# Gravity and Photon

1. Oct 4, 2007

### journeytospace

hi...I read from Hyperphysics site the relation between gravity and photon which says that

When the photon escapes the gravity field, it will have a different frequency

f ′= f (1 - GM/r*c²)...so when photon escapes r increases and so the value in the bracket also increases and hence frequency would be increasing which means it would be blue shifted when photon escapes gravitational field then why we say that light from a gravitational field would be red shifted ..is this right? Please clarify...

2. Oct 4, 2007

3. Oct 4, 2007

### meopemuk

There are two alternative ways to explain the red shift of photons in the gravitational field.

First explanation. One can imagine that the photon, like a massive particle, has both kinetic (K) and potential (V) energy, so that the total energy E = K + V remains constant while the photon is moving in the gravitational field. Then, when the photons moves away from Earth its potential energy V increases and kinetic energy K decreases, so that the frequency (assumed to be proportional to the kinetic energy) goes down as well. So, the red shift is the result of the photon's attraction to the massive body.

Second explanation. Photons are emitted in transitions between energy levels of atoms, nuclei, etc. For atoms deep in the gravitational field the separations between their energy levels decrease. Therefore, photons emitted by such atoms have lower energy. This energy doesn't change while the photon is traveling in the field. So, the red shift occurs because atoms emitting the photons are attracted to the massive body.

These two approaches were discussed in

L.B. Okun, K.G. Selivanov, V.L. Telegdi, "On the Interpretation of the Redshift in a Static Gravitational Field" http://www.arxiv.org/abs/physics/9907017

where it was concluded that the second explanation is actually correct.

I can also add the following argument against the first explanation. When the photon is registered by a detector it is absorbed completely. So, its total energy gets released in the detector. Therefore, its measured frequency should be proportional to the total energy E (which does not depend on the position of the photon in the gravitational field) rather than its kinetic energy K.

Eugene.

4. Oct 4, 2007

### pmb_phy

You've made a mistake in intrpreting the forumula f ′= f (1 - GM/r*c²). As r increases then the quantity GM/r*c² starts to decrease and not increase. As GM/r*c² decreases there is less and less to subtract from 1. Therefore the term (1 - GM/r*c²) is increasing and thus so is f'

Pete

5. Oct 5, 2007

### journeytospace

thank you...i mean value of the bracket as a whole increases and so frequency increases so there should be blue shift right...

6. Oct 7, 2007

### Jorrie

I think what the formula you quoted is referring to is a photon emitted at frequency f at Schwarzschild radial parameter r. It is observed by a distant observer (at r -> infinity, in free space) as frequency f ′, which is lower than f and hence redshifted. This equation does not track the observed frequency of a photon over changing observer distances.