e=mc^2

[alchemist]

what is the actual equation of e=mc^2? this is only the simplified equation, and i have forgotten the actual one already...

[chroot]

You probably mean

E2 = p2c2 + m2c4

- Warren

[mceddy2001]

what does the P stand for? and is that equation homogenous?

[chroot]

It's a lowercase p, and it stands for (linear) momentum. I don't know what you mean by "homogenous."

- Warren

[quartodeciman]

It all depends on whether you are a massist or an energist.

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A massist is willing to attribute mass values to anything, in any state of motion. For a massist, this m is really m0, a mass attributed to something in its rest frame of reference. For a massist

E2 = p2c2 + m02c4
p = mv

are always true in any inertial frame. For light quanta,

E = pc
p = mc

, because m0 = 0 for light quanta. But m = p/c = E/c2, a mass value dependent upon total energy of a quantum.
So E = mc2 is true for a light quantum as well as a particle with a non-zero rest mass.

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An energist is willing to attribute energy values to anything, in any state of motion. For an energist, m can only be attributed to something in its rest frame, so the subscript 0 is never needed. For an energist,

p2 = E2/c2 - m2c2

is always true in any inertial frame. The energy E must come from other physics. For light quanta, p = E/c is a given, so

p2 = p2 - m2c2

, so

m2c2 = 0

. Since c > 0,

m = 0 for a light quantum.
So, E = mc2/(1 - v2/c2)1/2 only in the case of a particle with non-zero rest mass.

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Most modern day physicists, especially high-energy physicists, tend to be energists rather than massists.

[quartodeciman]

I said:

So, E = mc2/(1 - v2/c2)1/2 only in the case of a particle with non-zero rest mass.

I should have said:

So, E = mc2/(1 - v2/c2)1/2 only in the case of a particle with non-zero mass.

[pmb]

Originally posted by alchemist
what is the actual equation of e=mc^2? this is only the simplified equation, and i have forgotten the actual one already...

The equation E = mc2 is the mass-energy equation relating the mass m of a particle to the free-particle energy E. The proof can be found here

www.geocities.com/physics_world/sr/mass_energy_equiv.htm

If the particle is a tardyon (i.e. a particle which travels at speeds less than light) then

m = m0/sqrt[1-(v/c)2]

Multiply both sides by c2

mc2 = m0c2/sqrt[1-(v/c)2]

Substitute in E = mc2 to get

E = m0c2/sqrt[1-(v/c)2]

This equation can be rewritten as

E2 - (pc)2 = (m0c2)2

Pete