Black hole event horizon and escape velocity

[iantresman]

Some time ago, I asked a question about Black holes and the velocity of light, which was kindly answered, and similar to another query I have.

The escape velocity of the Earth is 11.2 km/s, which I thought defined the speed, that an unpowered ballistic object requires, to break free from Earth's gravitational pull and reach an orbit at infinity. If I shoot an object skyward at just below escape velocity, it may still leave Earth, just not reach infinity... but this could be a significant distance from Earth.

I understood that a black hole's event horizon was a surface at which its escape velocity equalled the speed of light. Since light is travelling at the escape velocity, at the event horizon, won't it be able to leave its surface, and reach an "orbit at infinity".

In comparing the use of "escape velocity" with regard to Earth and black holes, it seems that the former allows objects to leave the surface, whereas a black hole event horizon is described as an impenetrable surface.

One solution is that an observer at infinity would see something different to an observer at twice the event horizon distance. But even if a photon can extend a tiny bit above the event horizon, then it could extend further. But can someone clarify my dilemma?

[korjik]

First, an event horizon isnt an impenetrable surface. That is a misunderstanding by most people. It is a surface that you wont even notice going one way, and cant ever reach going the other.

The analogy between light and escape velocity isnt a good one tho. Light dosent behave ballistically, so it dosent act the same. Properly, a photon at the event horizon is redshifted to an infinite wavelength, IIRC. That would be why it cant cross the horizon.

Another problem is that the geometry inside the horizon is different. A photon travelling radially outward just under the event horizon, will never reach the event horizon. So, a photon just cant extend a 'tiny bit' above the event horizon.

[WayneFrancis]

Some time ago, I asked a question about Black holes and the velocity of light, which was kindly answered, and similar to another query I have.

The escape velocity of the Earth is 11.2 km/s, which I thought defined the speed, that an unpowered ballistic object requires, to break free from Earth's gravitational pull and reach an orbit at infinity. If I shoot an object skyward at just below escape velocity, it may still leave Earth, just not reach infinity... but this could be a significant distance from Earth.

I understood that a black hole's event horizon was a surface at which its escape velocity equalled the speed of light. Since light is travelling at the escape velocity, at the event horizon, won't it be able to leave its surface, and reach an "orbit at infinity".

In comparing the use of "escape velocity" with regard to Earth and black holes, it seems that the former allows objects to leave the surface, whereas a black hole event horizon is described as an impenetrable surface.

One solution is that an observer at infinity would see something different to an observer at twice the event horizon distance. But even if a photon can extend a tiny bit above the event horizon, then it could extend further. But can someone clarify my dilemma?

IF the photon was produced at a location≥its frequency then the photon would escape to infinity but would be EXTREMELY red shifted by the time it actually got a good distance out.

At the Event Horizon you can think of it as walking on a tread mill where you are walking the same speed as the tread mill. Your centre of mass would not move. This is just a mental picture and the actual workings is a bit different. We don't expect there to be photons just sitting at the EH trying to get out. The nature of space/time as we know it would have even those photons curve back in towards the centre of the black hole.

The key here is a photon would have to be above the EH by its entire wavelength at the time it was created to escape and that means that it was never within the EH so there is no problem.