Quote by HallsofIvy
No, that is not true. A massive object can never have speed c, even for an "instantly short period" (whatever that means).

If I was to ask you what the vertical velocity of a free falling object relative to the ground was at its apogee, you would probably say 0 m/s. I would then ask you how you propose to measure this on the principle that we only believe what we can directly measure. You might propose that we measure its position 1 femto second before apogee and one femto second after apogee and calculate its average velocity and obtain zero. However, I would say that is not good enough and point out that we both accept its velocity is continually changing and I want to know its
exact velocity at the instant the object is
exactly at its apogee and I think you would have to concede that is impossible to measure because it requires a finite time interval to measure a velocity. Demanding to know the the instantaneous velocity at an exact location or point in time requires the concept, that I think passionflower was getting at, of a time interval with a duration of exactly zero time units and that is not practical or useful. This might seem like extreme nit picking, but it is important in relativity because as we all know, being able to get arbitrarily close to the speed of light is still a world away from actually travelling at the speed of light. Similarly, if we accept the idea that it is impossible to have a stationary observer exactly at the event horizon, (along with the notion that there is no such thing as an instant  See paper by Peter Lynds) then we have to accept that there is no practical way to measure the the velocity of an object exactly at the EH. Even though we can establish that the velocity of a free falling object can be as arbitrarily close to the speed of light as measured by an observer hovering arbitrarily close to the event horizon, this does not establish that the object reaches light speed exactly at the the EH.
Quote by HallsofIvy
The speed of an object crossing the event horizon has nothing to do with the escape velocity. Saying that the escape velocity of a black hole is c simply means that no massive object can escape from the black hole, going away from the black hole. That has nothing to do with crossing the event horizon going toward the black hole.

I think it is only fair to point out that generally the escape velocity at a given point in a gravitational field has everything to do with the free fall velocity of an object at that point falling
towards the massive body. The EH (and all points below it) may be an exception, because it is not possible to have a
stationary observer at or below the EH as mentioned above. Of course to prove this claim we need to prove that it is not possible to have a stationary observer at the EH.
To investigate this further we need to check the GR equations for free fall velocity and acceleration in the region of the EH. THe local free fall velocity is given by:
[tex] \frac{v}{c} = \sqrt{\frac{2GM}{rc^2}}[/tex]
Interestingly this equation for the escape velocity (or free falling velocity) of an object is the same in Newtonian and GR physics. It gives us the superficial result that the free fall velocity at the event horizon when [itex]r = 2GM/c^2[/itex] is the speed of light. However this is the velocity as measured by a stationary local observer at the EH and we have not established that it is possible to have such an observer. An alternative equation:
[tex] \frac{v}{c} = \sqrt{\frac{2GM}{rc^2}} \left( 1\frac{2GM}{rc^2} \right)[/tex]
is the coordinate velocity and this suggests that the free fall velocity at the event horizon is zero and does not support the idea that it is impossible to be stationary at the EH, but in turn it does not support the idea that it is possible to free fall at the speed of light at the EH. So the coordinate point of view gives stationary at the EH and the local point of view gives essentially undefined at the EH. Next, we have to look at the equations for gravitational acceleration to shed some light on this issue. The local acceleration is given by:
[tex] g = \frac{GM}{r^2\sqrt{12GM/rc^2} }[/tex]
supporting the view that it is impossible to be stationary at the EH because the acceleration becomes locally infinite there. However, the coordinate equation for the acceleration is:
[tex] g = \frac{GM}{r^2} \left(1\frac{2GM}{rc^2}\right)[/tex]
which gives the result that the coordinate gravitational acceleration at the EH is zero supporting the equation that the coordinate free fall velocity is also zero and in coordinate terms it is possible to have a stationary observer at the EH. It all depends upon whether you believe the local equations or the coordinate equations are a better reflection of reality, but it is only fair to state that the overwhelming conventional wisdom is that the local view is the better reflection of reality (infinite acceleration at the EH, no stationary observers at the EH and undetermined free fall velocity at the EH).