Lorentz interval

1. Feb 4, 2008

tomprice

a. Justify the notion that the Lorentz interval between two events is independent of the observer, without referring to the Lorentz transformations.

b. Derive the Lorentz transformations without resorting to the notion that the space-time interval between two events is independent of the observer.

Thank you very much in advance.

2. Feb 5, 2008

Fredrik

Staff Emeritus
a) I can't. Some thoughts about this: Any such derivation would have to start with some set of postulates. If we start with postulates that are sufficient to fully define special relativity (such as the claim that space-time can be represented by Minkowski space), the postulate(s) would imply that the Lorentz transformation holds, so you might as well start with the Lorentz transformation.

b) Yes. See e.g. post #8 of this thread for a derivation from Einstein's postulates. (I didn't explain every step carefully though, and this is only for 1+1 dimensions).

Edit: Note that Einstein's second postulate says that the speed of light is invariant, and that this is equivalent to saying that the Lorentz interval must be invariant when it's =0.

Last edited: Feb 5, 2008
3. Feb 5, 2008

DrGreg

Here is the outline of a proof using k-calculus and radar coordinates.

An inertial observer O is equipped with radar and emits a pulse at time e, and receives a radar echo at time r from a remote object P. For motion in a single dimension, the two values (e,r) (which are both measured by O's clock) uniquely determine any event and are called radar coordinates.

Similarly an observer O' can define radar coordinates (e',r') using the clock of O'.

See the attached spacetime diagram, in which it is assumed that O, O' and P lie in a straight line at all times.

Notice that a radar pulse emitted by O at time e (by O's clock) travels at exactly the same speed as a pulse emitted by O' at time e' (by the clock of O') (by Einstein's 2nd postulate), so both pulses arrive at P simultaneously and are echoed back. O' receives both pulses at time at time r' (by the clock of O'), and O receives both pulses at time at time r (by the clock of O).

In fact O emits a sequence of pulses at time-intervals of de. O' measures these intervals as de' = k de, where k is the Doppler factor between O and O'. We don't yet know the value of k, just that it is constant (provided O and O' are both inertial).

Next note that the Doppler factor for O relative to O' must, by Einstein's 1st postulate, be equal to the Doppler factor for O' relative to O. It follows that the pulses travelling past O' at intervals of dr' must be detected by O at intervals of dr = k dr'. k is the same value as above.

To summarise, we have:

de' = k de .......(1)

dr = k dr' .......(2)

Eliminating k between these two equations gives

de dr = de' dr'

In other words, de dr is invariant and we denote it by ds2.

ds2 = de dr = de' dr' ......(3)

It is not hard to show that O's radar coordinates (e,r) are related to O's standard time and distance coordinates (t,x) by

x = c(r - e)/2

t = (r + e)/2

(Technical note to those who are interested. It is only when we use t that we rely on Einstein's synchronisation convention. Results involving only r and e are independent of any synchronisation convention.)

Therefore

e = t - x/c ......(4)

r = t + x/c ......(5)

(with similar equations for the primed coordinates of O').

Substitute (4) and (5) into (3) to get the standard equation for the Lorentz interval.

Finally, substitute (4) and (5) into (1) and (2) to obtain

dt' = (k+k-1)dt/2 - (k-k-1)dx/(2c)

dx' = (k+k-1)dx/2 - (k-k-1)cdt/2

These equations are the standard Lorentz transform with gamma = (k+k-1)/2 and v = (k-k-1)/(k+k-1).

By considering the special case where x' = 0, you can verify that v is just the velocity of O' relative to O and that gamma is related to v by the usual formula.

The above proof does skate over a few of the details but I hope you can work out any missing parts yourself.

References

k-calculus:

* Bondi, Hermann (1980), Relativity and Common Sense, Dover Publications, New York, ISBN 0-486-24021-5

* http://www.geocities.com/autotheist/Bondi/intro.htm

* Geroch, Robert (1978), General Relativity from A to B, University of Chicago Press, Chicago, ISBN 0-226-28864-1

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4. Feb 5, 2008

1effect

DrGreg did a great job answering both a. and b.
Einstein was the first to answer b. here

5. Feb 5, 2008

1effect

Very nice! This was great!

6. Feb 5, 2008

7. Feb 5, 2008

bernhard.rothenstein

lorentz interval (Minkowski distance,space-time interval?)

In what concerns your first question I think that the celebrated light clock leads to
xx-cctt=dd
where d is the distance between the mirrors measured perpendicular to the direction of relative motion which is a relativistic invariant as a consequence of the first postulate only. Its proof does not involve light signals at all.
So
xx-cctt=x'x'-cct't'
the absence of c'c' in the right side being a consequence of the second postulate.
In what concerns your second question you find in the literatture of the subject titles like
"derivation of......without using the Lorentz transformations. So the addition law of parallel speeds can be derived from length contrtaction and time dilation. The addition law of parallel speeds leads to the Lorentz transformations for the space-time coordinates of the same event.

8. Feb 6, 2008

1effect

I don't think it does. The "light clock" moving at speed $$v$$ along the x-axis has as invariant the value:

$$c^2t^2-v^2t^2$$ since it is equal to $$d^2$$.

But the above is not the same as the Lorentz invariant in discussion: $$c^2t^2-x^2$$ . Besides, DrGreg has already given avery nice proof to point a.

You can't derive the invariance of the Lorentz interval from the "light clock" because the light clock demonstrates the invariance of a very particular case ($$c^2t^2-v^2t^2$$), not the general case.

But this is not what point b. in the original post is asking.

9. Feb 6, 2008

bernhard.rothenstein

space-time interval invariance

Thanks for your answer. The "light clock" does not move along the OX axis, it is its lower mirror which moves that way. The upper mirror moves a distance x=vt during the time t=r/c in which the light signal starting from lower mirror arrives at its location. So x=vr/c
the invariance of distances measured perpendicular to the direction of relative motion requiring that
d^2=(ct)^2-(vr/c)^2=(ct)^2-x^2=(ct')^2-(x')^2
where d is the distance between the mirrors.
I understood that the question was if the LT could be derived without using the invariance of the space-time interval. English is not my first language.
All I told above are only questions and not statements

10. Feb 6, 2008

1effect

I just explained to you why taking $$x=vt$$ in the case of the "light clock" does not demonstrate the invariance of the Lorentz interval in the general case. $$x$$ in the Lorentz interval $$c^2t^2-x^2$$ is an independent variable, setting $$x=vt$$ makes it...dependent of $$t$$, which is clearly the wrong thing to do.

Yes, that was the original question but you answered something that has no connection to the question.

Last edited: Feb 6, 2008
11. Feb 6, 2008

country boy

For the special case of the "Lorentz" intervals of light rays, it is easy enough to justify that they are independent of the observer. That is what the Michelson-Morley experiments showed. The Michelson-Morley result was that the speed of light, r/t=c, is universal. This gives r^2-(ct)^2=0 for the light interval, which must also be the same for everyone. The universality of c leads to the well-known conclusion that an expansion of light from a point will have the shape of a sphere for all observers. From this you can derive the Lorentz transformations. Then you can further show that the invariance also holds for non-zero intervals.

But this just says that the Lorentz transformations follow from the universality of c and that the invariant interval is another way of looking at it. All quite circular.

12. Feb 8, 2008

bernhard.rothenstein

space-time interval

I aggree that it is wrong to put V (the relative speed of I and I') but it is correct to put u the velocity of the mirrors relative to I.
So we should have
d^2=(ct)^2-(ut)^2=(ct')^2-(u't')^2 (1)
u' representing the speed of the mirror relative to I'.
For u'=0, u=V (the relative speed of the mirror relative to I. Under such conditions t'-0 becomes a proper time interval and the time dilation formula is derived.
It is easy to show that (1) is satisfied only if u and u' are related by the addtion law of parallel speeds.
As Countryboy mentioned the problem has a circular character (the invariance of c) but I think, mentioning that facts is a good pedagogical performance, extending the teaching power of the light clock which is not a wrist watch, involving wrist watches at each mirror.
I am preparing a paper in which I analyse the physics behind the derivations of the formulas that account for particulare relativistic effects and how many of them should be invoved a an one space or two space dimensions derivation of the Lorentz transformations. The Forum offers me a good opprtunity to test what I am working and so thanks to all who help me.

13. Feb 8, 2008

1effect

No, you are still wrong, the mistake is that you replace the independent variable
$$x$$ with $$ut$$. Therefore, the "light clock" cannot be used to prove the invariance of the general Lorentz interval $$(ct)^2-x^2$$

You are welcome. try to pay a little better attention to the criticism.

Last edited: Feb 8, 2008
14. Feb 9, 2008

bernhard.rothenstein

Please explain me in a few words what is wrong considering in a given inertial reference frame that x=ut and x'=u't' in I and in I' respectively?
Thanks

15. Feb 9, 2008

1effect

I already explained that to you several times: in the lorentz interval $$x$$ and $$t$$ are independent. By making them $$x=ut$$ you make them dependent. So, your proof loses generality , thus becoming incorrect.

16. Feb 14, 2008

DrGreg

Oops, the above equation was incorrect -- it is obviously dimensionally wrong (unless you assume c=1). The correct equation is

v = c(k-k-1)/(k+k-1)