Local Coordinates for Non-Uniformly Accelerated Observer

[LAHLH]

Hi,

I'm reading a paper about acceleration and the author states the local coordinates of the observer $$(\tau,x)$$ (for a non-uniformly accelerated observer) are specified (in relation to the inertial coordates (T,X)):

$$T(\tau)=\int^{\tau}\,\mathrm{d}\alpha\,[1+g(\alpha)x]\cosh{(\chi(\alpha))}$$
$$X(\tau)=\int^{\tau}\,\mathrm{d}\alpha\,[1+g(\alpha)x]\sinh{(\chi(\alpha))}$$

the author references another paper that then references another book by Moller (1969) but when I got that book I couldn't find the relevant section that explains these coordinates. So the above is how the inertial coordinates (T,X) are related to the local coordinates $$(\tau,x)$$.

If one is just interested in the trajectory of this accelerated (non-uniformly) observer in the inertial coords (T,X) then I can see that the trajectory is:

$$T_{*}(\tau)=\int^{\tau}\,\mathrm{d}\alpha\,\cosh{(\chi(\alpha))}$$
$$X_{*}(\tau)=\int^{\tau}\,\mathrm{d}\alpha\,\sinh{(\chi(\alpha))}$$

where $$d\chi(\tau)/d\tau=g(\tau)$$ and $$g(\tau)$$ is the time varying proper acceleration in all of the above.

So I can see where the trajectory of this observer in the inertial coords comes from (one just has to differentiate and take the four vector product of the four acceleration to see it has the correct norm etc. But I'm trying to get my head around these local coordinates, do they correspond to the Fermi-Walker tetrad somehow?

 

[LAHLH]

Actually it seems possible to invert these to write:

$$x=\frac{1}{g(\tau)}\left[\sqrt{\left(\frac{dT}{d\tau}\right)^2-\left(\frac{dX}{d\tau}\right)^2}-1\right]$$

I'm not sure what metric the author is using but assuming $$U^2=-1$$ along the accelerated trajectory then $$\sqrt{\left(\frac{dT}{d\tau}\right)^2-\left(\frac{dX}{d\tau}\right)^2}=1$$ then x=0 for the accelerated trajectory? so x is a good coordinate for a comoving reference frame with the observer?

------------

I also seem to find inverting to try and get $$\tau$$:

$$\chi(\tau)=\text{arctanh}(\frac{dX}{dT})$$

so given that $$d\chi/d\tau=g(\tau)$$:

$$g(\tau)=\frac{1}{1-\left(\frac{dX}{dT}\right)^2} =\gamma^2\text{??}$$