Observation of matter falling into Black Holes

[ObjectivelyRational]

How long would it take for an object, stationary with respect to a black hole (mass of the sun), to fall from 1 AU into it beyond its event horizon?
From our frame of reference can we observe over time, a black hole growing in size as matter falls beyond the event horizon?
From our frame of reference can we observe over time, a black hole growing in size as another black hole falls into it beyond both event horizons?
Are any answers of the above different for a black hole merger which has already been "observed"?

 

[PeterDonis]

How long would it take for an object, stationary with respect to a black hole (mass of the sun), to fall from 1 AU into it beyond its event horizon?

How long according to what observer?

According to the infalling observer, the infall time goes approximately (IIRC) as the 3/2 power of $R / 2M$, where $R$ is the radial coordinate from which the infall starts and $2M$ is the hole's Schwarzschild radius.

An observer who stays stationary far outside the hole never sees the infalling object cross the horizon at all, and any assignment of "time" made by that observer for the infall will depend on their choice of coordinates.

From our frame of reference can we observe over time, a black hole growing in size as matter falls beyond the event horizon?

Observations have nothing to do with "frame of reference"; they are invariant. If you remain outside the hole, you can't observe things falling past the horizon directly, but you can detect the increase in the hole's mass as matter falls in.

From our frame of reference can we observe over time, a black hole growing in size as another black hole falls into it beyond both event horizons?

One black hole doesn't "fall into" another black hole; they just merge. If you're far away from both holes, you won't detect any change in mass as they merge, unless the merger emits significant gravitational waves and you wait until those gravitational waves have passed you on the way out from the merger.

Are any answers of the above different for a black hole merger which has already been "observed"?

I'm not sure what you mean. If you've already observed the effects of a merger, what else would you expect to observe?

 

[ObjectivelyRational]

An observer who stays stationary far outside the hole never sees the infalling object cross the horizon at all

What do you mean by "never"? What is observed by the stationary observer?

If you remain outside the hole, you can't observe things falling past the horizon directly, but you can detect the increase in the hole's mass as matter falls in.

By "as it falls in" do you mean "as it falls toward" the event horizon or "as it crosses the event horizon"?

If you detect the increase of the hole's mass "as it falls in", does this constitute an indirect observation of an in falling object crossing the horizon?

 

[PeterDonis]

What do you mean by "never"? What is observed by the stationary observer?

The stationary observer observes (meaning, "sees in light signals coming from the infalling object") the infalling object moving slower and slower and the light from it redshifting more and more, but always above the horizon.

By "as it falls in" do you mean "as it falls toward" the event horizon or "as it crosses the event horizon"?

Both. See below.

If you detect the increase of the hole's mass "as it falls in", does this constitute an indirect observation of an in falling object crossing the horizon?

You detect the increase of the hole's mass, to a first approximation, as soon as it is closer to the hole's horizon than you are--more precisely, you do so if the object falls right past you. If the object is falling in on the opposite side of the hole from you, there is an additional delay due to light travel time before you detect the increased mass. But in any case the time at which you detect the increased mass is not directly related to the time when the infalling object crosses the horizon. Whether you want to call detecting the increased mass an "indirect observation" of the object crossing the horizon is a matter of words, not physics.

 

[PAllen]

Let's modify the example a bit. Suppose you are 1 AU from a 5 solar mass BH, and passing not far from you, a 1 solar mass white dwarf star heads directly toward the BH. Then, against the background stars via lensing, you will see a round enlarged merged BH disc within a fairly short time. You will continue to "theoretically see" light whose origin you can deduce is from the white dwarf before horizon crossing, but this light will soon be at wavelengths of kilometers, and energy of a hundreds of orders of magnitude less than CMB and apparent origin much less than Planck distance from the merged horizon (i.e. nothing resembling an image of the infalling white dwarf; just stray unobservable 'photons' ).

 

[ObjectivelyRational]

The stationary observer observes (meaning, "sees in light signals coming from the infalling object") the infalling object moving slower and slower and the light from it redshifting more and more, but always above the horizon.
Both. See below.
You detect the increase of the hole's mass, to a first approximation, as soon as it is closer to the hole's horizon than you are--more precisely, you do so if the object falls right past you. If the object is falling in on the opposite side of the hole from you, there is an additional delay due to light travel time before you detect the increased mass. But in any case the time at which you detect the increased mass is not directly related to the time when the infalling object crosses the horizon. Whether you want to call detecting the increased mass an "indirect observation" of the object crossing the horizon is a matter of words, not physics.

I think I get what is "seen" by an observer but perhaps I asked the wrong questions. I really wanted to ask about the relative passage of time as an object falls to just reach the event horizon of the black hole.

To identify it and distinguish it from the rest of the universe... other objects etc. is it safe to think of the black hole as "everything within a particular event horizon" or possibly preferable "everything not outside a particular event horizon"?

What is the relative passage of time for an object falling into a black hole and the passage of time for a far away observer, for all times the object is outside of the event horizon of the black hole and in the limit of it reaching the event horizon? i.e. How much time will pass for the far away observer in the time (measured at the falling object) it takes the object falling to just reach the event horizon?

 

[PeterDonis]

I really wanted to ask about the relative passage of time as an object falls to just reach the event horizon of the black hole.

There is no invariant way to define "relative passage of time" for observers that are not co-located in a curved spacetime. It depends on your choice of coordinates.

is it safe to think of the black hole as "everything within a particular event horizon" or possibly preferable "everything not outside a particular event horizon"?

If, as I assume, the difference is that the horizon itself is included in the latter but not the former, I would prefer the latter.

 

[PAllen]

What is the relative passage of time for an object falling into a black hole and the passage of time for a far away observer, for all times the object is outside of the event horizon of the black hole and in the limit of it reaching the event horizon? i.e. How much time will pass for the far away observer in the time (measured at the falling object) it takes the object falling to just reach the event horizon?

You choose. Specifically, consider a far away observer and an infalling body. Then any pair of events on their world lines connected by a spacelike geodesic may be considered to be simultaneous. Assuming the infaller crosses the horizon at some time per its clock, there is a specific moment on the far away world line such that a light signal emitted by the far away body will reach the infaller at the moment of horizon crossing. The relation between this far away event and horizon crossing is future light like. For any time later than this on the far away world line, it is plausible to consider this the moment of horizon crossing. Literally you choose. Even up to never, because there is never a time when the horizon crossing becomes in the past light cone for the far away observer (it will always be spacelike). Thus, for example, the faraway observer may consider the horizon crossing to have occurred any time between 3 pm Tuesday, or never, ever.

 

[Ibix]

Just to add to PAllen's comment, all ways of defining synchronisation involve the clocks exchanging light signals (or can be replaced by such a method, if you want to be picky). But once one clock reaches the horizon, you can't exchange light signals and therefore "at the same time" no longer has any definable meaning. So you can pick what you want.

 

[jbriggs444]

But once one clock reaches the horizon, you can't exchange light signals and therefore "at the same time" no longer has any definable meaning.

Defining a coordinate system is adequate to define a meaning for "at the same time". There is no need to exchange light signals.

The transmission and receipt of light signals do define the limits of valid coordinate assignments within a single coordinate patch. You are not allowed to assign the same time coordinate to two events where one lies within the future light cone of the other.

There is a point on the falling clock's world line and on the stationary clock's world line after which neither has any point on the other's world line in their causal future. No assignment of identical time coordinate values to event pairs drawn from the two world lines past those points can run afoul of that coordinate assignment rule.

 

[Ibix]

There is a point on the falling clock's world line and on the stationary clock's world line after which neither has any point on the other's world line in their causal future. No assignment of identical time coordinate values to event pairs drawn from the two world lines past those points can run afoul of that coordinate assignment rule.

True. The point I was trying to make is that you can't pick one definition as preferred by some physical process (exchanging light signals, for example), or even pick one out by symmetries, once one clock has crossed the horizon. If any definition is equally valid, they're all physically meaningless. As you say, you can simply pick one by fiat.

I hope that makes sense.

 

[ObjectivelyRational]

I think we are getting closer... my questions needed some honing in.

The (non-rotating) time dilation equation relates proper times between events for an observer within a gravitational field to coordinate time between those events for an observer at an arbitrarily large distance from the source of the gravitational field. Of course this assumes a stationary observer in the gravitational field.

Is there a way to integrate the that time dilation equation (of relative time) over a range of distances so that say a physicist falling toward the black hole could deduce the TOTAL time which has passed back on Earth while he has fallen from a specific point toward the event horizon.. IN THE LIMIT of but not including being at the event horizon.

 

[Ibix]

Is there a way to integrate the that time dilation equation (of relative time) over a range of distances so that say a physicist falling toward the black hole could deduce the TOTAL time which has passed back on Earth while he has fallen from a specific point toward the event horizon.. IN THE LIMIT of but not including being at the event horizon.

Of course. However, he has to assume a simultaneity criterion, unless he returns to Earth. There's an obvious one (as long as he doesn't cross the horizon), which uses equal Schwarzschild coordinate time. But it's still an arbitrary choice.

 

[PeterDonis]

The (non-rotating) time dilation equation relates proper times between events for an observer within a gravitational field to coordinate time between those events for an observer at an arbitrarily large distance from the source of the gravitational field

Only outside the horizon, since the "time dilation equation" you refer to is only valid there.

 

[PeterDonis]

Is there a way to integrate the that time dilation equation (of relative time) over a range of distances so that say a physicist falling toward the black hole could deduce the TOTAL time which has passed back on Earth while he has fallen from a specific point toward the event horizon

As has been pointed out, answering this requires a choice of simultaneity convention (or, equivalently, a choice of coordinates). Another way to put this is that, as far as physics is concerned, there is no such thing as "now". This is true even in special relativity. If two observers are spatially separated, there is no invariant answer for either one of them to the question of what the other one is doing "now". The only questions that have invariant answers are questions like "if I send a light signal to the other observer at this instant by my clock, what will his clock read when he receives the signal?", or "what time did the other observer's clock read when he sent the light signal I am just receiving at this instant?"

 

[jbriggs444]

As has been pointed out, answering this requires a choice of simultaneity convention (or, equivalently, a choice of coordinates). Another way to put this is that, as far as physics is concerned, there is no such thing as "now". This is true even in special relativity. If two observers are spatially separated, there is no invariant answer for either one of them to the question of what the other one is doing "now". The only questions that have invariant answers are questions like "if I send a light signal to the other observer at this instant by my clock, what will his clock read when he receives the signal?", or "what time did the other observer's clock read when he sent the light signal I am just receiving at this instant?"

So you want to adopt a nice symmetric simultaneity convention. The same one in which gravitational time dilation is conventionally expressed. For instance, have a hovering outer observer send a light signal to a hovering inner observer and assign a simultaneity convention such that the travel times of the upward and downward signal are identical.

You then want to extend this simultaneity convention down toward the event horizon.

Sure. Unless I am mistaken, that yields Schwarzschild coordinates and the total elapsed time for an outside observer is infinite.

 

[ObjectivelyRational]

Ok wrong question again I am sorry I do not know all the issues.

Suppose instead the following variation.

An indestructible physicist is orbiting a small black hole at a sufficient distance such that he is subject to 1 G just like his colleagues back home on Earth. He has chosen to orbit this small black hole because it is on a collision course with a completely isolated supermassive black hole whose spherical event horizon is many orders of magnitude larger than his orbit, such that it is essentially flat, and parallel to his circular orbit about his small black hole. Here I assume from his perspective he and his small black hole fall into the supermassive black hole at the same rate/time. The physicist knowing the distance between his small black hole (which is also the plane in which he orbits) and the isolated supermassive black hole's essentially (locally) flat event horizon, and the distance between the supermassive black hole and Earth, could he deduce a total amount of time which passes on Earth before his colleagues "observe" the "black hole merger" with say LIGO?

If he can deduce such a thing could he split (perhaps as a ratio?) that total time passed on Earth into the time passed on Earth corresponding to the time it takes for the black holes to approach each other to merge and the time it takes for the space-time distortion (gravity waves... whatever LIGO observes) caused by the merge to reach Earth?

 

[PeterDonis]

have a hovering outer observer send a light signal to a hovering inner observer and assign a simultaneity convention such that the travel times of the upward and downward signal are identical.

That's one way to choose such a convention, yes. But it's not the only way, and it's only "symmetric" for observers that are at rest relative to each other. But they aren't in the OP's scenario; one observer is free-falling inward. The convention you describe is not "symmetric" for that case, at least not if "symmetric" means "works the same for both observers".

Unless I am mistaken, that yields Schwarzschild coordinates

You aren't mistaken; you do indeed get Schwarzschild coordinates from the convention you describe.

 

[PeterDonis]

could he deduce a total amount of time which passes on Earth before his colleagues "observe" the "black hole merger" with say LIGO?

Time which passes on Earth between what event on Earth and his colleagues observing the merger?

 

[PeterDonis]

If he can deduce such a thing could he split (perhaps as a ratio?) that total time passed on Earth into the time passed on Earth corresponding to the time it takes for the black holes to approach each other to merge and the time it takes for the space-time distortion (gravity waves... whatever LIGO observes) caused by the merge to reach Earth?

Only once he has chosen a simultaneity convention (or, equivalently, coordinates). Go back and read what I said in post #1`5 again and again until it sinks in. Asking the same question in different ways won't change the answer.

 

[Nugatory]

could he deduce a total amount of time which passes on Earth before his colleagues "observe" the "black hole merger" with say LIGO?

You've just moved the simultaneity convention problem from the end of the time interval in question to the beginning. There's a event on the worldline of the Earth: "LIGO detects merger", that's the end of the time interval you're asking about. You're asking what a clock on Earth would will read at that event if it is zeroed now - that's what "time elapsed on Earth" means..

But what point on the Earth's worldline are you calling "Now"?

 

[PAllen]

Also, please reread my post #5, which was my attempt to answer a similar question as well as physically possible, without any discussions of arbitrary things like distant simultaneity.

 

[ObjectivelyRational]

Thanks to all I think I have my answer.I was trying to resolve misleading "pop science" claims that to outside observers matter falling towards a black hole never actually crosses the event horizon of a black hole. This is an overstatement. Confirmed by the comments I have received here.

In fact we only never "see" it actually cross. Any EM signals from it become increasingly red shifted which we see until we simply don't see any more signals (too low frequency and/or no more signals arriving). The progression of the redshift as well as the time at which we see nothing more are in and of themselves an observation confirming that the thing has crossed the event horizon. Apparently pop science really adheres to "seeing is believing" quite literally, when it claims we could never observe something fall into a black hole.

It also is contrary to LIGO observations which ARE observations of black hole mergers. Those observations rely on disturbances in space time ... which travel at the universal constant speed of light ... from a particular general location at some particular time in the past (in our general rest frame).

Please correct me if I am mistaken, and if not, please accept my apologies for going about getting to this answer in what has turned out to be the worst way possible.

 

[ObjectivelyRational]

So I decided to watch "Nova: Black Hole Apocalypse" on Netflix.
At 9:20, sometime after the narrator says: "Approach one, and time itself begins to change", Neil deGrasse Tyson says:

"Time will become so slow for you that you will watch the entire future of the universe unfold before your very eyes."

From the responses of this thread I assume this statement (ENTIRE FUTURE) would only be true if the space traveler was able to (somehow) prevent himself from falling in, and remained (for a relatively long period of time) in a region of extreme time dilation. Correct?

On the contrary, if the space traveler were orbiting the small black hole as it merged with the super massive one (as in my hypothetical above) the traveler could not see "the entire future" before he fell in because the process would complete in finite time... LIGO has observed black holes merge... not black holes forever merging. Correct?

 

[PAllen]

So I decided to watch "Nova: Black Hole Apocalypse" on Netflix.
At 9:20, sometime after the narrator says: "Approach one, and time itself begins to change", Neil deGrasse Tyson says:

"Time will become so slow for you that you will watch the entire future of the universe unfold before your very eyes."

From the responses of this thread I assume this statement (ENTIRE FUTURE) would only be true if the space traveler was able to (somehow) prevent himself from falling in, and remained (for a relatively long period of time) in a region of extreme time dilation. Correct?

On the contrary, if the space traveler were orbiting the small black hole as it merged with the super massive one (as in my hypothetical above) the traveler could not see "the entire future" before he fell in because the process would complete in finite time... LIGO has observed black holes merge... not black holes forever merging. Correct?

Pretty much correct. In fact, the whole notion of a near horizon hovering test body should probably be avoided in discussing GR . The test body concept is useful to the extent it approximates a realizable physical situation. A purported test body hovering very near a BH horizon would need thrust of e.g. a billion g (and a supermassive black hole wouldn’t help with this, as it does for reducing tidal gravity). No such situation is realizable because putting a source for the thrust would radically change the solution. Thus in GR, a test body hovering near a horizon is essentially a self contradictory concept, similar to asking what would happen to earth’s orbit if the sun suddenly disappeared. Neither situation can be validly represented in GR.

 

[Chris Miller]

If you remain outside the hole, you can't observe things falling past the horizon directly, but you can detect the increase in the hole's mass as matter falls in.

Interesting. So am I seeing the effect (mass increase) prior to the cause (falling past event horizon, which I never see)?

 

[Ibix]

Interesting. So am I seeing the effect (mass increase) prior to the cause (falling past event horizon, which I never see)?

No, the additional gravitational field you feel is a result of the matter you can see. It doesn't need to fall past the horizon to affect you gravitationally.

 

[Nugatory]

Interesting. So am I seeing the effect (mass increase) prior to the cause (falling past event horizon, which I never see)?

The Schwarzschild solution describes the gravitational effects in the vacuum outside of a spherically symmetrical mass distribution. The mass of the black hole is not determined by what’s passed through the event horizon, it’s determined by everything that’s in that spherically symmetric mass distribution, whether it’s fallen through the horizon yet or not.

 

[PAllen]

No, the additional gravitational field you feel is a result of the matter you can see. It doesn't need to fall past the horizon to affect you gravitationally.

To add to this, as soon as infalling matter gets closer to the horizon than you are, you detect increased mass 'beneath you'. However, it will take some time for you (and partners you have arranged around the BH) to detect the mass as spherically symmetric again (ringdown and gravitational wave emission will occur before this). Once the BH settles, you will see an enlarged radius of black 'hole' against the lensed image of background stars. This radius will correspond to a BH with the increased mass (what's left after that carried away by gravitational radiation during the merge/ringdown phase). The relic ultra redshifted photons you detect, in principle, from the infalling matter after stabilization, will all appear to come from less than a Planck distance from the enlarged horizon, i.e. not in any sense an image of the infalling body as it was before.

 

[ObjectivelyRational]

The relic ultra redshifted photons you detect, in principle, from the infalling matter after stabilization, will all appear to come from less than a Planck distance from the enlarged horizon, i.e. not in any sense an image of the infalling body as it was before

Is it safe to say:

1. For any specific portion of infalling matter directly in line of sight between you and the cog of the bh, "after stabilization" no photons emitted directly towards you from that specific portion of the infalling matter (straight line path) can reach you (it being beyond the event horizon). (or better no photons can even be directed toward you... or there is NO towards you)

2. Any contribution to the gravity wave you detect caused by that specific portion of infalling matter during stabilization (crossing the event horizon), are received by you (say at LIGO... at the speed of light) after any photons are received by you (at LIGO... at the speed of light) emitted directly to you from that specific portion of the infalling matter (line of sight, not around the black hole etc) which were emitted prior (in the frame of reference of the specific portion) to stabilization (prior to it crossing the event horizon).

i.e. There is no inversion of observed timing of events in the direct line of sight to the cog of the black hole.

?

 

[PeterDonis]

the cog of the bh

There is no "center of gravity" of the black hole. The "center" of the hole at $r = 0$ is not a point in space; it's a moment of time.

 

[jbriggs444]

1. For any specific portion of infalling matter directly in line of sight between you and the cog of the bh, "after stabilization" no photons emitted directly towards you from that specific portion of the infalling matter (straight line path) can reach you (it being beyond the event horizon). (or better no photons can even be directed toward you... or there is NO towards you)

As a practical matter, this is a somewhat reasonable heuristic statement.

There is a last moment on your remote worldline after which the likelihood of receiving any further photons from the infalling object amounts to an impossibility. If you want to consider this the time after which the black hole has stabilized, nobody can stop you.

 

[ObjectivelyRational]

There is no "center of gravity" of the black hole. The "center" of the hole at $r = 0$ is not a point in space; it's a moment of time.

Ah.. then would it be better to say in a direction directly toward the center of the solid angle subtended by the event horizon, on the cosmic background, as it appears to someone at LIGO?

perhaps there's a better alternative?

 

[PeterDonis]

would it be better to say in a direction directly toward the center of the solid angle subtended by the event horizon, on the cosmic background, as it appears to someone at LIGO?

That's a reasonable way to think about it, yes. That spatial direction is well-defined, although the question of what, exactly, it is pointing to inside the hole's horizon is not.

 

[PAllen]

Is it safe to say:

1. For any specific portion of infalling matter directly in line of sight between you and the cog of the bh, "after stabilization" no photons emitted directly towards you from that specific portion of the infalling matter (straight line path) can reach you (it being beyond the event horizon). (or better no photons can even be directed toward you... or there is NO towards you)

2. Any contribution to the gravity wave you detect caused by that specific portion of infalling matter during stabilization (crossing the event horizon), are received by you (say at LIGO... at the speed of light) after any photons are received by you (at LIGO... at the speed of light) emitted directly to you from that specific portion of the infalling matter (line of sight, not around the black hole etc) which were emitted prior (in the frame of reference of the specific portion) to stabilization (prior to it crossing the event horizon).

i.e. There is no inversion of observed timing of events in the direct line of sight to the cog of the black hole.

?

This is not true, in theory. The correct description is subtle.

1) You will continue (in theory) to receive em signal from said infalling matter forever, long after the gravitational radiation from its absorption by the BH was received. In practice, you won't because no detector can detect an em signal with a wavelength of e.g. 1 light year (or photon with a frequency of 1 cycle per year).

2) But there is, even in principle, no inversion of timing, as you describe it.

The explanation is that this long delayed relic photon was emitted by some particle of infalling matter when it was a tiny fraction of a Planck length from the stabilized horizon. The gravitational radiation you observed earlier was generated by spacetime distortions further away from the horizon than this. All gravitational radiation you observe is generated by phenomena occurring outside the horizon. Since it was generated further away from the horizon than the photon in (1), you receive it earlier.