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I find it best to start this question with a thought experiment. We have a rocket ship and Earth. For our purposes, Earth has no acceleration or velocity and begins at the origin. The rocket ship accelerates away from Earth with a constant acceleration $a$. This rocket ship begins at the origin ($x'_0=0$) with 0 velocity. In this thought experiment, we ignore higher order derivatives and gravity. If anything, we should derive the relativistic effects of gravity from the results of this thought experiment, as shown by the equivalence principle.

In my attempts to solve this so far, I have assumed that the speed of light is invariant in the accelerating reference frame (so that$x'=ct'$ here for a light pulse). Is this valid? Is acceleration (as shown in the thought experiment) absolute or relative (are both the Earth and the spaceship accelerating relative to each other or is the spaceship absolutely accelerating)? So far, here is what I have come up with in trying to solve this. Please tell me if I am wrong anywhere, what next steps I should take, and/or what conclusions I should come to:

  • The Lorentz factor gamma will still be a part of this accelerating coordinate transformation, but instead of being a function of $v$, it will be a function of $at$, as that is the velocity between the reference frames (remember that $v_0$ is 0 for my initial conditions, so I never considered it for simplicity, but feel free to include it as it is important in this transformation) so that gamma takes the form $1/\sqrt{1-(at)^2/c^2}$.

  • For the $x' $ transformation equation, $ x $ is unchanged, as it refers to the position of an event, and thus stays the same when acceleration is introduced to the transformation.

  • For the $x' $ transformation equation, $ vt $ is integrated (I am probably using an incorrect term here, I mean that acceleration is included in the velocity), and becomes $1/2at^2+v_0^2$ (where $v_0 $ is 0 for my initial conditions). as such, the $x' $ transformation equation is $\gamma((x_{event})-(v_{frame})t-1/2(a_{frame})t^2)$.

I attempted to find the $t' $ transformation equation with these parameters and knowing that $ x'=ct' $ while $ x=ct$, but I wasn't able to solve it. What are the spacetime coordinate transformations between reference frames when one is accelerating relative to the other? How do we find them?

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I have assumed that the speed of light is invariant in the accelerating reference frame (so that x'=ct' here for a light pulse). Is this valid?

There is no standard reference frame defined for a non-inertial observer like there is for inertial observers. There are many valid non-inertial reference frames where the speed of light is not equal to c. However, there is a class of non-inertial reference frames where the speed of light is equal to c, where the time and space coordinates are based on radar time and radar distance.

See Dolby and Gull: https://arxiv.org/abs/gr-qc/0104077

What are the spacetime coordinate transformations between reference frames when one is accelerating relative to the other? How do we find them?

The transforms are substantially more complicated than you were suggesting. The full transform is derived in the Dolby and Gull citation. In particular, you are interested in their radar coordinates in region U described in Figure 6 and the associated equations. As they derive (in natural units) $$t’=\frac{1}{2a}\log\left(\frac{x+t}{x-t} \right)$$ $$x’=\frac{1}{2a}\log\left( a^2 x^2-a^2 t^2\right)$$

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  • $\begingroup$ Alright, I looked at that article. Are you sure that it is technically accurate/trustworthy? I've been told that arXiv is generally untrustworthy due to reasons such as a lack of peer review. In addition, the equations reduce to 0 if a=0, regardless of coordinates. Is this because I am assuming that x_0=0 and v_0=0? Finally, the t' equation seems to imply that, due to the log function in the t' equation, as x increases, t' increases at a varying rate. This is confirmed in the article, where t' in the 'hypersurface of simultaneity' is a logarithmic function. As such, doesn't the shifting of $\endgroup$
    – Sciencemaster
    Nov 26, 2020 at 4:43
  • $\begingroup$ time proportional to space relative to an observer (shift of t' as x increases) have an exponential rate of change? This doesn't seem to make sense to me, but maybe that is just because I am used to non-accelerating special relativity. After all, space shifts at the 'same rate' as ct, and the observer is accelerating in space (the difference between x -> x' is exponential). $\endgroup$
    – Sciencemaster
    Nov 26, 2020 at 4:46
  • $\begingroup$ @Sciencemaster you are right that arxiv is not a reliable source. In fact, it should not be considered a source at all. It is simply an archival service. The source of the Dolby and Gull article is not arxiv, but rather the peer-reviewed Am.J.Phys. 69 (2001) 1257-1261. Arxiv is just a convenient place to access that source. Regarding the fact that the coordinates do not meet your expectations, you are welcome to do your own derivation. You can start with the basic radar time and distance formulas and choose your own parameterization for the accelerating observer. I encourage you to do so $\endgroup$
    – Dale
    Nov 26, 2020 at 4:58
  • $\begingroup$ Alright, thanks for clearing that up. In addition, looking back, I made an error: the article states that the 'hypersurface of simultaneity' is a function of tangent, not log (which still seems counterintuitive to me, but hopefully I can accept it once I derive it.) However, the concept of radar time is new to me. I attempted to search for it online, which yielded no results other than the same article (which is a red flag for the credibility of the article in my mind). At any rate, it just seems to be the average of the time between when a light beam is emitted and when it is received. $\endgroup$
    – Sciencemaster
    Nov 27, 2020 at 6:33
  • $\begingroup$ Another issue I take with this article that I would like to clear up is with the initial assertion that it wouldn't make sense for an event to be assigned multiple time coordinates in the accelerating reference frame and that an object 'sweeps back in time'. At first glance these seem to be very valid and intuitive properties of spacetime that need to be 'fixed'. However, it is important to note that this only occurs outside of an accelerating observer's light cone. More important to note however is that in the case where an observer accelerates to 'turn around', the observer... $\endgroup$
    – Sciencemaster
    Nov 28, 2020 at 6:13

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