B How Can We Observe Black Holes Growing?

[mef]

TL;DR Summary

A very simple question

IT IS WIDELY KNOWN that from the point of view of a distant, BUT FROM a FINITE distance, the time of falling into a black hole is INFINITE.

Then how does the BH absorb the surrounding matter?

It turns out that for any observer located at a finite distance R (the coordinate of the Schwarzschild metric) from the BH, no matter will ever reach the BH and it will never be able to absorb anything and increase in size.

 

[Dale]

Summary:: A very simple question

IT IS WIDELY KNOWN that from the point of view of a distant, BUT FROM a FINITE distance, the time of falling into a black hole is INFINITE.

Please don’t RANDOMLY SHOUT at us. It is rude and distracting

Then how does the BH absorb the surrounding matter?

In order to fall in, matter must lose its angular momentum. Typically this happens in an accretion disk, where there are lots of collisions that can redistribute angular momentum and energy so as to allow particles to fall in.

It turns out that for any observer located at a finite distance R (the coordinate of the Schwarzschild metric) from the BH, no matter will ever reach the BH and it will never be able to absorb anything and increase in size.

It turns out that physics is about more than the Schwarzschild coordinates.

 

[mef]

You are answering a completely different question.
What happens to particles with zero angular momentum (radial)?

 

[PeterDonis]

What happens to particles with zero angular momentum (radial)?

They fall in. They reach the horizon in a finite time by their own clock (and hit the singularity a finite time by their own clock after that).

Your misconception is a common one. I suggest reading this series of Insights articles:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

 

[mef]

Do you not understand what I wrote about the view, from the point of view of an outsider?

 

[Nugatory]

Do you not understand what I wrote about the view, from the point of view of an outsider?

We understand what you are asking, truly we do!

The “point of view of an outsider” is based on light from the infalling object reaching their eyes. As the object approaches the horizon it takes longer and longer for that light to reach the outsider and light from the horizon crossing never gets out at all; that’s what it means to say that the object gets closer to the horizon but never crosses it.

However, if we start with an object of mass $m$ and drop it into a black hole of mass $M$, we find that very quickly we have a black hole of mass $M+m$ - it’s just that we don’t get to see the object crossing the horizon.

Be aware that this answer contains some serious oversimplifications; the links posted by @PeterDonis would be a good start if you want a better understanding. There are also many older threads here as your question is a common one; the list of “Related Threads” below may be helpful, as well as https://www.physicsforums.com/threa...ching-a-bh-and-bh-growth.921644/#post-5814635

 

[PeroK]

Do you not understand what I wrote about the view, from the point of view of an outsider?

There are a lot of misconceptions online about an object falling into a black hole. These sources generally over-emphasise the role of a distant observer and what they see in terms of the light signals that reach them.

The argument they present, which you are repeating, is essentially that the object takes "infinite time" to reach the event horizon. This is false. There are other misconceptions, such as that the infalling object sees the whole history of the outside universe. That is also false.

One key point is that in the curved spacetime of a black hole, not all events may be observed by all observers. That the distant observer never observes an event does not mean that the event does not take place.

 

[Orodruin]

Do you not understand what I wrote about the view, from the point of view of an outsider?

Please understand that the gap of knowledge here is a big one. You are dealing with people who are very well versed in relativity and also very used to seeing precisely the type of misconceptions that you present. Instead of being dismissive and ungrateful to their replies because they do not fit with your own preconceptions, you should try to understand them.

 

[mef]

Please understand that the gap of knowledge here is a big one. You are dealing with people who are very well versed in relativity and also very used to seeing precisely the type of misconceptions that you present. Instead of being dismissive and ungrateful to their replies because they do not fit with your own preconceptions, you should try to understand them.

In addition to general words, no calculations.
OK.
Let's do it differently.
How much will the remote observer's clock show when, according to the local clock, matter reaches the horizon?

 

[mef]

There are a lot of misconceptions online about an object falling into a black hole. These sources generally over-emphasise the role of a distant observer and what they see in terms of the light signals that reach them.

The argument they present, which you are repeating, is essentially that the object takes "infinite time" to reach the event horizon. This is false. There are other misconceptions, such as that the infalling object sees the whole history of the outside universe. That is also false.

One key point is that in the curved spacetime of a black hole, not all events may be observed by all observers. That the distant observer never observes an event does not mean that the event does not take place.

I'm setting a very simple task... so that it becomes completely clear to you. How long does it take for a Newtonian apple to move from Earth to a black hole in the center of our galaxy.
From Newton's point of view

 

[Orodruin]

In addition to general words, no calculations.
OK.
Let's do it differently.
How much will the remote observer's clock show when, according to the local clock, matter reaches the horizon?

You have another misconception here: The idea that a global ”when”. The idea that there is a unique time of crossing the horizon as seen by a distant observer is a false one.

 

[PeroK]

I'm setting a very simple task... so that it becomes completely clear to you. How long does it take for a Newtonian apple to move from Earth to a black hole in the center of our galaxy.
From Newton's point of view

Newtonian gravity does not apply to black hole event horizons. As far as the apple is concerned, it reaches the event horizon in a finite time.

 

[Nugatory]

Let's write down the SPECIFIC time of the apple's fall from the Earth to the horizon of the black hole in the center of our galaxy. Take some mass of this BH. The attraction of the Earth and other gravity in the galaxy should be ignored.

By the “specific time” do you mean the proper time along the the infalling object’s worldline - that is, the time measured on the infaller’s clock between when it is released and it reaches the horizon?

If you mean something else, you will have to be more precise about exactly what the end points of the interval we’re measuring are. That’s what @Orodruin is getting at when he reminds you that there is no global “when” in curved spacetime.

 

[PeroK]

Newtonian gravity does not apply to black hole event horizons. As far as the apple is concerned, it reaches the event horizon in a finite time.

PS in terms of GR, there is a thread here about the time to fall into a black hole:

https://www.physicsforums.com/threads/how-long-does-it-take-to-fall-into-a-black-hole.1000235/

 

[Ibix]

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

Depends on how the remote observer defines "when". You are thinking three dimensionally, where you have little choice about what "when" means. In four dimensions, you have a choice. You can pick "when" so that the matter never crosses the horizon, or "when" so that it crosses almost as soon as it is released. It's a choice - not of physics, but of what you mean by the word "when". There's no universal meaning to it.

 

[mef]

Newtonian gravity does not apply to black hole event horizons. As far as the apple is concerned, it reaches the event horizon in a finite time.

I didn't talk about Newtonian gravity, only about his apple))

 

[mef]

You have another misconception here: The idea that a global ”when”. The idea that there is a unique time of crossing the horizon as seen by a distant observer is a false one.

there are strict initial conditions , but there is no definite answer? And then why do we need such a theory?)) How will you test students on the exam? He will write ANY answer... and he will ...this can also be))

maybe you have in terms of quantum mechanics? with such a probability, such an answer, with another probability, another answer))

 

[Ibix]

there are strict initial conditions

No there aren't. You keep failing to define "when". If you do that, there is an answer.

 

[mef]

No there aren't. You keep failing to define "when". If you do that, there is an answer.

t=0 start))

 

[PeroK]

I didn't talk about Newtonian gravity, only about his apple))

The GR calculations are in the thread I gave a link to.

 

[Ibix]

t=0 start))

You mean the Schwarzschild t coordinate? That isn't defined on the horizon so your question has no answer in those terms.

 

[PeroK]

t=0 start))

There is no "universal time in GR". Again, what you are syaing is: IF we assume universal time, then a black hole makes no sense. Which is true. Universal time (the same $t$ for everything) is not part of the theory of relativity.

 

[Orodruin]

there are strict initial conditions , but there is no definite answer? And then why do we need such a theory?)) How will you test students on the exam? He will write ANY answer... and he will ...this can also be))

No. You persist in applying your own misconceptions without any regards to being told that your question as posed is not well defined according to the theory. Obviously such a question would not be asked in an exam precisely because it is not a good question to ask.

 

[weirdoguy]

And then why do we need such a theory?

Let me guess - you didn't come here to learn, did you? You came here to force upon us all of your misconceptions just to tell us that GR is not correct.

 

[Dale]

You are answering a completely different question.
What happens to particles with zero angular momentum (radial)?

They fall in, just like particles with zero angular momentum about any spherical mass.

Do you not understand what I wrote about the view, from the point of view of an outsider?

Why should the particle care about the point of view of an outsider?

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

This is not a physical question, it is a mathematical question. And even as a mathematical question it is not fully specified since you have not indicated which coordinates you want to use to determine “when” things happen. If you use mathematically bad coordinates you will get a bad mathematical answer.

I'm setting a very simple task... so that it becomes completely clear to you. How long does it take for a Newtonian apple to move from Earth to a black hole in the center of our galaxy.

It is not a simple task because it is incompletely specified. You need to specify the coordinates. And even once you do so, this remains a mathematical exercise rather than a physical one

but there is no definite answer?

That is correct. You cannot have a definite answer to an indefinite question

 

[mef]

You mean the Schwarzschild t coordinate? That isn't defined on the horizon so your question has no answer in those terms.

This is defined at a finite point R (observer coordinate)
On the horizon, time is defined at least as a limit, just like in mathematical analysis. At the same time, some limits may diverge to infinity, but they exist!

but there are limits that do not exist, for example, sin(1/x) for x -> 0

 

[mef]

You have another misconception here: The idea that a global ”when”. The idea that there is a unique time of crossing the horizon as seen by a distant observer is a false one.

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

 

[PeroK]

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

Because he's taught relativity to hundreds of students and can see the mistakes you're making.

The only question for you is whether you have the desire to learn GR, which is hard work. Or, whether you prefer not to learn it, which is very easy to do.

You quoted some Real Analysis. 99.9% of the world's population do not understand real analysis and why $\sin \frac 1 x$ does not converge, as $x \rightarrow 0$. Real analysis is not easy to learn and many people think it's just some rubbish that mathematicians invented. It's easy for them to trash mathematics rather than learn it.

Ultimately, we've all put in the months of hard study to learn GR - and overcome the difficulties in understanding thinks like coordinate-dependence and invariance.

We can't force you to learn GR. But, you can't persuade us it's all wrong any more than you could persuade us that mathematics is all wrong. Because we've learned and understood it.

 

[Orodruin]

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

Because you are trying to construct a ”when” including both a stationary observer and an object falling through the event horizon by asking ”when” for a distant observer does an object pass the horizon. There is no such construction. The time coordinate t is just a coordinate. It does not define a universal sense of simultaneity. That no such concept exists should be clear already from special relativity. You yourself is the one standing in your own way by asserting that your ill defined questions are well defined.

On the horizon, time is defined at least as a limit, just like in mathematical analysis.

No, it is not and repeating the statement is not going to make it so. Time is undefined on the horizon because regular Schwarzschild coordinates are singular there. This also underlines that coordinate time is an exceedingly bad choice to define global simultaneity. It is somewhat useful for that outside the horizon due to its role as parametrising a timelike Killing flow. This however ends on the horizon.

 

[Orodruin]

and can see the mistakes you're making.

The hardest part of learning relativity is arguably letting go of hard ingrained preconceptions that are no longer valid.

 

[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.