Derivation of E=mc^2 in Wikipedia

[birulami]

http://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies" derives the kinetic energy of a rigid body at relativistic speed to be

E_k = m\gamma c^2 - m c^2​

The continue to say:

The mathematical by-product of this calculation is the mass-energy equivalence formula—the body at rest must have energy content equal to: E_{rest}=m c^2

Can anybody explain this reasoning? Just because the zero value of the kinetic energy at zero speed has the representation x - x does not mean that the rest energy must be x, right? Or is that the reasoning?

Thanks,
Harald.

 

[Fredrik]

The reasoning is just that the energy required to accelerate a particle from 0 to v is equal to the change in the quantity \gamma m c^2, so it makes sense to think of that quantity as representing a kind of energy.

I think you have to look at conservation of four-momentum to really justify the equivalence between mass and energy.

 

[Steely Dan]

There's an important distinction here; Wikipedia calculates the kinetic energy as what you listed above, however, the rest energy is not the kinetic energy, obviously, because the kinetic energy is obviously zero when v = 0.

So what they mean is that
E_t = E_k + E_m = m\gamma c^2 - m c^2 + m c^2 = m\gamma c^2

Or, equivalently,

E_r = E_m = E_t - E_k = (m\gamma c^2) - (m\gamma c^2 - m c^2) = m c^2Where E_m is the mass energy, E_t is the total energy, and E_r is the rest energy.

Sorry if that was confusing.

 

[birulami]

Don't worry, this is not confusing. But it somehow gets me where I started. I was after a simple derivation for E=mc^2, found a link to Wikipedia and now I am back with E_t=m\gamma c^2 asking where this comes. Hmm, yes, trivially it is E_t = E_k + E_m So I need to look closer to other recommendations in this other thread, I am afraid.

Thanks,
Harald.