https://gizmodo.com/why-doesnt-the-black-hole-image-look-like-the-one-from-1833949289

 

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https://gizmodo.com/why-doesnt-the-black-hole-image-look-like-the-one-from-1833949289

If you tilt your head and cross your eyes just right, you can see the number 42.

Edit: Heck...you don't even need to tilt your head.

 

This disturbs me. One huge assumption is that distance and acceleration (a function of both distance and time) are constant in highly curved space. My spidy-sense tells me there are orders-of-magnitude errors in the answers -- especially in problem 3.

Gravitational time dilation, special-relativistic time dilation are not physical processes, they are a difference between observers. The outside observer will see all sorts of weirdness in highly curved space but the person falling should only see effects consistent with the local space-time properties on the localized mass.

https://en.m.wikipedia.org/wiki/Penrose–Hawking_singularity_theorems

 

View attachment 174774

https://gizmodo.com/why-doesnt-the-black-hole-image-look-like-the-one-from-1833949289

Looks upside down.

 

The outside observer will see all sorts of weirdness in highly curved space but the person falling should only see effects consistent with the local space-time properties on the localized mass.

Yes! But localized space for the infalling observer inflates. Imagine: I drop a rock, then, a short time later, I jump in myself chasing the rock. As space inflates due to increasing gravitational distortions, the rock appears to accelerate away from me -- not because it is, but because space is getting "bigger".

This is, IMHO, exactly why the event horizon always appears below the infalling observer -- regardless of how long or how far he has fallen.

 

This is, IMHO, exactly why the event horizon always appears below the infalling observer -- regardless of how long or how far he has fallen.

Isn't that argument a little like Zeno's paradox?

 

Yes! But localized space for the infalling observer inflates. Imagine: I drop a rock, then, a short time later, I jump in myself chasing the rock. As space inflates due to increasing gravitational distortions, the rock appears to accelerate away from me -- not because it is, but because space is getting "bigger".

This is, IMHO, exactly why the event horizon always appears below the infalling observer -- regardless of how long or how far he has fallen.

The person (an infalling observer) actually falling in the event horizon will fall for a finite time to the 'singularity' and die.

https://arxiv.org/PS_cache/arxiv/pdf/0705/0705.1029v1.pdf

For a stellar mass black hole, this will be a fraction of a second, but for a supermassive black hole, this may be hours. As will be shown later, this maximum time applies to a faller who drops from rest at the event horizon and any one who starts falling from above the event horizon and free falls into the hole will experience less proper time on the journey from the event horizon to the singularity.

 

The person (an infalling observer) actually falling in the event horizon will fall for a finite time to the 'singularity' and die.

Of course! Assuming a singularity.

I make no such assumption.

 

Of course! Assuming a singularity.

I make no such assumption.

OK then, the person will hit the rock in the middle in a finite period of time and die.

 

OK then, the person will hit the rock in the middle in a finite period of time and die.

Well, actually, he'll die first. But that's not important.

What is important is how much time it takes the observer to cross the event horizon (and the author makes the assumption that he does). I claim the event horizon is always in his future. He never broaches it.

Edit: Oh! And there is no middle.

 

Well, actually, he'll die first. But that's not important.

What is important is how much time it takes the observer to cross the event horizon (and the author makes the assumption that he does). I claim the event horizon is always in his future. He never broaches it.

Yes, he'll die first. The author assumes the observer is impervious to all forces other than inelastic collision.

The black hole has a net mass, a net angular momentum now, a net charge now, not in the future. Gravitational time dilation effects can be tricky.

http://www1.phys.vt.edu/~jhs/faq/blackholes.html#q11

 

The black hole has a net mass, a net angular momentum now, a net charge now, not in the future.

Yup. And it's the distribution in space of that mass, momentum, and charge that changes as the observer falls.

Edit: I'd love to stay and chat...but I have a lot of work to do today and I gotta get to it...

 

Yup. And it's the distribution in space of that mass, momentum, and charge that changes as the observer falls.

None of that stops the observer from falling to the 'rock' at the center.

 

Edit: I'd love to stay and chat...but I have a lot of work to do today and I gotta get to it...

Same here... talk to you all tonight, hopefully. Have a nice (and productive) day.

 

None of that stops the observer from falling to the 'rock' at the center.

As you drop the mic.

kv

 

As you drop the mic.

kv

Not yet.
https://www.scientificamerican.com/article/the-reluctant-father-of-black-holes-2007-04/

Two Views of a Collapse

LET US START with an observer at rest a safe distance from the star. Let us also suppose that there is another observer attached to the surface of the star--co-moving with its collapse--who can send light signals back to his stationary colleague. The stationary observer will see the signals from his moving counterpart gradually shift to the red end of the electromagnetic spectrum. If the frequency of the signals is thought of as a clock, the stationary observer will say that the moving observer's clock is gradually slowing down.

Indeed, at the Schwarzschild radius the clock will slow down to zero. The stationary observer will argue that it took an infinite amount of time for the star to collapse to its Schwarzschild radius. What happens after that we cannot say, because, according to the stationary observer, there is no after. As far as this observer is concerned, the star is frozen at its Schwarzschild radius.

Indeed, until December 1967, when physicist John A. Wheeler of Princeton University coined the name black hole in a lecture he presented, these objects were often referred to in the literature as frozen stars. This frozen state is the real significance of the singularity in the Schwarzschild geometry. As Oppenheimer and Snyder observed in their paper, the collapsing star tends to close itself off from any communication with a distant observer; only its gravitational field persists. In other words, a black hole has been formed.

But what about observers riding with collapsing stars? These observers, Oppenheimer and Snyder pointed out, have a completely different sense of things. To them, the Schwarzschild radius has no special significance. They pass right through it and on to the center in a matter of hours, as measured by their watches. They would, however, be subject to monstrous tidal gravitational forces that would tear them to pieces.

 

Not yet.
https://www.scientificamerican.com/article/the-reluctant-father-of-black-holes-2007-04/

Only change I would've made to the previous citation:

They would, however, be subject to monstrous tidal gravitational forces that would tear them to pieces shreds.

 

Only change I would've made to the previous citation:

You would snap into pieces, shredding from top to bottom.

 

​

 

A bit of hope for the black hole fallen in despair.