Gravitational radiation shift

1. Principle of Equivalence

Experiments performed in a uniformly accelerating reference frame with acceleration a are indistinguishable from the same experiments performed in a non-accelerating reference frame which is situated in a gravitational field where the acceleration of gravity = g = a = intensity of gravity field.

The same experiments are indistinguishable from each other if they are performed in two reference frames: the first is a uniformly accelerated with acceleration a, and the second is not accelerated but present an accelerated field with the same acceleration a.

Thus if we measure the frequency ν of a photon in an accelerating reference frame where the acceleration is a, the same value will find in a non-accelerated frame present in a gravitational where the acceleration of gravity = g = a.

Therefore if we are in a non-accelerating reference (or inertial) frame within the gravitational field, the frequency of an accelerating source of photon will be shifted as it is in the Doppler effect

We want then to measure the shift in radiation frequency related to the relativistic Doppler shift due to an accelerating light source, by the gravitational field.

2. Relativistic Doppler effect
Moving source toward an observer at rest

2.1. Doppler effect

The observer is at rest and the source S is moving at a speed V_s toward this observer. The wave propagates at the speed V_win all directions and toward the observer of course. When this observer receives a wavefront, at this precise time, the source is not located at the distance Ls but at L's. For the observer, the period of the wave is T's. And at this precise time, the wave has travelled Ls = V_w. T , where T is the real period. T's is the apparent period.
We can write:
L's = Ls - V_s . T
Because Ls = V_w. T
and L's = V_w. T's
Thus :
V_w. T's = V_w. T - V_s . T = (V_w- V_s) . T Or:
1/T's = V_w/(V_w- V_s) (1/T)
Æ’' = V_w/(V_w- V_s) Æ’

Æ’_obs = Æ’ . V_w/(V_w- V_s)

Æ’_obs is the apparent frequency.

2.2. Relativistic Doppler effect

In the case of a photon
Vw = c ,
and a source moving at speed
Vs = v

We have:

Æ’_obs = Æ’ . c/(c - v)

In the relativistic case:
Æ’_rel-obs = ƒobs /γ
Where
γ = (1 - v²/c²)^-1/2

Therefore

Æ’_rel-obs = Æ’(1 + v/c) = Æ’(1 + at/c)

3. Gravitational radiation shift

The acceleration is g, so v = gt and t = h/c.
h is the length of the photon's path in the time t. Hence

ƒ_rel-obs = ƒ(1 + gh/c²), or

c/ƒ_rel-obs = c/ƒ(1 + gh/c²)

λ_rel-obs = λ/(1 + gh/c²)

The equation in terms of wavelengths is written as:

λ_rel-obs = λ/(1 + gh/c²)

In terms of frequencies, we have:
ƒ_rel-obs = ƒ(1 + gh/c²)

or

ƒ_rel-obs - ƒ = ƒ. gh/c²

That is

(ƒ_rel-obs - ƒ)/ƒ = gh/c²

The relative gravitational radiation shift is:

(ƒ_rel-obs - ƒ)/ƒ = gh/c²

Since the frequency is reduced, this effect is called the
gravitational red shift or the Einstein red shift.

In 1 km, with g = 9.8 m/s² and c = 3 x 10⁸ m/s,
the related relative gravitational radiation shift is about 10^{- 11} %.

4. Gravitational radiation shift and Schwarzschild radius

This shift is due to a lost of energy for the photon. If we attribute a mass m for the photon of initial energy hν_o and final energy at the height hν, we can write:

hν = hν_o + potential energy

This potential energy can be written as:

PE = - G m M/r

G is the gravitational constant
m mass of photon
M mass of sphere of gravity (earth)
r the height (position) where the photon has the energy hν

Therefore

hν = hν_o - G m M /r

The total energy of the photon having the mass m is:
hν_o = mc².

Hence

m = hν_o/c².

The expression of the potential energy becomes:

PE = - G m M /r = - G M hν_o/r c²

PE = - G m M /r = - G M hν_o/r c²

The above equation becomes:

hν = hν_o - G M hν_o/r c² =
hν_o[1 - G M /r c²]
Or
[hν - hν_o]/hν_o = - G M /r c²

(ν - ν_o)/ν_o = - G M /r c²

This result shows that the frequency is reduced.

Recall that the Schwarzschild radius r_s is:

r_s = 2GM/c²

Which also equal to the escape speed of the photon.

(ν - ν_o)/ν_o = - G M /r c² = - r_s/2r

(ν - ν_o)/ν_o = - r_s/2r

The Earth's Schwarzschild radius is about 9.0 mm, so:
- r_s/2r = - 1 at r = r_s/2 = 4.5 mm. In this conditions:
(ν - ν_o)/ν_o = - 1. That is ν = 0.
That is a photon traveling 4.5 mm in the earth gravitational field will disappear. It does make any sense!.