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Speed of light in NON-inertial frames

  1. Aug 15, 2010 #1
    I haven't studied this very much, but how do EM waves behave in noninertial frames? Do photons have an acceleration in order to maintain constant speed c in the noninertial frame?

    What happens to the Lorrent'z force in non-inertia frames? A charge moving in a magnetic field is equal to a magnetic field moving across a charge from an inertial observer; the velocity vector is relative. But what if a magnetic field source is accelerating across a charge?

    Anyone got a good read on this, or can type up an explanation?
     
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  3. Aug 15, 2010 #2

    Vanadium 50

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    How do you define a "speed of anything" in a non-inertial frame?

    Let's set aside light for a moment. You have an object at speed v. You are traveling at v'(t) - i.e. you are in a non-inertial frame. What's the (single) speed of this object in your frame?
     
  4. Aug 15, 2010 #3
    It's a mistake to think of the non-inertial frame as 'affecting' anything. It's merely a point of view.

    Picture yourself on a children's roundabout - you are an observer in a non-inertial frame.

    How do photons behave? - Exactly the same as they did before - the fact that you happen to be spinning round as you look at them makes no difference to them.
    Of course, what YOU see takes a bit of working out - but that's YOUR problem - not the photons.

    :smile:
     
  5. Aug 15, 2010 #4
    The phase velocity of EM waves in a gravitational field changes.
     
  6. Aug 15, 2010 #5
    in a small time interval, do you see photons moving away from you at C?
     
  7. Aug 15, 2010 #6

    JesseM

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    No, dx/dt for a photon can be different than c if x and t represent the coordinates of a non-inertial frame. A non-inertial frame is basically any arbitrary way of assigning position and time coordinates to different points in spacetime that doesn't happen to be an inertial frame--for example, even for a single non-inertial object, there are an infinite different number of ways of constructing a non-inertial "rest frame" for it.
     
  8. Aug 15, 2010 #7

    bcrowell

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    The speed of light is always found to be c by any local measurement made by any observer, regardless of whether the observer is accelerating or not.

    No. Relativistic velocities don't add linearly. If you combine a velocity of c with some other velocity v, the relativistic equation for combination of velocity always gives c, not c+v.
     
  9. Aug 15, 2010 #8

    Fredrik

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    I think an even more important question is, how do you define an accelerating coordinate system? I mean, a coordinate system is just a ("nice") function from a region of spacetime into [tex]\mathbb R^4[/tex], but when we talk about "inertial" or "accelerating" coordinate systems, there's a specific synchronization procedure that we're supposed to use to associate a coordinate system with the motion of an object. And the standard procedure ensures that the speed of a ray of light through the origin in the accelerating coordinate system is =c.

    However, if we try to measure that speed with an accelerating measuring device that was designed for local measurements (measurements in a region of spacetime that's so small that the effects of acceleration can be ignored), the result will be different from c. This doesn't mean that the speed of light is different from c in the accelerating coordinate system. It just means that the device isn't working properly when it's accelerating.
     
  10. Aug 15, 2010 #9
    So see if I get this right concerning your opinion on this matter:

    If I have a Born rigid spaceship undergoing constant acceleration and I measure the roundtrip time of light from the floor to the ceiling and back with a clock and an emitter on the floor and a mirror on the ceiling and find a different time from the case where the spaceship is inertial this is due, in your opinion, because the measuring setup is not working properly? And, out of curiosity, and again, in your opinion, is this setup also not working properly if we measure the roundtrip time from the top to the base of a tower standing on a massive planet?

    As I wrote in another thread the measured speed of light depends on two factors here:

    1. If the spaceship travels inertially.
    2. The height of the spaceship.

    That means for instance that if the spaceship is undergoing constant acceleration then in the limit, which is simply a special case, when the height of the ship approaches zero the measured result will be identical to the case where the spaceship travels inertially.
     
  11. Aug 15, 2010 #10
    For a short interval, in a very small space vicinity, speed of light is "c" in rotating frames.
    The science advisor "Demystifier" (H.Nikolic) has an excellent paper on this subject.
     
  12. Aug 15, 2010 #11
    In Einstein's thought experiment, shooting a laser horizontally in an elevator that is accelerating upwards will cause the light to "bend" as seen by the observer in the elevator.

    What I'm asking is, if I was in a train that is accelerating and I shoot a paintball gun head on, it would appear as I am "catching up" to the paintball since it stops accelerating after leaving the gun (ignore gravitational force) but *I* am accelerating, thus the speed of the paintball seems to decrease with respect to my eyes. The only way the speed will stay constant with respect to my face is if the paintball had an acceleration matching mine (and the train's which I'm standing on).

    What if you do this same experiment with a laser instead of a paintball gun? If the speed of light is constant, then would it appear as if the photons have an acceleration matching mine?

    P.S. I know I'm accelerating because my coffee will spill over if I hold it vertically, and I was once "at rest" but now I feel the acceleration.
     
  13. Aug 15, 2010 #12

    bcrowell

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    No, for the reason given in #7. The photons have velocity c according to any local measurement. In your frame they will get red-shifted as they fly, but they will always have velocity c.
     
  14. Aug 16, 2010 #13

    Fredrik

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    If we want to measure the roundtrip time, then there's nothing wrong with the setup. If we're trying to measure the speed of light at the floor, then there is.

    Same thing there. If we want to find the local speed of light at a specific height, this setup is less accurate the taller the tower is. If the setup "works" depends on how accurate we need the measurement to be.

    Yes, I think you have pretty accurately summarized my views on this. I admit that it's possible that I might have some detail wrong, because I haven't proved all of this rigorously to myself, but the calculations I've worked through (Hurkyl's calculation, my addition to it, and a calculation of the one-way travel time) all seem to support this view.
     
  15. Aug 16, 2010 #14
    You have to be careful by what you mean by speed. especially when you yourself are accelerating.
    In an accelerating frame, the instantaneous speed of light is constant. That means that if you measure it very,very quickly, over a short distance, you will get the universal constant c.
    But if you take your time over it and measure it over a significant distance, your own motion messes with the values of distance and time you get. If you then apply a naive calculation of v= distance/time with your wobbly measurements you will get a silly answer.

    It's exactly the same situation as if someone were to jog your arm and fiddle with your watch while you were measuring. In the lift, you are being joggled very smoothly and in a predicable way, but it's still a joggle.
     
  16. Aug 16, 2010 #15
    Speed is something that one can measure, it is distance over time. One can measure it in the limit or over a larger region and time. One can measure it when one is traveling inertially or accelerating. Why do you think the definition of speed or velocity should depend on whether someone is traveling inertially or not?

    One of the first things to know about relativity is that motion is relative, in general relativity the physical equations take the same form in all coordinate systems and what an observer measures is real and significant to him! You cannot say, well yes you measure it but it is not real because something is messed up, that is not science but sheer nonsense!

    Are you saying one cannot measure anything when you are accelerating except in the limit? Do I, standing on earth, when I measure something get silly answers because I am accelerating?

    Whatever an observer measures, independent whether he travels inertially or is accelerating, is of real scientific significance and is not silly or wobbly.
     
    Last edited: Aug 16, 2010
  17. Aug 16, 2010 #16

    JesseM

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    They only have a velocity of c in a locally inertial frame. If you use a coordinate system where your accelerating ship is at rest (like Rindler coordinates, where time is measured by accelerating clocks and distance is measured by rulers undergoing Born rigid acceleration) then light won't necessarily move at c.
     
  18. Aug 16, 2010 #17

    JesseM

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    No it isn't. For example, in Rindler coordinates the instantaneous coordinate velocity of a light beam approaches zero in the limit as it approaches the Rindler horizon. Again, only if you are using a locally inertial frame is the instantaneous velocity guaranteed to be c.
     
  19. Aug 16, 2010 #18

    DrGreg

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    It depends what you mean by "speed". If you mean "coordinate speed" i.e. rate of change of coordinate distance with respect to coordinate time, then JesseM is correct.

    But I suspect bcrowell and AJ Bentley didn't mean that. If you use a "local ruler" and a "local clock" then even accelerating observers measure an instantaneous speed of c (but not an average speed over a non-zero distance).
     
  20. Aug 16, 2010 #19

    JesseM

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    But in that case you aren't really measuring the speed in an "accelerating frame". Instead, you are getting an accelerating observer to measure the speed with an inertial ruler and clock that happen to be instantaneously at rest relative to himself (as opposed to using a local ruler and pair of clocks undergoing Born rigid acceleration, for example).
     
  21. Aug 16, 2010 #20
    Exactly right!
     
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