PHYS 631: General Relativity Homework #4

Prakash Gautam May 16, 2019

1. (Schutz 6.29) In polar coordinates, calculate the Riemann curvature tensor of the sphere of unit radius 2 2 2 whose metric is gθθ = r , gφφ = r sin θ, gθφ = 0. Solution: The metric for polar coordinate on the surface of unit sphere is

1 0  0 sin2 θ

The christoffel symbols are given by 1   Γµ = gµσ g + g − g νρ 2 νσ,ρ ρσ,ν νρ,σ The only non zero derivative of metric is with respect to θ so we get 1 1 Γθ = gθθ (−g ) = − sin 2θ φφ 2 φφ,θ 2 Similarly the other non zero Christoffel symbols are cos θ Γφ = Γφ = θφ φθ sin θ

And the Riemann tensor is given by

Rα = Γα Γσ − Γα Γσ − Γα + Γα βµν σµ βν σν βµ βµ,ν βν ,µ   R = g Γλ Γσ − Γλ Γσ − Γλ + Γλ αβµν αλ σµ βν σν βµ βµ,ν βν ,µ

Calculating   R = g Γφ Γσ − Γφ Γσ − Γφ + Γφ φθφθ φφ σφ θθ σθ θφ θφ,θ θθ,φ   = sin2 θ Γφ Γσ − Γφ Γσ − Γφ + Γφ σφ θθ σθ θφ θφ,θ θθ,φ   = sin2 θ −Γφ Γφ − Γφ φθ θφ θφ,θ  cos2 θ 1  = sin2 θ − + sin2 θ sin2 θ = sin2 θ

Now we can permute the coordinate with the symmetry property to obtain

2 2 2 Rφθθφ = − sin θ Rθφφθ = − sin θ Rθφθφ = sin θ

These are the non zero components of Riemann tensor. 

1 2. (Schutz 6.30) Calculate the Riemann curvature tensor of the cylinder. Solution: The line element ins the cylindrical coordinate system is ds2 = dr2 + r2dφ2 + dz2

So the metric in is 1 0 0 2 gµν = 0 r 0 0 0 1

The Christoffel symbols are given by 1   Γµ = gµσ g + g − g νρ 2 νσ,ρ ρσ,ν νρ,σ The only non zero derivative of metric is with respect to θ so we get 1 Γr = grr (−g ) = −r φφ 2 φφ,r Similarly the other non zero Christoffel symbols are 1 Γφ = Γφ = rφ φr r

And the Riemann tensor is given by Rα = Γα Γσ − Γα Γσ − Γα + Γα βµν σµ βν σν βµ βµ,ν βν ,µ   R = g Γλ Γσ − Γλ Γσ − Γλ + Γλ αβµν αλ σµ βν σν βµ βµ,ν βν ,µ Calculating   R = g Γφ Γσ − Γφ Γσ − Γφ + Γφ φrφr φφ σφ rr σr rφ rφ,r rr,φ   = r2 Γφ Γσ − Γφ Γσ − Γφ + Γφ σφ rr σr rφ rφ,r rr,φ   = r2 −Γφ Γφ − Γφ φr rφ rφ,r  1 1  = r2 − + r2 r2 = 0

Now we can permute the coordinates and with symmetry all the rest are zero too.

Rφrrφ = 0 Rrφφr = 0 Rrφrφ = 0

So all the components of Riemann tensor are zero, showing that the surface of cylinder is a flat surface. 

3. One way of describing the metric of a flat, homogeneous, expanding universe is: −1 0 0 0   0 a(t)2 0 0     0 0 a(t)2 0  0 0 0 a(t)2

where a(t) is a function of time only, and the coordinates are t µ x x =   y z

2 (a) Compute all non vanishing terms of the Riemann Tensor. Solution:

The Christoffel symbols are given by 1   Γµ = gµσ g + g − g νρ 2 νσ,ρ ρσ,ν νρ,σ The only non zero derivative of metric is with respect to t so we get 1 Γt = gtt (−g ) = aa˙ xx 2 xx,t These are true for y and z coordinates. 1 1 Γt = gtt (−g ) = aa˙ Γt = gtt (−g ) = aa˙ yy 2 yy,t zz 2 zz,t Similarly the other non zero Christoffel symbols are a˙ Γx = Γx = tx xt a These are also true for y and z.

a˙ a˙ Γy = Γy = Γz = Γz = ty yt a tz zt a And the Riemann tensor is given by

Rα = Γα Γσ − Γα Γσ − Γα + Γα βµν σµ βν σν βµ βµ,ν βν ,µ   R = g Γλ Γσ − Γλ Γσ − Γλ + Γλ αβµν αλ σµ βν σν βµ βµ,ν βν ,µ

Calculating

 x σ x σ x x  Rxtxt = gxx ΓσxΓtt − ΓσtΓtx − Γtx,t + Γtt,x

2  x σ x σ x x  = a ΓσxΓtt − ΓσtΓtx − Γtx,t + Γtt,x

2  x x x  = a −ΓxtΓtx − Γtx,t  a˙ 2 a˙ 2 a¨  = a2 − + + a2 a2 a = aa¨

Now we can permute the coordinate with the symmetry property to obtain

Rxttx = −aa¨ Rtxxt = −aa¨ Rtxtx = aa¨

Similarly the rest of the values can be calculated as

2 2 Ryxxy = −a a˙

The rest of them can be obtained by permuting the index using the (anti-)symmetry property.

2 Rzxxz = Rzyyz = Ryzzy = Rxzzx = Rxyyx = −a a˙



3 (b) Compute all Non-vanishing terms of the Ricci Tensor. Solution: The raised version of Riemann tensor is

α ασ R βγµ = g Rσβγµ

The first index non vanishing term is 0 x tt :0 xx yy ¨¨* zz : 0 Rttx = g Rtttx + g Rxttx + g ¨Ryttx + g Rzttx a¨ = a−2 (−aa¨) = − a Using the symmetry property and the elements of metric we get the rest of components of Riemann tensor as a¨ a¨ Rx = − Rx = ttx a txt a a¨ a¨ Ry = − Ry = tty a tyt a a¨ a¨ Rz = − Rz = ttz a tzt a

Now the components of Ricci tensor in terms of elements of Riemann Rensor are

µν ν Rαβ = g Rαµβ

Specifically for Rtt we get 0 tt t> xx x yy y zz z Rtt = g Rttt + g Rtxt + g Rtyt + g Rtzt = −a−2aa¨ − a−2aa¨ − a−2aa¨ = −3¨a/a

Similarly rest of the components can be calculated. They are

2 Rxx = Ryy = Rzz = aa¨ + 2˙a

These are the components of Ricci tensor 

(c) Compute Einstein Tensor. Solution: The components of Einstein tensor are given by 1 Gµν = Rµν − gµν R (1) 2 The Ricci scalar can be calculated by contracting the Ricci tensor as aa¨ +a ˙ 2 R = Rt + Rx + Ry + Rz = 6 (2) t x y z a2 Now the Einstein tensor simply is the substitution (2) into the (1). The first component of this tensor is 1 a¨ 1 6 aa¨ +a ˙ 2
a˙ 2 G = R − g R = −3 + = 3 tt tt 2 tt a 2 a2 a2 Similarly the rest of the components can be calculated. 1 1 6(aa¨ +a ˙ 2) G = R − g R = aa¨ + 2˙a2 − a2 · = −2aa¨ − a˙ 2 xx xx 2 xx 2 a2

4 2 Gxx = Gyy = Gzz = −2aa¨ − a˙ The raised version of Einstein tensor similarly are1.

3 a˙ 2 a˙ 2 + 2aa¨ Gtt = Gxx = Gyy = Gzz = − 2 a2 a4 These are the required components of Einstein tensor. 

2 4. (Schutz 6.35) Compute 20 independent components of Rαβµν for a manifold with line element ds = 2Φ 2 2Λ 2 2 2 2 2
−e dt + e dr + r dθ + sin θdφ , where Φ and Λ are arbitrary functions fo the coordinate r alone. Solution: Writing down the metric from the given expression for line element

2Φ 2Λ 2 2 2 gtt = −e ; grr = e ; gθθ = r ; gφφ = r sin θ The inverse metric is 1 1 gtt = −e−2Φ; grr = e−2Λ; gθθ = ; gφφ = r2 r2 sin2 θ The Christoffel symbols can be calculated by the expression 1   Γµ = gµσ g + g − g νρ 2 νσ,ρ ρσ,ν νρ,σ Evaluating the these we get

t t Γrt = Γtr = Φ,r r −2Λ 2 r −2Λ+2Φ r r −2Λ Γφφ = −re sin θ Γtt = −re Φ,r Γrr = Λ,r Γθθ = −re 1 1 Γθ = sin 2θ Γθ = Γθ = φφ 2 θr rθ r 1 cos θ Γφ = Γφ = Γφ = Γφ = φr rφ r θφ φθ sin θ The Riemann tensor is given by Rα = Γα + Γα − Γα Γσ − Γα Γσ βµν βµ,ν βν ,µ σµ βν σν βµ R = g (Γλ + Γλ − Γλ Γσ − Γλ Γσ ) αβµν λα βµ,ν βν ,µ σµ βν σν βµ

Explicitly for Rtrtr we get

t Rtrtr = gttRrtr  0  2Φ r r¨¨* r σ r σ = −e Γrt,r +¨Γrr,t + ΓσrΓrr − ΓσrΓrt

2Φ r r r r = −e [Φ,rr + ΓrrΓrr − ΓrrΓrt] 2Φ h 2 i = −e Φ,rr + (Φ,r) − Φ,rΛ,r

The rest of the components can be similarly calculated 2

2Φ Rtrrt = ((Λ,r − Φ,r)Φ,r − Φ,rr) e 2Φ Rtrtr = (− (Λ,r − Φ,r)Φ,r + Φ,rr) e

1This was solved mostly using Cadabra. https://www.physics.drexel.edu/~pgautam/courses/PHYS631/ einstein-tensor-expanding-universe.html 2I did this using Cadabra. The detail of this exercise is at https://www.physics.drexel.edu/~pgautam/courses/PHYS631/ HW4Schutz6.35.html

5 −2Λ+2Φ 2 Rtφtφ =re sin θΦ,r −2Λ+2Φ Rtθtθ =re Φ,r −2Λ+2Φ 2 Rtφφt = − re sin θΦ,r −2Λ+2Φ Rtθθt = − re Φ,r 2 Rrφrφ =rsin θΛ,r Rrθrθ =rΛ,r 2Φ Rrtrt = (− (Λ,r − Φ,r)Φ,r + Φ,rr) e 2 Rrφφr = − rsin θΛ,r Rrθθr = − rΛ,r 2Φ Rrttr = ((Λ,r − Φ,r)Φ,r − Φ,rr) e Rθrrθ = − rΛ,r Rθrθr =rΛ,r 1   R = r2 e2Λ sin (2θ) (tan θ)−1 − 2e2Λ cos (2θ) + cos (2θ) − 1 e−2Λ θφθφ 2 −2Λ+2Φ Rθtθt =re Φ,r 2 2Λ
−2Λ 2 Rθφφθ =r 1 − e e sin θ −2Λ+2Φ Rθttθ = − re Φ,r 2 Rφrrφ = − rsin θΛ,r 2 2Λ
−2Λ 2 Rφθθφ =r 1 − e e sin θ 2 Rφrφr =rsin θΛ,r 2 2Λ
−2Λ 2 Rφθφθ =r e − 1 e sin θ −2Λ+2Φ 2 Rφtφt =re sin θΦ,r −2Λ+2Φ 2 Rφttφ = − re sin θΦ,r

These are the non zero components of Riemann tensor. 

5. (Schutz 7.7) Consider the following four different metrics, as given by their line elements: i. ds2 = −dt2 + dx2 + dy2 + dz2 ; ii. ds2 = − (1 − 2M/r) dt2 + (1 − 2M/r)−1dr2 + r2(dθ2 + sin2 θdφ2) where M is a constant. 2 2 2  2 − 2 2 2 2 2  iii. ds = −dt + R (t) (1 − kr ) 1dr + r (dθ + sin θdφ ) , where k is a constant and R(t) is an arbitrary function of t alone.

(a) For each metric find as many conserved components pα of a freely falling particle’s four momentum as possible. Solution: The rate of change of momentum is given by dp 1 m β = g pν pα dτ 2 µα,β

The momentum pβ is conserved when gµα,β = 0. From the given metric the conserved quantities are

for i. : pt, px, py, pz

for ii. : pt, pφ

for iii. : pφ



6 (b) Write i. in the form ds2 = −dt2 + dr2 + r2 dθ2 + sin2 θdφ2
From this argue that ii. iii. are spherically symmetric. Does this increase the number of conserved components of pα? Solution: The coordinate transformation from Cartesian to polar is x = r sin θ cos φ =⇒ dx = sin θ + cos φdr + +r cos θ cos φdθ − r sin θ sin φdφ y = r sin θ sin φ =⇒ dy = sin θ + sin φdr + +r cos θ sin φdθ + r sin θ cos φdφ z = r cos θ =⇒ dz cos θdr = − sin θdθ Substituting these in the line element we get dl2 = − dt2 + dr2(sin2 θ sin2 φ sin2 θ cos2 φ + cos2)+ + dθ2 r2 cos2 θ cos2 φ + r2 cos2 θ cos2 φ + r2 sin2 θ
+ dφ2 r2 sin2 θ sin2 φ + r2 sin2 θ cos2 φ
= −dt2 + dr2 + r2 dθ2 + sin2 θdφ2

This is the required transformation in spherical form. 

π θ (c) It can be shown that for ii. and iii. a geodesic that begins with θ = 2 and p = 0- i.e., one which π θ begins tangent to the equatorial plane- always has θ = 2 and p = 0. For these cases use the equation ~p·~p = −m2 to solve for pr in terms of m, other conserved quantities, and known functions of position. Solution: Expanding the relation ~p · ~p = −m2 we get 2 t
2 r 2 θ
2 φ
2 −m = gtt p + grr (p ) + gθθ p + gφφ p Given θ = π/2 and pθ = 0 we get s −m2 − g (pt)2 + g (pφ)2 pr = tt φφ grr t φ r Since p and p are conserved substituting the corresponding metric values gαβ gives the quantity p s −m2 + 1−2M (pt)2 + r2 (pφ)2 for ii.; pr = r 1 − 2M/r s 2 1 − kr  2 2 for iii.; pr = −m2 + (pt) + (R(t)r)2 (pφ) R2(t)

r These are the required expression for p in terms of conserved quantities. 

(d) For iii., spherical symmetry implies that if a geodesic begins with pθ = pφ = 0, these remain zero. Use this to show that when k = 0, pr is a conserved quantity. Solution: The rate of change of momentum is given by dp 1 m β = g pν pα dτ 2 µα,β dp 1 m r = g (pt)2 + g (pr)2 + g (pθ)2 + g (pφ)2
dτ 2 tt,r rr,r θθ,r φφ θ φ But for k = 0, grr,r = 0 and gtt,r = 0 and given p = p = 0 we get dp m r = 0 dτ This proves that pr is a conserved quantity. 

7 6. What fractional energy does a photon lose if it goes from the surface of the earth to deep space? Solution: When the photon goes from the surface of earth to outer space, it must lose the gravitational potential energy that is has near the surface of earth. So the photon must lose this energy. For photon

0 2 (U ) g00 = −1

On surface of earth with weak field limit

g00 = −(1 − 2φ)

So near the surface of earth

U 0 ' 1 + φ

In far space metric Minkowski g00 = −1 so in far space

U 0 = 1

So ratio of energy 1 = 1 − φ 1 + φ

So change in energy is ∼ φ On the surface of earth the gravitational potential is

GM 6.672 × 10−11 × 6.0 × 1024 φ = − = − ≈ 7 × 10−10 c2r 6.4 × 106 × 9 × 1016

So the photon must lose this energy fractionally. 

8