How Can We Observe Black Holes Growing?

[PeroK]

Thank you all, but I would like to receive in response not reasoning, but CALCULATION...I think I can understand him

I'm happy to discuss the calculations in posts 75 and 77 here:

https://www.physicsforums.com/threa...into-a-black-hole.1012103/page-3#post-6599513

 

[Orodruin]

Thank you all, but I would like to receive in response not reasoning, but CALCULATION...I think I can understand him

Except that you cannot get a computation because your question is ill defined. As has been explained to you repeatedly, you have a significant amount of misconceptions that are mirrored in your questions. Until you resolve those and pose a properly defined question, all we can do is to point out that you in essence have asked what to do when the traffic light shows blue.

 

[Dale]

Thank you all, but I would like to receive in response not reasoning, but CALCULATION...I think I can understand him

Before you can get a calculation you need to specify the problem sufficiently that it can be calculated. You have not yet done that. No calculation is possible until you do.

 

[Orodruin]

Here is something to illustrate some of the ambiguities that you are facing without actually having the complications of Schwarzschild spacetime:

Consider the spacetime with the metric given by
$$
ds^2 = \alpha^2 x^2 dt^2 - dx^2.
$$
Consider a stationary observer in this spacetime (i.e., at constant ##x = x_0##) looking at a free-falling object released at time ##t=0##. The world line of such an object (parametrized by its proper time) is given by
$$
t = \frac 1\alpha \tanh^{-1}(s/x_0), \quad x = \sqrt{x_0^2 -s^2}.
$$
The coordinate ##x = 0## in this spacetime represents a horizon. Once the object reaches that horizon, no information from it will reach the stationary observer at ##x = x_0##, light emitted by the object will become increasingly redshifted when reaching the distant observer as the horizon is approached. The crossing occurs at infinite coordinate time ##t## as the inverse hyperbolic tangent approaches infinity as the argument approaches one, but the crossing occurs at finite proper time ##s=x_0##. I now ask you the same type of question you have been asking in this thread. What is the time a distant observer measures when the object crosses the horizon? Does the object really cross the horizon (time ##t## is infinite when this would happen)?

 

[sha1000]

I would like to join the question since I really can not understand how can reality differ from one observer to another.

Let's say that the distance between the stationary observer (A) and BH is R (Schawrzchild metric). And at t=0 observer (A) drops an apple (B).

At t = 0 both clocks (A and B) are synchronized. Then apple starts to fall.
As far as I understand there must be equations which permit to observer (A) calculate the time at apples clock (B). And vice versa.

So from the point of view of (A) apple never reaches the horizon. Not only, because he can't measure it but because this is his reality. Because he learned at school that everything "freeze" at the horizon

But you all are saying that in the apples reference frame (B) it does actually cross the horizon. But how so? Maybe I'm wrong, but something tells me that if we take time measured at (B) the moment it crosses the horizon; and from that we will try to calculate time at (A) (using initial distance R and all our knowledge about GR), we will get some infinite value. But observer (A) does need finite time to observe the growth of the BH.

 

[PeterDonis]

I really can not understand how can reality differ from one observer to another.

Reality does not differ from one observer to another.

But what part of reality a given observer can see can differ from one observer to another. The observer that stays outside the hole cannot see the part of reality that is at or below the hole's horizon. But that doesn't mean that part of reality isn't there. It just means the outside observer can't see it.

 

[Orodruin]

Let's say that the distance between the stationary observer (A) and BH is R (Schawrzchild metric). And at t=0 observer (clock A) drops an apple (B).

As has been indicated in this thread already, things are not this simple. First of all the Schwarzschild r coordinate is not a distance from the singularity of the black hole. It is related to the area of the sphere through ##A=4\pi r^2##. There is no well-defined distance to the singulary.

As far as I understand there must be equations which permit to observer (A) calculate the time at apples clock (B). And vice versa.

What time? Simultaneity is not well defined for objects that are not colocated so you need to be more precise for your assertion to be defined.

So from the point of view of (A) apple never reaches the horizon.

You need to be more careful with your statements. It is not clear what you mean by this.

because this is his reality

What do you mean by ”reality”?

Because he learned at school that everything "freeze" at the horizon

I certainly hope he did not learn that in school. If he did he should have gone to a better school.

But you all are saying that in the apples reference frame (B) it does actually cross the horizon.

You should drop the ”reference frame” from this statement. Anything that happens in relativity happens in all coordinate systems that appropriately cover the relevant part of spacetime. There can be no disambiguity between different coordinates. Only coordinates that cover or don’t cover the relevant events.

Maybe I'm wrong but something tells me that if we take time measured at (B) and from that we will try to calculate time at (A), we will get some infinite value.

As already stated, you cannot do this without assuming an arbitrary unphysical simultaneity convention.

 

[sha1000]

Reality does not differ from one observer to another.

But what part of reality a given observer can see can differ from one observer to another. The observer that stays outside the hole cannot see the part of reality that is at or below the hole's horizon. But that doesn't mean that part of reality isn't there. It just means the outside observer can't see it.

But this observer knows that it takes infinite time for an object to reach the horizon (from his point of view). He does not need to see it.

 

[sha1000]

As has been indicated in this thread already, things are not this simple. First of all the Schwarzschild r coordinate is not a distance from the singularity of the black hole. It is related to the area of the sphere through ##A=4\pi r^2##. There is no well-defined distance to the singulary.What time? Simultaneity is not well defined for objects that are not colocated so you need to be more precise for your assertion to be defined.You need to be more careful with your statements. It is not clear what you mean by this.What do you mean by ”reality”?I certainly hope he did not learn that in school. If he did he should have gone to a better school.You should drop the ”reference frame” from this statement. Anything that happens in relativity happens in all coordinate systems that appropriately cover the relevant part of spacetime. There can be no disambiguity between different coordinates. Only coordinates that cover or don’t cover the relevant events.As already stated, you cannot do this without assuming an arbitrary unphysical simultaneity convention.

Thank you for your response.

Indeed I always thought that times slows down near horizon and eventually reaches 0 (from the perspective of the external observer). Is it wrong?

 

[Orodruin]

Thank you for your response.

Indeed I always thought that times slows down near horizon and eventually reaches 0 (from the perspective of the external observer). Is it wrong?

This sounds like something from popular science. Popular scientific texts are not written to be unambiguous or to teach you science. It is not directly wrong but you do need to interpret it in a particular way and not overstep that interpretation.

More accurate would be to say that the observer has no way of uniquely defining the time on the clock of the infalling object simultaneous to some time on their own clock. All you can do is to compute the redshift of signals emitted by the object. This will make the object appear to have time running slower, but this is also true for moving objects in special relativity and for the case I presented in #39, which is still for the OP to handle.

 

[Orodruin]

But this observer knows that it takes infinite time for an object to reach the horizon (from his point of view). He does not need to see it.

No. This is based on a particular simultaneity convention based on Schwarzschild coordinates, which are not valid at the horizon. They are in no way or form equivalent to the observer’s point of view, which can only be based on measurements made locally.

 

[Dale]

But this observer knows that it takes infinite time for an object to reach the horizon (from his point of view).

This “infinite time” is not part of “reality”. It is simply a coordinate system where one of the labels goes to infinity. “Reality” doesn’t care about the labels you use.

Indeed I always thought that times slows down near horizon and eventually reaches 0 (from the perspective of the external observer). Is it wrong?

As a physical statement, yes it is wrong. A physical statement would be about proper time, which does not stop at the horizon. As a statement about Schwarzschild coordinate time it is correct, but not physical.

 

[sha1000]

This sounds like something from popular science. Popular scientific texts are not written to be unambiguous or to teach you science. It is not directly wrong but you do need to interpret it in a particular way and not overstep that interpretation.

More accurate would be to say that the observer has no way of uniquely defining the time on the clock of the infalling object simultaneous to some time on their own clock. All you can do is to compute the redshift of signals emitted by the object. This will make the object appear to have time running slower, but this is also true for moving objects in special relativity and for the case I presented in #39, which is still for the OP to handle.

Thank you again.

Ok. Let's reason this out by emission of signals. Again, we have stationary observer (A) placed at some distance R from the horizon.

His friend (B) decides to fly on the spaceship towards the BH. He takes with him a laser attached to a clock which emits a signal towards (A) every 10 seconds.

So our observer (A) starts to analyse signals. From his point of view the interval between signals grows as spaceship approaches the BH. At some point the interval between two signals will reach billion years, and then 100 of billions years but it will never actually disappear completely. So from his perspective the spaceship "never" reaches the horizon since he is still getting signals (even though the intervals are infinitely large).

But, if from the point of view of the spaceship it actually crosses the horizon (at some point). The signals will disappear.

If one can explain me this contradiction in terms of the signals I will then understand what you are trying to say.

 

[Orodruin]

Again, we have stationary observer (A) placed at some distance R from the horizon.

Again, R is not what you think it is.

So our observer (A) starts to analyse signals. From his point of view the interval between signals grows as spaceship approaches the BH. At some point the interval between two signals will reach billion years, and then 100 of billions years but it will never actually disappear completely. So from his perspective the spaceship "never" reaches the horizon since he is still getting signals (even though the intervals are infinitely large).

A will never see the spaceship cross the horizon no. Simply because A cannot see inside the horizon by definition of the horizon. This is also true about the situation in post #39.

As Peter said, just because A cannot see it happen does not mean it doesn’t happen.

But, if from the point of view of the spaceship it actually crosses the horizon (at some point).

Yes. The spaceship will cross the horizon (and hit the singularity) in finite proper time.

The signals will disappear.

What signals? The signals are sent to A, not to B.

 

[Dale]

So from his perspective the spaceship "never" reaches the horizon since he is still getting signals (even though the intervals are infinitely large).

The outside observers never receive a signal from the horizon crossing. This is not the same as the horizon crossing not happening. Many things happen that specific observers do not receive signals from.

You can talk about “perspectives” or you can talk about “reality”, but they are not the same thing. I would recommend not getting too caught up in perspectives.

 

[sha1000]

Again, R is not what you think it is.A will never see the spaceship cross the horizon no. Simply because A cannot see inside the horizon by definition of the horizon. This is also true about the situation in post #39.

As Peter said, just because A cannot see it happen does not mean it doesn’t happen.Yes. The spaceship will cross the horizon (and hit the singularity) in finite proper time.What signals? The signals are sent to A, not to B.

Indeed A is no longer observing the spaceship. All he observers are the signals which are emitted by the spaceship.

And what he can see is that the interval between those signals is increasing and at some point becomes infinitely large, but never completely disappears.

 

[Orodruin]

Indeed A is no longer observing the spaceship. All he observers are the signals which are emitted by the spaceship.

Huh? What would be the difference between these two?

 

[sha1000]

The outside observers never receive a signal from the horizon crossing. This is not the same as the horizon crossing not happening. Many things happen that I do not receive signals from.

So if I understand correctly the observer A will stop to receive the signals. And from this point can he conclude that his friend crossed the horizon?

 

[Dale]

So if I understand correctly the observer A will stop to receive the signals. And from this point can he conclude that his friend crossed the horizon?

He can conclude that his friend crossed the horizon well before that.

 

[Orodruin]

So if I understand correctly the observer A will stop to receive the signals. And from this point can he conclude that his friend crossed the horizon?

No. But he will never receive any signal from the horizon crossing time or later as based on the infalling observer’s clock.

I believe this has now taken a significant detour even if it is still mainly on topic… maybe it should be moved to a separate thread? OP should go back to post #39 and examine it closely.

 

[sha1000]

He can conclude that his friend crossed the horizon well before that.

Thank you. I wanted to know if GR tells us that there is a way for the external observer to conclude when object crosses the horizon. This is the answer I was looking for.

 

[sha1000]

No. But he will never receive any signal from the horizon crossing time or later as based on the infalling observer’s clock.

I believe this has now taken a significant detour even if it is still mainly on topic… maybe it should be moved to a separate thread? OP should go back to post #39 and examine it closely.

Yep. I think I get it better now. Thanks

 

[Orodruin]

Thank you. I wanted to know if GR tells us that there is a way for the external observer to conclude when object crosses the horizon. This is the answer I was looking for.

Careful with your statements. There is no ”when” an object passes the horizon for a distant observer as that would require an unphysical simultaneity convention.

 

[Dale]

Thank you. I wanted to know if GR tells us that there is a way for the external observer to conclude when object crosses the horizon. This is the answer I was looking for.

Careful. “When” something distant happens is a matter of coordinates, not physics.

 

[PeterDonis]

this observer knows that it takes infinite time for an object to reach the horizon (from his point of view).

For an appropriate definition of "time" and "his point of view", yes. But that doesn't mean what you think it means. See below.

He does not need to see it.

This statement is literally correct, but not in the way you think. The outside observer can calculate that the infalling object takes only a finite time by its clock to reach the horizon, and that it continues on inward. So indeed he does not need to see these things happen, to know that they do in fact happen.

The mistake you are making is to think that one particular aspect of the outside observer's "point of view"--the fact that his time coordinate goes to infinity at the horizon--is telling him something about the physics of the infalling object. It's not: it's telling him about the increasing distortion of his "point of view" (more precisely, of Schwarzschild coordinates) as the horizon is approached, culminating in "infinite distortion" at the horizon. It's as if the outside observer concluded, from the fact that a Mercator projection of the Earth's surface assigns an infinite coordinate to the North and South poles, that the North and South poles cannot exist. The problem is in the coordinates.

 

[PAllen]

Careful with your statements. There is no ”when” an object passes the horizon for a distant observer as that would require an unphysical simultaneity convention.

Hmm, convention yes. Unphysical is a matter of taste. I would call simultaneity defined from a family of free falling observers in a manner equivalent to how standard cosmological simultaneity is defined, a physically plausible convention. There is no reason a stationary observer is not allowed to adopt it, and then be able to state, per this convention, exactly when some body crosses the horizon.

 

[Orodruin]

physically plausible convention.

That does not make it physical. It is also not necessary to compute the time on the infalling observer’s clock as it passes the horizon. Furthermore, the Schwarzschild t coordinate is arguably more ”physically plausible” and gives a different answer.

 

[PAllen]

That does not make it physical. It is also not necessary to compute the time on the infalling observer’s clock as it passes the horizon. Furthermore, the Schwarzschild t coordinate is arguably more ”physically plausible” and gives a different answer.

Point remains there is no objective definition of physically plausible. It is a matter of aesthetics. To me, there are many physically plausible conventions possible. And if you want to talk about horizon crossing events in relation to an external stationary observer, you pick one that reaches the horizon.

 

[Orodruin]

Point remains there is no objective definition of physically plausible. It is a matter of aesthetics. To me, there are many physically plausible conventions possible. And if you want to talk about horizon crossing events in relation to an external stationary observer, you pick one that reaches the horizon.

But this is precisely the point. You can obtain any answer you want just by choosing coordinates. This makes the "what is the time for the external observer when the object crosses the horizon" an unphysical one. Sure, you can compute it based on some simultaneity convention, but really, why bother? The same thing goes for cosmology. You can compute the distance to an object "now" (according to standard cosmological choice of FLRW coordinates) but that really holds very little physical significance and it cannot be measured anyway.

 

[PAllen]

But this is precisely the point. You can obtain any answer you want just by choosing coordinates. This makes the "what is the time for the external observer when the object crosses the horizon" an unphysical one. Sure, you can compute it based on some simultaneity convention, but really, why bother? The same thing goes for cosmology. You can compute the distance to an object "now" (according to standard cosmological choice of FLRW coordinates) but that really holds very little physical significance and it cannot be measured anyway.

Of course, there is the invariant transition, for any external observer, of when no part of dropped probe's history is in its causal future anymore. That is, the time after which any light signal sent by the external observer will reach the singularity after the probe has already reached it.

For an 8 solar mass BH, and an external observer hovering at twice the SC radius, this transition is just 242 microseconds after dropping a probe. Any time after this (but not before) it is possible to consider that the probe has reached the singularity.

 

[Dale]

the time after which any light signal sent by the external observer will reach the singularity after the probe has already reached it.

That is a good criterion. Of course, it cannot be measured, just predicted. But with knowledge of the maximum thrust of the vehicle, you could place an upper bound on it.

 

[PAllen]

But this is precisely the point. You can obtain any answer you want just by choosing coordinates. This makes the "what is the time for the external observer when the object crosses the horizon" an unphysical one. Sure, you can compute it based on some simultaneity convention, but really, why bother? The same thing goes for cosmology. You can compute the distance to an object "now" (according to standard cosmological choice of FLRW coordinates) but that really holds very little physical significance and it cannot be measured anyway.

I find I still rebel against the notion that "not invariant" must mean "not physical". IMO, length contraction and time dilation are frame variant physical phenomena.

In the current context, consider yet another example of IMO physically plausible, coordinate dependent ways to state that an object has crossed the horizon at some time for a distant observer:

In GR, while there are no global inertial frames, there are local inertial frames. Then one can extent these to large scale coordinates by geodesically extending the basis (the result is Riemann Normal coordinates defined by a 4-velocity at an event). Then it is true that for such a coordinate system build from a momentarily stationary observers, such an extended coordinate system (though different from standard Schwarzschild coordinates) will never include a horizon crossing event. However, for an observer a light year from a stellar BH, a year after it formed, such coordinates built for an observer moving just 1 meter/second toward the BH, the horizon and singularity will included in the present. Thus, claims that the horizon never forms for distant observers are patently false, unless you somehow want to exclude any observer not completely 'stationary'.

 

[PeterDonis]

I find I still rebel against the notion that "not invariant" must mean "not physical". IMO, length contraction and time dilation are frame variant physical phenomena.

I have to disagree with this. I would say that "length contraction" and "time dilation", if they are to be considered physical phenomena, can and must be defined as invariants. An observer in a particular state of motion can measure invariants that can be appropriately described as the "length contracted length" of an object moving relative to that observer, or the "time dilated clock rate" of a clock moving relative to that observer. Those invariants are unproblematically physical things, and, with proper definition, will in fact correspond to what are usually called the "frame variant" quantities "length contraction" and "time dilation". And with a scenario properly set up along such lines, one can usefully discuss "length contraction" and "time dilation" in terms of invariants.

However, many discussions do not properly put into place or consider the background that needs to be set up to define such concepts in terms of invariants (and many participants in such discussions have no idea either that such a background can be put in place or that it needs to be). And not all coordinate-dependent quantities can even be given a useful correspondence with an invariant.

 

[PeterDonis]

one can extent these to large scale coordinates by geodesically extending the basis (the result is Riemann Normal coordinates defined by a 4-velocity at an event).

I assume you are allowing the metric in terms of such "extended" coordinates to vary from the Minkowski metric? Otherwise this construction will not work as soon as tidal gravity effects become non-negligible.

Also, even allowing the metric to vary, such a coordinate patch can only be extended through a region in which none of the geodesics being used cross. Perhaps this is what you have in mind when you say that no such coordinate patch constructed from the 4-velocity of a static observer will include a horizon crossing event (since radial spacelike geodesics in any such construction would cross at a point on the horizon).

 

[Orodruin]

IMO, length contraction and time dilation are frame variant physical phenomena.

These are concepts that I find confuse students for no real reason as they start obsessing over why the rod does not break in frames other than its rest frame from being pushed together or run in loops around the twin paradox. A bit similar to how we tend to not use relativistic mass.
Are they physical? It depends, as long as there is an operational way to measure them that gives the same result regardless of frame they will be, but they are always accompanied by the caveat of ”as measured by X” or something to that effect. We do not ascribe the measurement of some random observer as the length of an object (although we do define rest length as the length measured in the object’s rest frame).

For the horizon crossing, the same is true, you cannot define the time of horizon crossing from picking a random coordinate system. Only a time in that system, which to my understanding does not really solve the OP’s problem - which partially is wanting to use Schwarzschild t as the time coordinate.

Then one can extent these to large scale coordinates by geodesically extending the basis (the result is Riemann Normal coordinates defined by a 4-velocity at an event).

There generally is no guarantee that such a construction will actually cover the part of spacetime we are interested in. If they did we would not need to worry about using local coordinates at all. You could even end up in a situation where the simultaneities cross, making it possible for events with timelike separation to be time reveresed in the frame and I don’t know what to call that if not unphysical.