Finding a Local Inertial Frame

1. Jan 7, 2015

PeroK

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

I am trying to find a local inertial frame for the following metric:

$ds^2 = -(1+\Phi(x))dt^2 + (1-\Phi(x))dx^2$

I can get the transformed metric to equate to η at any point, but I can't get the first derivates wrt the transformed coordinates to vanish.

2. Relevant equations

Co-ordinate transformations.

3. The attempt at a solution

With a transformation of the form:

$t = \frac{\gamma}{a}(t' + vx') \ , \ x = \frac{\gamma}{b}(vt' + x')$

And setting $a= \frac{1}{\sqrt{1+\Phi(x_p)}}$ and $b= \frac{1}{\sqrt{1-\Phi(x_p)}}$ for a particular point $x_p$ I get

$g'_{αβ} = η$ as required

I've got my one degree of freedom, in terms of the arbitrary velocity $v$ to try to make the first derivative of $g'$ wrt $x'$ vanish. But, for example, I get:

$g'_{00} = \gamma^2(-\frac{(1+\Phi(x))}{a^2} + \frac{v^2(1-\Phi(x))}{b^2})$

Giving:

$\frac{\partial g'_{00}}{\partial x' } = - \frac{\gamma^3}{b} \Phi '(x) (\frac{1}{a^2} + \frac{v^2}{b^2})$

Which is not going away. The general argument is that one can always find ccordinates where all the first derivates vanish, but I don't see it in this case.

This was an example I set myself to try to see how the general process of finding a local IRF worked.

2. Jan 7, 2015

George Jones

Staff Emeritus
Are your primed coordinate curves geodesics? For example, if $x_p$ is the event labeled by $\left( x' , t' \right) = \left( 0 , 0 \right)$, are the curves $t' = 0$ and $x' = 0$ geodesics?

3. Jan 7, 2015

PeroK

The primed co-ordinates have the same origin as the unprimed. Should I look instead at primed co-ordinates centred on $x_p$?