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Photon angular moment in space

  1. May 10, 2008 #1
    In the present state of the knowledge, we can admit like true the formula:

    (1) v R(t) = C1
    v = frequency of a photon, R(t)=scale facteur, C1=constant ("Gravitation "Cosmology and, Stefen Weinberg, J. Wiley & Sons, page 416).

    As the linear moment of the photon is:
    p = hv/c
    one can one can write (1) so:

    (2) pR(t) = C2, where C2 is again a constant.

    As Stefen Weinberg written in his book that "R(t) can justly be called the radius of the universe ", the formula (2) show that the angular moment of the photon is constant and that the trajectory of the photon is a geodesic represented by a circumference with radius R(t).

    If the universe is a closed spherical space, with k = +1, admitting momentarily that R(t) is stationary, the trajectory of a photon in this space is a circle with the perimeter equals to the perimeter of a circle with the ray R(t), whatever is the position of the photon. And the angular moment of the photon is pR (t) = constant.
    Do now consider an Euclidien space, with k=0. What will be the trajectory of a photon with
    pR(t)=constante, assuming that R(t) is stationary? It would be, as in the previous case, with pR(t) = constant, that all photons have a trajectory on the surface of the sphere.
    But this it is not true, because the photons should be certainly present inside the space.

    I ask if there is some mistake in my arguments.
    Last edited: May 10, 2008
  2. jcsd
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