Black hole fun (not sun)

Here, right before your very eyes, I present you with a black hole of one solar mass which has a Schwarzschild radius of 3000 meters.

Here, also, we see a poor sad sack radially falling into the aforementioned black hole from a very great distance where spacetime was flat.

Now, the poor infalling slob notices something very strange. The Special Relativistic length contraction due to his speed exactly offsets the General Relativistic length increase between the reduced circumferences r. Ergo, his dr is the same as the reduced circumference dr.

Now, this would seen to imply that whether he travels through space, time or Jello he’s going to have to find some way to traverse that 3000 meters.

Right here, in this textbook, I see that the proper time to travel between the event horizon and the nasty place is equal to 6.57x10[sup]-6[/sup] x M/M[sub]sun[/sub] secs.

To my very great annoyance when I divide 3000m by 6.57x10[sup]-6[/sup] secs I get 4.57x10[sup]8[/sup] m/s which is not only faster than a speeding bullet, but is also faster than a speeding light. This, to put it mildly, is not good.

Now I know that the r and t change signs when the horizon is crossed, and therefore r becomes a time coordinate, but I still need enlightenment as to what is transpiring here.

For starters, could you specify which textbook you’re using for that proper time figure?

Actually, I lied. I found it on a website, but I do have Carroll and “Exploring Black Holes” by Wheeler, and I was pretty sure it was the same as I’d seen in Wheeler. And so it is: on page 3-22 of Wheeler’s book under, “Wristwatch Time from Horizon to Crunch” he has the exact same equation.

Tau = 4/3M, and when you do all the conversions to get to seconds he lists the same formula as above.

I realize that this is proper velocity, and proper velocity goes to infinity, but in this special case the distance as calculated from the reduced circumference (which I believe applies inside the EH) would seem to equal the infalling observers proper distance.

of course what you’re calculating is Δr/Δτ, even in special relativity dx/dτ (where x is a Minkowski space cooridinate) can be greater than c so it’s no biggie that Δr/Δτ in general relativity can be bigger than c.

ris the Schwarzchild observers coordinate, not the infalling observer’s coordinate. I’m not sure what you mean by proper distance here, proper distance between which two events? how is proper distance being defined here? What would proper distance/proper time even mean?

So r[sub]s[/sub] - r[sub]0[/sub] is also the distance the infalling observer must traverse. He doesn’t see a contracted distance.

Not quite sure I follow your logic.

The drfor a Schwarzchild observer and dr for an infalling observer absolutely cannot be the same. An infalling observer will pass any schawrchild observer they encounter with a non-zero velocity.

I also think as well that the concept of proper distance doesn’t make genreal sense in general relatvity (except when the termis used for a different purpose in the FLRW metric). Only cooridnate distance (which is a genrealisation of proper distance) makes sense annd I’m not sure that dividing a coordinate distance by proper time will give you anything physcially meaningful. Even in special relativity dividing one observer’s proper distance by another observers proper time doesn’t seem to mean much,

Let’s say we have an observer at r = 30 and he measures the distance between r = 30 and r = 29. He will find that the measured difference is greater than one by a factor of 1/(1-2M/r)[sup]1/2[/sup].

Now let’s say an observer is free falling from infinity. He will measure the distance between the two reduced circumferences r to be shorter by a factor of (1-v[sup]2[/sup])[sup]1/2[/sup].

And for this special case (free falling from infinity) the two factors are equal and opposite and will cancel and therefore the two drs will be the same.

I think where we get the greater than c velocity is from the infalling observer’s slow running clock.

BTW what do you mean by the Schwarzschild observer? A faraway observer or one suspended at a specific r or what?

Isn’t that sort of the reason a black hole is “black”?

Not exactly.

Near a gravitating mass time runs slow which means that the light emitted at the horizon has a lower frequency. (This is called red shift - the lower the frequency the redder the light. The higher the frequency the bluer the light)

Since time comes to a stop at the event horizon of a black hole the light is infinitely red shifted, and therefore has zero energy (E = hf) and is perfectly black.

Sorry, I should have said, for a faraway observer (us) time comes to a stop at the event horizon of a black hole.

Hi Ring,

I’ve got it now I think you’ve made relatively simple but easy mistake, it’s not the reduced circumference that is contracted due to Lorentz-Fitzgerald contraction (though of course the effect does lead to a contraction of the reduced circumference IYSWIM), it’s the stationary observer’s ‘dr coordinate’ that is contracted. So in the special case of an observer free-falling from infinity the contraction of the reduced circumference is infact the square of the Lorentz-Fitzgerald factor.

What I mean by asbolutely dr’s cannot be the same is that there are no co-moving stationary observers along the worldline of the infalling observer so they cannot share the same dr, as the two observers (stationary and infalling) have different bases in the tangent space at any given event on the worldine.

Yes when I said Schwarzchild observer I was being very wishy-washy and incorrect as I was conflating the faraway and the staionary observers.

I think we’re just confusing terminology. Here’s what John Archibald Wheeler says from