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Old December 22nd 05 posted to sci.physics.relativity,sci.physics,sci.skeptic,sci.philosophy.tech
Androcles
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Default Einsteinians Test Einstein about E=mc2


"newedana" wrote in message
oups.com...
The Origin of Unscientific Equation, E=Mc^2.

It starts from the special theory of relativity for mass,
m=m'(1-v^2/c^2)^-1/2. If you expand this to be a poly-nominal series,
it gives, m=m'[1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . . .].

Since v/c is negligibly small as in usual case, you can eliminate after
third term. Thus the simplified equation becomes, (m-m')c^2
=1/2m'v^2=E, and E=mc^2.



AMERICAN JOURNAL OF PHYSICS

_______________________________

VOLUME 33, NUMBER 1. Jan 1965

_______________________________



Evidence Against Emission Theories

J.G. Fox

Another proof, more closely related to the present discussion, may be made
by the following modification of a demonstration due to Langevin. Consider
a source which is at rest with respect to an observer O and which radiates a
simultaneous, oppositely directed pair of equal quanta, hu, e.g.,
annihilation radiation. While the total energy radiated is ?E = 2hu, the
total momentum radiated is zero, so the source remains at rest with respect
to O.

Now, consider this phenomenon from the point of view of an observer O' who
moves with respect to O with the constant velocity v = bc along the line
defined by the radiation. On account of the first-order Doppler effect O'
observes two quanta with the frequencies hu (1 + b) and hu (1 - b). He thus
concludes that a net amount of momentum hu(1+b)/c - hu (1-b)/c = 2hub/c is
emitted in the direction in which the source and O appear to move with
respect to him. From the conservation principle for momentum he concludes
that the source loses this same quantity of momentum. Now the velocity of
the source with respect to O' does not change since it remains at rest with
respect to O, as has been seen. Thus O' is forced to conclude that the mass
of the source has decreased by an amount Dm, where (Dm)u = 2hub/c. Thus,
Dm = DE/c2.

Androcles.



Do you think that this false equation can explain the Atomic Nuclear
Energy? Nonsense, absolutely! E=mc^2 is simply unscientific.

If you cannot neglect v/c, you should know another way of proving the
unscientific nature of E=mc^2.
Dr. Hansik Yoon disproved both deBroglei equation, ?=h/p, and one of
the key equations in the particle physics, E=h?.

From these two unscientific equations, another false equation, E=mc^2

is directly derived:

When deBroglei equation is applied to a photon(QM theorists defined
that photon has zero mass, so they defined arbitrarily, E=pc, pc=h?),
it becomes 1/?=h/mc, where ?=1/?, p=mc. Then the unscientific
equation, E=mc^2, is derived, combining with E=h?.

The real problem with E=Mc^2 is that this equation does not suggest any
information concerning the nuclear activities producing the nuclear
energy in the atomic nuclear structures.

This equation is nothing but the half of the kinetic energy of mass
moving at the speed of light.


Based on "Natural Science Founded on A New Atomic Model" by Hansik Yoon
( http://www.yoonsatom.net and http://yoonsphysics.blogspot.com/ )


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