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- Thread starter jimmycricket
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I do understand you question, and am not sure of the answer, but if you entered a BH, the universe would end for YOU, that's for sure

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Haha ok so assuming I can survive entering the black hole, is it possible that time would elapse so slowly there relative to the universe outside the black hole (apart from other similarly massive black holes) that the universe would essentially "run its course" during the time you spend in the black hole.I do understand you question, and am not sure of the answer, but if you entered a BH, the universe would end for YOU, that's for sure

Edit: Oops misread your post I thought you said you didnt understand the question

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The problems is not surviving since for a supermassive BH, there is nothing to survive out near the EH. Spaghetification doesn't occur w/ a SMBH until much closer to the singularity. You would not even notice crossing the EH. The problem is that you can't hover once inside, you HAVE to proceed to the singularity whether you like it or not. This does not, however, really answer your question, so keep looking.

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Chalnoth

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Unfortunately, once you've entered the black hole, you are drawn towards the center in a short, finite time (from your perspective). Even if you were (somehow) able to survive the trip towards the center, that trip wouldn't take enough time for you to observe much that goes on outside the black hole.Haha ok so assuming I can survive entering the black hole, is it possible that time would elapse so slowly there relative to the universe outside the black hole (apart from other similarly massive black holes) that the universe would essentially "run its course" during the time you spend in the black hole.

Edit: Oops misread your post I thought you said you didnt understand the question

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timmdeeg

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You would see the things far away extremely blue shifted due to gravitational time dilation (stars burning out) in case you would be hovering outside the black hole close to its event horizon. But it makes a big difference if you enter the black hole in free fall. Then (regardless if you are still outside or already inside the black hole), falling away from the stars outside causes redshift which is more or less canceled by the gravitational blueshift. I'm not sure about the net effect however.

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Chronos

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OK, so our indestructable observer has just fallen through what is the horizon as far as external observers are concerned.

He can observe only red shifted light coming from behind him, and that,s all.

If he could make sense of what he see behind him what does he see? - everything vastly sped up, though still looking very red.?

Galaxies and solar systems forming then dissipating in seconds?

Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

He can observe only red shifted light coming from behind him, and that,s all.

If he could make sense of what he see behind him what does he see? - everything vastly sped up, though still looking very red.?

Galaxies and solar systems forming then dissipating in seconds?

Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

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This would depend on the initial conditions. There is a continuum of different coordinate velocities you can pass the horizon with. And there are of course also different black hole sizes.OK, so our indestructable observer has just fallen through what is the horizon as far as external observers are concerned.

He can observe red shifted light behind him, and that's all.

Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

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Nugatory

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The proper time he experiences from horizon to singularity, assuming free fall, is something on the order of 15 microseconds for a sun-sized blackhole.Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

It scales linearly with the mass of the black hole.

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PAllen

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timmdeeg

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If just falling away was to be considered one should expect that the infalling observer would see the stars outside infinitely red shifted while crossing the horizon at the speed of light. On the other side the light of the stars gains energy entering the gravity well of the black hole. But I'm not sure if this reasoning is correct. Anyhow, how do you calculate the factor 2?The external universe viewed by an infalling observer inside the event horizon is redshifted by a factor approaching 2. It is not blueshifted because you are falling away from the external universe, and you can only view photons approaching you from behind.

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Chronos

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George Jones

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Actually, if the astronaut started falling from a great distance, the astronaut will see light red-shifted, i.e,, the distant outside universe will appear to run slow for the astronaut.

Suppose that observer A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler redshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength [itex]\lambda[/itex], then observer C, who falls from rest freely and radially from A, receives light that has wavelength

[tex]\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).[/tex]

The event horizon is at [itex]R = 2M[/itex], and the formula is valid for all [itex]R[/itex], i.e., for [itex]0 < R < \infty[/itex]. In particular, it is valid outside, at, and inside the event horizon.

See posts 5 and 7 in

https://www.physicsforums.com/showthread.php?p=861282#post861282

I have since done the calculations using Painleve-Gullstrand coordinates that are vaild even on the event horizon.

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timmdeeg

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Great, that's what I have been looking for, thanks George Jones and also Chronos.An earlier post of mine:

... The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength λ \lambda , then observer C, who falls from rest freely and radially from A, receives light that has wavelength

λ ′ =λ(1+2MR − − − − √ ).

\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).

So, the factor 2 is clarified.

Unfortunately my intuition is still a bit behind.

A sees B (hovering) extremely redshifted and C (in free fall) even more redshifted, because in the latter case the redshift consists of a gravitational and a kinematic component additional.

Passing by, C sees B extremely redshifted (purely Doppler) but sees A at z = 2 according to the formula you have shown. Would it be correct to say that also this redshift (z = 2) consists of a kinematic (not purely Doppler, because Doppler depends on a well defined relative velocity which in curved space-time is possible only locally) and a gravitational shift. So, it seems that the kinematic effect prevails over the gravitational effect. But why?

Kindly correct my reasoning.

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PAllen

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It won't help your intuition necessarily, but there a couple of universal approaches to red shift that don't involve gravitational redshift at all (which is strictly a non-essential, derived concept in GR, that is useful only for the special case of stationary spacetimes).

1) Establish the 4-momentum of the emitted light pulse (it will be a null vector, whose time component will be the energy if the vector is expressed in a local frame of the emitter). Parallel transport this 4-momentum along the null geodesic path (must specify path, because, of course, parallel transport is path dependent) to the receiver. Compute the dot product with the receiver world line's tangent vector (4-velocity) at the event of reception. The ratio of the energy specified in emitter local frame and this dot product will be the red/blue shift.

2) Equivalently (not obvious at all), you can parallel transport the emitter 4 velocity along the null geodesic to the reception event, express the transported 4-velocity in the local frame of the receiver world line at that event, and apply the pure SR Doppler formua (as if the emitter were local, with transported 4-velocity).

Both of these methods are universally applicable in any GR situation, no matter how complex. They explain why there is really only one fundamental phenomenon, not two. So called pure gravitational redshift for static observers in a static spacetime falls right out of either of these computational methods. They also cover any combination of non-static observers, in any spacetime, as noted.

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This doesn't sound right to me. There can't be a fixed, finite Doppler shift for a free-falling observer observing an external signal, because two different free-falling observers could have different states of motion.

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I think this is one of those questions that is very easy if you understand how to use Penrose diagrams, very hard otherwise. The question here is basically whether or not an event inside the event horizon has timelike infinity i+ in its past light cone. The answer is no if you look at the Penrose diagram.

By the way, it's a common misconception that the fate of all matter in the universe is that it will ultimately end up in black holes. Not true. In the distant future, due to the acceleration of cosmological expansion, there will be many cases where you have a single subatomic particle that is all alone inside its own cosmological horizon.

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PAllen

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I assumed what was meant was the classic free faller from infinity. This is unique - the limit of free falling at zero speed from a static observer, as the static observer approaches infinite distance from the BH.This doesn't sound right to me. There can't be a fixed, finite Doppler shift for a free-falling observer observing an external signal, because two different free-falling observers could have different states of motion.

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timmdeeg

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Thanks, PAllen, for your explanation. Yes, I think it is important to know that there is one phenomenon only, the observed frequency-shift, which is invariant.1) Establish the 4-momentum of the emitted light pulse (it will be a null vector, whose time component will be the energy if the vector is expressed in a local frame of the emitter). Parallel transport this 4-momentum along the null geodesic path (must specify path, because, of course, parallel transport is path dependent) to the receiver. Compute the dot product with the receiver world line's tangent vector (4-velocity) at the event of reception. The ratio of the energy specified in emitter local frame and this dot product will be the red/blue shift.

2) Equivalently (not obvious at all), you can parallel transport the emitter 4 velocity along the null geodesic to the reception event, express the transported 4-velocity in the local frame of the receiver world line at that event, and apply the pure SR Doppler formua (as if the emitter were local, with transported 4-velocity).

Both of these methods are universally applicable in any GR situation, no matter how complex. They explain why there is really only one fundamental phenomenon, not two. So called pure gravitational redshift for static observers in a static spacetime falls right out of either of these computational methods. They also cover any combination of non-static observers, in any spacetime, as noted.

Frankly speaking, I was inspired by authors like Peacok and Chodorowski,

http://arxiv.org/PS_cache/arxiv/pdf/0809/0809.4573v1.pdf

http://arxiv.org/PS_cache/arxiv/pdf/0911/0911.3536v3.pdf

who think of the cosmological redshift as being combined of kinematic and gravitational shifts. In the expanding FRW-model co-moving galaxies fall away from each other. Since the same is true in the Schwarzschild case regarding objects in radial free fall, I started reasoning why not thinking of said decomposition as well. Just to get a better feeling or intuition of the situation that the observer at R = 2M sees the far away exterior redshifted at z = 2 only, though moving locally with c.

But, of course, it is questionable whether the decomposition of the redshift makes any sense, because there is no invariant definition of a gravitational and a kinematic shift, which in combination yield the measured redshift.

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Chronos

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timmdeeg

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Oh yes, very interesting papers, thanks.

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PAllen

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For the second of these papers, the author tries to distinguish a classic result of Synge from a newer equivalent derivation by Narlikar, and seems to argue that Narlikar's result is not universally true. I find this to be a mis-representation.

The problem starts from the summary of Synge's result (I have his book; it is a favorite of mine). The author describes Synge's result as if there were two independent results of the parallel transport, a relative speed and recession speed. The authors do concede that both are derived locally (at receiver) from parallel transport of the emitter 4-velocity, so only one transported vector is involved. However, the key point is that after accounting for Synge's quite unusual notations, his resulting formula is

Thus, Synge's result, which the author admits is general, already establishes the universality of redshift in GR as 'pure Doppler' given acceptance of using source 4-velocity transported along the connecting null geodesic.

[Edit: To flesh this out a bit, Synge's equation (45) in Ch. III, section 7 is identical to equation (1) in the above wikipedia link once it is understood that in Synge's unusual notation:

1 + v

and

vR = γ (v cos θ sub o), with v on the RHS having the normal meaning as wikipedia.

The key is that Synge's spatial speeds have scaling by γ built into them by the particular geometric definitions he is using.]

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PAllen

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Ah, but starting from SR (let alone GR) any such question has no unique meaning. It all depends on the simultaneity convention adopted (equivalently, the foliation into time (per infaller) X space, chosen by the infaller). Only what you observe is devoid of convention - GR makes specific predictions for what you observe, and predicted observations don't depend on coordinate choices. But what is happening 'now' at a distance, is

For a given moment on the infaller's world line after horizon crossing but before reaching the singularity, the only constraint on what distant event is considered 'now' is that it can't be in either the past or the future light cone of that event. Thus, you can choose, for the now time of (e.g. earth) anything after the last observed time on earth, up until infinite future (since none of the exterior is in the future light cone of an interior event). Thus, if at some moment you see 3PM today timestamped on signal from earth, you can say 'now' there is anything from 3PM + epsilon, up until infinite future. Pick whatever you want, all choices in this range are valid.

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