BH event horizon apparent size

[pavelcherepan]

OK, so I was thinking about black holes for some reason last night and got a question.

 

Say, here on Earth the escape velocity at the surface level s around 11.2 km/s and any object crushing on Earth from outer space, even if initial relative velocity was 0, should end up hitting the surface at least at 11.2 km/s, if there were no air or the object is sufficiently massive so it's not slowed down by atmosphere as much. But if the object has a relative velocity to begin with, it can impact at a higher velocity.

 

Now, by the very definition of the Event Horizon of the black hole, the escape velocity at that region equals to c and, therefore, if you start at 0 velocity wrt the black hole, you should cross the even horizon at c. But what if you're moving towards the BH?

 

Obviously, you can't move faster than c (or even at c for that matter), which means that you'd have to cross event horizon sooner to account for all that extra velocity. Say, I'm flying towards the black hole having started at some relative velocity v and at one moment in time I'm passing the other observer, who is stationary wrt the black hole. If at that moment we both observe and measure the apparent size of the event horizon, will I see it as being larger/closer to me?

Edited June 12, 2017 by pavelcherepan

[imatfaal]

The thing about escape velocity is true - but to an extent coincidental. The black hole forms because space time is so warped by the gravity that the geometry does not allow any outward passages from within the event horizon. Remember that on a normal object the escape velocity is the speed that you need to be going to escape with no further effort - ie you will coast to infinity without ever being pulled back. On a normal object you could (given a very long ramp*) just walk up the gravity well ( it would just take a very long time) at a much slower speed than escape velocity. Inside the event horizon of a black hole no amount of effort can lift you away from the centre - because all directions lead to the centre; it is a geometrical trap - all roads lead to the same place and that is further in.

 

So your calculations and conclusions are based on a very shakey ground. A very large black hole will have an acceleration due to gravity at the event horizon of 10^5 metres per second - which is huge but not beyond imagination. I will dig out a nice primer on black holes. FYG the escape velocity of light being the reason for black holes is much older than General relativity (Mithell & Laplace 18C)

 

 

*Four thousand metres vertical distance is a very respectable day of cycling up hills - at this rate of vertical ascent it would take 50 days to reach low earth orbit

[Strange]

Part of your question is (I think) answered by the fact that velocities do not increase linearly in relativity and so you would need to use a relativistic formula for the acceleration caused by the force of gravity (assuming it is valid to use this Newtonian approximation in this scenario).

 

 

 

Say, I'm flying towards the black hole having started at some relative velocity v and at one moment in time I'm passing the other observer, who is stationary wrt the black hole. If at that moment we both observe and measure the apparent size of the event horizon, will I see it as being larger/closer to me?

 

This is a different question. My understanding is that as you approach a black hole, the event horizon is always ahead of you. So in this case you would "observe" (I'm not sure you can) or calculate the event horizon as being smaller than the stationary observer (or at least, when you pass the point where they think it is).

Inside the event horizon of a black hole no amount of effort can lift you away from the centre - because all directions lead to the centre; it is a geometrical trap - all roads lead to the same place and that is further in.

 

Excellent answer! The only thing I would add is that inside the event horizon, the radial direction to the singularity becomes your future. And we all know there is no escaping that...

[pavelcherepan]

 

 

On a normal object you could (given a very long ramp*) just walk up the gravity well ( it would just take a very long time) at a much slower speed than escape velocity.

 

Suppose we could build a ramp that starts at the event horizon of the black hole, say we use extremely powerful engines to prevent from falling in. We should still be able to ride a bicycle up from the black hole and eventually to the point where it's gravitational influence is negligible?

 

 

 

This is a different question. My understanding is that as you approach a black hole, the event horizon is always ahead of you. So in this case you would "observe" (I'm not sure you can) or calculate the event horizon as being smaller than the stationary observer (or at least, when you pass the point where they think it is).

 

Strange, can you please clarify what you mean by "event horizon is always ahead of you"? Why would you observe it as being smaller?

[Strange]

 

Suppose we could build a ramp that starts at the event horizon of the black hole, say we use extremely powerful engines to prevent from falling in. We should still be able to ride a bicycle up from the black hole and eventually to the point where it's gravitational influence is negligible?

 

If you did this "at" (just outside) the event horizon, then you could (in principle) do this.

 

If you tried it inside the even horizon, then whatever you did, your ramp would head towards the centre.

 

 

 

Strange, can you please clarify what you mean by "event horizon is always ahead of you"? Why would you observe it as being smaller?

 

 

Not really. This is something I read (from someone who does understand the mathematics). I just checked and I slightly mis-remembered it.

 

If you are falling towards a black hole, the event horizon is invariant (the same for all observers) and so you will fall through that. But then there is another "personal" even horizon ahead of you. This separates you from all the events that you will never observe because you will hit the singularity before the light from them can reach you.

I am going to quote this from another forum, as I think it is a good description of what happens

Any light we see inside the event horizon was emitted by objects falling towards the singularity. When any object falls into the singularity, it leaves behind a "supply" of photons inside the event horizon: those emitted near the event horizon wind down their radial coordinate only slowly initially; those emitted just above the singularity hit the singularity almost immediately. So when we fall below the event horizon, we are presented with photons, coming from ahead, emitted by objects that preceded us into the black hole. If we fall in immediately after the preceding object, we'll be able to watch almost all of its trip to the singularity during our own trip to the singularity. If we fall in some time afterwards, we'll catch up to signals emitted by the preceding object during only the initial part of its fall: it will take us our whole fall to observer photons from (say) just the first fraction of a second of the preceding object's fall.

So there's an event horizon of sorts, marking off a region of spacetime from which we can never receive signals before our own appointment with the singularity: but it slopes through time, cutting off long-ago events close to the inside of the event horizon and more recent events close to the singularity. But at any given moment of our own fall, we will (at least in potentiality) have signals from all around, from objects falling ahead of us, alongside us, and behind us.

 

This all connects seamlessly to the external view:

As we fall towards the event horizon, we encounter photons from objects that fell into the black hole ahead of us. The longer ago the object fell in, the more delayed and sparser are these photons, bringing us old views of the object's last moments above the event horizon, which we can watch over some extended period during our fall. As we pass through the horizon, we (potentially) see images of everything that went through the horizon ahead of us: but the sheaf of photons in the vicinity of the event horizon disperses with time, so the images of objects that passed through long ago will eventually wear down to their last photon and disappear. Inside the horizon, we encounter images from photons emitted inside the horizon. The more recently the object preceded us through the horizon, the more complete the history we can observer during our own remaining time before the singularity.

 

Grant Hutchison

https://forum.cosmoquest.org/showthread.php?109938-What-is-the-event-horizon-of-a-freefaller-into-a-black-hole&p=1823342#post1823342

 

And, another interesting thing is that the "black disk" you would see if you looked at a black hole is larger than the event horizon. It is actually the photon sphere:

http://rantonels.github.io/starless/