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Does a single photon have 'color'

  1. Dec 7, 2004 #1
    Does a single photon have 'color'? If so, what physical attribute(s) of the photon embody the Hubble velocity color shift?
  2. jcsd
  3. Dec 7, 2004 #2


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    Yes, the photon's frequency defines its color.

    - Warren
  4. Dec 7, 2004 #3
    what physical attribute(s) of the photon embody the Hubble velocity color shift?
  5. Dec 7, 2004 #4


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    Is that a homework question? Can you think of a way/reason a photon's color would change?
  6. Dec 8, 2004 #5
    I'm way past homework. (There is a hidden agenda behind the question.) It doesn't seem to pass any test of reason that the emitter of the photon 'knows' who is going to be doing the detecting. So it would seem to follow that, for a given type of emitter (element), the attributes of the photon are always the same. That leaves us with the detector. The only attribute of the photon that I can see that would cause the detector to see a shift in color would be the velocity of the photon - where the Hubble shift says the emitter and detector have some relative velocity difference.
  7. Dec 8, 2004 #6


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    The energy and frequency of a photon are tied together, if one varies, so does the other, in direct proportion. Since a photon's energy can vary depending on circumstances so can its frequency. Energy is not invariant; different frames will see the same photon with different energy, hence with different color. This is the mechanism of the relativistic doppler effect, which is what causes the color shifting of the remote galaxies which are moving away from us.
  8. Dec 8, 2004 #7


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    Welcome to Physics Forums arrell!

    All observers measure the speed of light (in a vaccuum) to be the same ... c. This is a prediction from Einstein's theory of Relativity, and there are no (AFAIK) good experimental or observational results that are inconsistent with this theory.

    While photons from distant galaxies in images taken by the Hubble Space Telescope are detected as 'red-shifted' (their wavelengths are greater than comparable photons emitted in a lab near you) - an effect caused by the expansion of the universe - you don't have to go so far to see red-shifted photons. For example, helioseismology relies on the (local) Doppler effect (this involves both redshifting and blueshifting).

    Another example, this time the classic Pound and Rebka experiment ... the photons from the emitter were received by the detector (or not!) at a different frequency, but there was no relative motion! This is called 'classic' because it first demonstrated the 'gravitational redshift' predicted by Einstein.
  9. Dec 8, 2004 #8
    Thanks for answering ! I'm a little confused, however. I thought that the speed of light (in a vacuum) was a postulate of Einstein's SR theory rather than a prediction. And Einstein seems to go on to do formulation based on light from the rest frame 'chasing' a point in the moving frame so that the 'closure' velocity is c-v. In my question, I guess I was looking at the detection of the photon being interesting in a case where there is a detector at rest wrt the emitter and then there is my detector which is moving at velocity v away from the emitter. And I guess there is no reason not to be able to constrain the consideration to be that when the photon arrives, we'll have the two detectors at that moment be equidistant from the emitter. The only physical difference that I can see for the two point detectors is that one has a velocity v, so it seems that this would be what has to make the detection show a lower energy for the photon - which would be consistent with the photon 'chasing' the detector. (I thought Pound/Rebka was based on GRT?)
  10. Dec 8, 2004 #9


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    The relative speed of light to any unaccelerated frame is c; this is a postulate of SR. Of course it is good physics to verify that that postulate agrees with reality! That is why Nereid referred to the experimental status. Since both frames in your example see the photon moving at c (there is no v-c involved) you have to adapt your conditions to that fact.
  11. Dec 8, 2004 #10
    From Einstein's paper,

    t(B) - t(A) = r(AB) / (c-v)

    which would seem to say that he considers that the light (photon) from the rest system is 'approaching' moving point B at less than c ???
    Last edited: Dec 8, 2004
  12. Dec 10, 2004 #11
    Apologies. I reversed the A and B subscripts. Should have been:
    t(A) - t(B) = r(AB) / (c-v)
    While we are this stage of Einstein's formulation, there is something that puzzles me.
    When the photon reaches point B, it would seem like we might consider that the communication of the photon from the rest frame to the moving frame has taken place. As the equation above is written, the photon is approaching B at velocity c-v. If we regard the photon as having made the transition to this frame and now consider that this frame has the properties where A and B are at rest with respect to one another, at the instant the photon is reflected from B, it seems like the velocity must be c-v in the opposite direction.
    so t'(A) - t(B) = r(AB) / (c-v)
    Or, I don't see any problem with still characterizing this frame as moving at velocity v as Einstein apparently does. However, if at the time the photon arrives at B, we assume that B (and A) are still regarded as moving at velocity v, it seems like the equation for the reverse path must be written:
    t'(A) - t(B) = r(AB) / (c - v - v(B) + v(A)) = r(AB) / (c-v)
    It feels pretty satisfying that two different ways of characterizing the frame yield the same resulting equation.
    However, when Einstein formulates the second equation, he has to be using some principle(s) that I'm not aware of.
    In the case where the frame is considered now at rest when the photon reaches B, in order to get his equation, the reflection must add velocity 2v to the reflection:
    t'(A) - t(B) = r(AB) / (c-v + v + v) = r(AB) / (c+v).
    Or in the other case, where he considers the frame as still moving at v, the reflection must add velocity v to reflected velocity of the photon:
    t'(A) - t(B) = r(AB) / (c-v + v + v(B)) = r(AB) / (c+v).
    What I need help with is that I don't see how the reflection can add any velocity to the reflected motion of the photon much less two different amounts depending on whether you now regard the frame as at rest or still moving. ???
  13. Dec 10, 2004 #12


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    Moving this to Special & General Relativity ...
  14. Dec 10, 2004 #13


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    Making some logical assumptions about the coordinate systems being used (which should really be spelled out explicitly), the equation

    t(B) - t(A) = r(AB) / (c-v)

    does not say that the photon moves at a velocity c-v in any frame.

    What it says that if a photon is approaching another moving object, the time it takes in the frame where the object is moving is given by solving

    d = c*t +A (a light beam starting out at d=A at t=0)
    d' = v*t+B (an object starting at d=B at t=0 with velocity v)

    for the time when d=d'. This means c*t+A = v*t+B, or (c-v)*t = B-A

    When you go to the frame of the moving object, you find (of course) theat the velocity of the photon in the moving frame is equal to c. You finding that in the moving frame, the light beam and the object do not start out at the same time 't'. that the distance is lorentz contracted, and that time is dilated.


    I'm going to explain this more fully, in the hope that it will get through. (I'm guessing that this is the same person I talked to before that was confused about this issue, though of course it's a resaonably common confusion)


    We have two frames of reference. One is stationary, one is moving at velocity v. The origins of the two frames coincide. We have an object, who is also moving at velocity v (the same velocity that the two frames are moving).

    in frame 1, a light beam starts out at t=A, v=c

    In frame 1, the equation of the light beam is

    x = A + c*t

    To convert this to frame 2, we use the Lorentz transform

    x' = gamma*(x - v*t) t' = gamma*(t - v*x/c^2)

    where gamma = 1/sqrt(1-v^2/c^2)

    since x=A and t=0, we find

    in frame 2, a light beam starts out at

    x' = gamma*A, and t' = -gamma*v*A/c^2.

    Since the velocity of light is a constant, 'c', in any frame, the equation for the motion of the light beam in frame 2 is

    1) x = gamma*A + c*(t' + gamma*v*A/c^2)

    (the only equation for an object moving at 'c' that passes through the given point).

    Similarly, in frame 2, we find that an object starts out at

    x' = gamma *B and t' = -gamma*v*B/c^2

    The relativistic velocity equation will give a velocity for the object in frame 2 as zero, because I am making the assumption that our frame is comoving with this object (but has an origin that's displaced).

    Thus the equation of motion for the moving object as a function of time in frame 2 is

    2) x' = gamma*B

    So, the two objects will meet when eq 1) equals eq 2)

    gamma*A + gamma*A*v/c

    gamma*A + c*(t' + gamma*v*A/c^2) = gamma*B

    which boils down to gamma*A(1+v/c) + c*t' = gamma*B

    or t' = gamma*(B-A)/c - gamma*A*v/c^2

    which is quite consistent, it says that the distance is contracted by gamma, and you have a time offset of -gamma*A*v/c^2 due to the relativity of simultaneity.
    Last edited: Dec 10, 2004
  15. Dec 10, 2004 #14
    your courtesy in taking the time to answer is very much appreciated !! It's going to take me some time to work thru. It should be interesting. Again, thanks !
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