# Timelike geodesic

1. Jan 7, 2017

### binbagsss

1. The problem statement, all variables and given/known data

The question is to find $A$ and $B$ such that the specified curve (we are given a certain parameterisation , see below) is a timelike geodesic , where we have $|s| < 1$

I am just stuck on the last bit really.

So since the geodesic is affinely paramterised $dL/ds=0$ and so I can set $L=constant$, $L$ the Lagrangian of a freely-falling particle.

Let $L$ be this constant.

And with the specified metric and parameterised curve, which are all given to us, this gives:

$B^2(\frac{A^2-s^2}{1-s^2}) = L$

This is all fine.

MY QUESTION IS...

2. Relevant equations

see above

3. The attempt at a solution

MY QUESTION IS...

From this I conclude that (since a null curve is given by $L=0$, a space-like by $L < 0$ and a time-like by $L>0$, since the metric signature in the question is ( +, - ) ) that we require $|A|<1$ since we have $|s| < 1$ , and $B\neq 0$, however the solution gives:

we need $A=\pm 1$ and $B\neq 0$.
I don't understand where $A=\pm 1$ comes from, I thought we just need it such that $L > 0$ and $A=\pm 1$ does this

2. Jan 7, 2017

### Orodruin

Staff Emeritus
You have not given us the full problem. You have forgotten to specify the given curve and metric.

3. Jan 9, 2017

### binbagsss

I know, Im pretty sure theyre not needed, it is just the final conclusion described above that I am stuck on. but I will post them now.

curve $t= A tanh^{-1} s$ , $x=B(1-s^{2})^{1/2}$
metric : $ds^2=x^2 dt^2 - dx^2$

Last edited: Jan 9, 2017
4. Jan 9, 2017

### Orodruin

Staff Emeritus
Did you try inserting the curve into the geodesic equations for the given metric?

5. Jan 11, 2017

### binbagsss

the question was completed using the euler-lagrange equations. One replaced with setting $L$ to a constant as above, the other the E-L equation for $t$ which gave no new constraints on $A$ and $B$ .

It is just the conclusion as I say in OP that I am on stuck on.