# The experience of the infalling body of an EH

## Main Question or Discussion Point

Alice is free-falling towards the event horizon (EH) of a black hole (BH) large enough such that local tidal forces are negligible. Bob is sufficiently ("infinitely") far away such that he is unaffected by the BH and inertial. Bob's view of Alice is well established: she will appear to slow down and red-shift as the approaches the EH, eventually "freezing" and "dimming to nothingness" before she ever appears to cross it.

The question is, what is Alice's view of the outside world? I've looked through many sources (GR textbooks, Googling, etc) and so far have been unable to find an answer to this. A few people seem to think that Bob would appear blue-shifted, but there is some contention as to whether this blue-shifting would be asymptotically infinite as the EH is approached. Is there a well-established answer?

Related Special and General Relativity News on Phys.org
Alice sees the universe go into extreme fast forward because of the effects of time dilation. i think.

tiny-tim
Homework Helper
hi rjbeery!

you're confusing time dilation with red-shifting (or blue-shifting)

time dilation is what you measure, red-shift is what you see

if you move away from me at speed v = tanhα, then we actually see each other's clock going slow by the red-shift factor e = √((1 - v)/(1 + v)),

but we make allowance for the red-shift effect (it would happen even without relativity, and does happen with sound waves, but using the speed of sound instead of c), and so each of us regards the other's clock as going slow only by the lesser time dilation factor sechα = √(1 - v2/c2)

if Alice hovered just above the event horizon, the gravitational blue-shift (inverse of her own gravitational time dilation, (1 - 2GM/rc2)) would make her see Bob's light blue-shifted, but if she was free-falling (at the same place), this would be countered by the red-shift due to her high speed

as she approaches the event horizon, her speed approaches the speed of light (she actually crosses the horizon at the same speed as a ray of light), so i think (i haven't actually checked the maths you can do that! ) the 1/(1 - 2GM/rc2) blue-shift would be more than offset by the e redshift

Thanks Erk3452, but if someone could give a description, ideally mathematical, of how GR predicts Bob would appear to Alice it would be fantastic. I'm looking for something definitive, possibly with a reference? I imagine this question has been asked and answered many times before, so it surprises me that a definitive response seems to be elusive...
(edit: I posted this before seeing your response, tiny-tim. What you wrote "seems right", but how certain are you?)

Last edited:
pervect
Staff Emeritus
It's fairly well established what you'll see when falling to the event horizon of a static, non-rotating black hole, though it's hard to find a textbook reference.

On the WWW, http://casa.colorado.edu/~ajsh/quiz.html, there is an answer in the form of a quiz question and its answer.

question 5:

As you fall freely into a black hole, you see the entire future of the Universe played out before your eyes. True or false?

Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.

In order to watch the history of the Universe unfold, you would have to remain outside the horizon, the Schwarzschild surface. One way to watch all the history of the Universe would be to stay just above the horizon, firing your rockets like crazy just to stay put. The Universe would then appear not only speeded up, but also highly blueshifted (probably roasting you in gamma rays), and concentrated in a tiny piece of the sky just above you.
I didn't see any direct working out of this problem in either MTW or in Kip Thorne's book on black holes, the first two places I'd look. Which doesn't mean it's not there, the amount of time I had to look was limited. I do recall doing the calculations myself on this, once upon a time, though.

I hope this does answer your question - with infinite blueshift, one should conclude that one would see the entire history of the universe play out.

tiny-tim
Homework Helper
hi rjbeery!

(just got up :zzz: …)
(edit: I posted this before seeing your response, tiny-tim. What you wrote "seems right", but how certain are you?)
for hovering, as i said, it's obviously a blue-shift

for free-falling (which you specified), i think the red-shift eventually exceeds the blue-shift (but only of course for a fraction of a second before entering the event horizon) …

but i'm leaving you to do the maths!

George Jones
Staff Emeritus
Gold Member
i think the red-shift eventually exceeds the blue-shift (but only of course for a fraction of a second before entering the event horizon) …
I am not sure what you mean by this. I have used Painleve-Gullstrand coordinates, which are valid outside, on, and inside the event horizon, to do the calculation, and the red-shift increases below the event horizon and approaches infinite red-shift as the freely falling observer approaches the singularity.

If Bob, who hovers at great distance from the black hole, radially emits light of frequency $\omega$, then Alice, who falls from rest freely and radially from Bob, receives light that has frequency

$$\omega' = \omega \left( 1+\sqrt{\frac{2M}{R}}\right) ^{-1}.$$

This is valid for $0 < R < \infty$.

For the definition of Painleve-Gullstrand coordinates (but not for the red-shift calculation), see

Thanks to everyone for the feedback. I spent some time in the library stacks this weekend and discovered that the answer is just what tiny-tim suggested: blueshifting occurs at the EH but is almost completely compensated for by the redshifting due to the high velocity of the infalling body. The result is an external world whose frequency is cut in half..!

My next question then becomes what does the infalling body experience (regarding the outside world-view) as she approaches the singularity. This is something I should be able to work out on my own and post a bit later (I'm not very familiar with LaTex)...but don't let that stop you from taking a crack at it!

Last edited:
tiny-tim
Homework Helper
hi rjbeery!
I spent some time in the library stacks this weekend and discovered that the answer is just what tiny-tim suggested: blueshifting occurs at the EH but is almost completely compensated for by the redshifting due to the high velocity of the infalling body.
no, actually my guess was that the redshift far exceeds the blueshift

but until i see the calculations (and i'm too lazy to do them myself ), i'm more inclined to believe George (according to whom also I'm wrong) than some library book …

can you identify the book, and perhaps quote a passage from it?
I am not sure what you mean by this. I have used Painleve-Gullstrand coordinates, which are valid outside, on, and inside the event horizon, to do the calculation, and the red-shift increases below the event horizon and approaches infinite red-shift as the freely falling observer approaches the singularity.
Hi George!

I'm surprised that the equations are valid inside the event horizon …

I thought that anything with positive rest-mass overtook radially-infalling light inside the horizon, and so someone looking radially out wouldn't see anything?

George Jones
Staff Emeritus
Gold Member
no, actually my guess was that the redshift far exceeds the blueshift
and this agrees with the the book found by rjbeery states.
Thanks to everyone for the feedback. I spent some time in the library stacks this weekend and discovered that the answer is just what tiny-tim suggested: blueshifting occurs at the EH but is almost completely compensated for by the redshifting due to the high velocity of the infalling body. The result is an external world whose frequency is cut in half..!
This happens at the event horizon, i.e., as the event horizon is approached, the total shift approaches a red-shift factor of 2. This agrees with what I posted above. So I think everyone agrees that up to the event horizon, red-shift wins! For the situation outside the event, I gave a derivation in post#5 (correction in post #7) of

I have since worked out a more careful derivation that is also valid inside the event horizon, but I have not posted this.
I'm surprised that the equations are valid inside the event horizon …

I thought that anything with positive rest-mass overtook radially-infalling light inside the horizon, and so someone looking radially out wouldn't see anything?
No, I don't think so, but I don't have time right now to elaborate.
My next question then becomes what does the infalling body experience (regarding the outside world-view) as she approaches the singularity. This is something I should be able to work out on my own and post a bit later (I'm not very familiar with LaTex)...but don't let that stop you from taking a crack at it!
I didn't give a derivation, but my post above considers this.

Last edited:
The Wiki article explains this well and I believe it

As the rain observer approaches the singularity, $r \rightarrow 0\,\!$ , and $\cos \boldsymbol{\Phi}_r \rightarrow \sqrt{\frac{r}{2M}}\,\!$ . Most of the stars are squeezed to a narrow band at the 90° viewing angle. The observer sees a magnificent bright ring of stars bisecting the dark sky.
Article

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

Thanks to everyone for participating. FYI, my references were "Black Holes" by Raine Thomas and "The Mathematical Theory of Black Holes" by Chandrasekhar.

tiny-tim
Homework Helper
Last edited by a moderator:
Yes, Derek Raine and Edwin Thomas. It's a book wonderfully suited to my level of intellectual laziness.

DrGreg
Gold Member
I thought that anything with positive rest-mass overtook radially-infalling light inside the horizon, and so someone looking radially out wouldn't see anything?
The equivalence principle still applies inside the event horizon as anywhere else. Within a small enough region around the observer, special relativity still applies as a local approximation, so the local speed of light relative to the observer is still c; you can't overtake a local photon. As you get close to the central singularity, the region where the equivalence principle approximation applies shrinks (presumably to zero, for a given margin of error you're prepared to accept as "approximate".)

DrGreg