On Oct 23, 2009, at 4:26 AM, Mauro Lacy wrote:
Horace Heffner wrote:
This is to examine the feasibility that gravity has a role in fusion
at some distance. The Coulomb force between two particles is:
Fc = Cc * q1 * q2 / r^2
where Cc is the Coulomb constant 8.99x10^9 m/F, the charge q1 or q2
of a particle is typically +-1.602x10^-19 C, and r is the particle
separation.
The gravitational force between two masses is:
Fg = Gc * m1 * m2 / r^2
How do you know that those formulas are valid at those scales?
At what scales? No scale is specified.
Newton's law is only an aproximation. It assumes point masses.
The above should work fine over the volume of any portion of a
wavefunction. It's Coulomb's law, and the gravitational equivalent,
not Newton's.
So, to ve
valid, that formula has contourn conditions. Namely, that r must be
greater than the radius of the two masses.
Not true.
Because in Reality there are
no point masses.
Irrelevant. Mean forces can be summed over the wavefunctions.
Newton's law ceases to be valid when the point of equilibrium(the
point
of zero gravity) between two "point masses" lie on the inside of
one of
the "point" masses.
Not true. It appears you are confusing Newton's laws with Coulomb's law.
If this were not the case, the force would tend to
infinite at small scales(when r tends to zero), which again is
something
that does not make sense.
When the centers of charge of two wavefunctions overlap, the net
force is zero, which is just fine.
So, it's perfectly possible to think that "in between"(when r is
approaching 0), gravity could behave in a manner completely different
than at scales when r is clearly greater than the radius of the
"point"
masses.
No, gravity and charge behave normally, they are just distributed in
space.
It could behave exponentially, to a point, and reach an
equilibrium afterwards. Or it can become repulsive, when r is less
than
a given value.
Where is the evidence for this? If you are referring to spin coupling
then, again, the electromagnetic coupling overwhelms the gravimagnetic.
On the other hand, the same happens with the Coulomb force. Why are
you
inclined to talk about the Coloumb force at those scales, when the
electron orbiting then nucleus clearly violates it?
Show the violation.
The Coloumb force
again has contourn conditions, and could cease to be valid(indeed, it
ceases to be) when r tends to zero. The Coloumb force also assumes
point
charges, which again is something that does not exist in Reality.
Again, at small distances the Coulomb force is valid but takes on a
statistical nature, as does the gravitational force between chunks of
the wavefunction. The effective charge in a volume is equal to the
probability of the charge being found there times q. The equivalent
is true of the mass. Similar ratios, all greater than 10^30, apply.
Gravity is totally unimportant.
where Gc is the gravitational constant 6.673x10^-11 m^3/(kg s^2), m1
and m2 are particle masses, and r is the particle separation. Given
the ratio of neutrons to protons is typically around 1, the largest
mass to charge nucleus is tritium, which has 2 neutrons and only one
proton, and a mass of 5.00736x10-27 kg.
The best ratio brgcf of gravitational force to Coulomb force is thus:
brgcf = Fg/Fc = (Gc * m1 * m2) / (Cc * q1 * q2)
which is clearly independent of distance assuming mass and charge
occupy the same volume. The best ratio is given by:
brgcf = Gc * (5.00736x10-27 kg)^2 / (Cc * (1.602x10^-19 C)^2)
brgcf = 7.25186x10^-36
A similarly small ratio is obtained when comparing spin coupling
gravimagnetic vs magnetic forces. It thus appears gravitation plays
no significant role in fusion or in any atomic mechanics at any
distance. This even applies when only neutrons are involved, because
the electromagnetic spin coupling dwarfs both the gravitation force
and the gravimagnetic force. The force of gravity must only be large
in the interaction of extremely small and thus energetic neutral
bosons, e.g. a photon ball early in the big bang.
Comments?
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
