Re: recoiling photons evidence?

"George Dishman" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> "ralph sansbury" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
> >
> >     Yes and someone called these tachyons(and it is perfectly
> > reasonable to suppose orbiting charged particles at such
speeds
> > inside the electron and proton without assuming them to
increase
> > to infinite mass and undefinedness as c is approached from
> > above).
>
> The mass of the photon remains zero. The factor called
> reativistic mass can be useful for massive particles as
> long as you remain aware of its limitations but is doesn't
> work for photons as it becomes undefined (not infinite).

   Yes if you add the premise to the list of photon premises that
the change in the velocity of the proton as it moves from rest to
the speed of light is instantaneous unlike any other particle.

>
> > But to get back to the subject of reasons aside from
> > momentum conservation to expect aspacecraft to be shoved back
by
> > the emission of 8Watt,GHz photons from its transmitter.
> >    The magnetic force argument would make this possible and
this
> > argument it seems to me is more convincing than the photon
> > momentum argument.
>
> Since you seem to be saying that both would imply the
> craft will experience a force, they are not in conflict
> and no choice is needed.

> > > given the somewhat arbitrary undefined self
> > contradictory nature of the photon.
>
> You mean the contradictions you invent by using 19th
> century concept that cannot be applied to massless
> particles and trying to do just that?

 No the contradictions that entailed by the  old fashioned
concept and that can only be  avoided by adding  arbitrary
conventions.

> >    That is the pushing of the oscillating electrons in the
center
> > feed wire of a dish antenna against the reflected
oscillations in
> > the dish might push against the dish.
>
> If the craft is pushed by this, how is momentum conserved?
> For photons, the balancing momentum is simply carried off
> by the particles. In your description, the eletrons in the
> feed and the dish are both attached to the craft.

Yes I was thinking about this when I wrote it but was trying
to be accomodating and thinking that something about
the different shapes of the feed wire and the flat dish would
somehow make this possible.  But the bottom line is that  the
force of
repulsion is between the feed and dish and both are attached to
the craft. The magnetic argument works for a source and mirror
where the mirror is free to move independently of the source.
So I am sorry nature and logic are not accomodating and the
photon conservation of momentum argument and analogy with
bullets and exploding gases from a gun is not reliable because of
the difference between the photon particle properties and the
bullet
particle properties.

>
> >   Now the question is is this enough to explain the anomalous
> > acceleration?
> No. The measured acceleration corresponds to 56W of power and
> is towards the Sun while the beam is 8W and pushes the craft
> away from the Sun. The anomaly is treated as 63W equivalent
> since 8W worth of push is used overcoming the beam reaction.

   The speed of the spacecraft is greater than the speed of
planets at the
same distance so it may be that the possible charge polarization
proportional
to the motion of objects that can explain the gravitational
attraction to the sun
of planets and other objects is greater than that implied by the
force of gravity
which does not take this explicitly into account. A more detailed
argument for
this is given at http://www.bestweb.net/~sansbury


   "Compare this to the gravitational force between two protons
one meter apart which is (6.67)(10-11) times [(1.67)(10-27)]2
which if set equal to the force between electrostatic dipoles of
unknown length s, (9)(109)(es)2 implies s=(.9)(10-18).
   It is important to note that this value is consistent with
that implied by Einstein's equation for the transformation of
energy into mass and vice versa, E=mc2. The energy supposedly
latent in the mass, m, and which can be partially or wholly
transformed into energy eg including the kinetic and potential
energy of other particles etc., this energy may be viewed as the
potential energy plus the kinetic energy of a particle,m*,
orbiting another particle,m**, inside of the larger particle with
mass,m.

   That is, the energy of oppositely charged particles in an
orbital system can be written independently of the masses of the
orbiting particle and the central particle as -(9)(10-9)(2)e2
/R(x).
   Suppose the charge  of one particle is -e and the charge of
the other is +2e.  The total charge of this orbital system is +e,
the charge of a proton.  If the signs are reversed, the charge of
the orbital system is the charge of an electron.
   The total mass of this system can be denoted, m(x). Suppose
m(x) is the mass of a proton or of an electron. The deflection of
such particles in an electric field or a magnetic field can be
used to measure this mass of the particle. Thus the mass of the
proton is about 1836 times that of the electron because it is
deflected that much less by the same deflecting force. Etc..
   The rest energy of such a  particle then is m(x)c2
=-(9)(109)(2)e2 /R(x).Hence the rest mass of the
proton,m(x)=1.67(10-27)kg  =
 -(9)(109)(2)e2/c2R(x) which, if we ignore the minus sign,
determines the value of R(x)=
(18(2.58)/9(1.67))(109-38-16+27=3.08(10-18) for x=proton.
   It would seem then that  the smaller mass electron is larger
in volume than the greater mass proton!
   The problem with this argument is that we have ignored the
masses inside the orbital systems which also contribute to the
energy of the orbital system mc2=E. The implication of the
argument is that the central mass of the electron and that of the
proton are smaller than the total mass measured and that we are
measuring these masses and their binding energies as given by the
orbital system formula above
   If the initially assumed circular orbit inside electrons and
protons is now assumed to be an elliptical orbit with the
eccentricity needed to produce a 10-18 meter displacement between
centers of opposite charge, we see that Einstein's equation E =
mc2 implies, or at least is consistent with, an electrostatic
dipole theory of gravity.
    We are assuming that the charge displaced is 'e'  when in
fact it might be some multiple of 'e' greater than one or less
than one and it would be necessary to reassess "s".  The mass of
protons are known from their deflection when propelled by an
electric field  through a  magnetic field in mass spectrometers;
that is from the degree of charge polarization inside the nuclei
due to the electric field propelling them and the strength of the
magnetic field relative to the degree of  charge polarization in
the nuclei due to gravity and the gravitational strength of the
Earth.

  Now consider how many atoms there are in the Earth and how many
protons plus neutrons in the average atom eg a total of 28
protons-neutrons if all silicon on average. (56 if all iron , 12
if all oxygen etc..) There are   6.02 times 1026 atoms of silicon
in 28kg so if the mass of the Earth has (5.98 times 1024)/28
times 6.02 times 1026 atoms and each of these times has  28 (=
14protons plus 14 neutrons) yields 3.6 times 1051 dipoles. Hence
the force between half these dipoles concentrated at a point R/2
meters from the surface and a single dipole at the Earth's
surface is (9)(109)(3.6/2)(1051)times
[(6.37/2)(106)(1.6)(10-19)(.9)(10-18)]2 divided by
[(6.37/2)(106)]4. This reduces to (3.32)(10^60-38-36-12)
=(3.32)(10-26) newtons compared to (1.6)(10-26) newtons as
calculated above in the usual way with the gravitational
constant.
  Note that this dipole length in each proton neutron is due to
the 465m/sec spin of the earth,v, and the inhibiting effect of
the forces due to all the other dipoles. Thus the dipole
associated with n protons-neutrons is
(v/c)(ns/(S+s))=(n)(10^(3-8))(s/(S+s))=(10^-18) which implies
that s/S is only 10^-13 while n/10^51 is for n=1 up to n=10^38 is
a much smaller number.  This is attributable  to the fact that
more distant dipoles have a less inhibiting effect and most
dipoles are at greater distances.
   The exact process is a matter for further research. But the
equivalence of the electrostatic dipole representation of the
gravitational force and the Newtonian representation is
unequivocal."