Photon angular moment in space

[Luis Carvalho]

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.

 

[lippyka]

There is no mistake in your argument; however, the conclusion is incorrect. The reason is that the assumption that R(t) is stationary is not valid in the case of an Euclidean space. In general, R(t) is not constant and can vary over time, depending on the expansion or contraction of the universe. Therefore, the trajectory of a photon in an Euclidean space will not necessarily be a circumference with radius R(t).