Finding the shapes of all timelike geodesics

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Homework Statement


Consider the two-dimensional spacetime with the line element
dS2 = -X2dT2+dX2.
Find the shapes X(T) of all timelike geodesics in this spacetime.

2. The attempt at a solution
I have the solution to this problem but I don't understand one step. For timelike worldlines
dS2 = -dt2 = 0 (where dt is the proper time)
We also have that the Lagrangian is L = (X2(dT/dσ)2 - (dX/dσ)2)1/2
The Euler-Lagrange equation gives us that ∂L/(∂(dT/dσ)) = const.
In the solutions, it is stated that this constant is identically equal to "e", and I do not understand why this is.

Could anyone explain this or point me in the right direction? Thanks.
 
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gnulinger said:
For timelike worldlines
dS2 = -dt2 = 0 (where dt is the proper time)

No, that's null worldlines. For timelike dS2 < 0.
 
clamtrox said:
No, that's null worldlines. For timelike dS2 < 0.

Oops. You're right.

Anyway, here is the part that I don't understand (taken from the solutions manual):
http://i.imgur.com/YL66F.png
 
Do you understand why this quantity related to the Killing vector is conserved? Each Killing vector corresponds to a coordinate which can be transformed to remain constant, ie. they correspond to conserved quantities. Timelike Killing vector gives you conservation of energy, so e in this case is energy density or something like that. You can find the theory behind Killing vectors in most GR books.

A less tricky way of doing the same calculation would be just to integrate the geodesic equations directly. So start like you usually do from the Lagrangian and find Christoffel symbols from it. It's just a bit more work, as geodesic equations are 2nd order while the Killing vector method gives you directly first order equations.
 
I understand that the quantity is conserved, but I thought that the solutions manual was stating that the constant was the number "e," as opposed to just some arbitrary label of a constant. I was wondering if there was some strange math fact that led to this.

Thank you for the help.
 
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