Re: Path in Schwarzschild

Dear sal:

"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:46eb6a7c$0$24906$ec3e2dad@xxxxxxxxxxxxxxxxxxxxxxxxx
....
One particular question I have is:

At r = 2 G M c^2, does the massive body behave
exactly like a photon, in so much that both get
"stuck" at the event horizon?

No massive body is stuck at the horizon. Only a perfectly
aligned photon is stuck at the horizon, such that an observer at
"infinity" will wait an infinte time to see it. This includes
*the image* of an infaller crossing the event horizon. Only the
image, not the body itself.

I am wondering this because at this radius,
the massive body's internal time rate drops to 0.

No, because any mass infalling is travelling less than c on the
"outside" of an event horizon.

That's probably not an ideal way to think of it.
OTOH, in the rest frame of the falling body,
the horizon blows past it at C, which does
rather suggest it's acting a lot like a photon, eh?

No, because the infaller never sees the event horizon. The event
horizon for the infaller is always "ahead". He never really
notices that the Universe he has left is no longer a place that
he can have effect on. The Schwarzchild event horizon is for a
statiotnary observer at infinity.

Probably need to use Kruskal-Szekeres coordinates
to answer any such question, and then converting the
result back to anything comprehensible is likely to be
tricky.

As to whether things get "stuck" at the horizon or
not ... that's an interesting question but it seems to
hinge on the definition of the word "stuck". A finite
amount of proper time passes for an infalling
body by the time it goes through the horizon.

And a finite time for a distant observer too. Just don't expect
he'll ever see light emitted at the instant of crossing.

Asking "what time is it back on Earth at that
moment" is slippery, because in the wildly
curved space around a black hole, there is no
universal time reference you can use to define
"that moment" anywhere except locally.

David A. Smith


.

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