Re: GR: 'Line-of-sight' distance to a black hole is infinite?

On Jun 27, 11:21 pm, Roland PJ <rolan...@xxxxxxxxx> wrote:
Last question before my R&R.

The intergral( s dr ) diverges towards r = 2M, where r is Schwarzchild
radius, and s is proper distance.

Not in the way you have currently written it, it does not. Are you
trying to write the coordinate distance between a point outside, but
near, the event horizon and one far away? What path does it take? Is
it a massive particle?


Now, it seems to me that l = integral( s dr ) is the distance that we
would 'see' from earth, or more concretely, the distance corresponding
to the apparent size, or parallax, of objects in our line-of-sight to
the black hole.

It only diverges if the object is _inside_ or on the event horizon.
Integrals remain finite outside the black hole in Schwarzschild
coordinates. The distance between an external observer and any point
_outside_ of r = 2M is finite.


So, I had the impression previously that we would 'see' infalling
objects to slow down and essentially come to a halt at r = 2M (the
'frozen star' view).

No. Infinitely redshifted. Play around with null paths, and you will
see what I mean.

[...]


Then my next question was, how can an infalling object cover the
apparently infinite proper distance in a finite time. So, this is due
to SR effects, right? The distances measured by the infalling observer
are contracted by Lorenz as she really does achieve v = c at the event
horizon, and contracted exactly correctly to provide an observer
distance which is finite.

When will people learn that SR has nothing to do with black holes?

The infalling object _doesn't_ cover the distance in finite observer
time. It _does_ cover the distance in finite proper time.


Is this all accurate?
Thanks in advance for my further tuition
Roland


.

Relevant Pages

  • Re: GR: Line-of-sight distance to a black hole is infinite?
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