# In special relativity, is there a prescription to find the coordinate transformation to a general (for instance accelerated) observer? And in GR?

Suppose we are in an inertial frame, so that the metric in our coordinates is just the Minkowski metric, \begin{align} \text{d}s^2 = -\text{d}t^2 + \text{d}x^2 + \text{d}y^2 + \text{d}z^2. \end{align} And suppose that some other (possibly non-inertial) observer $\mathcal O$ is described by a worldline $x^\mu(\tau)$ in our coordinates.

Given this worldline, does there exist a general prescription for how to relate the coordinate system $(t',x',y',z')$ of the observer $\mathcal O$ to our own coordinate system $(t,x,y,z)$?

If the worldine of $\mathcal O$ is a straight line, then clearly the transformation from $(t,x,y,z)$ to $(t',x',y',z')$ is given by a Poincare transformation, but I'm interested precisely in worldlines for which this is not the case.

As an example, suppose that we work in $1+1$ dimensions and that the worldine of $\mathcal O$ is given by $x^\mu(\tau) = (\frac{1}{\alpha}\sinh(\alpha\tau),\frac{1}{\alpha}\cosh(\alpha\tau))$. This describes an observer with constant acceleration (understood in the appropriate sense). How do we find out what the world looks like to him, and in particular, what line element $(\text{d}s')^2$ he sees?

The question is not necessarily specific to special relativity, of course. More generally, given a worldline in some coordinate system on any spacetime (not just flat spacetime), is there a general prescription to find the coordinate transformation to the observer corresponding to this worldline? (Assuming that there actually exists an observer corresponding to the worldline.)

• This is not in general a reasonable thing to expect to be able to do. In SR, these coordinate transformations will only work locally. The coordinates can only be extended to a distance $x\sim c^2/a$ away from the observer's world-line, where $a$ is the acceleration of the observer. At greater distances, the coordinate transformation isn't one-to-one. In GR, coordinate systems do not describe observers.
– user4552
Jan 19, 2018 at 22:46
• @BenCrowell You write that 'these' coordinate transformations will only work locally. So it seems you are implying that natural candidates for these transformations actually do exist? I would be more than glad to know how to find those. I understand that they won't in general be global. Jan 20, 2018 at 0:49
• Something else to consider: an accelerating observer (instantaneously Rindler) will see space-time as having the same event horizon that the Rindler observer does, for that instant. That is, assuming that the higher derivatives in the path aren't somehow relevant. Jan 20, 2018 at 0:58
• @Sjorszini: So it seems you are implying that natural candidates for these transformations actually do exist? There is only one natural coordinate transformation to consider locally, and that's a Lorentz transformation into the frame that is instantaneously comoving with the observer.
– user4552
Feb 25, 2018 at 19:04
• – user4552
Feb 25, 2018 at 19:04

In a very broad sense, a reference frame (observer) could be defined as a $(1,1)$-tensor field $\mathcal{R}$ on the spacetime $M$ with $Tr(\mathcal{R})=1$. To simplify the matter, we could start with a vector field $\Gamma$ and a one-form $\theta$ such that $\theta(\Gamma)=1$ and set $\mathcal{R}=\Gamma\otimes\theta$ (in this way we are fixing a parametrization for the integral curves of $\Gamma$ describing the worldlines of the observer). Roughly speaking, the integral curves of $\Gamma$ defines the "time", while, if $\theta$ is integrable ($\theta\wedge\mathrm{d}\theta=0$), its kernel defines a 3-dimensional distribution the leaves of which may be interpreted as instantaneous spatial slices. Clearly, you can find a set of coordinates which is adapted to the foliation induced by $\theta$ and its complementary foliation (the integral curves of $\Gamma$) only locally. This means that, in general, there is no rule to associate with a reference frame (observer) a global coordinate system. The situation is different for inertial reference frames in Minkowski spacetime because in that case $\mathcal{R}$ splits into the tensor product of a linear vector field with a linear one-form, and there is always a global coordinate system associated with such objects.