- #71
PeterDonis
Mentor
- 46,863
- 23,330
JDoolin said:It comes out so whether you do the calculation by (X and T) or by (x and [itex]\tau[/itex]), you get the very same result for the acceleration.
This looks OK to me as a calculation of the acceleration (or rather a verification that the calculation gives the same results in the inertial frame and the accelerated frame). I assume you would then just take the derivative with respect to x to get dg/dx?