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jambaugh
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Mar2-12, 02:24 PM
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Moving interferometer

Quote Quote by Edi View Post
... The thought experiment is here. You want numbers? ok. The laser and the detector is moving at 0.8 c relative to the planet in the same direction and they are 1000 km apart.
OK, good. Now you say they are 1000km apart, is that as seen by an observer on the planet or as seen by an observer moving with the laser and detector?

It makes a quantitative difference though not qualitative: Let's say the 1000km is as measured by someone moving with the laser and detector...

according to relativity from the perspective of the person moving with the laser-detector they are all stationary and it is the planet which is whizzing past. If the observer synch's up clocks on the laser and detector (two clocks there) and the laser clock hits t=0 just as it emits a pulse of light, then the detector clock will record the time of arrival as t=1000km/300000(km/sec) = 1/300th of a second later. All reasonable and straightforward right?

The strange then then is that an observer on the planet will also see the pulse as having left when the laser clock reads t=0 and arrive when the detector clock reads t = 1/300 sec. However:

a.) The planetary observer still sees the pulse as having moved at speed c = 300000km/s.
b.) The planetary observer will see the distance between laser and detector as shorter:
(by a factor of [itex] 1/\cosh(\beta)[/itex] where [itex]\tanh(\beta) = 0.8=4/5[/itex]
That comes to [itex]\beta = 1.0986123, \cosh(\beta)=5/3 ,\frac{ 1}{\cosh(\beta)} =3/5 = 0.6[/itex]
(I suggested 0.8 c because it gives "nice" numbers, and yes those are hyperbolic trig functions. [itex] \cosh = 5/3, \sinh = 4/3, \cosh^2 - \sinh^2 = 3/3 = 1[/itex])

And the final strangeness is...
c.) The planetary observer sees the laser clock and detector clock as being out of synch with each other by a factor of
...hmmm calculating...
The transformation between coordinate systems is (for differences in coordinates):
[itex] \Delta x' = \Delta x \cosh\beta + c\Delta t \sinh \beta[/itex]
[itex] c\Delta t' = \Delta x \sinh\beta + c\Delta t \cosh\beta [/itex]
Mapping the t=0 events for both emitter and detector to primed = planet coordinates:
[itex](1000km,0) \mapsto (1000\cosh\beta, 1000\sinh\beta)= (\Delta x',c\Delta t')[/itex]
yes, so the change in synchronization will be 1000km/c = 1/300 seconds times [itex]\sinh(\beta) = 4/3[/itex], so that's 4/900 ths of a second out of sync.
That is why, due to the moving of the laser and detector, the observer sees the events where they each read t=0, as farther apart (but also separated in time) but still sees the objects themselves at a given (t') time as closer together via length contraction.

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OK, I described the events as relativity describes them for your 0.8c moving laser-detector pair. You didn't ask a specific question in your 2nd post, or state a problem, just gave a scenario.

The beta, [itex]\beta[/itex] parameter I used was a boost parameter. When you draw the world line of a moving object the velocity is the slope of that line v =dx/dt. Which is c times dx/cdt = c times % of c speed in common space-time units (here km). In Euclidean geometry the slope of a line is the tangent of the angle. In space-time you have hyperbolic geometry and the velocity/c is the hyperbolic tangent of a "boost parameter" or "pseudo-angle". You can then resolve Lorentz transformations as hyperbolic "pseudo-rotations" using the formulas I used above.

Note when you try to add velocities it is like adding slopes for lines. In Euclidean (elliptic) relativity you "add" slopes by adding angles. (think about stacking two wedges). (Note that if space-time obeyed "Euclidean relativity" you could by accelerating just rotate around and your time would move backwards relative to mine).

In Minkowski or Einstein (hyperbolic) relativity you "add" slopes (velocities) by adding the boost parameters which are pseudo-angles. What I call beta, But be careful! some texts use beta for the value of v/c. I prefer u = v/c.

In Galilean (parabolic) relativity (what we intuitively use) you add "parabolic" angles but that comes to just adding velocities.

You can define "parabolic trig" as: paracos(v) = 1, parasin(v) = paratan(v) = v. Note these are the small angle, or small pseudo-angle limits of the regular or hyperbolic trig cases. Parabolic trig/relatiivty is the boundary between elliptic and hyperbolic cases and what one sees when common t units are much bigger than common x units. (1 second = 299,792.458 kilometers).

[edit: Note that even if we had "Euclidean" relativity where velocities could be arbitrarily high, and objects could even move backward in time, we'd have to have a constant c to convert t units to x units in that unified space-time. Imagine trying to figure rotations if we measured x in furlongs and y in fathoms.]