[David Cooper]

In case someone has an answer as to whether the speed of light inwards across the EV is higher than the speed outwards across it (zero), it would also be useful to get some picture of how high/low that inward speed is (i.e. whether it's c, 2c or some other value between 0 and 2c).

[AndroidNeox (OP)]

The change to an infinite blue shift happens at the singularity in the middle; not at the EH.

That's wrong. The blueshift from any point in space to the event horizon is infinite just as the redshift up from the event horizon to any point in space above the EH is infinite. The blueshift downward is equal to the redshift upward. That's a simple consequence of conservation of energy.

[AndroidNeox (OP)]

The solution is that the light never quite reaches the event horizon, but slows down to a near halt. The same applies to matter falling in - during the entire lifetime of the universe, that material never reaches the event horizon (unless the event horizon moves further out if the black hole expands). The calculations that suggest that objects can cross the event horizon (and that light crosses it too) are based on the idea that clocks never really slow down, so objects are imagined to go on falling through the event horizon from their own point of view and continue on down to a singularity, but this could only happen after more than an infinite amount of time has gone by for the rest of the universe, which means there are no singularities yet and that all the stuff falling towards a black hole centre is frozen in place at the same distance from the centre as it was when it stopped at where the event horizon was at the time it stopped there. The black holes should evaporate away (due to Hawking radiation) before that material has a chance to move any further in, so in reality most of it can never reach the singularity, but perhaps the light can - as the black hole evaporates, the event horizon will migrate inwards as the energy density just outside the event horizon is reduced, so the light will get a chance to move a bit further in. At some stage when the event horizon disappears and the light is free to to go straight through what was the centre of the black hole, it will be shoot out the other side, liberated, never at any stage being infinitely blue-shifted.

I agree with your interpretation (except for one point). I think you've correctly understood what happens.

I personally interpret the situation a bit differently. Since Relativity is based on the universality of c, I interpret the time delay of  light passing through a gravity well as space dilation... essentially, the rubber sheet model. If space didn't dilate exactly the same amount as time then c wouldn't be universal when measured within every frame of reference. 

[AndroidNeox (OP)]

The gravitational potential between those two points is finite, but enough for escape speed to be c. 

No, this is wrong. The event horizon is defined as the point where the gravitational potential energy barrier is infinite and that is why the escape velocity is c. To accelerate a test mass to c requires infinite energy. It's not a finite energy barrier unless there's no event horizon.

[AndroidNeox (OP)]

But not true for the observer falling in.

This is not correct. The falling observer measures finite time in their fall to the event horizon. That does not in any way imply that the fall ever ends. Time for the falling observer (even ignoring special relativistic effects) slows asymptotically approaching zero time passage as the observer approaches the event horizon. Just as the area under a decaying exponential curve is finite while the curve is infinitely long, the forever slowing observer never reaches zero.

[AndroidNeox (OP)]

The light does in fact get there, and beyond.  Rocks really do fall into black holes, with nothing unusual about the event.  No Hawking radiation observed by the rock for instance.  It just doesn't fall into the black hole in the frame of this distant observer is all.

As with all of Relativity, the observations of all observers are consistent. However, there is no possible way to make the infinite value of the time observed by any observer (not only distant) stationary with respect to the black hole match the finite time you suggest. If the falling object reaches the event horizon in finite time then that event must be observable outside of the event horizon within finite time. But, as you say, it is not observable.

The Shapiro delay is infinite. That means light cannot travel to an event horizon in finite time. Unless you are suggesting that Shapiro is wrong?

[Halc]

The blueshift from any point in space to the event horizon is infinite just as the redshift up from the event horizon to any point in space above the EH is infinite. The blueshift downward is equal to the redshift upward. That's a simple consequence of conservation of energy.

Energy is not conserved over different reference frames, so that property is being applied in an invalid manner here.  In fact, yes, the guy at the EH appears stopped (infinite redshift) to some outside observer, but that guy on the EH sees the distant observer finite redshifted.  There is light reaching the falling guy from the distant point at exactly one moment, and light from earlier times already fell in, and light from later times hasn't yet got there.
The two are moving apart from each other, so just like Joe and Bob in different galaxies see each other both redshifted, so do these guys, one falling into a black hole.

[Halc]

The gravitational potential between those two points is finite, but enough for escape speed to be c.

No, this is wrong. The event horizon is defined as the point where the gravitational potential energy barrier is infinite and that is why the escape velocity is c. To accelerate a test mass to c requires infinite energy. It's not a finite energy barrier unless there's no event horizon.

Well, it would take infinite acceleration to keep something outside of the EH, so I'm willing to concede this.  It doesn't violate energy conservation, so it works.
Escape velocity might be c there, but only for light pointed directly out.  At 1.5x the radius of the EH, light orbits the black hole.  It is moving at more than escape velocity, but it doesn't escape.  Light orbiting below that point will not orbit, but be pulled in.  So clearly there are two different escape velocities that are the same for minor objects like Earth, but quite different for relativistic anomalies like black holes:  One is the velocity needed going straight up, and the other is the velocity needed tangential.  For a black hole, the latter is c at 1.5 the radius of the event horizon, so the value is greater than c anywhere between the two.

This can be observed by seeing a black hole eat material from a nearby star.  The material falling in comes from well outside the event horizon and you'd think it would pick up enough speed to orbit like a comet, but it actually gets sucked in by the bent space and spirals in just like the light does.

[AndroidNeox (OP)]

Energy is not conserved over different reference frames, so that property is being applied in an invalid manner here.

No, that's incorrect. All frames under consideration in this problem are stationary with respect to the black hole. All light is traveling only radially. Energy is conserved.

[Halc]

Energy is not conserved over different reference frames, so that property is being applied in an invalid manner here.

No, that's incorrect. All frames under consideration in this problem are stationary with respect to the black hole. All light is traveling only radially. Energy is conserved.

OK, I was speaking of the frame of an observer falling in.  The frame you speak of (inertial, stationary with the BH, at the event horizon) is an invalid reference frame, but an observer in a frame close by (one meter outside the EH) will observe an arbitrarily large blue shift of the distant observer, so I agree in that sense.

[AndroidNeox (OP)]

Energy is not conserved over different reference frames, so that property is being applied in an invalid manner here.

No, that's incorrect. All frames under consideration in this problem are stationary with respect to the black hole. All light is traveling only radially. Energy is conserved.

OK, I was speaking of the frame of an observer falling in.  The frame you speak of (inertial, stationary with the BH, at the event horizon) is an invalid reference frame, but an observer in a frame close by (one meter outside the EH) will observe an arbitrarily large blue shift of the distant observer, so I agree in that sense.

No, it's not invalid. The thought experiment is a perfectly valid relativistic thought experiment. Nothing in the experiment entails anything being stationary at the event horizon. Maybe you should read the original post and work through the thought experiment.

[Halc]

No, it's not invalid. The thought experiment is a perfectly valid relativistic thought experiment.

You seem to contradict yourself between these two statements:

All frames under consideration in this problem are stationary with respect to the black hole.

Nothing in the experiment entails anything being stationary at the event horizon.

Red and blue shifting of light is a frame-dependent thing, not an actual thing light does.  You need to specify a frame, but your contradicting statements don't tell me which one to use.

Maybe you should read the original post and work through the thought experiment.

The OP doesn't specify any frame.  It says these things:

The blueshift a light beam will undergo when traveling from any point in space to an event horizon is infinite.
...
By the time the front of the laser beam intersects the event horizon, the beam is infinitely blueshifted.

These are frame dependent claims.  I can look at light from a distant galaxy and the light is red shifted to one observer moving away from that galaxy, and blue shifted to another observer in the same place, but moving towards the galaxy.  The light is not actually shifted in either case, but appears red or blue relative to different frames.

Ditto for the black hole.  Light is finite red shifted for a falling observe at (or even beyond) the EH of a black hole.  For an observer stationary to the distant light source, the light appears blue shifted due to the time dilation of the gravity well, and yes, this approaches infinite dilation as the point approaches the horizon, but at the horizon itself is no more a valid inertial frame than is the inertial frame of a photon. Such a frame violates the principle of relativity.

[yor_on]

The light from the lasers platform is indeed blue shifted, but only as measured at the Event horizon. Passing that EV it is part of the singularity and might be described as 'infinitely blue shifted', although that makes no sense from the position of making a measurement. Outside, or at, the EV it can't be infinitely blue shifted.

The light the platform possibly can measure on will be reflected light, and that, as it comes back from the vicinity of the black hole, must be red shifted. You always need to be clear about what frame of reference you use when setting up a thought experiment like this. 

When it comes to the time dilation then you should be correct about the platform finding it to take a very long time to get any reflections of that light, if it now was possible to measure. But from the frame of reference of the infalling 'photons' the time it will take to reach the Event Horizon is defined by 'c', locally measured, so it should have no trouble reaching the EV in a 'finite time'. And as soon as you're outside a 'singularity' of some sort you can't speak of something being 'infinitely slow', because that will fall under the physics we know.
=

This is presuming we could use a frame of reference for a photon  :)
You can't really, but it doesn't matter for this, 'c' is 'c', and it's always a local measurement. And something assumed to 'propagate', as a 'photon', must then have some sort of locality in Space and Time. We can simplify it by imagining you to throw a rock at the Black Hole from the platform. Then let someone on that rock constantly measure 'c' as he 'falls' into a geodesic towards the Black Hole. He will reach it too in a, to him, finite time.

[yor_on]

" I'm in the tiny minority of people who believe that there is NOTHING beyond the event horizon, that there is no BEYOND the horizon ... i.e., that the event horizon itself is the closest thing to a "center" that a black hole has. "

I've seen that definition too Mike and it makes sense. Once you passed a event horizon, if we define that as being inside the region from where light can't leave, it seems to me that there will be no real definition of what a center should be. You're already inside a region where physics breaks down.

[Halc]

The light from the lasers platform is indeed blue shifted, but only as measured at the Event horizon. Passing that EV it is part of the singularity and might be described as 'infinitely blue shifted', although that makes no sense from the position of making a measurement. Outside, or at, the EV it can't be infinitely blue shifted.

The light the platform possibly can measure on will be reflected light, and that, as it comes back from the vicinity of the black hole, must be red shifted. You always need to be clear about what frame of reference you use when setting up a thought experiment like this. 

And yet you don't specify one.  If I put a mirror stationary (relative to the platform) near a black hole, the observer on the platform will see his own reflected light coming back the same frequency as it left.  Not so if the mirror is falling.  But you didn't specify above, which is the gist of most of my prior post.
You can get arbitrarily high blue or red shift, but the only infinite shift I can think of is that from a laser falling into the black hole as seen from the platform outside.

[yor_on]

What?

are you telling me that  " If I put a mirror stationary (relative to the platform) near a black hole, the observer on the platform will see his own reflected light coming back the same frequency as it left. "

Sorry
Fail

[yor_on]

A frequency is the number of occurrences of a repeating event per unit of time. That means that the platform watching the red shifted light also will find it to be at another frequency. Light is 'observer dependent'. meaning that you can see a light quanta as being constant, no matter if it seems to red or blue shift from another frame of reference.

[yor_on]

The point being, you have a good mind Halc, I enjoy reading you.
 Adapt and overcome.

you will get there.

[Halc]

A frequency is the number of occurrences of a repeating event per unit of time. That means that the platform watching the red shifted light also will find it to be at another frequency.

Assuming it is red shifted, yes.  But you haven't justified your assertion that it is red shifted.

Light is 'observer dependent'. meaning that you can see a light quanta as being constant, no matter if it seems to red or blue shift from another frame of reference.

Correct again, but we're looking at the reflected beam from the frame of the platform from which it was sent, reflected by a mirror stationary relative to that platform.  Regardless of any gravity well in which the mirror might be, that is going to result in no red or blue shift as seen by our platform observer.  If it were otherwise, you run into contradictions.

[yor_on]

Want to show me the contradictions?