# 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 '18 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. – Sjorszini Jan 20 '18 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. – Sean E. Lake Jan 20 '18 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 '18 at 19:04
• – user4552 Feb 25 '18 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.