0
$\begingroup$

Let's assume an observer orbits close to a black hole, he is not alone, massive cosmic rays, like electrons and protons and other kind of space dust comes from the outer space and may hit him.

Since these cosmic rays are falling down the gravitational potential well of the black hole they get accelerated and gain energy.

Does this mean an observer orbiting near a black hole will experience cosmic rays with higher energies than usual?

Since at the event horizon infinite energy would be needed to escape to a distant point, does this mean that massive cosmic rays will get accelerated to infinite energy (and tear apart everything they hit on before it could reach the singularity)?

$\endgroup$
  • $\begingroup$ Do said cosmic rays need to escape? $\endgroup$ – Quill May 19 '16 at 7:33
0
$\begingroup$

An orbiting observer is a bit problematic because there are no stable orbits for $r \le 3r_s$, so let's instead consider an observer hovering at some distance $r$ from the black hole. In that case as $r \rightarrow r_s$ the blue shift does indeed $\rightarrow\infty$ and the observer would indeed be roasted.

But this shouldn't be surprising. The acceleration required to hover at a distance $r$ from the black hole goes to infinity as the observer approaches the horizon, so in effect the infinite blue shift corresponds to an infinite acceleration. If you are falling freely into the black hole then the blue shift does not become infinite. In fact a freely falling observer sees the light redshifted not blueshifted.

$\endgroup$
  • $\begingroup$ Does the infalling observer see the light red shifted from all directions or just above? If the light is red shifted does that means that the cosmic rays are actually of a weaker energy too? $\endgroup$ – Calmarius May 19 '16 at 9:00
  • $\begingroup$ @Calmarius: The calculation is only easy for radial trajectories so I can't comment on how the red shift depends on angle. Light coming from directions ahead of the faller would be lensed, so I think the calculation would end up being exceedingly complex. $\endgroup$ – John Rennie May 19 '16 at 9:03
  • $\begingroup$ I'm guessing that by cosmic rays you mean high energy particles (mostly protons). If so their kinetic energy would be reduced in a comparable way, though geodesics of massive particles are more complicated than geodesics of light, so I would have to go away and work out the details. $\endgroup$ – John Rennie May 19 '16 at 9:05

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.