I'm learning about accelerating reference frames (to eventually get grasp of general relativity too).

I've just read about the Rindler coordinates and this one caught my eye

Note that Rindler observers with smaller constant x coordinate are accelerating harder to keep up! This may seem surprising because in Newtonian physics, observers who maintain constant relative distance must share the same acceleration. But in relativistic physics, we see that the trailing endpoint of a rod which is accelerated by some external force (parallel to its symmetry axis) must accelerate a bit harder than the leading endpoint, or else it must ultimately break. This is a manifestation of Lorentz contraction. As the rod accelerates its velocity increases and its length decreases. Since it is getting shorter, the back end must accelerate harder than the front.

So does this mean, that in Einstein's elevator (that's used to introduce general relativity, and is supposed to accelerate uniformly) the acceleration felt by the observer on the top is smaller than the acceleration felt by the observer at bottom?


I did some research and calculations:

To summarize: The relativistic rocket won't break apart, uniform acceleration along it is possible. But the observers will measure different accelerations due to the gravitational time dilation.

In more detail:

Let's assume the observer at the bottom measures $\alpha$ acceleration. So for an inertial observer outside (who draws the Minkowski chart) this accelerating observer's motion will be hyperbolic. The semi mayor axis of this hyperbola will be $c^2/\alpha$.

Let's say the length of the relativistic rocket is $h$ this is measured before the launch, and the mechanical stresses during the travel try to keep this value constant in the reference frame of the rocket, otherwise the rocket would break apart (we assume it don't break apart). As the rocket accelerates the plane of simultaneity rotates from the viewpoint of the inertial observer. So the two ends of the rocket won't trace two identical hyperbolas. But the two ends always connect two points on the two hyperbolas whose slope is the same (you can see this on the Rindler chart). So all parts of the rocket travel with the same speed at the local frame simultaneously, so the rocket's acceleration will be uniform and it won't break apart.

But the two hyperbolas are different. The bottom traces a hyperbola whose semi-mayor axis is $c^2/\alpha$, the top traces a hyperbola whose semi-mayor axis is $c^2/\alpha + h$. The acceleration that corresponds to the second hyperbola is $1 + \alpha h / c^2$ times smaller than $\alpha$.

This a bit paradoxical situation, because I stated that the acceleration is uniform along the rocket, now I state it's different due to the different hyperbolas.

This paradox can be resolved if I introduce gravitational time dilation, so I assume that the clock of the observer at the top ages faster with the rate I mentioned above. So the top observer measures less acceleration this way.

There is an event horizon at the Rindler-horizon where $h = -c^2/\alpha$, there the time stands still. This is somewhat analogous with the black hole's event ho+rizon.

The gravitational time dilation formula mentioned in the Wikipedia and the one mentioned in the comment is the same, but that exponentional formula never reaches zero, that would mean the Rindler-horizon does not exist... Which would be a bit odd. So I still need some research.

Update: The Wikipedia article has been fixed since last update. So the general formula to the gravitational time dilation is $e^{\int^h_0 g(h)dh / c^2}$, where $g(h)$ is the measured gravitational acceleration at the given level. For Rindler observers $g(h) = c^2/(H+h)$ where $H = c^2/\alpha$. Doing the integral gives $e^{ln(H+h) - ln(H)} = (H+h)/H$. Substituting $H$ back I will get the original $1 + \alpha h / c^2$ I mentioned earlier.

  • $\begingroup$ Calmarius wrote: "Let's say the height of the relativistic rocket is $h$." What is meant by "height of a rocket"; how is this quantity to be measured; does it have some particular relation to distance" (a.k.a. "[en.wikipedia.org/wiki/Length]") between two "ends" which are and remain at rest to each other? -- "But the two ends always connect two points on the two hyperbolas whose slope is the same (you can see this on the Rindler chart)." Please elaborate; because there's no mentioning of "slope" at [en.wikipedia.org/wiki/Rindler_coordinates] $\endgroup$ – user12262 Jun 24 '13 at 19:08
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    $\begingroup$ The length of the rocket remains constant in the reference frame of the rocket (edited). If the hyperbolas share the same center and asymptotes, they are basically magnified versions of each other. If you draw a ray from the common center the rays will intersect all hyperbolas at the points where the slope (call it whatever you want) is the same. The planes of simultaneity of the accelerating body intersect at the center. Thus the velocity is same everywhere along the object, as it accelerates (as seen from outside). I've just thought this can be simply seen on the Rindler chart. $\endgroup$ – Calmarius Jun 25 '13 at 8:19
  • $\begingroup$ Thanks for the response + effort. What's bugging me shows here: "Thus the velocity is same everywhere along the object, as it accelerates (as seen from outside)." -- Is this meant as an exact statement, or only as approximation? Is there at least one inertial system whose members find exactly simultaneously that the rocket ends moved at exactly equal velocity? And: "The length of the rocket remains constant in the reference frame of the rocket (edited)." -- How is "height" to be determined by members of this particular "reference frame" (such as the two rocket ends in particular)? $\endgroup$ – user12262 Jun 25 '13 at 19:46
  • $\begingroup$ Full disclosure: I asked this question after having submitted an answer here. $\endgroup$ – user12262 Jun 25 '13 at 19:47
  • $\begingroup$ @user12262 Well this velocity thing is quite tricky, and I probably used the wrong words. And it's quite hard to describe anyway. I meant proper velocity on both ends here, but despite its name, it's not invariant, you will need a "map frame" to measure proper velocity against. That's why someone uses the word "celerity" instead... But the gravitational/acceleration time dilation makes this tricky. $\endgroup$ – Calmarius Jun 26 '13 at 19:07

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