# Spacetime/Proper Length

Tags:
1. Nov 4, 2013

### Mikeal

In special relativity, length dilation is defined as follows:
X' = X0√(1 - V2/c2), where X' is the apparent/dilated length and X0 is the "proper length"
Therefore proper length: X0 = X'/√(1 - V2/c2), where c > V > -c

I read a book on the spacetime approach to relativity using the relationship: S = √(ct)2 - X2 for ct > X and: S = √X2 - (ct)2 - for X > ct

The spacetime approach produces the same results as special relativity, with the exception of "proper length" which it defines as X0 = X'√(1 - c2/V2) where ∞ > V > c and -c > V > -∞

In all other cases including time dilation, proper time, energy and momentum relationships etc., the spacetime and special relativity equations agree.

What is the physical significance of the different results in the proper distance relationships?

2. Nov 4, 2013

### Mentz114

The term 'proper length' in the spacetime approach is not the same thing as your X0. The proper length (S) is the spacetime distance along a worldline between two events. See the Wiki article for instance
http://en.wikipedia.org/wiki/Proper_length

3. Nov 5, 2013

### Naty1

You'll also find the following definition and explanation a convenient way to see the difference between the two definitions ...length r versus spacetime interval [distance] s.

http://en.wikipedia.org/wiki/Spacetime_interval#Spacetime_intervals

The invarient interval shown
s2 = [delta r]2 - c2[delta t]2

becomes s = r when the observation time interval, delta t, is zero, that is for example, in the rest frame of the object.

4. Nov 5, 2013

### Mikeal

Thanks the Wikapedia article explains the difference.
What confused me was that in special relativity, time dilation is defined as:
t' = t0/√(1 - V2/c2), therefore proper time: t0 = t'√(1 - V2/c2), where c > V > -c
The spacetime approach yields the same result. I guess this is because in both cases c > V > -c, whereas "proper length" is defined for c > V > -c and proper distance is defined for ∞ > V > c and -c > V > -∞

5. Nov 5, 2013

### Mentz114

I don't understand why you refer to two different things 'special relativity' and 'the spacetime approach'. There is only one special theory of relativity and it is a theory of spacetime. I don't think you've grasped the difference between the two usages of 'proper length'.

One is the length of and object measured in its rest frame, the other is a spacetime distance along a worldline, which is what a clock reads. They are completely different things.

6. Nov 5, 2013

### Mikeal

Agreed. I understand that there is only one special theory of relativity. I was simply referring to two different approaches to arrive at the time and length dilation equations. I started with a text-book that didn't mention spacetime per se and developed the equations using moving trains and trackside observers etc. I then read a book on spacetime that discussed hyperbolic spacetime diagrams and developed the relativity relationships from these. The results came out to be the same in all cases, except for proper length/proper distance which I agree now are two separate things. However, there appears to be no such difference in the temporal realm. Both approaches yield the same result (i.e. proper time).

7. Nov 5, 2013

### PAllen

You can see them as equivalent, at least in pure SR:

- The proper distance between two arbitrary events (with spacelike separation) is the same as the proper (rest) length of a ruler moving such that it connects them simultaneously (at constant time) in its own rest frame.

- The proper time between to arbitrary events (with timelike separation) is the time measured by a clock connecting them (thus holding constant position in its own frame).

8. Nov 15, 2013

### Mikeal

I now understand that in most instances "proper length" and "proper distance" are different.
However, the above wikipedia link starts by stating that, "proper distance is an invariant measure of the distance between two spacelike-separated events"

I take this to mean that proper distance equals the spacetime distance between them, irrespective of whether the events are simultaneous or not.

It then states that, "the proper distance between two spacelike-separated events is the distance between the two events, as measured in an inertial frame of reference in which the events are simultaneous. It is based on the invariant spacetime interval and is denoted by

\Delta\sigma=\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - c^2 \Delta t^2}"

This seems contradictory as generally speaking, space-like separated events are not simultaneous (i.e Delta t is not zero), although they can be.

Is it true or not that space-like events have to be simulatneous to be a measure proper distance? If it is true, why is Delta t not zero in the above spacetime equation?

9. Nov 15, 2013

### PAllen

I emphasize a key word. Specifically, thought it can be computed in any inertial frame in SR, with common formula, what it measures is the distance between the events in a frame in which they are simultaneous. There is always such frame for events with spacelike separation.
There is no contradiction. That formula will give the same result in any frame - that's what invariant means. Further, if you apply a Lorentz transform such the events are simultaneous, you are left with Euclidean distance. Since it is invariant, the value computed in any frame is this same thing: distance as measured in a frame where they are simultaneous.

10. Nov 15, 2013

### PAllen

Try the following:

Perform a Lorentz transform of (0,0) and (0,1) where my convention is (t,x). After the transform, the t coordinates will be different. However, if you apply the formula for interval distance, you will still get 1 as the answer.

11. Nov 15, 2013

### Mikeal

I agree with the invariance part as that is the point of computing spacetime distances as opposed to t,x coordinates. I also agree that for a given spacetime distance of a space-like event pair there is a frame in which t will be determined to be zero (events simultaneous). But there will also be frames where the spacetime distance is the same but t is non-zero (events not simultaneous). As proper distance is equal to the spacetime distance, for a given event pair, is it fair to say that the proper distance in non-simultaneous frames is the same as the proper distance in a simulatneous frame?

12. Nov 15, 2013

### PAllen

Yes, because proper distance means distance as measured in a frame where the events are simultaneous, irrespective of what frame is used to compute it.

13. Nov 15, 2013

### Mikeal

And so taking it to its logical conclusion, the proper distance between two space-like events is equal the the spacetime distance between them. This is true even when the spacetime distance is measured by an observer in a frame that doesn't determine the events to be simultaneous?

14. Nov 15, 2013

### PAllen

Yes.

15. Nov 15, 2013

### Mikeal

Good. Thankyou for taking the time to help me with this issue.