Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif

[Robin van Spaandonk]

In reply to  Horace Heffner's message of Sat, 11 Oct 2008 17:49:52 -0800:
Hi,
[snip]
>This is because the electric field about an infinite plane of uniform  
>charge is given by:
>
>    E = a rho/(2 * epsilon_0)
>
>so it is just a matter of applying the gravimagnetic isomorphism to  
>obtain the result.  In both formulations rho includes the sign of  
[snip]


....however in reality, the plane is not infinite. In fact if you look at real
galactic jets, the jet usually extends much farther out into space than the
diameter of the accretion disc. 

Therefore, consider a point e.g. 10% in from the edge of an accretion disk and
some distance away from it. An inscribed circle in the plane of the accretion
disk centered on the normal projection of the point onto the plane thereof, and
with a radius of 10% of that of the accretion disc will have perpendicular
gravity vector components that cancel one another, while the parallel components
(toward the disc) all reinforce one another. IOW if that small (non-concentric)
circle were all there were, then the point mass would indeed experience an
attractive force normal to the disc. However it isn't all there is. The rest of
the accretion disc is there too, and it is largely to one side of the small
"virtual" disk, hence its gravitational component will shift the direction of
the overall vector toward the centre of the accretion disk.
(and that's without taking the mass of the black hole itself into account).

(The "virtual" disc is inside the real one, has a smaller radius, and it's outer
edge just touches the outer edge of the real disc - see attached gif file).

Regards,

Robin van Spaandonk <[EMAIL PROTECTED]>

<<attachment: discs.gif>>