# Thread: What is the event horizon of a freefaller into a black hole?

1. Originally Posted by thothicabob
Your question about the radius 'event horizon' being 'invariant': it IS invariant for a given black hole of a given mass, but as the mass increases, so does its radius (this follows from the formula. ...
I believe per Grants post above that he does not agree with this. It may be invariant for an observer at infinity ( and I would assume in a non gravitational well??? ) and not moving relative to the BH.

Otherwise the radius will change along with appropriate GR effects as will any other distance ( radius ).

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Originally Posted by Strange
Also, I am not entirely sure that passing through the event horizon would be completely undetectable. I assume the curvature of spacetime at that point means that your view of the universe outside would be distorted in a specific way at that point (I think, it would be visible as entirely behind a plane tangent to your point of entry but I may have that wrong).
For an infaller, relativistic aberration shoves the event horizon into the forward view, so that as you cross the event horizon it subtends only about 84 degrees ahead. But, yes, the infaller can easily detect the event horizon by observation and an understanding of GR. He also requires observation and an understanding of GR to detect this personal event horizon, which I now wish to hell I had never mentioned.

Grant Hutchison

3. Originally Posted by grant hutchison
For pity's sake, what is your question? Telling us you're baffled isn't a question.

Grant Hutchison
I would prefer not to get into that here ... without thinking about it a bit myself ... but LOTS of questions arise because of this ...

4. even Keeping Mass constant it is still not invariant.

Originally Posted by thothicabob
or to make it simple: "My open question at this point is about the Schwarzschild radius ? r=2GM/c^2...with r not being invariant leaves me baffled. This is my open question. "

the radius changes as the black hole acquires more mass. that seems 'not invariant' to me...?

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Originally Posted by tommac
I was careful to use the words "apparent signifigance" to connote that the freefaller has no way of measuring ( please correct me if I am wrong ) any effects of the "true" EH.
Originally Posted by tommac
Originally Posted by Strange
Also, I am not entirely sure that passing through the event horizon would be completely undetectable. I assume the curvature of spacetime at that point means that your view of the universe outside would be distorted in a specific way at that point (I think, it would be visible as entirely behind a plane tangent to your point of entry but I may have that wrong).
From what I understand this is not true. If it is true can someone post some information on how a freefaller could detect/measure some effect of passing through the EH?
You're wrong. The freefaller can easily detect the real event horizon by observation and an understanding of GR. With a sextant and a knowledge of the sky he can pin down the moment he crosses the event horizon with arbitrary accuracy. He likewise needs observation and an understanding of GR to detect this personal horizon which you seem to have latched on to so alarmingly.

Grant Hutchison
Last edited by grant hutchison; 2010-Dec-01 at 04:46 PM. Reason: Nested quote from Strange, for context

6. just to review the complicated thread:

1) A freefaller can see light that is inside the EH only after crossing the EH himself.
2) A freefaller has a apparent EH which is different from the BH EH ... this "personal EH" is a point in space time to which he can never communicate with
3) The radius of an EH is not invariant even when mass is kept constant.
4) the SR of a black hole is only for an observer at infinity that is not moving relative to the BH.

Does everyone agree on this summary? Anything else to add?

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Originally Posted by thothicabob
... so, in a sense, the horizon for a 'real' black hole is likely not very 'invariant' as it's likely accumulating mass all the time ...
The use of "invariant" here is a technical term, meaning "the same for all observers". Things like the location of an absolute event horizon, the rest mass of an object, and the spacetime interval between events are invariant in that sense. Length and time intervals are not invariant, because different observers disagree.
So although the black hole's mass and Schwarzschild radius may change with time, the whole spacetime structure of the event horizon is invariant: all observers, everywhere, agree which events are inside the horizon and which events are outside the horizon.

Grant Hutchison

8. umm, if the mass is constant, the radius is constant. it may not 'appear' that way relative to an infaller, especially after he's fallen in, but that's due to changed geometry of spacetime within the black hole (look up schwarzchild geometry for more on that). in fact, i think the event horizon would begin to appear as if it's receding from the infaller at c once he's inside, if he could 'see' it. but irregardless, the radius of a black hole is determined by its mass - that's what the formula says - and therefore it's not invariant it accumulates more mass (such as hapless infallers and careless observers who get too close). the relationship of the radius to the mass, i suppose, could be said to be 'invariant', if you want to be picky/clever tho.

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Originally Posted by Strange
I think the point is, while you are still outside the EH, you will only see those photons from the (other infalling) object before it passes through the EH. These photons will be limited in number (and red shifted) because the object does pass through the EH at some point. If you stayed outside the EH for long enough, that object would fade from view (and you would never see it again even if you then fell in behind it[*]). But if you fall through the EH after the object (while it is still visible) then at some point (i.e when you cross the EH) you will start seeing photons from the object emitted after it crossed the EH., This will be seamless - you will never lose site of it.

Did I get that right, Grant?
You certainly got it the way I've been saying it.

Grant Hutchison

10. Relative to who? To the freefaller ? To the person at infinity? To some standard measuring stick ???

Originally Posted by thothicabob
umm, if the mass is constant, the radius is constant. it may not 'appear' that way relative to an infaller, especially after he's fallen in, but that's due to changed geometry of spacetime within the black hole (look up schwarzchild geometry for more on that). in fact, i think the event horizon would begin to appear as if it's receding from the infaller at c once he's inside, if he could 'see' it. but irregardless, the radius of a black hole is determined by its mass - that's what the formula says - and therefore it's not invariant it accumulates more mass (such as hapless infallers and careless observers who get too close). the relationship of the radius to the mass, i suppose, could be said to be 'invariant', if you want to be picky/clever tho.

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Originally Posted by tommac
1) A freefaller can see light that is inside the EH only after crossing the EH himself.
That's right.
Originally Posted by tommac
2) A freefaller has a apparent EH which is different from the BH EH ... this "personal EH" is a point in space time to which he can never communicate with
It's not a point in spacetime. It's a hypersurface that separates the observer from the set of events he will never see, because he will hit the singularity after light from those events hits the singularity. The observer has to understand that his future is limited by a singularity in order to understand that he has this personal event horizon.
Originally Posted by tommac
3) The radius of an EH is not invariant even when mass is kept constant.
4) the SR of a black hole is only for an observer at infinity that is not moving relative to the BH.
The Schwarzschild radius is a property of the Schwarzschild metric, which is a set of extended coordinates specific to a distant observer at rest relative to the black hole. Other observers may observe other things.

Grant Hutchison

12. What do you mean by the quote below? The point is that you will lose site of it once it gets past your "personal" event horizon. The rest of that quote was right.

Originally Posted by Strange
This will be seamless - you will never lose site of it.

Did I get that right, Grant?

*Although, presumably, if you used your jetpack to fall faster than the other object, it could become visible again.

13. Originally Posted by tommac
Relative to who? To the freefaller ? To the person at infinity? To some standard measuring stick ???
That depends. :P

Actually, grant and strange are doing a much more concise and accurate job, so i'll just try to halt any further infalling and assume the role of a distant observer at infinity for a while...

14. Originally Posted by grant hutchison
The observer has to understand that his future is limited by a singularity in order to understand that he has this personal event horizon.
Sorry man ... I know that this has been a rough thread ... but can you explain this last line?

Why is that understanding needed? From the freefallers perspective the flashlight that fell in before him is accelerating away from him at some point that flashlight will be accelerating away from him faster than the speed of light and at that point the photons being released ( in his direction ) from the flashlight will also be accelerating away from him slowly.

Sure the singularity is in his future ... but why does he need to "understand" that ...

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Originally Posted by tommac
What do you mean by the quote below? The point is that you will lose site of it once it gets past your "personal" event horizon. The rest of that quote was right.
The whole of the quote was right.
Strange was talking about crossing the black hole's event horizon: you never lose sight of an object that is falling through the horizon ahead of you. You also never lose sight of an object because it crosses your personal event horizon. In principle, you can still receive photons from it right until you hit the singularity yourself: what you can't see is the later part of that object's worldline, because light from that part of its worldline hits the singularity before you do.

Grant Hutchison

16. Originally Posted by tommac
What do you mean by the quote below? The point is that you will lose site of it once it gets past your "personal" event horizon. The rest of that quote was right.
Good catch - I was trying to be unambiguous. What I meant was: if you lose sight of it before passing through the event horiozon, you won't see it again afterwards; if you don't lose sight of it before then you will continue to be able to see it as you pass through the EH. What won't happen is that you lose sight of it and then start seeing it again once you pass through the EH (assuming you are both freefalling).

17. Originally Posted by tommac
Sorry man ... I know that this has been a rough thread ... but can you explain this last line?
Why don't try to read the thread again, from the start, keeping in mind what you understand now? Grant has explained it before.

18. Originally Posted by tommac
Why is that understanding needed?
You can't "see" the event horizon; the only way you know it is there is to calculate it.

19. Originally Posted by slang
Why don't try to read the thread again, from the start, keeping in mind what you understand now? Grant has explained it before.
That's not a bad idea. I might do that myself

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Originally Posted by tommac
Sorry man ... I know that this has been a rough thread ... but can you explain this last line?

Why is that understanding needed? From the freefallers perspective the flashlight that fell in before him is accelerating away from him at some point that flashlight will be accelerating away from him faster than the speed of light and at that point the photons being released ( in his direction ) from the flashlight will also be accelerating away from him slowly.

Sure the singularity is in his future ... but why does he need to "understand" that ...
But all he sees, during his entire fall, is a flashlight becoming progressively more redshifted. In principle, light signals can continue to reach him from that flashlight, at longer intervals, redder and fainter, right until he hits the singularity himself. The flashlight leaves a trail of photons behind it, all the way to its own impact in the singularity. Those photons fall into the singularity themselves, but the observer is able to run down that train of infalling photons and see some proportion of them before he gets to the singularity himself.
So unless the observer understands GR, and understands that there is a singularity in his future, and understands the significance of this progressive redshift, he can imagine that he'll be able to just run along this stretched photon train, merrily observing this slow-motion flashlight for ever. He only has a personal event horizon because there's a singularity ahead.

How can the infaller know that he has a personal event horizon? He needs to understand GR and make observations.
How can the infaller detect that he has crossed the black hole's event horizon? He needs to understand GR and make observations.

Grant Hutchison

21. When the flashlight is travelling away from him faster than the speed of light ... he doesnt see that flashlight right? and doesnt see the photons emitted from it ( even at infinite redshift ) ???? I understand that this would only be for a VERY short time OR it would need to be a very Super massive BH for any of this to come into play ...

But ... at some point the difference in velocitys between the freefaller and the flashlight will exceed the speed of light. And the difference between the freefaller and the photon will actuallly get bigger even though the photon is pointed directly at him.

Is this true or not true ...

Originally Posted by grant hutchison
But all he sees, during his entire fall, is a flashlight becoming progressively more redshifted. In principle, light signals can continue to reach him from that flashlight, at longer intervals, redder and fainter, right until he hits the singularity himself. The flashlight leaves a trail of photons behind it, all the way to its own impact in the singularity. Those photons fall into the singularity themselves, but the observer is able to run down that train of infalling photons and see some proportion of them before he gets to the singularity himself.
So unless the observer understands GR, and understands that there is a singularity in his future, and understands the significance of this progressive redshift, he can imagine that he'll be able to just run along this stretched photon train, merrily observing this slow-motion flashlight for ever. He only has a personal event horizon because there's a singularity ahead.

How can the infaller know that he has a personal event horizon? He needs to understand GR and make observations.
How can the infaller detect that he has crossed the black hole's event horizon? He needs to understand GR and make observations.

Grant Hutchison

22. Originally Posted by Strange
You can't "see" the event horizon; the only way you know it is there is to calculate it.
Haha. .... my head hurts ... and my wording sucks ...

I think the thread was talking about the personal event horizon ... I understand that we would only be able to calculate the real BH EH ... but I would still think that we could measure and observe things about our personal event horizon without knowing / understanding that we are about to crash into the singul a..... ahhahhhahahha .. SPLAT

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Originally Posted by tommac
When the flashlight is travelling away from him faster than the speed of light ... he doesnt see that flashlight right? and doesnt see the photons emitted from it ( even at infinite redshift ) ???? I understand that this would only be for a VERY short time OR it would need to be a very Super massive BH for any of this to come into play ...

But ... at some point the difference in velocitys between the freefaller and the flashlight will exceed the speed of light. And the difference between the freefaller and the photon will actuallly get bigger even though the photon is pointed directly at him.
The fact there are photons the infaller can't see is already dealt with: those are the photons that will follow the flashlight into the singularity before the infaller gets to them. All the photons that the flashlight emits outwards are falling into the singularity more slowly than the flashlight (since they depart from the flashlight at a relative velocity of c). So the flashlight necessarily leaves a continuous stream of photons in its wake: those near the event horizon wind down their radial coordinate comparatively slowly; those near the singularity move inwards very quickly. So the stream of photons thins out as the bottom end disappears into the singularity in the wake of the flashlight, while the top end dawdles inwards from just under the event horizon; but there's never a sudden gap or cut-off, just a continuous stretching and attenuation.
Now, wherever the infaller is, he will be moving inwards more quickly than the flashlight photons in his vicinity (because he is falling just as quickly as the flashlight was at that point, and the photons are moving just as quickly as the photons originally emitted outwards at that point). So there are always (in principle) photons for the infaller to see, just progressively more stretched out and attenuated. For the infaller, running down that fading trail of photons, the flashlight never disappears; it just fades and redshifts progressively. Even in the moment before he strikes the singularity, the infaller has the chance to observe one last photon from the flashlight.

Originally Posted by tommac
Is this true or not true ...
Whether or not the flashlight and the infaller ever achieve a separation speed greater than lightspeed (and I'm not entirely convinced that's true) is irrelevant, because the infaller is still travelling through "old" photons, left in the wake of the flashlight.

Grant Hutchison

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Originally Posted by Jeff Root
Originally Posted by WayneFrancis
Originally Posted by tommac
...
-As we fall towards the BH our reference point gradually changes.
In fact we will even be able to see light that is coming from
inside the EH...
which is false, 100% unadulteratedly false.
I don't agree. I say that it is exactly correct. Really.

I'm willing to argue that if you want, but I'm also willing to
agree right now that the statement is ambiguous. The way I
interpret the statement, it is clearly true. You may prefer to
interpret it in a way that makes it untrue.
Originally Posted by WayneFrancis
Please...show me I'm wrong. Show me how any photons can
ever be emitted from an event inside a black hole's event
horizon and have any observer outside the EH see those
photons while they are still outside the event horizon.
No, I agree: Photons emitted inside an event horizon can't
be observed outside the event horizon. Tom would agree
with that.

The questions are: 1)Where is the event horizon located?
and 2) Where is the observer located?

The observer is falling into the black hole. So the observer
necessarily crosses the event horizon. As argued many times
by many posters, an outside, non-falling observer will not see
the falling observer cross the event horizon. However, we can
calculate what time a falling observer's clock will read when
he crosses what distant observers consider to be the event
horizon. The falling observer can set his alarm clock to buzz
at the moment he will cross that horizon.

The falling observer watches a line of dropped clocks falling
into the black hole below him. He sees the clocks redshifting
and dimming as they accelerate away from him. However, the
redshifting and dimming is very slight, because he's accelerating
downward, too, almost but not quite keeping up with them.

The clocks below him are more redshifted and dimmed the
closer they are to the center of the black hole, and the farther
they are from him.

I'm not sure where this scenario is going. I asserted in a previous
post (#34) that you would never see anything ahead of you that
you could not previously see. Grant objected to the beginning of
the paragraph in which I said that. I'm not sure: Does something
special happen just after you cross the event horizon? Is there a
pileup of light there, trying to escape, from the all the things that
fell in before you? Do you see this piled-up light just after you
cross the event horizon? If so, you would see the clocks disappear
below you as you approach the event horizon, then see them again
after crossing. So the event horizon would be a special place for
the free-faller, after all the times we've been told it is not.

Otherwise, I would argue that the field of vision does not suddenly
extend when crossing the event horizon. If it is limited above the
horizon, it is even more limited below. But the fact that the faller
can see anything at all below him indicates that his horizon is not
at the same location as the horizon distant observers see.

I think the scenario with the line of falling clocks can be useful in
further analysis, but I'm not up to doing it all myself.

-- Jeff, in Minneapolis

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Originally Posted by Strange
Originally Posted by Jeff Root
I'm virtually certain that the description of photons moving away
from the black hole, making a U-turn, and falling in, like the loops
of a bow, is incorrect.
I think the confusion may be about where the event horizon is in
these descriptions; the event horizon is the bounding surface of the
bow or the curved lines in the diagram. You may be right that the
exact nature of the curves in the diagram is wrong, but it seems to
be a simpler representation of the diagram on the page Grant linked
The diagram (under the heading "Schwarzschild spacetime diagram")
does not show anything like loops of a bow. Nothing in the diagram
moves away from the black hole inside the event horizon. There are
lines (yellow) that can give the impression that they go outward if
you don't study them carefully. Notice the direction of the arrowhead.
All the lines inside the event horizon go toward the center.

-- Jeff, in Minneapolis

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Originally Posted by caveman1917
Originally Posted by Jeff Root
If a photon inside the photon sphere
(at 1.5 Schwarzschild radii) is moving downward, it falls in.
(my bold)

I think you meant tangential, the photon sphere is the minimum
distance for stable orbits. If it's moving downward it falls in no
matter how far it starts.
No, I said what I meant.

I didn't say anything about a photon that is in the photon sphere,
"orbiting" the black hole. That is what makes the photon sphere
so interesting, of course, and what gives it its name, but I did not
refer to that aspect of the photon sphere.

A photon which is moving downward toward a black hole will
fall in if the photon's path crosses inside the photon sphere.
If the path does not reach the photon sphere, then the path will
just be bent around the black hole, and the photon will move
away from the black hole again after the encounter. It has been
gravitationally lensed.

In order for light to be exactly in the photon sphere, it needs to
be emitted horizontally by an object which is falling through the
photon sphere.

The last stable orbit is at 3 Schwarzschild radii. That's as close to
a black hole as massive objects can orbit without being drawn in.
Anything orbiting closer than that needs a rocket or something to
keep it in orbit.

-- Jeff, in Minneapolis

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Originally Posted by publius
The notion of "horizons" in GR is, like all things, a pretty subtle thing. It's hard to pin things down, as they never hardly mean what you think they should mean. As Grant mentioned, we have the notion of apparent horizons, and absolute horizons (the term event horizon is itself somewhat subtle and can apply to several types of horizons, depending on just what you mean, and there are things like Cauchy horizons, Killing horizons, etc, etc). Absolute horizons are geometric properties of the global spacetime, and to define them, the spacetime must be asymptotically flat turns out. There must exist the notion of a future null (light-like) infinity, allowing one to thus speak of photons which can "escape to infinity" and those that cannot, defining the absolute horizon as the boundary between those.

Note that we're talking spacetime here. We must know the whole thing, which means we must have a global solution, and know the future absolutely. If we don't know that (and we really don't in general), then absolute horizons are technically unknown exactly.
I've been meaning to make a comment arising from this, but forgot in the recent flurry. (Recent flurry in more ways than one: during the couple of hours I was out of this thread earlier, I had the unusual experience of having to dig myself back into my street. While I was out in the car, a snowplough went past the end of the road and constructed a wall of snow completely blocking access. In the midst of a heavy dump of blowing snow and attendant rotten visibility, I had to get out the snow shovel and cut a slot through this barrier, all the while uncomfortably aware that my car was currrently blocking a narrow, steep road which was still alarmingly slippy after the plough had been through. I kept expecting to hear another vehicle sliding into mine.)

Anyway. This global aspect of the absolute horizon makes it do some interesting things. Imagine a Schwarzschild black hole accreting a spherical shell of matter. In the last moment before the infalling shell arrives at the "existing" event horizon, a photon emitted from just above that horizon will start to propagate outwards, but will encounter the infalling shell just as the total density enclosed by the shell gets high enough to form a new horizon, farther out, surrounding a black hole that is now more massive by the mass of the accreted shell. So the photon is doomed never to escape, even though it may have "felt" like it could escape initially. That means the photon was never outside the absolute horizon. The absolute horizon actually rises to meet the incoming shell. Because it's a global property of spacetime, the absolute horizon anticipates the arrival of new mass, and reaches out towards it. That's the only way in which it can always mark the boundary between what escapes and what does not.

Grant Hutchison

28. This is the part that I missed. Now the picture is complete! Thanks your your patience and willingness to share your wealth of knowledge!

Originally Posted by grant hutchison
Now, wherever the infaller is, he will be moving inwards more quickly than the flashlight photons in his vicinity (because he is falling just as quickly as the flashlight was at that point, and the photons are moving just as quickly as the photons originally emitted outwards at that point). So there are always (in principle) photons for the infaller to see, just progressively more stretched out and attenuated.

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Originally Posted by grant hutchison
Originally Posted by tommac
This has NEVER been in question ... what has been in question
is does the "true" EH have any apparent signifigance for the free
faller. The free faller doesnt know that he is falling. He is in his
own frame and can not detect that he has crossed the "true"
event horizon. Nobody is or has disagreed that light somehow
escapes the EH ... it is clear that the freefaller falls past the EH
and catches up to the light.
That's the first time it's actually been clear that you
understand that. You write so vaguely, and you gallop off in
so many directions, we have to keep pinning you down to
make you us a clear form of words.
I understood that he understood that all along. The problem
is that the horizon is in a different place for the faller than it is
for the Schwarzschild observer, so when you say something
about an event that takes place relative to the horizon, it is not
clear where you mean. Your statements have been unclear for
the same reason that Tom's have been. I think you and others,
probably including me, have been giving Tom Schwarzschild
solutions when he needs Rindler solutions... or something....

Originally Posted by grant hutchison
Originally Posted by tommac
This being said ... it is also agreed even by Grant ... that there
is an apparent EH for the freefaller. This is the EH that the
freefaller can mention.
What's with the "even by Grant"? I'm not exactly out on a limb here.
No, but people keep claiming that Tom is out on a limb, and here
you are, the most trusted of trustworthy authorities, agreeing
with what he said. Maybe he isn't out on a limb at that.

Originally Posted by grant hutchison
The freefaller can "mention" any event horizon he cares to mention.
I might not be able to translate it into English, but I understood
what he meant.

Originally Posted by grant hutchison
He is able to detect the presence of both, if he understands GR and
makes observations.
The question is, can he directly observe the event of his reaching
and passing through the event horizon? We have been told many,
many, many times that nothing special happens there. Now you are
telling us that something special *does* happen there: Namely, that
the infaller will suddenly see things that he could not previously see,
at and inside the event horizon.

Originally Posted by grant hutchison
Originally Posted by tommac
Now grant mentioned that the "true" EH is invariant but only in
the sense of it being a horizon.
Is there another way for a horizon to be invariant, apart from
in the sense of being a horizon"?
Yes: The location of the horizon could be invarient.

-- Jeff, in Minneapolis

30. Originally Posted by Jeff Root
Tom has more problem expressing himself clearly than most of
us have. Richard ("publius") has more problem expressing himself
clearly than most of us have, too, and he's probably a genius. Or
he does a pretty good genius impersonation. They're both a little
bit annoying because of their communication problems, but I can
usually figure out what they mean, and when I can't, I can wait
until the intent is clarified a few posts down the thread. I have
learned quite a lot from both of them.

-- Jeff, in Minneapolis
My only response to that is don't call me Shirley, but I can call you Betty, and Betty when you call me, you can call me Al.

-Richard

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