Does light lose energy in transit?

Question

Consider a photon is an energetic particle and therefore has a gravitational field. When a photon passes a molecule or particle of dust in space it will pull the dust towards it and deform the dust. This will lose energy in the normal Newtonian way, and the energy is supplied by the photon.

Assuming that the speed of light is constant this implies that the photon will become less energetic and therefore red-shift over its travels, simply as a consequence of interactions with near objects. Is the effect real and/or measurable?

Answer 1

Yes the effect is real, potentially at least, but no it's not measurable.

As an aside, the redshift of light by its gravitational interaction with (homogeneous and isotropic) dust is exactly what the FLRW metric predicts, but this clearly isn't what the question means.

The gravitational field of a beam of light is calculated in the paper On the Gravitational Field Produced by Light by Tolman, Ehrenfest and Podolsky (the link is a 1.5MB PDF). A test particle feels a force normal to the light ray (there is no force parallel to the light ray), so some of the light's energy would be transferred to the test particle. In principle, when a light ray passes a dust particle we'd get something like this:

Some of the light's energy would be transferred to the particle and it would be red shifted as a result. The change is too tiny ever to be measurable, but in principle it would be there.

However this isn't likely to happen outside of a physics lab. For starlight shining through a dust cloud the interaction would actually look like:

because the light is symmetric aound the dust particle the particle feels no net force and is not accelerated. Therefore no energy is transferred to it and the light is not red shifted. As a result there is no cosmological red shift due gravitational interactions with dust.