# Derivation of E=mc^2 in Wikipedia

1. Jan 11, 2009

### birulami

Wikipedia 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:
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.

2. Jan 11, 2009

### Fredrik

Staff Emeritus
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.

3. Jan 12, 2009

### 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,

$$E_r = E_m = E_t - E_k = (m\gamma c^2) - (m\gamma c^2 - m c^2) = m c^2$$

Where $$E_m$$ is the mass energy, $$E_t$$ is the total energy, and $$E_r$$ is the rest energy.

Sorry if that was confusing.

4. Jan 12, 2009

### 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.