## Concerning E=mc2

 Quote by Pengwuino To avoid confusion, no, $m^2 / s^2$ is not a unit of velocity. It simply is what it is.
Are you sure? Is there any difference between saying "that car is traveling at 4 $m/s$" and "that car is traveling at 16 $m^2/s^2$" (other than the latter being quite unconventional and needlessly complicated but, ultimately, the same thing as the former)?

 One might say it's a unit of Energy per mass.
Where is that coming from?

 Quote by Warp Are you sure? Is there any difference between saying "that car is traveling at 4 $m/s$" and "that car is traveling at 16 $m^2/s^2$" (other than the latter being quite unconventional and needlessly complicated but, ultimately, the same thing as the former)?
It's not the same. Why would it be? The latter is a square of the former.

 Quote by Warp Are you sure? Is there any difference between saying "that car is traveling at 4 $m/s$" and "that car is traveling at 16 $m^2/s^2$" (other than the latter being quite unconventional and needlessly complicated but, ultimately, the same thing as the former)?

Yes there is quite a difference. The first way is correct, the second way is wrong.

Would you say "the length of this table is 9 m^2"? No you would not, m^2 is a unit of area not length. You can only say "the length of this table is 3m".

 Quote by Neferkamichael What rule or mechanism or whatever allows for the squaring of the speed of light in this equation.
Assuming the energy of a body is linear correlated to its mass (as used by Newton):

$E = k \cdot m$

Than the change of its energy is

$dE = k \cdot dm = F \cdot ds = \left( {m \cdot \frac{{dv}}{{dt}} + v \cdot \frac{{dm}}{{dt}}} \right) \cdot ds = m \cdot v \cdot dv + v^2 \cdot dm$

Integration of the resulting differential equation

$\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k - v^2 }}$

$m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{k}} }}$

The constant of integration m0 is the mass of the body at rest and as this equation has real solutions for v²<k only the factor k must be the square of a maximum velocity that can not be reached or exceeded by the body. Experiments show that this speed limit is the speed of light in vacuum. Thus there can be no proportionality between energy and classical inertial mass except:

$E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$
 Center O Bass , thank you for the insightful information.

 Quote by Warp 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 = mc2 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 ex. 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)ex

p'- = -(hf-/c)ex

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 ex = -m'vf ex + (hf+/c)ex - (hf-/c)ex

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/c2

E’ = Δm’c2

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 = mc2 as was to be shown.

 Quote by Boston_Guy Therefore E = m2 as was to be shown.
Now that's a weird conclusion!

 Quote by Dead Boss Now that's a weird conclusion!
Oops! My bad! I forgot the c after the m. Must be the painkillers.