If you mean “greater” and not “less” then yes. But approaching infinity as r approaches 2M - that is, as the object approaches the event horizon. Thus, any object crosses the event horizon at c.
What happens to the formula after the crossing of the event horizon is unknown, since the formula goes to imaginary numbers.
That is very intuitive, but it really doesn’t make any sense. I’ll show you why.
Multiplication and division are inverses.
if x = yz, then y = x/z and z = x/y.
If 5/0 = ∞, and 4/0 = ∞, and ∞/0 = ∞ then ∞*0 = 4,5, or ∞. Therefor 4 = 5 = ∞. This will work for any positive number. π/0 = ∞, e/0 = ∞, therefore π = e
In other words, we’ve shown any positive number is equal to any other positive number.
In other words if x/0 = ∞ then numbers have no meaning. Any number can be shown to equal any other number.
I know it’s very intuitive to think x/0 = ∞. It got me too, however the important thing is there no amount, not even an infinite amount of 0s that will get you 2M unless M is 0. It’s like trying to build a real wall out of imaginary bricks. No matter how hard or how many you imagine, you will never get a wall.
You’re tricking me into thinking about math man 
Just think of it in term of limits. If it is true that as r approaches 2M, a approaches infinity then what happens when r reaches 2M?
Either it can never reach 2M, or the math is wrong. There is no finite operation that can produce infinity in a finite amount of iterations. If it will repeat infinite times, that means it will repeat without end, meaning it never completes. Since all the iterations can be numbered, and there will always more iterations, then it never hits iteration infinity.
As always with these things I would love to be proved wrong, because whatever it was would be pretty cool.
Another argument that it is not infinite is if the acceleration is infinite, or without end, then it can’t decrease because that would be an end. In other words what is ∞ - 2? If you get a number, it wasn’t without end because what if you add 3 to that number?
∞ - 2 + 3? Why it’s ∞ + 1. So I guess ∞ wasn’t the highest number after all.
If it was unending, or infinite, it would never end. Planets, the universe, the galaxy, your car, even your dog, all would be falling, unable to even move even a hair or finger any direction but the hole. The sky would x-ray and gamma ray shifted one way, and completely black the other.
We know that it does reach 2M… And if there is another equation for the gravitational acceleration at event horizon, I’d like to find out.
It’s gravitational acceleration at the event horizon. Not everywhere.
a=1/sqrt(r-2M)
If r = 2m
then
a=1/sqrt(r-r)
then
a=1/sqrt(0)
then
a=1/0
How do I divide 1 into zero pieces to solve for a?
Both Chronos and I already said that the surface gravity of a black hole goes inversely to its mass, and is thus clearly finite (for a Schwarzschild black hole, it’s equal to 1/4M). What I suspect is happening with Wald’s formula – which, as you already noted, clearly can’t hold for the whole space, since it yields imaginary values beyond the horizon – is that it essentially falls victim to the coordinate singularity of the Schwarzschild solution at r = 2M. This is however not a physical thing, but an artifact of the mathematics; there are other choices of coordinates that are free from this singularity (unlike the singularity at r = 0, which can’t be gotten rid of this way).
It does not hold for the whole space, but does it hold for r>2M? Because if it does, and r is very close to 2M, the acceleration gets pretty big, and certainly higher than the other formula.
It seems to me that it would have to be so: if the only way to leave the black hole were to go at c, then obviously, the object is traveling c in the other direction. Or, okay, it could be really, really close to c and constantly accelerating.
I don’t think this picture is really useful: the problem is really that the radial direction beyond the horizon becomes timelike, so no acceleration is gonna get you out anymore than it’s gonna get you to yesterday. (Nevertheless, completing the analogy, moving faster than c could get you there.)
The other problem with the escape speed explanation is that you don’t actually need escape speed to escape an object, if you have continuing propulsion. But even with continuing propulsion, you can’t escape a black hole.
And let’s stop harassing Terr about the division-by-zero thing. There is a way to define acceleration for which the formula he gives is correct, and by that formula, in the limit as you approach the horizon from the outside, the acceleration approaches infinity. He was perhaps a bit sloppy in the way he phrased it, but I thought the meaning was clear enough.
What I am very confused about, Chronos, is that Wheeler et al, in the “Exploring Black Holes” text, concur that any object that crosses the event horizon, crosses it at c (whether it starts from rest at infinity or from any other position). Thus, one would think that the subjective time for the object’s frame contracts to 0, and the trip from horizon to singularity would take, subjectively, 0 time for the object. Yet in the same text, Wheeler is stating that it takes non-0 subjective time “horizon to crunch” and even calculates it for various black hole masses. How can it be non-0?
It’s non-zero from the perspective of the object. From the outside, we can’t measure it. Similarly, an object looking outward will see the universe recede from it at C at the event horizon, and beyond the horizon, it can’t measure them any more because it can’t seen them.
Note that I said “subjective” - that is, from the perspective of the object. According to the theory of relativity (correct me if I’m wrong) for an object moving at c, no time passes. That is, in the photon’s reference frame, it crosses the known universe, from one side to the other, in zero time. Correct? If so, how is it that the infalling object, from horizon to crunch, would take non-zero subjective time to cross?
Because from the outside perspective, it takes an infinite time for an infalling object to cross the horizon.