
2
∫ [e−2ky
(−(dx0dτ
)2+(dx1dτ
)2+(dx2dτ
)2+(dx3dτ
)2)+(dydτ
)2]dτ
First let’s consider variations x1 7→ x1 + δx1. Under such variations the only change in the integrand is
(dx1dτ
)2 7→ (dx1dτ
+d(δx1)
dτ
)2=(dx1dτ
)2+ 2
dx1
dτ
d(δx1)
dτ+O
((δx1)2
)Therefore, up to first-order in δx1 the variation of the integral is
δI =1
2
∫ [2e−2ky
dx1
dτ
d(δx1)
dτ
]dτ
Integrate by parts with u = e−2ky dx1
dτ and dv = d(δx1)dτ giving us
δI = e−2kydx1
dτδx1|boundary −
∫ [d2x1
dτ2e−2ky − 2ke−2ky
dx1
dτ
dy
dτ
]δx1dτ
Demanding that the variation of x1 vanish at the boundary, and setting the variation of the integral equalto zero we get the following equation
d2x1
dτ2− 2k
dx1
dτ
dy
dτ= 0
which implies the non-zero Γ1ij are
Γ114 = Γ1
41 = −k
where we use the convention that y = x4. Generalizing this calculation for other i’s we get
Γii4 = Γi4i = −k, i = 0, 1, 2, 3
Note that the minus sign in front of(dx0
dτ
)2does 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
(dydτ
)2 7→ (dydτ
+d(δy)
dτ
)2=(dydτ
)2+ 2
dy
dτ
d(δy)
dτ+O
((δy)2
)The other terms involve the exponential factor, whose variation is
e−2ky 7→ e−2k(y+δy) = e−2ky(1− 2k δy +O(δy2))
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Therefore the variation of the integral is (dropping terms O(δy2))
δI =1
2
∫ [− 2k e−2kyδy
(−(dx0dτ
)2+(dx1dτ
)2+(dx2dτ
)2+(dx3dτ
)2)+ 2
dy
dτ
d(δy)
dτ
]dτ
Integrating the last term by parts and dropping the boundary term we arrive at
δI =
∫ [− k e−2ky
(−(dx0dτ
)2+(dx1dτ
)2+(dx2dτ
)2+(dx3dτ
)2)− d2y
dτ2
]δy dτ
δI = 0⇒ k e−2ky(−(dx0dτ
)2+(dx1dτ
)2+(dx2dτ
)2+(dx3dτ
)2)+d2y
dτ2= 0
Now we read off the following connection coefficients
Γ400 = −ke−2ky
Γ4ii = ke−2ky, 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 toturn 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, sincethere are only so many non-zero connection coefficients, however for more complicated metrics this techniquecomes in quite handy.
The definition of the Riemann tensor is:
Rρ σµν = ∂µΓρνσ − ∂νΓρµσ + 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 thedefinition of the Riemann tensor can be recast as the matrix equation
Rµν = ∂µΓν − ∂νΓµ + ΓµΓν − ΓνΓµ
The components of the Γ matrices can be written as
(Γ0)ρσ = −k δ0ρδ4σ − ke−2kyδ4ρδ0σ
(Γi)ρσ = −k δiρδ4σ + ke−2kyδ4ρδiσ; i = 1, 2, 3
(Γ4)ρσ = −k3∑i=0
δρiδσi
Also note that because of the antisymmetry of the Riemann tensor in the last two indices, it suffices toconsider just the case where µ < ν.
Let’s do some example calculations:
R01 = ∂0Γ1 − ∂1Γ0 + Γ0Γ1 − Γ1Γ0
Because none of the connection coefficients depend on x0 or x1 the derivative terms drop out and we have
Repeating the previous calculation for the other components we find
R0404 = −k2;R4
004 = −k2e−2ky
Ri4i4 = −k2;R4ii4 = k2e−2ky; i = 1, 2, 3
All other terms not related to one of these by a symmetry vanish.Now we calculate the Ricci tensor by the definition, but this is aided by the observation that Rλµλν is
equal to zero unless µ = ν. Therefore all of the off-diagonal terms are zero and our work is almost done.
R00 = Rλ0λ0 = R1010 +R2
020 +R3030 +R4
040
Rewrite this in terms of the components that we have calculated:
(c) A metric defines a maximally symmetric space if it satisfies Carroll eq. 3.191, which for a five-dimensionalmanifold reads:
Rρσµν =R
20(gρµgσν − gρνgσµ)
Luckily the metric is diagonal and so we only need to worry about the terms Rµνµν and the symmetry ofboth 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−2ky(−k2e−2ky) = −k2e−4ky
Compare this withR
20(giigjj − gijgji) =
−20k2
20e−2kye−2ky = −k2e−4ky
Now consider i = 1, 2, 3
R0i0i = g0kRki0i = g00R
0i0i = −e−2ky(−k2e−2ky) = k2e−4ky
Compare withR
20(giig00 − gi0g0i) =
−20k2
20e−2ky(−e−2ky) = k2e−4ky
Now we just take care of the terms involving 4’s, first considering i = 1, 2, 3
Ri4i4 = gikRk4i4 = giiR
i4i4 = (e−2ky)(−k2) = −k2e−2ky
Compare withR
20(giig44 − gi4g4i) =
−20k2
20e−2ky(1) = −k2e−2ky
...Just one last case to check
R0404 = g0kRk404 = g00R
0404 = −e−2ky(−k2) = k2e−2ky
Compare withR
20(g00g44 − g04g40) =
−20k2
20(−e−2ky)(1) = k2e−2ky
All cases check out, so this metric does indeed define a maximally symmetric space.
![[LS]Mold TR_E_1208](https://static.documents.pub/doc/80x56/563dba1a550346aa9aa2bbc5/lsmold-tre1208.jpg)
