Introduction to General Relativity. (G4040 - Fall 2012). Solutions: Problem set 5.
Prof. A. Beloborodov. TA: Alex Chen. 1. (Schutz, 6.9: 12) Formally we can write,.
Introduction to General Relativity (G4040 - Fall 2012) Solutions: Problem set 5 Prof. A. Beloborodov TA: Alex Chen 1. (Schutz, 6.9: 12) Formally we can write, d dφ d d = =a . dλ dλ dφ dφ If the geodesic equation is satisfied by a curve parameterized by λ, then the geodesic equation for the new parameter φ ≡ aλ + b is, a2
dxµ dxβ d2 xα + a2 Γα = 0. µβ 2 dφ dφ dφ
We can divide by a2 to recover the original form of the equation proving that φ is in fact an affine parameter. 2. (Schutz, 6.9: 13) (a) A or B being parallel transported along U means that U α ∇α Aβ = U α (Aβ,α + Γβαγ Aγ ) = 0,
U α ∇α B β = U α (B β,α + Γβαγ B γ ) = 0.
Now let’s see what differential equation is obeyed by the inner product of A and B, (gαβ Aα B β ),γ = gαβ,γ Aα B β + gαβ Aα,γ B β + gαβ Aα B β,γ . We want to use somehow the parallel transport condition, so let’s multiple by U γ , σ β α γ β σ U γ (gαβ Aα B β ),γ = gαβ,γ U γ Aα B β − gαβ U γ Γα γσ A B − gαβ A U Γγσ B .
No notice that we can factorize U γ Aα B β , being careful to change correctly dummy indices, U γ (gαβ Aα B β ),γ = U γ Aα B β (gαβ,γ − gσβ Γσγα − gασ Γσγβ ) , U γ (gαβ Aα B β ),γ = U γ Aα B β ∇γ gαβ = 0. The last equality follows if we have a connection that is metric compatible, which is our case. This proves that the inner product of parallel transported vectors is constant along the curve. In other words, the derivative of the inner product along the curve is zero.
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(b) Let’s observe that the geodesic equation is in fact telling as that the tangent vector is parallel transported along the curve. The geodesic equation for a vector U µ = dxµ /dτ is dU µ + Γµαβ U α U β = 0, dτ ∂U µ ∂xγ + Γµαβ U α U β = 0. ∂xγ ∂τ The geodesic path xγ (τ ) is in fact only a function of τ , so in the previous equation we can factorize U γ getting, U γ ∇γ U µ = 0, proving that the norm of the tangent vector of a geodesic is constant along the geodesic. If a geodesic is somewhere spacelike it will remain spacelike, similar for null or timelike geodesics. 3. (Schutz, 6.9: 14) From equation 6.8 and the previous problem, we conclude that for geodesics the proper distance between 2 points, on the curve, is a constant times the difference of the parameter between those 2 points. (Since by assumption the curve is a geodesic we can assume that λ is itself an affine parameter). 4. (a) For the 2-Sphere the only non-zero Christoffel symbols are: Γθφφ
=
− sin θ cos θ,
Γφθφ
=
Γφφθ = cot θ.
The divergence of a vector field Aµ = (cos θ, sin θ cos φ) is, µ ∇α Aα = Aα,α + Γα αµ A ,
which is, ∇α Aα = − sin θ − sin θ sin φ + cot θ cos θ, (b) The geodesic equations are: dxρ dxσ d2 xµ + Γµρσ = 0. 2 dλ dλ dλ Therefore the geodesic equations on the 2-sphere become: 00
0
θ = (φ )2 sin θ cos θ, 00
0
0
φ = −θ φ cot θ, so we see that the curve {φ =constant,θ(λ) =constant×λ} satisfy the geodesic equation. (c) The geodesic equations for θ = θ0 constant, reduce to 0
0 = (φ )2 sin θ0 cos θ0 , 00
φ = 0. If sin θ0 cos θ0 6= 0 the only solution for φ is φ(τ ) = φ0 constant. Of course this is a “degenerate” geodesic joining a point with itself. On the contrary, if sin θ0 cos θ0 = 0, θ0 can be 0 or π/2 or π. For θ0 = {0, π} we have degenerate solutions, for θ0 = π/2 we get that the equator is a geodesic, which is true.
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