Crossing an event horizon

[GameDaddy]

If by some remote chance you are not torn asunder by the variable tidal gravitational pull of the black hole yourself... You still won't see them as they cross the event horizon. At that point gravity is sucking in the light as well, allowing no light to escape... no light = no vision. Mo light = color vision.

And they'll be dead already. Just so much compressed jelly. You might get a nice measurable little x-ray burst out of them though. And if you orbit just right you get out of the gravity well before reaching the event horizon yourself being close enough to see the luckless traveller during his/her last moments.

The gravitational tidal force are off the charts at the event horizon of a black hole.

Thank you. I've just figured something else new out...

 

[Umbran]

If by some remote chance you are not torn asunder by the variable tidal gravitational pull of the black hole yourself... You still won't see them as they cross the event horizon. At that point gravity is sucking in the light as well, allowing no light to escape... .

There's more to it than that. The relativistic effects muck with clocks, as we've noted above. If you're off at a distance, you see her clock slow to pretty much a stop as she falls - from the perspective of the rest of the universe, it takes an infinite amount of time for her to reach the event horizon.

The gravitational tidal force are off the charts at the event horizon of a black hole.

Not necessarily. For some, even most, that may be true, but not all.

Specifically, what you say is more true the smaller the black hole is. For large black holes (like the supermassive ones found in the hearts of galaxies) the event horizon can have only tiny tidal forces, less than that which the Earth is exerting on you at this moment, small enough to not damage materials entering the hole.

Now, inside the black hole, near the singularity itself, that's another matter.

 

[GameDaddy]

There's more to it than that. The relativistic effects muck with clocks, as we've noted above. If you're off at a distance, you see her clock slow to pretty much a stop as she falls - from the perspective of the rest of the universe, it takes an infinite amount of time for her to reach the event horizon.

Not necessarily. For some, even most, that may be true, but not all.

Specifically, what you say is more true the smaller the black hole is. For large black holes (like the supermassive ones found in the hearts of galaxies) the event horizon can have only tiny tidal forces, less than that which the Earth is exerting on you at this moment, small enough to not damage materials entering the hole.

Now, inside the black hole, near the singularity itself, that's another matter.

How is that possible? At the event horizon the gravitational pull of the black hole is so great, even light can no longer escape. If there is any variation in the density of the black hole it's going to create a gravitational sheer effect. You saying that the black holes in the center of the galaxy are perfectly spherical, and uniformly dense?

Given that the black hole sucks in mass at a variable rate, how is that mass evenly distributed in the singularity?

 

[jonesy]

If you are falling towards the horizon with the person you are observing, then neither of you experiences a slowdown of time. Relative to each other. But the stars around you would appear to speed up.

The farther away you are from the person you are observing, and the horizon he is falling towards, the greater the slowdown from your perspective, to the person falling in.

The farther in he is, the slower he seems to fall. At some point you will experience the interesting notion that he doesn't seem to be moving at all, because he will be moving inches by aeons. You will never in your own lifetime see him fall all the way through.

From his perspective the person observing him, and all the space around him, would speed up. The farther in he'd be, the faster everything would go.

 

[GameDaddy]

The farther in he is, the slower he seems to fall. At some point you will experience the interesting notion that he doesn't seem to be moving at all, because he will be moving inches by aeons. You will never in your own lifetime see him fall all the way through.

Only if you are observing him from a distance. What happens if you are closer?

Closer still.

and even closer?

Unless someone can provide some mathematics or an illustration to prove it is physically impossible to witness an event horizon transition I'll be in the school of: now you see him [He crosses the event horizon, you don't], Now you don't see him anymore... He simply vanishes... ummm like forever.

P.S. one would have to be careful when that close in to an event horizon, as I doubt they are perfectly spherical as well...

 

[jonesy]

Only if you are observing him from a distance. What happens if you are closer?

Closer still.

and even closer?

Unless someone can provide some mathematics or an illustration to prove it is physically impossible to witness an event horizon transition I'll be in the school of: now you see him [He crosses the event horizon, you don't], Now you don't see hime anymore... He simply vanishes... ummm like forever.

Did you not read what I just said? I didn't say it was impossible. I said in your own lifetime.

If we go with your original assumption of indestructible immortals, then sure, why not. But it's going to be a looooong wait for the observer if he watches the whole thing from beginning to end.

Edit: oh, it was Shag who said it. D'oh.

 

[Umbran]

image snipped

The scenario you show in this image is complicated, as you have two travelers who both approach the event horizon. It'll take me a bit to write that up properly. But, I can handle the next one quickly.

How is that possible? At the event horizon the gravitational pull of the black hole is so great, even light can no longer escape. If there is any variation in the density of the black hole it's going to create a gravitational sheer effect. You saying that the black holes in the center of the galaxy are perfectly spherical, and uniformly dense?

Given that the black hole sucks in mass at a variable rate, how is that mass evenly distributed in the singularity?

Perhaps you don't understand what I mean by "singularity".

A black hole forms when no force known can prevent collapse. Not even light gets away. Nothing stops the collapse. Nothing.

I repeat that - nothing. The thing shrinks down to the event horizon, and it keeps collapsing. The event horizon isn't a physical thing, it is not the surface of the collapsed object. It is just a demarcation line. The mass within continues to collapse, because there is no force in the universe that can hold it up, until it takes up zero volume. The density, therefore, is infinite - or more properly, it is undefined. There is no worry about the distribution of mass being uneven, as the mass isn't taking up any space!

The singularity is the point (and I mean that in the literal, mathematical sense) where we divide by zero, and the normal rules of physics no longer apply.

There is a volume of space between the event horizon and the singularity - all paths within the event horizon eventually lead to the singularity.

 

[Pbartender]

The density, therefore, is infinite - or more properly, it is undefined. There is no worry about the distribution of mass being uneven, as the mass isn't taking up any space!

The singularity is the point (and I mean that in the literal, mathematical sense) where we divide by zero, and the normal rules of physics no longer apply.

Or to look at it from another point of view...

It's perfectly spherical, perfectly dense, and perfectly even in a way that only nothing can be.

 

[Umbran]

Okay, so about that picture.

We have two travellers going near to the event horizon - one is falling straight in, the other is traveling so he skims just over the surface, and the question is how close does the skimmer have to be to see the other fall in.

Now, we have a problem. In order to skim by the black hole hole and not fall in, you have to be traveling quickly. The closer you want to go and still not fall in, the closer to the speed of light you need to go. To skim just above that surface, you have to be just below the speed of light.

The skimmer cannot actually be at the surface - that would require him to move at the speed of light, which he cannot, as he has mass. Only massless things can move that fast.

So, the person falling is at the event horizon, and the skimmer is above it. This is now no different from the case where one person is in a ship high above, and watching the other fall in. It doesn't matter how far above the observer is - if the observer isn't also falling in, the same clock-slowing will occur.

The math is kind of wonky that way. An inch is as good as a yard is as good as an astronomical unit.

 

[Umbran]

It's perfectly spherical, perfectly dense, and perfectly even in a way that only nothing can be.

So long as it is not spinning, anyway.