I would like to know whether my physical interpretation of some dynamics in accelerated frames is correct. In a frame with acceleration $a$ we have the metric $$ds^2 = (1+ax)^2 dt^2 - dx^2$$

The Christoffel symbols we find are $$\Gamma^x_{tt} = a(1+ax)$$ $$\Gamma^t_{xt} = \Gamma^t_{tx} = a/(1+ax)$$ and all others are $0$.

If we parametrise the curve of a massive particle by $\gamma(\tau)$ then if if this particle is at constant $x$ we find that $\ddot\gamma = 0$ (as we have to impose $\dot\gamma^\mu\dot\gamma_\mu = 1$ which implies that $\dot\gamma^t$ is linear in $\tau$ and we used that $\dot\gamma^x = 0$). Therefore the particle's acceleration is $$\ddot\gamma^\mu + \Gamma^\mu_{\sigma\nu}\dot\gamma^\sigma\dot\gamma^\nu = (0,\frac{a}{1+ax})$$ is this right?

And in the case of a free falling particle the acceleration is $0$ and therefore $$\ddot\gamma^\mu = -\Gamma^\mu_{\sigma\nu}\dot\gamma^\sigma\dot\gamma^\nu$$ which in the case of $a<<1$ and a slow moving particle is $$\ddot\gamma^\mu \approx (0,-a)$$

Thanks in advance for the help.

PS: If you want to know how I find the metric in the accelerated frame: starting from flat space-time (2 dimensional to keep the notation simple) we have: $$ds^2 = dT^2 - dX^2$$ And from this we can go to Rindler coordinates (where we have $T = \rho\sinh(\omega)$ and $X = \rho\cosh(\omega)$) in which the metric is $$ds^2 = \rho^2d\omega^2 - d\rho^2$$ A massive particle keeping $\rho$ constant is a particle that is constantly accelerating with acceleration $\rho^{-1}$. Therefore if we choose an acceleration $a$ we can get the normal frame of such an accelerating observer (with initial position $X = 1/a$ at $T = 0$) by choosing $$x = \rho - 1/a$$ $$t = \omega/a$$ which yields the metric $$ds^2 = (1+ax)^2 dt^2 - dx^2$$

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    $\begingroup$ Just as a general comment, not meant to be an answer, frames of reference do not normally play any important role in GR; GR has only local frames of reference, not global ones; and coordinates do not correspond to frames of reference. What you are doing here is probably also not GR, it's probably Minkowski space described in unusual coordinates. (I haven't checked that your metric is flat.) Are your coordinates supposed to be the same as Rindler coordinates? $\endgroup$ – Ben Crowell Jan 28 at 22:58
  • $\begingroup$ Well, these coordinates can be used in GR to describe what sees an observer in constant acceleration, and it's what we did in class. The coordinates are not flat: the Cristoffel symbols are not 0. It is clear that the coordinates are not global, but we can use them to get an understanding of some situations through the equivalence principle. The main example is the use of Rindler coordinates to study what happens close to the horizon of a black hole. In a similar way these coordinates can be used to describe a situation with a constant gravitational field. $\endgroup$ – albe_rola Jan 31 at 10:18

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