How Do You Find a Local Inertial Frame for a Given Metric?

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Homework Statement



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.

Homework Equations



Co-ordinate transformations.

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.
 
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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?
 
George Jones said:
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?

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