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Can Photon Accelerate in Vacuum?

  1. Apr 1, 2004 #1
    It cannot because the speed of photon is a constant in vacuum. So it also cannot deccelerate in vacuum.
  2. jcsd
  3. Apr 1, 2004 #2
    Thank you Capt. Obvious? :smile:

    Although it is only the magnitude of the velocity that is constant. Since it is possible to change the direction of the photon, you could say that it can be "accelerated", but the acceleration vector is always perpendicular to the velocity vector.
  4. Apr 1, 2004 #3


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    In circular motion.
  5. Apr 1, 2004 #4
    Or whenever the acceleration changes only the direction of the velocity vector and not its magnitude.
  6. Apr 1, 2004 #5
    Linear Acceleration Of A Photon

    There seems to be some question as to whether or not a photon can be linearly accelerated.

    For a photon:

    E = Pc

    P = mc

    E = mc^2 = hf

    m = hf/c^2

    Consider the gravitational attraction of two photons separated by a distance R.

    F = G m1 m2/R^2


    m1 a = G m1 m2/R^2


    a = G m2/R^2


    a = G hf/c^2R^2

    Thus, if there can be action at a distance between two photons, and the Newtonian gravitational force law is correct, and the relativistic equation for energy is correct, and the quantum mechanical relation for photon energy is correct, then a photon can be linearly accelerated.

    And if the Newtonian Gravitational Force Law is incorrect, but photons can experience real gravity, a photon will accelerate as it heads towards a sun, because it is being pulled. Indeed, if a photon can experience any force, then it can be accelerated.
  7. Apr 1, 2004 #6


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    This sort of "acceleration" is observed as red and blue shifted light. Since the speed of light can vary its wavelength must change.
  8. Apr 1, 2004 #7
    Is this the point where the dependent of kinetic energy as a function of velocity

    [tex] K.E. = \frac {1}{2}mv^2[/tex]

    breaks away to depend on wavelength or frequency?

    [tex] Energy = h \nu [/tex]
  9. Apr 1, 2004 #8

    Thanks you very much.
  10. Apr 1, 2004 #9


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    Recall that force is defined as F := dp / dτ, not F := m a. Since |p| = hf/c for a photon, we can differentiate p2 to get:

    2 p . dp / dτ = (h/c) df / dτ
    or, in other words,

    p.F = (h/2c) df / dτ

    P.S. I'm using m for rest mass, F and p are 4-vectors, and . for the (Minowski) dot product.

    P.P.S. Argh, you replied before I could delete it! I wanted to look up a detail or two. :frown: In particular, I think |p| = hf/c might only be valid for 3-momentum, not 4-momentum. It's been a while since I've done any of this in any detail! Maybe I should've wussed out and done it with 3-vectors!
    Last edited: Apr 1, 2004
  11. Apr 1, 2004 #10

    Thanks again.
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