Falling into a black hole

[PeterDonis]

Another possibility is (my favorite) that an evaporating “black” hole does not have a horizon at all.

This is the possibility I described at the end of post #10. However, I don't think that spacetime geometry (in which there is no event horizon and no singularity) is what the OP wanted to discuss.

 

[A.T.]

For that particular question the video below by our member @A.T. may be even more helpful.

@martix That video is based on a book, which also visualizes the finite proper-time vs infinite coordinate-time issue, for falling into a black hole:

https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/page/n201/mode/2up

But you should have a look at the previous chapters to understand those diagrams intuitively:

https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/page/n149/mode/2up

 

[PeterDonis]

This is not necessarily true.

The use of the term "coordinate time" may be confusing you. Your claim can be rephrased in coordinate-independent terms as follows: a distant observer will continue to see light signals from someone who fell into the evaporating black hole, showing them getting closer and closer to the horizon, even after the distant observer has seen the black hole evaporate.

And rephrased that way, the claim can be seen to be obviously false just by looking at the Penrose diagram of Hawking's original model. You don't even need to know the exact metric; the Penrose diagram, showing the causal structure, is enough.

 

[PeterDonis]

Recently, there was a very interesting discussion in another thread about an article
https://arxiv.org/abs/1102.2609
which claims (see fig.3) that infaling do not cross the horizon at all.

That thread is here:

https://www.physicsforums.com/threa...horizon-of-an-evaporating-black-hole.1062617/

I do not think the claim made in Fig. 3 of that paper is valid, for reasons I gave in the thread.

 

[Tomas Vencl]

Your claim can be rephrased in coordinate-independent terms as follows: a distant observer will continue to see light signals from someone who fell into the evaporating black hole, showing them getting closer and closer to the horizon, even after the distant observer has seen the black hole evaporate.

I do not think so, my claim:

This is not necessarily true. …

was reaction on this comment:

Specifically, in an evaporating spacetime it does not take an infinite amount of coordinate time to cross the horizon in Schwarzschild-like coordinates

I think that my claim can be rephrased for example that there is no horizon to cross (so there is no finite time to cross it, obviously), or, as another example, in some models with firewall also falling observer do not cross the horizon (similarly to mentioned fig.3 at arxiv article).

 

[Dale]

May be, but fact is that exact metric for evaporating “black hole” is unknown, so claiming that “in an evaporating spacetime it does not take an infinite amount of coordinate time to cross the horizon in Schwarzschild-like coordinates” is at least based on aproximations and rather speculative.

It is true that I don't know the exact metric for an evaporating black hole. However, there are a few things that we can indeed say based on first principles.

Since it is a black hole there is an event horizon where timelike worldlines can enter but not exit, this is what defines a black hole instead of a white hole. When evaporation is finished there is no more event horizon, this is what defines the evaporation. Without loss of generality, assign the time of the evaporation event to be $t_E$. Meaning, after $t_E$ there is no event horizon. All coordinate charts, by definition, are smooth and one-to-one, including the distant observer's Schwarzschild-like chart. So the crossing of the event horizon cannot be assigned a time coordinate $t_C>t_E$. Therefore, indeed, it does not take an infinite amount of coordinate time to cross the horizon.

 

[Tomas Vencl]

It is true that I don't know the exact metric for an evaporating black hole. However, there are a few things that we can indeed say based on first principles.

Since it is a black hole there is an event horizon where timelike worldlines can enter but not exit, this is what defines a black hole instead of a white hole. When evaporation is finished there is no more event horizon, this is what defines the evaporation. Without loss of generality, assign the time of the evaporation event to be $t_E$. Meaning, after $t_E$ there is no event horizon. All coordinate charts, by definition, are smooth and one-to-one, including the distant observer's Schwarzschild-like chart. So the crossing of the event horizon cannot be assigned a time coordinate $t_C>t_E$. Therefore, indeed, it does not take an infinite amount of coordinate time to cross the horizon.

Sure, but you are not sure (since we do not have exact metric) that your assumptions (event horizon + evaporation) are even together physically possible . So this could be similar like counting of number of angels on tip of the needle.
I can imagine that in case of evaporation instead of real black hole there can be different object (not so black hole :-) from distant perspective indistinguishable (but yes, this is speculation already).

 

[Dale]

Sure, but you are not sure (since we do not have exact metric) that your assumptions (event horizon + evaporation) are even together physically possible . So this could be similar like counting of number of angels on tip of the needle.
I can imagine that in case of evaporation instead of real black hole there can be different object (not so black hole :-) from distant perspective indistinguishable (but yes, this is speculation already).

They are not my assumptions, they are the assumptions of the question.

 

[Tomas Vencl]

They are not my assumptions, they are the assumptions of the question.

OK, perhaps it would be fair to also inform the inquirer that the assumptions may not be realistic (which I clumsily tried without any successs to do in my second comment).

 

[PeterDonis]

I think that my claim can be rephrased for example that there is no horizon to cross

Then you are not talking about Hawking's original model, or indeed any model with an event horizon. You are talking about a model like the Bardeen black hole that I referenced previously. But, as I have already said, I don't think that kind of model is what the OP of this thread is asking about.

 

[PeterDonis]

in some models with firewall also falling observer do not cross the horizon (similarly to mentioned fig.3 at arxiv article).

As I said in the other thread where that paper was discussed, I don't think Fig. 3 in that paper is valid for a "firewall" model. It claims to be describing an evaporating black hole, but it looks more like a diagram (in weird coordinates) of maximally extended Schwarzschild spacetime, with both a black hole and a white hole region, and some kind of weird thing in the middle. Whatever it is, I don't think it's relevant to the discussion in this thread.

 

[PeterDonis]

OK, perhaps it would be fair to also inform the inquirer that the assumptions may not be realistic (which I clumsily tried without any successs to do in my second comment).

I think most of the posts in this thread before you came into it were doing that.

 

[Tomas Vencl]

Then you are not talking about Hawking's original model, or indeed any model with an event horizon. You are talking about a model like the Bardeen black hole that I referenced previously.………….…..I think most of the posts in this thread before you came into it were doing that.

Yes,you are right. I am leaving the discussion for now. Thank you and Dale for your ttime.

 

[Grinkle]

@A.T. I have never seen that discussion of what lead Einstein to GR before - its very digestible, thanks, and has me hooked enough to download and read the entire text.