11
$\begingroup$

After having worked for a while with the Schwarzschild geometry, I have realized something I hadn't seen before and that I found slightly disturbing. Consider the 4-dimensional Schwarzschild metric:

$$ ds^2 = -\left(1- \frac{2M}{r}\right)dt^2 + \left(1- \frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2 ~, $$

and suppose two observers are held at fixed spatial positions $(R_1, 0, 0)$ and $(R_2,0,0)$, with $R_2 > R_1 > 2M$ (using some non-gravitational system, say rockets burning fuel). This is a typical situation to study gravitational redshift when the inner observer sends signals separated by $\Delta \tau_1$ (as measured by him, that is, $\tau_1$ is the proper time along his worldline). The famous result is:

$$ \Delta \tau_2 = \sqrt{\frac{1-\frac{2M}{R_2}}{1-\frac{2M}{R_1}}} \Delta \tau_1 ~~, \hspace{1cm} \nu_2 = \sqrt{\frac{1-\frac{2M}{R_1}}{1-\frac{2M}{R_2}}} \nu_1 ~, $$

where we think in terms of period of these signals in the last expression. If the inner observer approaches $R_1 \to 2M$, we get an infinite redshift from the outer observer's viewpoint. My question is the following: what happens if insted of the previous situation the outer observer sends signal to the inner one? Obviously, we obtain a blueshift (the analysis is the same), and that blueshift becomes arbitrarily large when $R_1 \to 2M$, since it is still true that:

$$ \nu_1 = \sqrt{\frac{1-\frac{2M}{R_2}}{1-\frac{2M}{R_1}}} \nu_2 ~, $$

and we are keeping $\nu_2$ constant now. This seems puzzling to me for several reasons:

  • It seems that low energy excitations (say photons for example) sent by the outer observer to the inner one can become arbitrarily energetic when received by the inner observer. So in some sense we need to know the details of the high-energy physics to work in a neighbourhood of the horizon.
  • At first I thought that keeping an observer at fixed $r$ close to the horizon without moving in space was unphysical in some sense (maybe the gravitational tidal forces become so large that it is not possible to do that). But, in principle, for large mass the horizon at $r = 2 M$ is not a place where gravitational forces become that strong (in a Newtonian approximation, $F \sim M / r^2 \sim 1/M $, something that I think I remember seeing somewhere justified more rigorously in the framework of GR by noting that the Riemann tensor components in a local inertial frame do not diverge at $r = 2M$).

I thought that maybe the above definition of frequency is tricky, so I also tried to compute the energy of a photon with four-momentum $k$ with the formula $E_{obs} = - k \cdot U_{obs}$, $U_{obs}$ being the observer's four-velocity. This also diverges for a photon sent with a finite energy from infinity (actually, you obtain the same expression as before), so nothing new. Is this a real problem or am I overestimating the consequences of this infinite blueshift? [For the sake of completeness, my fear of an infinite blueshift comes from some discussions I've read about stability of horizons. In particular, when considering rotating or charged black holes one has an inner horizon which spoils some nice features one expect physical solutions to have, like predictability. I found convincing in that case the fact that perturbations sent from outside the black hole become arbitrarily blueshifted when they reach the inner horizon, so maybe this horizon is unstable and it is only appearing as a consequence of the high degree of symmetry of our exact solutions - thus they do not violate strong cosmic censorship]. Any ideas or thoughts on this issue?

$\endgroup$
  • $\begingroup$ This can all take place in a neighbourhood of the horizon small enough that the curvature is negligible, so the physics is in fact identical in flat space. If my friend sends a photon towards me, and I subsequently accelerate towards him with my ultra powerful rockets, I will meet the photon highly blueshifted. This is precisely the same effect! $\endgroup$ – Holographer Apr 4 '18 at 18:51
  • $\begingroup$ Instead of using a rocket to accelerate, use the normal force from the surface of a planet (so the Schwarzschild solution is only valid outside the planet). Now you've just done the Pound-Rebka experiment! $\endgroup$ – Holographer Apr 4 '18 at 18:53
12
$\begingroup$

You are quite correct that the blue shift becomes arbitrarily large as the inner observer approached the event horizon, but you are also correct that it becomes impossible to hover at fixed distance as you approach the horizon. The calculation of the force needed to hover is described in What is the weight equation through general relativity? The force required is:

$$ F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} $$

and this diverges as $r \to r_s$

There is no paradox or inconsistency here. The blue shift tends to infinity as you approach the horizon regardless of the starting point, but tends to infinity does not means it tends to the same value.

The apparent energy density is a result of the infinite outward acceleration. If you are falling into the black hole you would not see the light building up to infinite energy densities at the event horizon.

$\endgroup$
  • $\begingroup$ That was exactly what I was looking for, apart from a wonderful example of how even knowing the underlying theory (I studied those ideas when trying to understand the classical meaning of surface gravity) you always get surprised by the subtleties of GR. Thank you! $\endgroup$ – Alex V. Apr 4 '18 at 15:31
  • $\begingroup$ Is there perhaps also the issue that if you are hovering very close to $r_s$ and the light is coming past you and into the black hole then $r_s$ increases... $\endgroup$ – Rob Jeffries Apr 4 '18 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.