Proving Free-Fall in Rindler Spacetime is Finite

[Antarres]

Okay, so, while discussing Rindler space with my professor, I was asked to prove that for a free-falling observer, proper time for passing through the Rindler horizon is finite. That is at least how the question is phrased.
So, the professor obviously assumes that it is clear and trivial to me what this means, but I'm having doubts. That is, there was no mention of gravity in the whole discussion(it is just hyperbolic motion standard treatment), so I assumed that free-fall means that the observer moving without any forces acting on it, hence it is moving with constant velocity(this is analogous to how free fall is treated in GR, just with abscence of curvature). In that case it is trivial to show that he will pass the Rindler horizon in finite proper time. However, it makes no sense to me that an observer who is accelerating with acceleration of constant magnitude(assuming this is what free-fall means, something similar to homogeneous gravity field), would pass through the horizon in finite proper time, since his accelerated motion is what defines this horizon.
Am I missing something here? I feel like this should be simple, and maybe I'm overthinking, or maybe what I have to prove makes no sense, since that professor has habit of phrasing questions like that vaguely(it's some sort of homework, but not real homework, since we're not expected to bring it for grades, it's just like a question to think about I guess).

 

[PeterDonis]

I assumed that free-fall means that the observer moving without any forces acting on it

Yes.

hence it is moving with constant velocity

Not in the standard Rindler coordinates used for Rindler spacetime, since those coordinates are not inertial. A free-falling object will accelerate "downward" (i.e., towards the Rindler horizon) in these coordinates.

You could, of course, switch to using standard inertial coordinates--indeed, it looks like this is what you have done when you say:

In that case it is trivial to show that he will pass the Rindler horizon in finite proper time.

But in that case...

it makes no sense to me that an observer who is accelerating with acceleration of constant magnitude(assuming this is what free-fall means

...you are confusing yourself by trying to work in two different coordinate charts at once. A freely falling object is not "accelerating with acceleration of constant magnitude" in standard inertial coordinates. It only is in non-inertial Rindler coordinates.

Further:

would pass through the horizon in finite proper time, since his accelerated motion is what defines this horizon

No, the accelerated motion of the Rindler observer--the one who has nonzero proper acceleration and follows a hyperbolic trajectory in standard inertial coordinates--is what defines the Rindler horizon (as the asymptotes of the hyperbola).

this is analogous to how free fall is treated in GR

Only if you do it correctly. See above.

 

[PAllen]

Your first guess is correct and your second guess is just wrong. There is no GR here, and free fall in SR means no proper acceleration, i.e. constant velocity in an inertial frame. Note, however, such a free fall motion will not be constant coordinate speed in Rindler coordinates. I assume your professor meant for you to describe inertial motion in Rindler coordinates and then, using the Rindler metric, verify that the horizon is reached in finite proper time. This is reasonably simple, but not trivial.

 

[A.T.]

I was asked to prove that for a free-falling observer, proper time for passing through the Rindler horizon is finite.

For the "free-falling observer" there is no Rindler horizon, so the question is confusing. Replace "free-falling observer" with "free-falling clock" to make sense of it.

 

[Orodruin]

For the "free-falling observer" there is no Rindler horizon, so the question is confusing. Replace "free-falling observer" with "free-falling clock" to make sense of it.

You could also make it clearer by stating "proper time for passing through the Rindler horizon of the accelerating observer is finite".

 

[Antarres]

Thank you for the answers. I assumed the first approach makes more sense, but I wasn't sure, so I wanted to see other opinions. I think I'll be able to solve it on my own then. It doesn't look complicated to me, though it's probably a bit technical.
@PAllen I said 'trivial', since that's how I usually call problems which are purely technical and don't require certain non-trivial steps in calculation. So to me anything that is straightforward completely, that you can do without much thinking technically, is trivial. That's what I meant, and I think this exercise will turn into a trivial one now that this has been cleared up. Thank you all very much.