How Can We Observe Black Holes Growing?

[Nugatory]

What makes you think I'm asking about the global when?

We think that's what you're doing because it is what you are doing. Any statement about what the observer's clock reads WHEN the object passes through the event horizon carries a hidden and incorrect assumption about there being a some sort of global WHEN.

This is what I was getting at in post #13 above: How have you chosen the point on the observer's world line that happens at the same time that the world line of the infalling object intersects the event horizon (or the observer finds that the black hole has increased by the mass of t infalling object)?

 

[Dale]

What makes you think I'm asking about the global when?
I'm asking about when only with the observer system

Right here, for example:

How much will the remote observer's clock show when, according to the local clock, matter reaches the horizon?

This requires a coordinate system/ synchronization convention to answer. You seem to think that specifying an observer uniquely specifies a coordinate system and it does not. And even once you do specify, the answer is coordinate dependent, so it is math not physics

 

[Vanadium 50]

no calculations...How much will the remote observer's clock show when...

You want us to answer a quantitative question without calculations?

 

[PAllen]

Let's do it differently.
How much will the remote observer's clock show when, according to the local clock, matter reaches the horizon?

As has been said so many time, it depends on which events on the infaller's world line are considered to be the same time as on a stationary world line some distance from the horizon. Schwarzschild t coordinate as a definition simply doesn't work because it isn't defined at the horizon and it behaves like an axial distance rather than a timelike coordinate inside the horizon.

However, there are many well behaved coordinate systems, such that a line of constant time and angular coordinates represents a smooth spacelike path through the horizon and up to the singularity. Each such different choice of well behaved coordinates will give a different answer to the above question. However, for any particular choice of well behaved coordinates there is a definite answer. For example, in:

https://www.physicsforums.com/posts/6601010/

I discuss platform adapted Lemaitre style coordinates. If these are used to define 'when', then:

If you have clock on a platform hovering at r=2R, where R is the Schwarzschild radius, and r is the Schwarzschild radial coordinate, then the platform clock will read $R(\sqrt 2 + \pi/\sqrt 2)$ when the falling test body reaches the horizon, and it will read $R\pi\sqrt 2$ when the test body reaches the singularity (counting from the platform clock being zero when it drops a test body; R is in units of light seconds).

For a typical stellar BH, e.g. 8 solar mass, the time on a platform clock at twice the Schwarzschild radius r coordinate, from when it drops a probe to when the probe reaches the singularity, for this definition of “when”, is just 355 microseconds.

 

[mef]

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

 

[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.