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Does photon have a rest frame?

  1. Jan 28, 2012 #1
    Lets say for a moment that everything that travels at speed below c has a rest frame..
    All kinds of waves are produced from entities which has a rest frame...
    for example, Water waves produced from water surface which has a rest frame, similarly sound waves from physical objects, light waves from atoms and so on...

    If photon is created at the atom level by the transition of electrons from one energy level to another, then it should have rest frame, since electron has rest frame...
    I mean the speed of electron can never reach 'c', so the photon which is produced from the electrons must be initially accelerated to the speed of 'c'... If photon has no acceleration, then it can never reach 'c'... So photon initially travels at a speed below 'c' before it is accelerated... Hence it has a rest frame?
  2. jcsd
  3. Jan 28, 2012 #2

    Doc Al

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    You assume that photons start out with zero speed and then are accelerated. That's not so.
  4. Jan 28, 2012 #3


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    read the FAQs in the cosmology section
  5. Jan 28, 2012 #4
    Can u please tell me which question you are referring to?
    There are 12 questions in there and none is related to photons...
  6. Jan 28, 2012 #5
    Im not assuming.. I just wonder how something could just come out of the rest particles with speed of light without any acceleration whatsoever....
  7. Jan 28, 2012 #6

    Doc Al

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    We have a short FAQ on this in this forum (Relativity): Rest frame of a photon
  8. Jan 28, 2012 #7

    Doc Al

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    Again you tacitly assume that it starts from rest. It doesn't. Photons only exist at speed c. (Just because the particle emitting the photon might be at rest in some frame has no implication for the photon's speed.)
  9. Jan 28, 2012 #8
    It is certainly a good question.

    And not easy to answer.
    For instance consider the wavefunction of a photon in time.

    I differ with most here who probably would discourage people from asking such questions in the first place.
  10. Jan 28, 2012 #9
    Photons in SR always travel at c. They are not accelerated.

    Photons are not localizable, therefore you cannot define a rest frame for them.

    A photon is not a tiny ball living inside an electron and then ejected and accelerated up to c, as you seem to believe.
  11. Jan 28, 2012 #10
    Since you pretend to know it better why don't you give use the truth of the matter.

    We have an electron, then a photon is emitted. What happens?

    The floor is yours........
  12. Jan 28, 2012 #11


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    Consider some graphics originally taken from http://en.wikipedia.org/wiki/Polarizer

    The first is labeled in nanometers. This is meant to represent a 532 nanometer photon flying by you (your rest frame - you are watching the photon fly past you) to the right at the speed of light. The photon has been forced into a "linear" form by the filters it has gone through.

    As it passes you the amplitude of the photon is increasing, then dropping back to zero then going negative as it passes you. The amplitude is 0 at position 0 and is back to zero by the time it reaches 532 nanometers to the right.


    The next graphic is meant to represent the photon from the photons point of view (the photons rest frame - you are flying along with the photon and watching it over time). What you see is the photon again being forced into a specific orientation by the filters.

    As time passes, the amplitude of the photon is increasing, then dropping back to zero then going negative, then going back to zero. At time 0, it has an amplitude of 0, by time 1.7 zeptoseconds the amplitude has returned to 0.

  13. Jan 28, 2012 #12
    Remember that there is also a 'spiral' component of the wave. I have not seen anyone turning that in some relativistic Thomas precession.
  14. Jan 29, 2012 #13


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    It is not possible to transform to the 'rest frame' of anything travelling a the speed of light by a Lorentz transformation, obviously.

    But there are coordinates in which the energy of radiation is seen as static. The energy momentum tensor of light propagating in the k direction is given by

    Tαβ = σ2 kαkβ where k is a null vector, kμkμ = 0.

    This is in the usual Minkwski coords, t,x,y,z.

    If we trasform this EMT to Brinkmann coords.

    u = (x-t)/2
    v= (x+t)2

    The EMT transorms to that of static energy, T00 = ρ, all other components zero.

    This works in reverse too, and static matter in Minkowski coords becomes radiation in Brinkmann coords. I think this is called an 'ultra-boost'.
    Last edited: Jan 29, 2012
  15. Jan 30, 2012 #14
    As Mentz114 says, it is not possible to transform the observer to the 'rest frame' of the photon. But it is still possible to approach that frame as closely as we like. And beyond a certain speed, the observer's picture of the photon stops changing and looks like this:

    1. Because the photon's light source is moving away from the observer at a speed immeasurably close to that of light, the Doppler effect red-shift's the photon's wave packet (wherein its position becomes immeasurably uncertain) and its wavelength, λ, to immeasurably large values. Hence, its wave number, w=1/λ, is immeasurably small, and its momentum, p=hw, is likewise.

    2. Since, its speed is the same in all frames, both its frequency is f = w/c, and therefore, both f and its energy, hf = hw/c, are immeasurably small.

    3. Conclusion: Long before the observer can ever "match the speed" of the photon, he knows what the answer will be: What photon?
  16. Jan 30, 2012 #15
    Correction: f=wc, of course!
  17. Jan 30, 2012 #16
    Essentially a photon just came into being (and traveling at c), while the equivalent amount of energy and momentum is taken away from the electron (perhaps dropping down to a lower atomic energy level).

    As far as we know, this is instantaneous and probabilistic at the single electron-photon level.
  18. Jan 30, 2012 #17
    But we all know that photons' speed is reduced when it is travelling through a denser medium... So for instance, imagine we are travelling just over an ocean close to the water level in a very fast(really fast) jet, and a light source is kept inside the ocean.. We can see the light path from the jet... Now since in air Photons' speed is higher than in water... So now if we manage to travel very fast in air, then we can actually reach the speed of photon('c' in water)....
    Remember we didnt reach value of 'c' for air, we reached value of 'c' for water In AIR....
    So from the jet , we can now see the rest frame of photon...

    Dont u think?
    Instead of just denying me right out, just think about it a little bit....
  19. Jan 30, 2012 #18
    Its finally so good to see a person who agrees with me... Thank you a lot....
  20. Jan 30, 2012 #19


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    No, it isn't:

    https://www.physicsforums.com/showthread.php?t=511177 [Broken]
    Last edited by a moderator: May 5, 2017
  21. Jan 30, 2012 #20
    Last edited by a moderator: May 5, 2017
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