newedana wrote:
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.

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

Equations don't explain; theories explain. The correct way to do this
is as follows:

Let m be the rest mass of a particle. The relativistic energy of the
particle is given as

E = \gamma mc^2,

where \gamma = (1-v^2/c^2)^-1/2 and v is the speed of the particle in
some inertial frame.

Now, \gamma expands in a binomial series as

1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . .

So, substituting in, we get

E = [1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . .]mc^2

or

E = mc^2+(1/2)mv^2 + higher-order terms in v^2

Now, assume the particle comes to rest, i.e., v = 0. How shall we
interpret E = mc^2? It's not an energy associated with speed and it's
not dependent on any model of the particle. We shall interpret it as a
latent energy. How latent energy is to be transformed into kinetic or
some other form of energy is separate question.

.....

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.

That's because it's not an equation dependent on a particular atomic
theory or model. Furthermore, it's been shown to apply rigorously at
the subatomic level.