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Speed of massless objects

  1. Jun 24, 2014 #1
    Objects with mass always travel less than c with respect to other objects with mass. Objects without mass always travel at c with respect to objects with mass. But what speed do objects without mass travel with respect to each other? Do two massless photons travel at c with respect to the other's frame of reference even if they are going in the same direction? If so, do they perceive themselves as going in the same direction at different speeds or in opposite directions with respect to objects with mass?
     
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  3. Jun 24, 2014 #2

    ZapperZ

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    Last edited: Jun 24, 2014
  4. Jun 24, 2014 #3
    If we apply the Lorentz transformations to an object traveling at the speed of light, we discover that the entire universe around it has flattened in the direction of travel. So the actual time in flight of the photon, as "measured" by the photon, would be zero.

    So photons don't have time to see other photons or any other scenery.

    For photons, it's not the journey, it's all about the destination.
     
  5. Jun 24, 2014 #4

    Nugatory

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    No, if we apply the Lorentz transformations to an object traveling at the speed of light (more precisely, to a frame in which light is at rest), we get garbage. The derivation of the transforms starts with the constraint that light travels at speed c which is not zero in both the primed and unprimed frames, so the transforms simply don't apply.
     
  6. Jun 25, 2014 #5
    Of course, you could always resort to limits as v->c, but you don't need to.

    At the speed of light, light still travels at the speed of light. The equations handle it. The potential problem with one photon not seeing another photon traveling at c is resolved by the lack of any photon having any time to look.

    The result is not only computable, it is also useful. If a neutrino is truly able to change during flight, it cannot have zero mass.
     
  7. Jun 25, 2014 #6
    I have always assumed the photon 'sees' things as Scott described. At least, since I wrapped my head around time dilation. And have wondered about the point of view of a gravity wave.

    Wondered too, if gravity wave is not a contradiction in terms.

    Regarding : https://www.physicsforums.com/showthread.php?t=511170

    The story I was told is that Einstein developed relativity by imaging himself in the rest frame of a photon. From this tale hangs a very large cat.
     
  8. Jun 25, 2014 #7

    ZapperZ

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    Well then, try to develop a dynamical description in that frame. How does one do that? Your sense of time and space are no longer what you think. One dimension is compacted into zero, and you have no way of defining time to measure anything. Are there anything changing in that frame?

    Einstein might have imagined traveling at the speed of light BEFORE he fully developed SR. But in all the illustrations and description of what it would be like, the description always says about how the world LOOKS LIKE in front and back (i.e. everything is blueshifted in the direction of motion, which it is all redshifted in the opposite direction). But just saying that AUTOMATICALLY IMPLIES that one is only moving close to c. Otherwise, one can't see the blueshifted and redshifted light!

    Zz.
     
  9. Jun 25, 2014 #8

    UltrafastPED

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    The Einstein story is repeated in many of his biographies; the gist is that at about age 16 he thought about what one sees if travelling on the crest of a wave: a co-moving series of waves, all at rest wrt to your position. As this would hold for any type of wave, what would happen with light? But a static electromagnetic wave violates Maxwell's equations - so there must be a contradiction somewhere in this brief thought experiment.

    It was at least ten years later, after all of his student days were completed, that he worked out where the contradiction is - and invented Special Relativity at the same time.

    PS: Gravity waves can best be thought of as periodic changes in the tidal forces; they are not at all exotic, just very small and hard to find.
     
  10. Jun 25, 2014 #9

    Dale

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    No, they don't. Such a transformation is not invertible, which is a key requirement of the math.
     
  11. Jun 25, 2014 #10

    PAllen

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    Do they, even in any silly sense? We have, for example:

    x' = [itex]\gamma[/itex](x-[itex]\beta[/itex]t)

    Tell me the value of [itex]\gamma[/itex] for going from a my frame to a photon frame. It is not ∞, it is undefined.

    The FAQ is correct - it is not possible within SR to define a frame for c - period.
     
  12. Jun 26, 2014 #11
    Yes. Exactly. Photon emits. Photon strikes object. No time in between.

    Again, I concur. At C, time stops. However, at the time of the thought experiment, Einstein could not know that, because he hadn't discovered it yet. This led him into error. In discovering and correcting that error, he made his name.



    EXCEPT when you are discussing the point of view of an electromagnetic wave.

    And point of order, aren't those really Heavysides' Equations? Didn't Maxwell have a few more things on his mind, and in his notes?

    Yes. And it's a good thing for the rest of humanity he didn't spend all that time in church, or the synagogue. Although I'm sure the clergy is disappointed, the rest of us have some great insight, second hand.

    If it's as simple as the turn of the tides, I'm not surprised to have missed it. I was looking for something a little more exotic. For instance, the observational effects of gravitational ripples caused by two black holes circling, getting closer and closer, whirling faster and faster, UNTIL... gee, I hope I'm not there, then!!!! Or a good reason for the fact that gravity has an apparently infinite velocity and therefore must be a heck of a lot more important and interesting than that sluggard, mere light.
     
    Last edited: Jun 26, 2014
  13. Jun 26, 2014 #12

    Bill_K

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    For other reasons, not because special relativity forbids it. Neutrino mixing requires an off-diagonal term in the Hamiltonian. This means the masses of the eigenstates cannot be equal, i.e. at least ONE of them must be nonzero.
     
  14. Jun 26, 2014 #13
    If there are three ways of computing a value and two of them involve dividing by zero, they don't invalidate the third path. And just because you can't compute everything doesn't invalidate the computations of what you can compute.

    Since the elapsed time of travel works out to 0, I'm not too worried about about all those things that involve velocity, acceleration, or any other time-dependent measurement.

    Basically, the photon sees its entire path collapse to a point so it is touching both its departure point and its destination. That condition lasts for 0 seconds.
     
  15. Jun 26, 2014 #14

    PAllen

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    You stated you could apply the Lorentz transform ("If we apply the Lorentz transformations to an object traveling at the speed of light"). I showed you could not. The Lorentz transform has two substantive equations, for x' and for t'. Neither can be applied for a boost to c.

    You made a false claim - end of discussion.
     
  16. Jun 26, 2014 #15

    Bill_K

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    You're not doing yourself any good by clinging to this idea, .Scott, it's a complete misunderstanding, and it's just telling us that you haven't yet come to grips with Minkowski geometry.

    Because it doesn't have anything to do with what a "photon sees", or what a Lorentz transformation looks like, it's just about the geometry of spacetime. Along any null curve, the invariant interval is zero. And so along that curve we switch to a different measure, an affine parameter, and use that in place of proper distance for describing the separation of points.
     
  17. Jun 26, 2014 #16
    So [itex]x' = \gamma(x-\beta[/itex]t) is Lorentz
    but [itex]x'\sqrt{1-\beta^2} = (x-\beta t)[/itex] isn't!
     
    Last edited: Jun 26, 2014
  18. Jun 26, 2014 #17

    Dale

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    It doesn't matter how you write it. It is not a valid coordinate transform at v=c because it is not invertible.
     
    Last edited: Jun 26, 2014
  19. Jun 27, 2014 #18

    PAllen

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    Now solve for x'. It is undefined, period. I put in x, what value of x' comes out? Face it, your claim that you can apply LT for boost of c is just wrong, pure and simple.

    [Edit: this ties back to Dalespam's point about non-invertible being a fundamental problem. Suppose you take the literal interpretation of your rearrangement of LT. Then, used in one direction, it claims the whole x' line corresponds to one event, for a given t. Used the other direction it claims the whole x line corresponds to one event for a given t'. That is a logical contradiction, similar to all the grade school paradoxes you can 'derive' by dividing by zero in disguised way. It is just nonsense to claim the LT applies to a boost by c.
     
    Last edited: Jun 27, 2014
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