Re: Please explain where you get e=mc^2 from 1905 paper

franklinhu@xxxxxxxxx says...

The last part says:

"Neglecting magnitudes of fourth and higher orders we may place

K0-K1 = 1/2(L/c^2)v^2

From this equation it directly follows that:--

If a body gives off the energy L in the form of radiation, its mass
diminishes by L/c^2."

Now just how does this in any way directly follow???? He might as well
have said " ... and then a miracle happens and ...".

It's not that big of a jump.

(1) Initially, you have an object moving
at speed v in some frame, where v is low compared with the speed
of light, so the nonrelativistic formula for kinetic energy may be used:
kinetic energy = 1/2 M v^2.

(2) The object emits energy in the form of light waves in such a way
that its total energy changes by an amount approximately equal to L,
and its kinetic energy changes by an amount approximately equal to
1/2 L/c^2 v^2.

(3) Afterwards, the object is still traveling at speed v. So its
kinetic energy is still given by KE = 1/2 M' v^2, where M' is its
new mass after losing energy. So the change in its kinetic energy
is given by 1/2 (M - M') v^2.

(4) So one way of calculating the change in kinetic energy gives
1/2 (M - M') v^2, while another way of calculating the change
in kinetic energy gives 1/2 L/c^2 v^2. Putting them together,
we get (M - M') = L/c^2.

(5) But L is approximately the change in total energy of the object.
So (change in mass) = (change in energy)/c^2. Or turning that around,

(change in energy) = (change in mass)*c^2.

--
Daryl McCullough
Ithaca, NY

.

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