B What Happens Inside the Event Horizon of a Black Hole?

[Stephanus]

Dear PF Forum,
What happens inside Event Horizon?

1. Will clock stop inside EH?
An object crosses the EH of a black hole around 1 billions solar mass which its Schwarzschild Radius is 3 billions km. It takes light to cross that distance (in 'normal' space) 30 thousands seconds.
The time for an object travels at, say, 99.99% the speed of light to travel that distance (in 'normal' space) according to a rest observer, after adjusting to time dilation is not 30 thousands second, but around 428 seconds.
Faster than that it could take just 1 second.
So what happen if a clock crosses the EH of a black hole, will it hit the singularity instantneusly? Or it will hit the singularity in 30 thousands seconds?

2. What is the speed of an object that crosses the EH? It's in a speed of light or it's an irrelevant question?
@tiny-tim said in https://www.physicsforums.com/threads/acceleration-inside-an-event-horizon.402258/#post-2711192, I can't quote it because the thread was closed.

Everything (including light) crosses the event horizon at the same speed (which we may as well call the speed of light).

But an object with mass can't travel at the speed of light Or the law of physics break down inside EH?

Thank you very much.

 

[PeterDonis]

Will clock stop inside EH?

No.

An object crosses the EH of a black hole around 1 billions solar mass which its Schwarzschild Radius is 3 billions km. It takes light to cross that distance (in 'normal' space) 30 thousands seconds.
The time for an object travels at, say, 99.99% the speed of light to travel that distance (in 'normal' space) according to a rest observer, after adjusting to time dilation is not 30 thousands second, but around 428 seconds.

This is all muddled; you are mixing up coordinate time and proper time, and you are trying to apply special relativity in a situation where general relativity is needed.

For a simple estimate of the proper time it takes for an object to fall from the horizon of a black hole to the singularity, your first number--the time it takes light to travel a distance equal to the Schwarzschild radius of the hole--is actually correct. So for a billion solar mass black hole, it will take an object approximately 30,000 seconds, by its own clock, to fall from the horizon to the singularity. The details of why this is true are too complicated for a B-level thread, but that's the answer.

What is the speed of an object that crosses the EH?

Speed relative to what?

Relative to the object, the horizon itself is moving outward at the speed of light. But that's because the horizon itself is made up of outgoing light rays.

The object at the horizon has no "speed" relative to an observer very far away; in general relativity the concept of "relative speed" has no meaning for spatially separated objects.

 

[Stephanus]

...So for a billion solar mass black hole, it will take an object approximately 30,000 seconds, by its own clock, to fall from the horizon to the singularity...

So are you saying that...

Scenario 1:
A "rest" observer (C) sees point A and B in space (at "rest" also). A-B distance is 3 billions kms. Then a clock travels from A to B 99.99%c, when it reaches A, A record the clock value, and when the clock reaches B, B records its value. Later when A-B reports to C, it will show that the clock only takes 428 seconds.

Scenario 2
And in different scenario
C sees two point A and B.
But B is a black hole here,
So a clock supposedly travels from A to B and at B the clock will read 30 thousands seconds as opposed to 428 seconds as in scenario 1?

If it's true then I will be satisfied with yes/no question because...

The details of why this is true are too complicated for a B-level thread, but that's the answer.

Speed relative to what?

Yes, that's the wrong question.

Relative to the object, the horizon itself is moving outward at the speed of light.

Can you please elaborate a little?
What do you mean by "outward" here? Toward the singularity or toward a coordinate outside the singularity?
Relative to the object (that crosses the EH)
A: The horizon is moving outside the black hole at c or
B: The horizon is moving inside the black hole at c?

What "horizon" do you mean here? Event horizon or space?
As in one of Michio Kaku video, he says, perhaps something like this, I don't know the exact word.
"I'm stuck in my chair because space is pushing me downward at 11 km per second"

But that's because the horizon itself is made up of outgoing light rays.

I'm sorry. What "horizon" do you mean? EH? I think EH is position $\frac{2GM}{c^2}$, not a something made up from light rays?

The object at the horizon has no "speed" relative to an observer very far away; in general relativity the concept of "relative speed" has no meaning for spatially separated objects.

Thanks FYI

 

[PeterDonis]

Scenario 1:
A "rest" observer (C) sees point A and B in space (at "rest" also). A-B distance is 3 billions kms.

Where are A and B in relation to the black hole? If A is outside the horizon and B is inside the horizon, your scenario won't work; B cannot be at rest relative to A.

You appear to be trying to apply intuitions derived from special relativity to a situation where SR does not apply. That won't work.

C sees two point A and B.
But B is a black hole here

A black hole is not a point, and it can't be "seen" in the ordinary sense.

So a clock supposedly travels from A to B and at B the clock will read 30 thousands seconds as opposed to
428 seconds as in scenario 1?

Same problem here as above; if B is inside the horizon of a black hole, B cannot be at rest relative to A.

Also, remember that we were talking about the time for an observer to fall from the horizon to the singularity, by the observer's own clock. Nothing has to travel from the observer to or from anywhere else to make that measurement. The clock just free-falls along with the observer, and he reads off what it says.

What do you mean by "outward" here?

Radially outward. If the observer is falling radially inward (which I am assuming is the case), then "outward" is the direction opposite to the direction in which the observer is falling.

Physically, the observer can define "outward" by picking a very distant object, say a star, that is directly overhead when he starts to fall. Then the direction in which he sees light coming in from that star is always "radially outward" as he falls.

Toward the singularity or toward a coordinate outside the singularity?

Neither. "Toward the singularity" is toward the future; it's a timelike direction, not a spacelike direction. "Toward a coordinate outside the singularity" doesn't make sense; coordinates don't have locations.

Relative to the object (that crosses the EH)
A: The horizon is moving outside the black hole at c or
B: The horizon is moving inside the black hole at c?

Neither. The horizon is the spacetime boundary of the black hole. Whether it is "moving" is relative; that's why I was careful to say that relative to the object falling inward across the horizon, the horizon is moving radially outward at c. To be more precise (and getting into I-level territory instead of B-level territory): if we set up a local inertial frame covering a small patch of spacetime centered on the event where the object crosses the horizon, then in that local inertial frame, the horizon is a set of light rays that are all moving at c in whatever spatial direction, in that frame, corresponds to "radially outward".

What "horizon" do you mean here?

The only horizon we've mentioned: the event horizon.

As in one of Michio Kaku video

Forget anything you think you have learned from such sources. You should not be trying to learn science from pop science sources. Find a textbook. A good one to start with is Sean Carroll's lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019

"I'm stuck in my chair because space is pushing me downward at 11 km per second"

This has nothing whatever to do with the topic of this thread.

I think EH is position $\frac{2GM}{c^2}$,

You think incorrectly. The event horizon is at the radial coordinate $2M$ (or $2GM / c^2$ in conventional units), but this value of the radial coordinate does not label a "position". Only radial coordinates greater than $2M$ can be viewed as labeling positions. The reason for that is too complex to explain in a B-level thread; to illustrate that, I'll give you the reason in I-level technical language: curves of constant $r$ are integral curves of a Killing vector field on Schwarzschild spacetime, but this KVF is only timelike outside the horizon; it's null on the horizon and spacelike inside it. Only in the region where the KVF is timelike can the curves of constant $r$ be viewed as labeling "positions".

At this point I am closing the thread because you are trying to reason with incorrect intuitions that can't be corrected (more than I already have) at the B level. You need to improve your background knowledge to the point where you can discuss this topic at the I level.