# Accelerating frame in special relativity

1. Jul 25, 2007

### jostpuur

Let (t,x) be coordinates of some point P in some inertial frame O. Suppose that at (0,0) there is a particle that has velocity u and acceleration a in the frame O. If the origo of a frame O' is attached to this particle, then what are coordinates (t',x') of the point P in the frame O'?

Or does that acceleration have any effect on the coordinates? It would seem reasonable, if correct answer is that one must use coordinates that transform in the usual way similarly as without acceleration, but always choose the value of u that is correct instantaneously.

Last edited: Jul 25, 2007
2. Jul 25, 2007

### Ich

That is reasonable, but there is also a definition of accelerated coordinates that is instructive as an outlook on GR. To use it, you
1. translate your coordinates such that the coordinates of your former origin satisfy v=t/x and x²-t²=1/a². Source:http://www.mathpages.com/home/kmath422/kmath422.htm
2. Check whether the event P is in the Rindler wedge 0 < x, -x < t < x; if so, transform to the accelerated coordinates, if not, it lies behind the Rindler Horizon of the particle and is out of reach. Source:http://en.wikipedia.org/wiki/Rindler_coordinates
If you're talking about a point in space rather than an event, it will generally fall back behind the particle an finally come to rest at the horizon, in strong analogy to the event horizon of a black hole.

3. Jul 25, 2007

### pervect

Staff Emeritus
The OP might also want to take a look at a couple of past threads on the topic or that wandered into the area of the topic besides the above references (which are good, esp. the mathpages url which is probably simpler than the wikipedia entry).

There are a few important things to point out:

1) "The" accelerated coordinate system of an accelerated observer isn't necessarily uniquely defined. The one I discuss in these links is a standard approach from a standard textbook, and takes the reasonable choice of using the inertial coordinates of an observer who co-moves with the origin of the coordinate system for the coordinates of an accelerating observer.

2) The coordinate system I describe above doesn't cover all of space-time. (See the remarks by Ich about the Rindler wedge) - it's only a local coordinate system.

4. Jul 27, 2007

### jostpuur

Okey, the idea of co-moving frame looks good.

The individual equations in Wikipedia's article of Rindler coordinates didn't seem very complicated, although I didn't understand what precisely is achieved with them.

So the co-moving frame is the most natural choice, but there exists also other ones? That seems strange, what kind of coordinates are they?

5. Jul 27, 2007

### George Jones

Staff Emeritus
The radar method gives one natural way of constructing these coorinates. I outlined such a construction here

Note the mistake in my construction:

$$t = \frac{e^{2at'} + 1}{e^{2at'} - 1}x$$

should be

$$t = \frac{e^{2at'} - 1}{e^{2at'} + 1}x.$$

6. Jul 27, 2007

### pervect

Staff Emeritus
GR allows arbitrary coordinates. Coordinates are just a set of numbers that give a label to some particular point, on the Earth for instance one might use lattitude and longitude. http://www.eftaylor.com/pub/chapter2.pdf with it's example of a rowboat with nails in it might be useful as an example of how coordinates, and "frames" are secondary to GR - what is important is the geometry, not the underlying coordinates, which are just numbers assigned to the events of space-time, i.e. the "nails" in the rowboat.

Because GR allows arbitrary coordinates, it needs a mechanism for turning coordinates into distances. This is the metric. The metric gives the distance (or, more precisely, the invariant Lorentz interval, which can serve as measure of either distance or time) between any two points, given their coordinates.

Note that the Earth's surface has a metric as well, and it serves the same purpose - it coverts $\Delta$ latititude and $\Delta$ longitude into distances (for small $\Delta$).