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Momentum of a photon heading towards a spherical mass

  1. Feb 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A distant observer is at rest relative to a spherical mass and at a distance where the effects of gravity are negligible. The distant observer sends a photon radially towards the mass. At the distant observer, the photon's frequency is f. What is the momentum relative to the distant observer of the photon when it is distance r from the mass? Assume that r > the radius of the mass.

    2. Relevant equations
    h = planck's constant
    f = frequency of the photon at some distance from the mass relative to the distant observer's frame of reference
    fobs = frequency of the photon at the distant observer's position relative to the distant observer
    v = velocity of the photon at some distance from the mass relative to the distant observer's frame of reference
    l = wavelength of the photon relative to the distant observer's frame of reference
    r = radial distance from the mass to the photon
    c = the speed of light in free space
    p = momentum of the photon relative to the distant observer's frame of reference
    Rs = Schwarzschild radius of the point mass = 2GM/c2 where G = gravitational constant and M = mass of the spherical mass.

    3. The attempt at a solution

    Here are two attempts with two different answers. They make assumptions which may not be correct.

    Attempt 1

    When the distant observer sends out the photon, it has a momentum of -hfobs/c. If the photon were to hit the mass and its energy totally absorbed by the mass and converted into kinetic energy, then the momentum of the mass would be -hfobs/c due to conservation of momentum. Thus the momentum of the photon relative to the distant observer would be -hfobs/c just prior to the collision, and it would be -hfobs/c at all times regardless of its distance r to the radial mass.

    Attempt 2

    p = -h/l
    l = v/f
    v = c(1-Rs/r) - from the Schwarzschild metric
    f = fobs
    therefore p = -hfobs/c(1-Rs/r)

    The reason that f(r) = fobs is that an observer at r will see a blueshifted photon (f'(r) = blueshifted(fobs) ) but their clock is slower so they'll see more cycles per their second. The distant observer will see fewer cycles per their second so that the frequency at r relative to the distant observer = f = redshifted(f') = redshifted(blueshifted(fobs)) = fobs.
     
  2. jcsd
  3. Feb 20, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    The photon interacts with the gravitational field (or curvature of spacetime if you prefer the GR view) before it hits the sphere. There is no reason to assume that this interaction does not exchange momentum (and it does).

    Well, you should get the same result for an observer at the place of the photon.

    I'm not sure if the momentum of a photon far away is such a well-defined concept, but I guess your attempt 2 result is the most meaningful value.
     
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