Re: value of g in black holes

[Vic Decarlo <vdecarlo@DEPAUW.EDU>]

Carl:
    I don't see an easy way of getting at that square root factor; my
derivation required nearly a full page of calculations. Certainly the
factor
makes sense in the limit R--> infinity - very far from the black hole,
g is identical to the Newtonian prediction (in these units). Ultimately,

the square root term comes from the Schwarzschild metric which
tells us about the degree of spacetime curvature (i.e., the deviation
from the flat spacetime of special relativity) near the black hole.
    Maybe someone with a little more expertise in General Relativity
than me can see a "simple" way of deriving that square root factor.

Vic DeCarlo
DePauw University


"Carl E. Mungan" wrote:

> Vic Decarlo wrote:
>
> >     I'm teaching a course on black holes this semester using
> > the new excellent new intro text by Taylor and Wheeler
> > ("Exploring Black Holes").  One of the exercises involves
> > calculating the local acceleration of gravity for someone
> > standing on a shell of reduced circumference R surrounding
> > a nonspinning, spherical black hole of mass M.  The answer
> > turns out to be
> >
> >                 g = (M/R^2) / sqrt(1 - 2M/R)
>
> Well, as I said my formula was probably wrong :-)
> Now educate me: can you give me a simple derivation of the sqrt
> factor in the above expression?
> --
> Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
>        U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
> mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/