# Derivation of Proper Time and Proper Space

1. May 23, 2010

I have tried to derive the formula for the proper time from the two equations of the Lorentz Transformation. The formula is as follows (see Wikipedia: http://nl.wikipedia.org/wiki/Eigentijd): [Broken]

tau = t*sqrt (1 - v**2/c**2)

The two equations of the Lorentz Transformation are as follows (see Wikipedia):

x '= x / sqrt (1 - v**2/c**2) - v*t / sqrt (1 - v**2/c**2) (1)

and

t '= - ((v*x) / c**2) / sqrt (1 - v**2/c**2)) + t / sqrt (1 - v**2/c**2) (2)

Now x 'and t' represent points. x' represents a location (point in space) and t' is a time (point in time). One may write the two equations of the Lorentz Transformation for intervals as follows:

Delta x ' = Delta x / sqrt (1 - v**2/c**2) - v * Delta t / sqrt (1 - v**2/c**2) (3)

and

Delta t '= - ((v * Delta x) / c**2) / sqrt (1 - v**2/c**2)) + Delta t / sqrt (1 - v**2/c**2) (4)

We derive the formula for proper time by assuming, that Delta x '= 0. This is the case when Delta x = v * Delta t. Substituting this into formula (4) we obtain:

Delta t '= Delta t * sqrt (1 - v**2/c**2)

or

tau = t * sqrt (1 - v**2/c**2)

Note, that tau refers to a time interval.

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Completely analogous to the derivation of the formula for the proper time, I derived the formula for the proper space (proper length). The formula is as follows (Wikipedia: http://en.wikipedia.org/wiki/Proper_length): [Broken]

sigma = x * sqrt (1 - v**2/c**2) .

We derive the formula for proper space by assuming, that Delta t '= 0. This is the case when Delta t = (v * Delta x) / c**2. Substituting this into formula (3) we obtain:

Delta x '= Delta x * sqrt (1 - v**2/c**2)

or

sigma = x * sqrt (1 - v**2/c**2)

Note, that sigma refers to a space interval (one dimensional).

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My question is: Are these derivations correct?

Last edited by a moderator: May 4, 2017
2. May 25, 2010

Is there anybody who might want to answer my question?

3. May 25, 2010

### Cyosis

I noticed that you kept the meaning of sigma and tau deliberately vague. As in sigma is a space interval and tau is a time interval. You need to arrive at a result where proper length and coordinate length and proper time and coordinate time are related to each other. For example your conclusion for proper length is: a space interval is equal to a different space interval times some factor, but there is no mention of proper length anywhere. Now my question is what is the proper length in that result?

Last edited: May 25, 2010
4. May 25, 2010

Proper length:

Delta x' = Delta x * sqrt (1 - v**2/c**2)

or

L' = L * sqrt (1 - v**2/c**2)

with L = Delta x and L' = Delta x'.

Proper space (along the x-axis only):

sigma = x' = x * sqrt (1 - v**2/c**2) .

Note, that sigma, x and x' can be negative, whereas L and L' cannot.

5. May 25, 2010

### Cyosis

It does I think. You're saying that sigma is the proper length as I suspected. When you look at your formula it says that sigma is always smaller than x. Which is wrong, because the proper length is the longest length of the object.

I am short on time I will elaborate further later.

Last edited: May 25, 2010
6. May 25, 2010

A moving ruler (measuring stick) is observed shorter in the direction of the movement!

Might this be of any help?

7. May 25, 2010

### Cyosis

Yes this is correct. However your derived equation, given that sigma is the proper length, says the opposite.

8. May 25, 2010

### Cyosis

We are doing relativity here which makes it very important that you are very clear on what your formulae and notations mean. You also need to be clear on the different type of reference frames. Your derivation exists of writing down the Lorentz transformations then doing some algebraic manipulations. But you never really say what you're doing.

First of all this is not an assumption. Proper time has a very clear definition. Proper time is the time between two events in the inertial frame where these events happen at the same location.

Let us investigate an inertial frame $S$ with coordinates $(x,t)$ and an inertial frame $S'$ with coordinates $(x',t')$, which are moving relative to each other with a speed $v$. If we measure an event at the coordinates $(x_1',t_1')$ and a second event at $(x_1', t_2')$ in frame $S'$ then the proper time is $\Delta t'=t_2'-t_1'$, because the two events happened at the same position in frame $S'$. Due to relativity of simultaneity these two events will happen at $x_1$ and $x_2$, with $x_1 \neq x_2$ in frame $S$.

Now if we measure the time between an event with coordinates $(x_1,t_1)$ and an event with coordinates $(x_1,t_2)$ in frame $S$ then $\Delta t =t_2-t_1$ would be the proper time. As it should be otherwise there would be some absolute inertial frame which goes directly against relativity.

Again there is no assumption in the definition of proper length. The idea is similar to that of proper time.

Consider two inertial frames again, namely S and S', which are again moving at a speed v relative to each other. Now let us place a rigid object, with length $\Delta x'$ in S' along the x'-axis, which is at rest in S'. Here $\Delta x'$ is the proper length. We want to find its length in frame S. To measure its length we will need to mark its end points simultaneously in S. We call these markings events and in frame S these events happen at the same time. Therefore we can set $\Delta t=0$. If you carry out the derivation correctly you will find that $\Delta x'>\Delta x$.

What you have done is mix frames. You mark the endpoints simultaneously in frame $S'$ and therefore set $\Delta t'=0$. So far so good. You then calculate $\Delta x'$ and call this the proper length. This is wrong because $S$ is the object's rest frame.

Last edited: May 25, 2010
9. May 25, 2010

I suppose $t_1' < t_2'$?

10. May 25, 2010

I suppose $S'$ is moving with velocity $v$ with respect to $S$ or measured from $S$?

11. May 25, 2010

I suppose $t_1 < t_2$?

12. May 25, 2010

### Cyosis

Sure, although it doesn't really matter as long as one precedes the other in all inertial frames.

13. May 25, 2010

To Cyosis

Your exposition is very clear. Thank you.

14. May 25, 2010

It looks to me, that this definition of proper time implies, that $\Delta x' = 0$. This is exactly what I did in deriving

Delta t '= Delta t * sqrt (1 - v**2/c**2)

Last edited: May 25, 2010
15. May 25, 2010

### Cyosis

16. May 25, 2010

### matheinste

The proper time between two events depends on the path taken between these events. Your definition is OK for an inertial path. All ideal clocks, whatever their motion, measure proper along their own worldlines, and its value is frame invariant, unlike coordinate time. In the inertial frame in which a clock is at rest, its worldline coincides with the time axis of that frame and so in this case, your scenario, proper time is equal to coordinate time.

Matheinste.

17. May 25, 2010

### yuiop

18. May 26, 2010