Order of Kilopi
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.Originally Posted by thothicabob
Otherwise the radius will change along with appropriate GR effects as will any other distance ( radius ).
Order of Kilopi
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
Order of Kilopi
I would prefer not to get into that here ... without thinking about it a bit myself ... but LOTS of questions arise because of this ...
Order of Kilopi
even Keeping Mass constant it is still not invariant.
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...?
Order of Kilopi
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.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?
Grant Hutchison
Last edited by grant hutchison; 2010-Dec-01 at 04:46 PM. Reason: Nested quote from Strange, for context
Order of Kilopi
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?
Order of Kilopi
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
Established Member
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.
Order of Kilopi
You certainly got it the way I've been saying it.
Grant Hutchison
Order of Kilopi
Relative to who? To the freefaller ? To the person at infinity? To some standard measuring stick ???
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.
Order of Kilopi
That's right.
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.
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
Order of Kilopi
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.
Established Member
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...
Order of Kilopi
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 ...
Order of Kilopi
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
Order of Kilopi
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).
Moderator
Why don't try to read the thread again, from the start, keeping in mind what you understand now? Grant has explained it before.
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Order of Kilopi
You can't "see" the event horizon; the only way you know it is there is to calculate it.
Order of Kilopi
That's not a bad idea. I might do that myself
Order of Kilopi
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
Order of Kilopi
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 ...
Order of Kilopi
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
Order of Kilopi
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.
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
Order of Kilopi
I don't agree. I say that it is exactly correct. Really.which is false, 100% unadulteratedly false.
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.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
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
Order of Kilopi
The diagram (under the heading "Schwarzschild spacetime diagram")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
to earlier: http://casa.colorado.edu/~ajsh/schwp.html#spacetime
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
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
Order of Kilopi
No, I said what I meant.(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.
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
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
Order of Kilopi
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.)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.
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
Order of Kilopi
This is the part that I missed. Now the picture is complete! Thanks your your patience and willingness to share your wealth of knowledge!
Order of Kilopi
I understood that he understood that all along. The problemThat'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.
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....
No, but people keep claiming that Tom is out on a limb, and hereWhat's with the "even by Grant"? I'm not exactly out on a limb here.
you are, the most trusted of trustworthy authorities, agreeing
with what he said. Maybe he isn't out on a limb at that.
I might not be able to translate it into English, but I understood
what he meant.
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.
Yes: The location of the horizon could be invarient.Is there another way for a horizon to be invariant, apart from
in the sense of being a horizon"?
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
Order of Kilopi
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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