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Relative velocity of light in special theory of realtivity?

  1. Apr 1, 2013 #1
    relative velocity of light in special theory of realtivity??

    hello everyone..

    as per the theory of special relativity,
    Postulate2:Any ray of light moves in the 'stationary' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.

    In other words, the speed c of light is constant as seen from any and every inertial frame of reference.
    But let us suppose we have a train that is presently at rest. Let there be a source of light on the top end of the train with a photosensitive screen straight at the bottom of the source. At time t=0, the train starts to move with velocity v (high velocity) in +ve x-direction, and the source is switched on for a short period of time. As seen by an observer inside the train the photosensitive screen will have mark made by light, and the path traveled by light as perceived by this observer is a straight vertical line. The same event is observed by an observer at rest with respect to he ground, will find that light struck the screen but it traveled a slant path and not a straight line. So as seen by the observer at rest the path traveled by light is slant, which could be possible only if the vector addition of velocities was done, thus making the velocity of light not the same.
    Further to set this velocity same as c length and time are subject to differ in two different frames, leading to length contraction and time dilation. But this whole idea is based on basic vector addition resultant observed by the rest observer.
    So how is c invariant when it gets subjected to (relative) vector addition?

    Further, let's assume instead of a normal source we have a very high resolution laser beam. Here we have two possibilities, first being laser is adjusted to fall straight at the photosensitive screen as per the rest frame. In this case the moving observer should not see light falling(because laser will fall straight down and by the time it will reach the lower end of the train, the screen and the train would have got displaced from their initial position, thus no contact) or making any impact on the screen but the rest observer should as the light is falling straight. Other case being laser adjusted as per the moving frame. (This situation has turned pretty confusing! i can't explain the outcomes, but looks like they seem to show light having relative velocity...)
    So please explain how is c invariant in these circumstances?
    (To ensure light is unaffected by the motion of the source, i considered laser moving straight even though the source is moving..)
     
  2. jcsd
  3. Apr 1, 2013 #2

    Doc Al

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    The velocity (direction) of a light ray is frame dependent, but its speed is invariant.
     
  4. Apr 1, 2013 #3
    Postulate2:Any ray of light moves in the 'stationary' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.
    But the statement says Velocity...
    Even wiki says so:
    The Principle of Invariant Light Speed – "... light is always propagated in empty space with a definite velocity [speed] c which is independent of the state of motion of the emitting body." (from the preface) ..

    Also i can easily argue ..as per the situation i have described above, the magnitude of velocity will be √ ( c2 +v2), which is more than c for v≠0. But the whole idea of length contraction and time dilation is derived in to set this right, yet light travels the same way in a slant path with same speed c.??
     
  5. Apr 1, 2013 #4

    Nugatory

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    That's not what the second postulate says, that's what a common mistranslation of the second postulate from the original German into English says. I suspect that the square-bracketed "speed" in the wikipedia article is the author's attempt to make it clear that he/she is not making the usual English-language distinction between "speed" and "velocity" here.

    Addition of velocities for slant paths is a bit trickier, but if you google around for "relativistic addition of velocities" you'll find your answer. Intuitively, the x direction is contracted but not the y direction when you calculate the distance traveled by the flash of light - but the time it takes to travel that distance is dilated such that it all comes out to a speed of c.
     
  6. Apr 1, 2013 #5

    Doc Al

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    Those statements may use the term velocity, but they mean speed.

    Ah, but you must add velocities using the relativistic addition of velocity formula. Using the Galilean addition of velocity formula gives the wrong answer, as you see.
    Length contraction, time dilation, and the relativity of simultaneity all combine (in the Lorentz transformations and its derivatives) to ensure that the speed of light remains the same in every frame.
     
  7. Apr 1, 2013 #6

    jtbell

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    In the early 1900s when Einstein developed SR, physicists did not consistently distinguish between the terms "speed" and "velocity" the way we do now. The convention of considering "velocity" as a vector and "speed" as its magnitude, came later.
     
  8. Apr 1, 2013 #7
    So it is not velocity that remains constant it is speed...
     
  9. Apr 1, 2013 #8

    Doc Al

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    Right.
     
  10. Apr 1, 2013 #9
    Incorrect, the speed of light cannot be considered constant in any system where time is not constant since calculation of speed requires time to be constant.
     
  11. Apr 1, 2013 #10

    Nugatory

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    If you go back and look at the original question, it was asked in the context of Special Relativity so we're talking inertial frames in flat spacetime, and there the speed of light can reasonably be considered constant - it's a postulate, for crissakes.
     
  12. Apr 1, 2013 #11
    My apologies it is difficult for me to look at something within a context that is inherently flawed. Carry on.
     
  13. Apr 1, 2013 #12

    Doc Al

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    Yeah, OK. :rolleyes:
     
  14. Apr 5, 2013 #13
    let me continue with this query ..

    If I have a photosensitive screen inside the train and one just below it in it the rest frame, at t=0 when the motion of the train is to start. Now I adjust the laser to direct its beam straight at the screen when the whole system is at rest, that is the laser container (or the source box instead)points vertically downward. Now the train starts to move and a pulse of laser light is given out, on which of the two screen light spot will be observed?

    ans: To answer this I assume that the direction of laser is not affected by the motion of the train. So the light falls straight down on the screen at rest. So the observer in the moving frame concludes that light traveled a slant path(in the backward direction to its own motion), suggesting the laser be seen to point in that slant direction. but we had adjusted the laser container straight down. But the moving observer needs to see it slanting to explain not getting a mark on his on screen. But the laser points straight downward and that's what he should observe too..
    This situation looks contradictory ...
    To resolve this contradiction all i can say is that the direction of laser is affected by the motion of the train....which goes against the second postulate!
    so what should it really be like??
     
  15. Apr 5, 2013 #14

    Dale

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    Are you asking about the time while the train is accelerating, or about the inertial motion parts before and after the acceleration?
     
  16. Apr 5, 2013 #15
    Inertial motion.
     
  17. Apr 5, 2013 #16

    Dale

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    Then, by the PoR, if the beam hits the inside screen when the laser and screen are at rest then it will hit the inside screen while the laser and screen are in inertial motion.
     
  18. Apr 5, 2013 #17
    I agree to you on this completely. But if the light were to hit the screen inside then it would have to travel in a slant path as seen by an observer outside.....while the outside observer sees it straight(because that's how it was initially adjusted) then how is it going to hit the screen inside the train........it should come straight out.
     
  19. Apr 5, 2013 #18

    Dale

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    No, it was initially adjusted so that an observer at rest wrt it will see it as straight, and that remains. While it was being initially adjusted, an observer moving relative to it would see it at a slant, and that also remains.
     
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