Warp said:
The proper way of expressing the above would be "the energy that any matter has is equal Could someone post a short summary on how the "E=mc^2" equation is derived?
That's a great idea. Here is the essense of Einstein's E = mc
2 equation.
Consider a body, at rest in the inertial frame S, which emits two photons, each of frequency f, one in the +x direction, the other in the –x direction. The total energy of the two photons is E = 2hf. For energy to be conserved there must be a decrease in the energy of the body. This implies that the body previously contained energy. There must have been an internal change that resulted in a physical state having lower value of energy. Since the total momentum of the photons is zero the emitting body must remain at rest otherwise total the total momentum of the system would not be conserved.
Now consider the process from the inertial fram S' which is in standard configuration with respect to S and is moving in the +x direction with speed
v. In S’ the body is moving in the –x’ direction with velocity
v’ = -
v e
x. An observer in S’ observed the body emit two photons. One photon is emitted in the +x’ direction and the other in the –x’ direction. The photons have frequencies f+ and f- respectively. The velocity of the body remains unchanged. Due to Doppler shift the photon moving in the +x direction is red shifted from f and the photon moving in the –x direction is blue shifted f. The shifted frequencies, i.e. f
+ and f
- are related to f by
f
+ = sqrt{[1 - (v/c)]/[1 + (v/c)]}f
f
- = sqrt{[1 + (v/c)]/[1 - (v/c)]}f
The total momentum in S’ before the photons are emitted is the initial momentum of the body given by
p'
i = m'
ive_
x
where m’
i is the initial mass of the body as measured in S’. If m’
f is the final mass of the body as measured in S’ then, since the velocity of the body remains unchanged, the velocity remains unchanged so that the momentum of the body after emission is
p'
f = m'
fve_
x
The momenta of the photons in S’ is given by
p'
+ = (hf
+/c)e
x
p'
- = -(hf
-/c)e
x
The total energy of the photons as measured in S’ is
E = hf
+ + hf
+
Conservation of momentum requires
p'
i = p'
f + p'
+ + p'
-
Substituting the values above gives
-m'
iv e
x = -m'v
f e
x + (hf
+/c)e
x - (hf
-/c)e
x
Upon equating the components on each side we get, upon rearranging terms and substituting the values in Eq. (19)
Δm'v = 2γβhf/c = Δm’v = 2γhfv/c^2
It can be shown that
E’ = 2γhf
Substituting the expression for E’ into the expression for the change in momentum
Δm’v = 2γβhf/c = E’v/c
2
E’ = Δm’c
2
From this Einstein deduced (he used L where I use E')
If a body gives off the energy L in the form of radiation its mass diminishes by L/c[sup2[/sup]. The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference, so that we are let to the more general conclusion that the mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9x1020, the energy being measured in ergs, and the mass in grammes.
Therefore E = mc
2 as was to be shown.
