What is the general transformation formula for uniform proper acceleration?

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• snoopies622
snoopies622
TL;DR Summary
Given one inertial frame and one frame accelerating with constant proper acceleration, how to transform the coordinates from one reference frame to the other?
I'm not perfectly clear about how Rindler coordinates work. Here's what i do understand:

Suppose I'm in an inertial reference frame and i define the location of the events around me with (x,t), (ignoring the y and z directions here) and a spaceship approaches me from afar and then flies away according to the right half of the hyperbola defined by

$x^2 - c^2t^2 = r^2$

Then the person on the spaceship will feel a constant acceleration in the positive x direction $\alpha=c^2/r$. If $\tau$ is the time according to the clock on the spaceship, then I can re-define the hyperbola above with

x = r cosh ( $\tau \alpha$ /c )

ct = r sinh ( $\tau \alpha$ /c )

If an event takes place on this hyperbola, then given the x and t that I observe, i can conclude that the clock on the spaceship at that event reads

$\tau = c/ \alpha$ inverse cosh (x/r)
or $\tau = c/ \alpha$ inverse sinh (ct/r)

and that the X coordinate of the event as defined by the spaceship observer must be zero.

What i don't know is a general transformation formula. That is, given the x,t coordinates of an event NOT on the hyperbola, how do I translate them into X,T as observed by the person on the spaceship?

Thanks.

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The Wikipedia article is pretty good, and you may recognise some names in the edit history. If you can't make sense of it, ask.

Thanks, i think i get it now. For me this is another case where changing x to xi and the hyperbolas to circles is very intuitively helpful.

Ibix said:
If you can't make sense of it, ask.

Question: the Wikipedia article says that, given a spacetime point defined by X,T in the inertial coordinate system, the time in the accelerated frame is

t = $\alpha ^{-1}$ arctanh(T/X)

which of course is undefined at X=0. I can see why this would be the case when T>=0 since a beam of light coming from the positive T axis never touches the accelerating spaceship hyperbola, but it appears that such a beam would indeed intersect the hyperbola if T<0. Why then would that point of origin (X=0, T is negative) also be undefined to the accelerating observer?

Have you seen the picture of the Rindler wedge at the article? The coordinate system doesn't cover ##X=0##. It's valid where ##|T|<X##

snoopies622 said:
TL;DR Summary: Given one inertial frame and one frame accelerating with constant proper acceleration, how to transform the coordinates from one reference frame to the other?

I'm not perfectly clear about how Rindler coordinates work. .........................
.................................. What i don't know is a general transformation formula. That is, given the x,t coordinates of an event NOT on the hyperbola, how do I translate them into X,T as observed by the person on the spaceship?

A couple of things to note that make Rindler coordinates much less confusing:

With reference to the above chart from the Wikipedia article ,, the diagonal line marked t=1 represents the line of simultaneity of the accelerating Rindler observers with equal Rindler coordinate time t=1. They do not have equal proper times. The t coordinate used in the Rindler coordinate system is not the proper time of the individual accelerating Rindler observers. An accelerating observer (A) on the x=1 parabola will see a second accelerating observer (B) on say the x=0.6 parabola as stationary with respect to himself at all times yet he will observe the natural clock carried by B to be running much slower than his own clock. If the two accelerating observers synchronise their clocks using the Einstein clock synchronisation convention, they will be out of sync again a very short while after. To create a sensible coordinate system in the accelerating Rindler reference frame, the proper times of the spatially separated observers have to be artificially adjusted so that they all run at the same rate. as a single clock somewhere on the positive X axis that is chosen to be the master reference clock of the accelerating observers. This master clock is often chosen to be the clock of the accelerating observer that passes through X=1 for convenience. The diagonal line t=1 also represents a line of equal velocity where all accelerating observers on that line have a velocity as measured in the inertial reference frame of $x^2 - c^2t^2 = r^2$. At any time T (except zero) according to the inertial observer the accelerating observers have different instantaneous velocities, with the ones nearest the origin moving the fastest.

The general transformation formulas that you requested are given by equation (2a) in the Wikipedia article but Wikipedia uses X,T for the inertial coordinates and x,t for the accelerating coordinates which is the opposite of the notation you introduced in your first post.

Hope that helps.

Note credit to Dr Greg for creating and making the Rindler chart freely available on Wikipedia

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vanhees71, PeterDonis and snoopies622
Thanks KDP, that helps a lot!

Doing a few calculations, i'm struck by this asymmetry: From the perspective of the inertial observer (A), the accelerating spaceship (B) has a speed which asymptotically approaches the speed of light, while from B's reference frame, A's speed asymptotically approaches zero. I haven't yet expressed these speeds as rapidities, but of course different speeds will mean different rapidities. But . . isn't rapidity frame invariant? I know that two inertial frames will observe each other with the same (but opposite direction) speed, and always thought it must be the same with accelerating frames.

snoopies622 said:
from B's reference frame, A's speed asymptotically approaches zero
A's coordinate speed does, yes. But since B's frame is a non-inertial frame, you cannot interpret coordinate speeds the way you would in an inertial frame. If you do the math, you will see that the coordinate speed of a light ray in B's frame asymptotically approaches zero as well.

snoopies622 said:
I haven't yet expressed these speeds as rapidities
You can't express coordinate speeds in B's frame as rapidities. That only works for inertial frames.

snoopies622
snoopies622 said:
isn't rapidity frame invariant?
No, of course not, any more than coordinate speed is.

vanhees71
snoopies622 said:
I know that two inertial frames will observe each other with the same (but opposite direction) speed, and always thought it must be the same with accelerating frames.
As should be evident, this is not correct for non-inertial frames.

vanhees71
snoopies622 said:
Doing a few calculations, i'm struck by this asymmetry: From the perspective of the inertial observer (A), the accelerating spaceship (B) has a speed which asymptotically approaches the speed of light, while from B's reference frame, A's speed asymptotically approaches zero.
This is the equivalence principle mandating "gravitational" time dilation. As the inertial ship falls below the accelerating one time dilation gets ever more severe. This is the beginning of the issues around defining the speed of something remote from you in curved coordinates (and later, curved spacetime).

If you consider a large family of inertial buoys at mutual rest, and have the accelerating observer measure each buoy's speed as it passes you will find that these speeds tend to ##c##.
snoopies622 said:
But . . isn't rapidity frame invariant?
Rapidity is analogous to an angle in Euclidean geometry. Yes, the angle between two lines is frame invariant, as is the rapidity (and velocity) between two worldlines. But the way you are using rapidity here you mean the rapidity between some worldline and your frame's timelike axis. That's naturally frame dependant because it depends on your choice of timelike axis.

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PeterDonis and snoopies622
Thank you both. Fascinating subject.

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