Re: value of g in black holes

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

> -----Original Message-----
> From: John Mallinckrodt [mailto:ajmallinckro@CSUPOMONA.EDU]
> Sent: Thursday, October 18, 2001 3:03 PM
> To: PHYS-L@lists.nau.edu
> Subject: Re: value of g in black holes

>  After all the value of g is normally
> associated with observations that would be made by a "stationary"
> observer--i.e., one who stays at constant R.
>
> At any rate I am not surprised to learn that the relativistically
> correct value diverges as one approaches the event horizon
> regardless of M.  I do wish, however, that I understood a little
> better what that *meant*!

It surprised me at first, since I'm used to trying to explain that the
metric singularity at the event horizion is a coordinate singularity and not
physical, as in the tidal forces are nice and finite at the event horizon.

A moments thought made me realize that its not surprising and the key is
what John M says in the first quoted paragraph above.

(g) is the same size as the acceleration that the stationary observer needs
to have to keep from being in a free-fall frame.  And at the horizon that
acceleration is infinite.  Or another way to say it if you consider a
succession of constant r shells closer and closer to the event horizon,
approaching (mathematically) from the outside.  A rocket that wants to stay
stationary will have to fire its engines (or else fall-in).  the closer you
get to the horizon the "harder" your engines will have to fire, the larger
acceleration your rocket must have relative to the free-fall frame.
 (e.g. for a rocket to hover 1 meter above the earth's surface requires it
to accelerate 9.8 m/s^2 relative to the free fall frame at that point.)

This acceleration must increase to infinity at the event horizon, as we know
that at the event horizon there is no acceleration that will prevent you
from falling in.

Joel R.


> John Mallinckrodt      mailto:ajm@csupomona.edu
> Cal Poly Pomona          http://www.csupomona.edu/~ajm