# Accelerated frames in SR

1. Aug 23, 2007

### pellman

Do accelerated observers also see the same constant speed of light as inertial observers?

How do we transform to an accelerated frame? For simplicity, I mean a constant acceleration, constant in direction and magnitude. By acceleration I mean the second derivative with respect to lab time not with respect to the proper time. (Just point me to a resource if you like, online or book, either is fine.) .

btw, I am aware that a constant force does not lead to a constant acceleration at relativistic energies, and that a constant acceleration cannot be maintained indefinitely. But transforming to a frame in which the particle travels at a constant velocity (or at rest) for the constant force problem is probably much more difficult than for constant acceleration. But either situation interests me.

2. Aug 23, 2007

### jcsd

No accelrated observers do not necessarily see a constant speed of light. The 2nd postulate of SR (the constancy of the speed of light) applies specifically to inertial (non-acclerated) reference frames and it does not extend into non-inertial reference frames.

The big problem is though that whilst inertial frames (in SR) lend themselves automatically to global spacetime coordinates, non-inetrial reference frames do not, so there's automatically the problem of defining speeds (for example accelrated frames can be badly behaved). Nevertheless you can define acclerated frames where it is sensible to talk about the speed of light and it can be not c and non-constant.

Actually it's easier (and of course more relatsitc) to deal with acceleration that is constant wrt to proper time. John Baez deals with the equations for accelarted motion in SR here: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

To create an accelerated frame you integrate over the momentarily co-moving inertial frames.

Last edited by a moderator: May 3, 2017
3. Aug 23, 2007

### pervect

Staff Emeritus
Accelerated observers will still see the speed of light as being constant and independent of their acceleration near the observer.

Accelerated observers will basically see the speed of light as a function of "altitude", i.e. distance measured in the direction of the acceleration. The speed of light will increase the altitude increases, and decrease as the altitude decreases. At the origin, however, where the observer is located, the speed of light will be equal to 'c'.

There is a quite definite limit on the size of the coordinate system, as jcsd mentions. A distance of c^2/a below the observer, the speed of light will drop to zero as measured in the coordinate system of the accelerated observer. (I've taken some liberties here to single out a particular coordinate system with some desirable properties as "the" coordinate system of an accelerated observer.) This forms an event horizon for this coordinate system, called the "Rindler horizon", which is quite analogous to to event horizon of a black hole.

4. Aug 23, 2007

### pellman

Thanks a bunch. I'll tell you where I'm going with this just in case you have further insight.

A general solution to the linear potential (i.e., "constant force") for the Schrodinger equation has the interesting property that making the substitution x --> x - (1/2)Ft^2/m (where F comes from V(x) = -Fx, the "force") does not reduce to the free particle solution that we get by letting F=0. However, measurables quantities do exhibit this. That is, expectation values for the linear potential case are the same as what you would get by measuring an appropriate free particle solution from an accelerating frame.

Well, I have a solution to a relativistic quantum linear potential and I wanted to investigate this same comparison between the solution and the free particle solution as seen from a suitably accelerated frame. But it is looking much more complicated.

Last edited: Aug 23, 2007
5. Aug 24, 2007

### JDługosz

I'm in the process of editing that and adding all the illustrations (see discussions in the newsgroups). You can see work-in-progress here. (At the moment, ignore Figure 4.)

Feedback welcome from this group, too!

Last edited by a moderator: May 3, 2017
6. Aug 24, 2007

### pervect

Staff Emeritus
OK, I'm glad to see this FAQ is being updated - the original was a bit confusing.

Here are some comments I have:

ROTFL - OK, this is a very early version. Ok, let's get serious.

Here's my suggestion on some more detail to add to the next section (the wording might need more tweaking). Basically, I think it could use some expanding. (Maybe I overdid it, I dunno).

Also, you need to decide on consistency of calling the moving observer either a "moving observer" or a "co-moving observer". A minor point, but potentially confusing.

Next up - I think you want to mention why you are drawing hyperbola's a little earlier:

Something like:

Question: how relevant is the space-time with the metric d\tau^2 = e^{2x}dt^2 - dx^2?

This was in the original FAQ, but it seems to me to be a bit of a digression. What's more interesting is the Rindler metric, i.e

d\tau^2 = -(1+gx)^2 dt^2 + dx^2

We occasionally get into arguments over what name to use to describe this, but the point is that the above metric is a vacuum space-time, and that's what we want.

Last edited: Aug 24, 2007
7. Aug 25, 2007

### JDługosz

I was wondering about that myself. It seems to be a total digression from the problem, except to point out that in such a situation ships would remain the same distance. But too much trouble to explain it! I think the real enlightening part is to relate this to how rigid objects must contract in length.

Oh, how do you think the drawings look?

8. Aug 25, 2007

### pervect

Staff Emeritus
The diagrams look good to me. I think the 5 different versions of diagram 2 may be a bit of overkill, personally. Of course diagram 4 needs work, but you already know that .

9. Aug 26, 2007

### JDługosz

For #2, I was thinking that client-side javascript could let you flip through them. The HTML would be like you see, but the script would stack them up if scripting is enabled.

#4 I'm having some issues with. The text wasn't clear but now I understand that the hyperbola depends on the initial separation. What bugs me is the units: the physical units are arbitrary (and not even shown). So what does the number "60" mean?

10. Aug 26, 2007

### pervect

Staff Emeritus
All of the hyperbolae must share the same asymptotes in the constant distance case. Thus the curves in figure 4 should never cross. Another equivalent way of saying this - the tangents to the hyperbolae all instersect at the same point.

x^2 - t^2 = c^2 / a, where a is the proper acceleration. (c^2/a) has dimensions of meters, and has some physical significance as the distance to Rindler horizon. So, for instance, a proper acceleration of 1 light year / year^2 implies that the constant K is 1 light year.

Basically, when you pick a distance unit, you pick a proper acceleration at that distance of c^2/a. There are no dimensionful constants in acceleration * distance = c^2 except c.

It turns out that the lines x = (some constant) * t are all minkowski-orthogonal to the worldlines of the hyperolae, so that that the lines of x = (some constant) * t are lines of simultaneity. You probably need to prove this more rigorously, though.

11. Aug 26, 2007

### robphy

I think you mean
the CENTERS OF the hyperbolae all intersect at the same point.
That is, these hyperbolae are concentric.

12. Aug 26, 2007

### pervect

Staff Emeritus
Given that the two observers are the two ends of a rod in Born rigid motion, I believe it is correct to say that the asymptotes are the same. In fact, more generally, the asymptote of the hyperbolae of all the points on the rod intersect at one unique point.

All lines of simultaneity also intersect at this same point. As t goes to infinity, the boost factor goes to infinity, and the line of simultaneity becomes the same line as the tangent to the curve.

Thus the asymptotes of the hyperbola (the tangent at t=infinity) intersect at the same point as the lines of simultaneity do at t<infinity. I've seen this point called (informally) the "pivot point" i.e. in

http://www.mathpages.com/home/kmath422/kmath422.htm.

I haven't seen any formal name for this point, it's just a very important point for hyperbolic motion.

I'd have to go back to the textbooks to offer a formal proof, though - I'm not sure offhand what the best way of demonstrating this is to the reader.

13. Aug 26, 2007

### pervect

Staff Emeritus
Here's a partial argument, though it's not bulletproof (but it's simple). A light ray emitted from what I call the "pviot point" will never catch up to any point on the accelerating rod. It would definitely be inconsistent if the light ray emitted from the pivot point reached the front of the spaceship (far away from the pivot point) without first passing through the back. It turns out, conveniently enough, that a light ray emitted from the pivot point never quite catches up to any point on the rod, though it gets very close to catching up as t goes to infinity. (The above argument isn't quite enough to demonstrate the stationariness of the pivot point, however).

Last edited: Aug 26, 2007
14. Aug 28, 2007

### JDługosz

I know, I need to continue working on it. The original manuscript just said "k>1" but did not relate k with the earlier K, the initial separation. The scale on the graph is [0..1] so I wasn't thinking that k represents a x distance!