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Moving interferometer

  1. Mar 2, 2012 #1


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    If one makes a interferometer with a "arm" giant enough or moves it fast enough..
    Forget that, basically you could have just a laser and detector floating in space, not connected, at huge distance, with the same speed and direction relative to a, say, planet in background.
    If the distance is large enough and/ or the system is moving fast enough, shouldn't the photon miss the detector by the time it gets there (as the detector has moved away from the position the laser was aimed at ) ?
    If we take away the planet in the background and there is only the laser and the detector in whole of space, then they are basically standing still. There is no way to measure their speed because there is nothing to measure it against (unless there is Eather .. which we still cannot really measure). If they are standing still, laser would surely hit the detector.

    I understand the system where the photon is replaced with .. something. A particle. System where the particle would simply have kinetic energy in another direction.. but can the same be for a photon? It would kinda have to move faster than c in order to cover that increased distance when summing the two vectors..

    Any help clarifying this in my mind would be appreciated.
  2. jcsd
  3. Mar 2, 2012 #2


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    Try to clarify your question. a.) often that will clarify you understanding and b.) it would help facilitate others attempts to shed some light.

    Create a specific though experiment, something like "suppose you have an interferometer 1000km long..." or "suppose an interferometer is moving at 0.8c..."

    Finally note that interferometers do not measure interference from one photon. They see interference patterns from many repeated one photon experiments by looking at the distribution of final positions. (And you can, if you like do this with a large number of equivalent distinct interferometers each doing a 1 photon experiment.)
  4. Mar 2, 2012 #3


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    Ok, as I said, Interferometer really has nothing to do with it. As we can make a simpler ar more to the point experiment. .. and I dont understand what else do you want me to do. 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.
  5. Mar 2, 2012 #4


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    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.

    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.]
  6. Mar 2, 2012 #5
    Moving laser has a moving mirror, and a second moving mirror at some distance from the first moving mirror. Some photons miss a mirror because the mirror is moving. The laser amplifies that kind of light that stays between the mirrors. A laser moving north and pointing east, shoots to the north east.
  7. Mar 2, 2012 #6


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    Ahhh..hmmm... Edi, where you thinking of the planet moving perpendicular to the laser/detector setup?

    In this case you don't need to worry about loss of simultaneity or even length contraction but you do need to take into account the slowing of the laser clock as seen by the planetary observer.

    The laser, say it's 1 meter long, initiates a pulse of light from the back at say t=0 which leaves the front at t = 1/3 x 10^-8 seconds.

    Let's say the t=0 time is the same t'=0 as the planet observer uses. He sees the light leave the laser at [itex] t' = \cosh(\beta) t = 5/3 \cdot 1/3 \times 10^-8s[/itex], which means it traveled a distance longer than a meter, specifically 5/3 meters. This is because in this time the front of the laser has moved laterally a distance of 0.8c times the time interval which is 5/3 * 4/5 = 4/3. You have a right triangle with the laser 3/3 m long and moving 4/3 of a meter laterally before the light leaves the front. The light follows the hypotenuse of this right triangle traveling 5/3 of a meter total.

    Draw yourself a picture (actually a pair of pictures) and compare the numbers.

    In short, just as with a gun fired from a train, the ejected particle has (as seen by the observer moving relative to the device) the same component of lateral velocity as the device firing it. The reason this doesn't imply faster than c motion is because the velocity in the direction of aim is less (as seen by the moving observer). The lessening of that velocity corresponds to the relative slowing of the clock measuring the time of flight on the device so everyone sees consistent outcomes.

    As the planetary observer sees it. The motion in the direction of aim is slower than c but the laser's clock mistakes if for c since it is running slow. As the laser observer sees it, his clock is just fine and the pulse really is moving at speed c.

    Now as far as the distant detector is concerned. Just extend the end of the laser all the way to the detector and it works exactly the same. Make the laser 1000km long.
  8. Mar 3, 2012 #7


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    This is not correct. Let's simply work this out from the Lorentz Transforms.

    If a pulse of light is moving at c in the +y' direction then the equation for its worldline is:

    If we substitute in the Lorentz transform expressions and simplify we get:
    [tex]\frac{v}{c} x + \sqrt{1-\frac{v^2}{c^2}} y = c t[/tex]

    Noting that:
    [tex]\left(\frac{v}{c}\right)^2 + \left(\sqrt{1-\frac{v^2}{c^2}} \right)^2 = 1[/tex]
    we immediately see that the speed is c in both frames.
  9. Mar 3, 2012 #8


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    ok, this is the picture as simple as it gets.
    3 images, one after another, as 3 steps in time.

    https://docs.google.com/file/d/0B96oKqGy81PsalFidjhHM0xSLWVyQm41Wmx5REtYQQ/edit?pli=1 [Broken]

    https://docs.google.com/file/d/0B96oKqGy81PsVDE3Ums3VnJSbXFwMzRQdURnRWptdw/edit?pli=1 [Broken]

    https://docs.google.com/file/d/0B96oKqGy81PsalFSNHJESC1SSnVUbDd0VE1wLXhSUQ/edit?pli=1 [Broken]

    see, if that was a ball or something, it would move sideways with the system.. but what about a photon.. (?)
    Last edited by a moderator: May 5, 2017
  10. Mar 3, 2012 #9


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    Light also moves sideways, as I showed in post 7, but it's speed is still c as also shown.
  11. Mar 4, 2012 #10


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    Ball or photon, it doesn't matter. Your picture is incorrect in that you are not resolving the photon/bullet as it travels down the length of the laser/barrel. It will not (as seen from the planet) move parallel to the laser's axis. This is not an SR issue, it is a general motion issue.

    Picture this... first forget all about Einstein's relativity. Picture two trains running side by side at the same speed. (put them on the moon so we can neglect the wind). You're on one train, shooting at a target on the other train exactly across from you. You don't have to lead the target you just aim right at it. You hit it.

    Now suppose you shoot a second shot just as both trains are about to enter a tunnel. Your bullet will leave the gun right before you enter and would have hit the target right after. Instead is grazes the side of the mountain between the two tunnels. You see, there's lateral motion on the bullet equal to the lateral motion on you and the target. In fact if you stick your barrel out too far the barrel will hit the mountain while the bullet is still in it traveling up its length and you'll ruin a perfectly good rifle.

    You try a third shot, this time as your train passes under an overhanging platform where I too take a shot. You don't lead your target but I, being stationary must lead. We both shoot at the same time and hit the same target at the same time. (I have a slightly higher power gun so the speeds perp. to the train are the same given my lead angle).

    Now visualize both our barrels superimposed. Mine is crossing yours with an angle and relative motions so that our bullets trace out the exact same path as seen from above.

    In the diagram, the blue frame is my "stationary" frame. The red is yours on the train. The point here is the bullet leaving your gun is traveling in the same direction as mine even though your aiming in a different direction.

    Your error is in thinking the bullet you fire will suddenly loose its component of motion in the direction the train travels just as it leaves the barrel. It doesn't. From the ground (or my overhanging vantage) the bullet travels at an angle forward from perpendicular to the train because the barrel moves forward (perp to its aim) while the bullet is traveling down its length.

    Understand that point first. Then consider what Einstein's special relativity has to say.

    Light does exactly the same thing and if it were not for SR I'd need to shoot light a wee bit faster than you in order to get my pulse to hit the target at the same time as yours. Relativity resolves this by (apparently to me) having your clock run slower.

    As I see it you measure the photon bullet at speed c with no motion lateral to your barrel and I see it having speed c with that extra component of motion traveling farther in a longer period of time than you measure with your slow clock.

    Everything works out so everyone always sees light travel at speed c, and there's no way to say who is actually moving and who is actually standing still.

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  12. Mar 5, 2012 #11


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    Ok, I understood already everything about the ordinary bullet scenario .. but I was confuses about light.
    Now I see that time dilation, length contraction and what not will always make c .. = c

    Thank you.
  13. Mar 5, 2012 #12


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    You're quite welcome.
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