In a message dated 8/24/2011 12:11:32 P.M. Eastern Daylight Time,
Spinozalens@aol.com writes:
Do Extremal Black Holes have Inertia?
Here is I think is an interesting question. Based on the Kerr Newman
metric
we can write a generalized equation for a black hole radius as; .
R= [G/c^2]*[m+sqrt[m^2-Q^2/G-(c^2/G^2)*a^2]]
Where Q is electric charge and a is angular momentum.
This gives us two black hole radiuses , the gravitational radius R(+)
which corresponds the positive square root and the null radius R(-)
which
corresponds to the negative square root. It believed that nature
enforces the
relationship
M^2 = > Q^2/G- (c^2/G^2)*a^2
to avoid a naked singularity. However, the case where the null radius
equals the gravitational radius we call an Extremal Black Hole. String
theory
is able to calculate rigorously the horizon degrees of freedom for
Extremal
black holes only. No one knows if the Extremal state for black holes is a
physically achievable state. Certainly not black holes produced by the
gravitational collapse of spent stars, but perhaps micro black holes ,
akin to
subatomic particles , might exist in Extremal states. For these states
the
quantization of spin should be in effect. Using the equations above we
can
produce two theoretical possibilities. for Extremal black holes. Of
course there are unlimited examples.
*************************
Case 1
m= 1.22 E 5 KG
Q= 1 ( charge of positron)
s= (1/2)*hbar
R(+) =R(-) = 9.08 E-28 Meters
*****************************************
Case 2
m= 1.54E-8 KG
Q=0
s= (1/2)*hbar
R= 1.14E -35 meters
**********************************************
As an aside an interesting proposal by Copperstock and Faraoni suggest
that
micro black hole radius be restricted the Compton wavelength giving an
extended Planck scale they call M_plex where solving for m in the first
equation above gives us;
M_plex= m_plk*sqrt[2*(1 +s^2)/(2-alpha*Q^2)]
Where alpha is the fine structure constant.
In any case this explanation gets us to my question. For any black hole
the
gravity field equals
g= R(+) - R(-)*c^2/(2*R(+))
So given that for an Extremal black hole
R(+)=R(-) we see that
g=0
An Extremal black hole has no gravity field, in essence no gravitational
mass. So the question is does it have inertial mass. Absent inertial mass,
the Extremal BH would follow NULL geodesics the same way photons do.
What happens to a Extremal black hole when it enters a gravity field?
Imagine the Extremal BH in case 1 above entering the gravity field of the
earth.
The earth wouldn't see any gravity field from the Extremal black hole,
but would the Extremal black hole see the gravity field of the earth.
A much better way to think about this , in terms of General Relativity, is
that the Extremal black hole creates no space time curvature but the
earth
does. Therefore the Extremal black hole should follows a NULL geodesic
path
created by the earth , hence effectively Extremal black holes are
massless,
that is they have zero inertial mass. Note that this would seem to be
required by the equivalence principle since we must have
M_grav= k*M_inertia
Where k is any positive number and is set to unity by standard units.