Gravitational red- or rather blue-shift when approaching Schwarzschild radius

[birulami]

I found http://physicspages.com/2013/05/05/schwarzschild-metric-gravitational-redshift/: $$\frac{\lambda_R}{\lambda_E} = \sqrt{\frac{1-2GM/r_R}{1-2GM/r_E}}$$ where the indexes R and E are for receiver and emitter respectively, and the speed of light is normalized to 1.

Most other sources on the net I found only show the limit for $r_R\to\infty$, so I was happy to see the explicit formula without the limit, because I wanted to see the frequency shift when the photon approaches the Schwarzschild radius of the mass $M$. This looks like $\lambda_R$ then tends to zero, meaning the photon's frequency and energy go to infinity.

But for which observer does this hold? I take it it is the one near the Schwarzschild radius, not someone looking from the outside?

 

[mfb]

This formula assumes both R and E are outside the event horizon.

Note that a formula for R at infinity is sufficient to find the more general formula - assume we have a mirror at infinity and follow the path E -> infinity -> R.

 

[birulami]

This formula assumes both R and E are outside the event horizon.

Yes, that is what I assume, with $r_R$ being just very slightly larger than the Schwarzschild radius (assuming a situation where the event horizon is at the Schwarzschild radius). So it is true that the energy of the photon is on its way to infinity at $r_R$ then?

 

[PeterDonis]

for which observer does this hold?

For an observer that is stationary at some $r_R$ above the horizon. The closer $r_R$ is to the horizon, the greater the energy and frequency the observer will measure photons coming inward to him to have.

 

[pervect]

Yes, that is what I assume, with $r_R$ being just very slightly larger than the Schwarzschild radius (assuming a situation where the event horizon is at the Schwarzschild radius). So it is true that the energy of the photon is on its way to infinity at $r_R$ then?

The energy of a photon depends on the observer. For the particular observer you mention, called a static observer, the energy and frequency of the photon are indeed blueshifted relative to the observer at infinity. There are no static observers at the event horizion of a black hole, but as you approach the limit of getting closer and closer to the event horizon, the energy measured by a static observer increases without bound.

Energies (or the sum of energies) observed by static observers are not a conserved quantity in GR though. In Newtonian mechanics, for instance, we'd say "the photon gains energy by falling in the gravitational field", so we wouldn't expect static observers to have a conserved energy. While our expectations of finding some additional "energy in the gravitational field" or "potential energy" to add or subtract wind up being eventually dashed, this is getting a bit-off topic. The most relative point is that if we went this route (considering some additive potential energy) and we tried to come up with a conserved quantity, we'd wind up trying to argue that the positive infinity canceled the negative infinity. Thus it's fortunate we can avoid taking this route.

It does turns out that under the proper conditions (a static metric, or more generally what's known as a stationary metric), the energy-at-infinity is both a constant of motion of the photon (it doesn't change as the photon falls in) , and the sum of the energies-at-infinity gives a measure of energy that is conserved. This is sometimes called the Komar energy.