This email did not make it last time I tried so here goes again.
The Plank mass m_p is given by:
m_p = (h-bar c/G)^0.5 = (h c/(2 Pi G)) ^0.5 (1)
The Plank charge q_p is given by:
q_p = (4 Pi h_bar c e_0)^0.5 = (2 h c e_0)^0.5 (2)
Applying the gravimagnetic isomorphism to (2) above, the Plank
gravimagnetic charge q_p_g is given by:
q_p_g = (2 h_g c e_g_0)^0.5 (3)
Where:
e_g_0 = 1/(4 Pi G). (4)
However, the Plank charge should also simply be i times the Plank mass
q_p_g = i m_p (5)
where i is the imaginary number i = (-1)^0.5. We thus have
q_p_g = i (h c/(2 Pi G)) ^0.5 (6)
Setting (6) and (3) equal we have:
i (h c/(2 Pi G))^0.5 = (2 h_g c e_g_0)^0.5 (7)
and solving for the gravimagnetic Plank constant h_g we have:
- (h c/(2 Pi G)) = (2 h_g c e_g_0) (8)
- (h c/(2 Pi G)) = ( h_g c/(2 Pi G)) (9)
- h = h_g (10)
Plank's constant has a 2002 CODATA recommended value of h = 6.626
0693e-34 J s
with an uncertainty of 0.000 0011e-34 J s. Under the gravimagnetic
isomorphism established in http://mtaonline.net/~hheffner/GR-and-
QM.pdf, the gravimagnetic Plank's constant h_g is given by:
h_g = - h = -6.626 0693e-34 J s
Since both constants have the same units, this provides to some
degree a unifying relation between gravimagnetism and
electromagnetism at the quantum level. Graviphotons carry the same
energy and momenta as photons at a given wavelength.
Given the gravimagnetic expression for graviphoton momentum,
p_g = h_g/lambda
We see the momentum carried by the graviphoton is negative. This
means impact with a graviphoton provides a thrust in the direction
from which the graviphoton came. A graviphoton rocket would have to
emit graviphotons in the direction in which it accelerates.
Given the gravimagnetic analog to Plank's equation
E_g = h_g nu
the energy carried by a graviphoton is negative. This is a curious
thing. Perhaps a mechanism exists to simultaneously emit photons in
one direction and graviphotons in the other and thereby provide
energy free propulsion.
Horace Heffner
