Theory of relativity/Rindler coordinates

Rindler coordinates(1) are coordinates appropriate for an observer undergoing constant proper acceleration (a constant g-force felt) in an otherwise flat spacetime. Given an unprimed inertial frame set of coordinates, one assigns the accelerated frame observer primed coordinates, "Rindler coordinates", given by

$ct=(\frac{c^2}{\alpha }+x^{\prime })\sinh \left(\frac{\alpha t^{\prime }}{c}\right)$

$x=(\frac{c^2}{\alpha }+x^{\prime })\cosh \left(\frac{\alpha t^{\prime }}{c}\right)-\frac{c^2}{\alpha }$

where $\alpha$ is the constant proper acceleration.

Developement

First Rindler solved the equation of motion for a constant proper acceleration problem. If we go to an inertial frame for which a test mass is instantaneously at rest, the four-vector of acceleration will relate to the coordinate acceleration by

$||A^{\prime }{\text{'}}^{\mu }||=(0,{𝐚}^{″})$

, but the length of the four-acceleration is an invariant we're calling the proper acceleration $\alpha$, so the inertial frame of instantaneous rest coordinate acceleration relates to the proper acceleration by

$\alpha =\sqrt{|{\eta }_{\mu \nu }{A}^{\prime }{\text{'}}^{\mu }{A}^{\prime }{\text{'}}^{\nu }|}=|a^{″}|$

The coordinate acceleration according to the inertial frame of instantaneous rest relates to the ordinary force by Newton's second law $f^{″}=m{a}^{″}$ so we have

$\alpha =\frac{|f^{″}|}{m}$

The x component of the ordinary force for a particle of velocity $𝐮$ with respect to the unprimed frame transforms between the two according to

$f_{x^{″}}=\frac{f_x-\frac{v}{c^2}(𝐮\cdot 𝐟)}{1-\frac{u_{x}v}{c^2}}$

And we are considering force along the direction of motion only so

$f^{″}=\frac{f-\frac{vuf}{c^2}}{1-\frac{uv}{c^2}}$

$f^{″}=f$

So our equation for proper acceleration becomes

$\alpha =\frac{f}{m}$

which means that the equation of motion for the case of constant proper acceleration is given by a constant ordinary force constant mass problem given according to whatever inertial frame we use. So we need to solve the equation of motion for a constant ordinary force, constant mass problem.

$m\alpha =\frac{dp}{dt}$

$m\alpha =m\frac{d\left(\gamma u\right)}{dt}$

$\alpha =\frac{(ud\gamma +\gamma du)}{dt}$

$\alpha =\frac{({\gamma }^3\frac{u^2}{c^2}du+\gamma du)}{dt}$

$\alpha dt=\gamma ({\gamma }^2\frac{u^2}{c^2}+1)du$

$\alpha dt=\gamma \left(\frac{\frac{u^2}{c^2}+1-\frac{u^2}{c^2}}{1-\frac{u^2}{c^2}}\right)du$

$\alpha dt=\gamma \left(\frac{1}{1-\frac{u^2}{c^2}}\right)du$

$\alpha dt=\frac{du}{{(1-\frac{u^2}{c^2})}^{\frac{3}{2}}}$

Integrating and choosing initial conditions so that its initially at zero velocity yields

$u=\frac{\alpha t}{\sqrt{1+{\left(\frac{\alpha t}{c}\right)}^2}}$

Using $\frac{dt}{d\tau }=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$ one can then solve for the time as a function of the proper time

$ct=\frac{c^2}{\alpha }\sinh \left(\frac{\alpha \tau }{c}\right)$

and using

$\frac{dx}{dt}=\frac{\alpha t}{\sqrt{1+{\left(\frac{\alpha t}{c}\right)}^2}}$

one can solve for the position as a function of time

${(1+\frac{\alpha x}{c^2})}^2-{\left(\frac{\alpha t}{c}\right)}^2=1$

and then the position as a function of proper time

$x=\frac{c^2}{\alpha }\cosh \left(\frac{\alpha \tau }{c}\right)-\frac{c^2}{\alpha }$

Having both the position and time now as a function of proper time

$ct=\frac{c^2}{\alpha }\sinh \left(\frac{\alpha \tau }{c}\right)$

$x=\frac{c^2}{\alpha }\cosh \left(\frac{\alpha \tau }{c}\right)-\frac{c^2}{\alpha }$

We merely make a natural choice of primed coordinates for which $x^{\prime }=0$ is described by those equations of motion. Rindler's choice was

$ct=(\frac{c^2}{\alpha }+x^{\prime })\sinh \left(\frac{\alpha t^{\prime }}{c}\right)$

$x=(\frac{c^2}{\alpha }+x^{\prime })\cosh \left(\frac{\alpha t^{\prime }}{c}\right)-\frac{c^2}{\alpha }$

which does just that.

Rindler Metric

Rindler's spacetime is then just an accelerated frame transformation of the Minkowski metric of special relativity. Doing the transformation

$ct=(\frac{c^2}{\alpha }+x^{\prime })\sinh \left(\frac{\alpha t^{\prime }}{c}\right)$

$x=(\frac{c^2}{\alpha }+x^{\prime })\cosh \left(\frac{\alpha t^{\prime }}{c}\right)-\frac{c^2}{\alpha }$

transforms the Minkowski line element

$d{s}^2=d(ct{)}^2-(d{x}^2+d{y}^2+d{z}^2)$

into

$d{s}^2={(1+\frac{\alpha x^{\prime }}{c^2})}^{2}dct{\text{'}}^2-(dx{\text{'}}^2+d{y}^2+d{z}^2)$

Generalizing beyond Rindler's

The Lorentz transformation can be written

$ct=\gamma c{t}^{\prime }+\gamma \beta x^{\prime }$

$x=\gamma x^{\prime }+\gamma \beta c{t}^{\prime }$

$y=y^{\prime }$

$z=z^{\prime }$

If we let $\beta$ and $\gamma$ be functions of the primed time, we can do a transformation given by (2)

$ct=\int \gamma dc{t}^{\prime }+\gamma \beta x^{\prime }$

$x=\gamma x^{\prime }+\int \gamma \beta dc{t}^{\prime }$

$y=y^{\prime }$

$z=z^{\prime }$

where we are doing anti-derivatives with respect to the primed coordinate time.

It turns out that Rindler's choice of coordinates for an observer undergoing constant proper acceleration is the special case of this transformation where $\beta$ and $\gamma$ are what you would get as a function of the primed time if they were describing constant proper acceleration. Doing this more general transformation for arbitrary time dependent acceleration the line element transforms to

$d{s}^2={(1+\frac{\alpha x^{\prime }}{c^2})}^{2}dct{\text{'}}^2-(dx{\text{'}}^2+dy{\text{'}}^2+dz{\text{'}}^2)$

$\alpha =c{\gamma }^2\frac{d\beta }{d{t}^{\prime }}$

so for arbitrarily time dependent acceleration, with these transformations you still get the Rindler spacetime, only $\alpha$ is now any function of $t^{\prime }$ instead of a constant.

Further, it turns out that for ${\alpha }_{x^{\prime }}$, ${\alpha }_{y^{\prime }}$, and ${\alpha }_{z^{\prime }}$ being any three arbitrary functions of $t^{\prime }$ that the line element given by

$d{s}^2={(1+\frac{{\alpha }_{x^{\prime }}{x}^{\prime }}{c^2}+\frac{{\alpha }_{y^{\prime }}{y}^{\prime }}{c^2}+\frac{{\alpha }_{z^{\prime }}{z}^{\prime }}{c^2})}^{2}dct{\text{'}}^2-(dx{\text{'}}^2+dy{\text{'}}^2+dz{\text{'}}^2)$

Is a zero Riemann tensor exact vacuum solution and so represents a more general accelerated frame transformation of the Minkowski metric of special relativity.

Rindler Horizon

The path of constant proper acceleration as observed by an inertial frame

The path of constant proper acceleration as observed by an inertial frame

The red curve of the image shows the path of something undergoing constant proper acceleration with respect to the unprimed inertial frame coordinates which we found was given by

${(1+\frac{\alpha x}{c^2})}^2-{\left(\frac{\alpha t}{c}\right)}^2=1$

Note it asymptotically approaches the line

$ct=x+\frac{c^2}{\alpha }$

The path of a light speed signal lays at 45 degrees. Therefore information from any event in the shaded region will never intersect the path of the accelerated observer so long as he maintains his constant proper acceleration. As such that asymptote constitutes an event horizon from the perspective of the accelerated observer beyond which he can not be reached by any information there. So let's find out where in the accelerated observer's coordinates this asymptote is. Refer to the transformation to Rindler coordinates

$ct=(\frac{c^2}{\alpha }+x^{\prime })\sinh \left(\frac{\alpha t^{\prime }}{c}\right)$

$x=(\frac{c^2}{\alpha }+x^{\prime })\cosh \left(\frac{\alpha t^{\prime }}{c}\right)-\frac{c^2}{\alpha }$

First lets move the constant on the second

$ct=(\frac{c^2}{\alpha }+x^{\prime })\sinh \left(\frac{\alpha t^{\prime }}{c}\right)$

$x+\frac{c^2}{\alpha }=(\frac{c^2}{\alpha }+x^{\prime })\cosh \left(\frac{\alpha t^{\prime }}{c}\right)$

Next lets replace the $x+\frac{c^2}{\alpha }$ from the equation for the asymptote.

$ct=(\frac{c^2}{\alpha }+x^{\prime })\sinh \left(\frac{\alpha t^{\prime }}{c}\right)$

$ct=(\frac{c^2}{\alpha }+x^{\prime })\cosh \left(\frac{\alpha t^{\prime }}{c}\right)$

Then square both sides

$c{t}^2={(\frac{c^2}{\alpha }+x^{\prime })}^2{\sinh }^2\left(\frac{\alpha t^{\prime }}{c}\right)$

$c{t}^2={(\frac{c^2}{\alpha }+x^{\prime })}^2{\cosh }^2\left(\frac{\alpha t^{\prime }}{c}\right)$

Now subtract the top equation from the bottom equation

$0={(\frac{c^2}{\alpha }+x^{\prime })}^2({\cosh }^2\left(\frac{\alpha t^{\prime }}{c}\right)-{\sinh }^2\left(\frac{\alpha t^{\prime }}{c}\right))$

Hyperbolic trig identity

$0={(\frac{c^2}{\alpha }+x^{\prime })}^2$

Now the solution to this is

$x^{\prime }=-\frac{c^2}{\alpha }$

So we see that the event horizon for the accelerated observer is observed by him as a constant distance coordinate behind him. This time-time element of the contravariant metric tensor in his coordinates is

$g^{00}=\frac{1}{{(1+\frac{\alpha x^{\prime }}{c^2})}^2}$

Note that the event horizon corresponds to a singularity in this element of the metric tensor for his coordinates.

Comparing to Schwarzschild

In comparing the physics an observer of constant proper acceleration observes to that a remote observer from an uncharged nonrotating black hole using Schwarzschild coordinates observes, it is best to choose the remote behavior of the accelerated observer's spatial coordinate scaled by the transformation

$1+\frac{2\alpha x^{″}}{c^2}={(1+\frac{\alpha x^{\prime }}{c^2})}^2$

With this choice of distance coordinate for the accelerated observer, the line element becomes

$d{s}^2=(1+\frac{2\alpha x^{″}}{c^2})dct{\text{'}}^2-\frac{d{x}^{\prime }{\text{'}}^2}{(1+\frac{2\alpha x^{″}}{c^2})}-d{y}^2-d{z}^2$

which is now comperable to the Schwarzschild metric in Schwarzschild coordinates expressed by

$d{s}^2=(1-\frac{2GM}{r{c}^2})dc{t}^2-\frac{d{r}^2}{(1-\frac{2GM}{r{c}^2})}-r^{2}d{\theta }^2-r^2{\sin }^2\theta d{\varphi }^2$

References

(1) Rindler, W., 1969, Essential Relativity: Special, General, and Cosmological, Van Nostrand, New York

(2) Relativity equation 3.3.8