Concerning E=mc2

Warp

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

Where is that coming from?

Dead Boss

It's not the same. Why would it be? The latter is a square of the former.

Matterwave

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".

DrStupid

Assuming the energy of a body is linear correlated to its mass (as used by Newton):

[itex]E = k \cdot m[/itex]

Than the change of its energy is

[itex]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[/itex]

Integration of the resulting differential equation

[itex]\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k - v^2 }}[/itex]

leads to

[itex]m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{k}} }}[/itex]

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:

[itex]E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/itex]

Neferkamichael

Center O Bass , thank you for the insightful information.

Boston_Guy

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 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/c²

E’ = Δm’c²

From this Einstein deduced (he used L where I use E')
Therefore E = mc² as was to be shown.

Dead Boss

Now that's a weird conclusion!

Boston_Guy

Oops! My bad! I forgot the c after the m. Must be the painkillers.