G4040_Sol_Set
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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 dλ = dφ dλ d dφ = a d 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, a 2 d 2 x α dφ 2 + a 2 Γ α μβ dx μ dφ dx β dφ =0. We can divide by a 2 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 A β = U α (A β ,α +Γ β αγ A γ )=0, U α ∇ α B β = U α (B β ,α +Γ β αγ B γ )=0. Now let’s see what differential equation is obeyed by the inner product of A and B, (g αβ A α B β ) ,γ = g αβ,γ A α B β + g αβ A α ,γ B β + g αβ A α B β ,γ . We want to use somehow the parallel transport condition, so let’s multiple by U γ , U γ (g αβ A α B β ) ,γ = g αβ,γ U γ A α B β - g αβ U γ Γ α γσ A σ B β - g αβ A α U γ Γ β γσ B σ . No notice that we can factorize U γ A α B β , being careful to change correctly dummy indices, U γ (g αβ A α B β ) ,γ = U γ A α B β (g αβ,γ - g σβ Γ σ γα - g ασ Γ σ γβ ) ,U γ (g αβ A α B β ) ,γ = U γ A α B β ∇ γ g αβ =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. 1
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Transcript of G4040_Sol_Set
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Introduction to General Relativity
(G4040 - Fall 2012)
Solutions: Problem set 5Prof. A. Beloborodov
TA: Alex Chen
1. (Schutz, 6.9: 12) Formally we can write,
d
d=d
d
d
d= a
d
d.
If the geodesic equation is satisfied by a curve parameterized by , then the geodesic equation for thenew parameter a+ b is,
a2d2x
d2+ a2
dx
d
dx
d= 0.
We can divide by a2 to recover the original form of the equation proving that is in fact an affineparameter.
2. (Schutz, 6.9: 13)
(a) A or B being parallel transported along U means that
UA = U(A, + A) = 0, UB = U(B, + B) = 0.
Now lets see what differential equation is obeyed by the inner product of A and B,
(gAB), = g,A
B + gA,B
+ gAB, .
We want to use somehow the parallel transport condition, so lets multiple by U ,
U(gAB), = g,U
AB gUAB gAUB.
No notice that we can factorize UAB , being careful to change correctly dummy indices,
U(gAB), = U
AB(g, g g), U(gA
B), = UABg = 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. Inother words, the derivative of the inner product along the curve is zero.
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(b) Lets observe that the geodesic equation is in fact telling as that the tangent vector is paralleltransported along the curve. The geodesic equation for a vector U = dx/d is
dU
d+ U
U = 0,
U
xx
+ U
U = 0.
The geodesic path x() is in fact only a function of , so in the previous equation we can factorizeU getting,
UU = 0,
proving that the norm of the tangent vector of a geodesic is constant along the geodesic. If ageodesic 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 theproper distance between 2 points, on the curve, is a constant times the difference of the parameterbetween those 2 points. (Since by assumption the curve is a geodesic we can assume that is itself anaffine parameter).
4. (a) For the 2-Sphere the only non-zero Christoffel symbols are:
= sin cos ,
= = cot .
The divergence of a vector field A = (cos , sin cos) is,
A = A, + A,
which is,
A = sin sin sin+ cot cos ,
(b) The geodesic equations are:
d2x
d2+
dx
d
dx
d= 0.
Therefore the geodesic equations on the 2-sphere become:
= ()2 sin cos ,
=
cot ,
so we see that the curve { =constant,() =constant} satisfy the geodesic equation.(c) The geodesic equations for = 0 constant, reduce to
0 = ()2 sin 0 cos 0,
= 0.
If sin 0 cos 0 6= 0 the only solution for is () = 0 constant. Of course this is a degenerategeodesic joining a point with itself. On the contrary, if sin 0 cos 0 = 0, 0 can be 0 or /2 or .For 0 = {0, } we have degenerate solutions, for 0 = /2 we get that the equator is a geodesic,which is true.
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