# E=mc^2, but m can change

1. Jul 8, 2008

### Crazy Tosser

So, by the theory of relativity: $$m=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$$

But then, we have $$E=mc^2$$.

So if you have (relative to YOU) a very fast moving body, when it radiates, the radiation is actually of higher energy than it would be if the body was static?

2. Jul 8, 2008

### bassplayer142

The E in E=mc^2 is only the rest energy of the particle. It does not include any Kinetic Energy.

3. Jul 8, 2008

### Crazy Tosser

KE, as in $$mv^2/2$$? No, it doesn't. But my question was just that, do relativistic effects on the mass modify the energies of the emitted waves?

4. Jul 9, 2008

### DopplerDog

No. There is a relativistic effect on emitted radiation which is known as the relativistic Doppler shift, but it doesn't have anything to do with the mass of the object doing the emitting (ignoring gravitation).

5. Jul 9, 2008

### WarPhalange

No, we don't. We have $$E=m_{0}c^2$$

6. Jul 9, 2008

### jablonsky27

$$E=m_{0}c^2$$ is the energy of a particle at rest.
$$E=mc^2$$ is the total energy of a particle.
So,
$$mc^2$$ = $$m_{0}c^2$$ + relativistic kinetic energy.

7. Jul 9, 2008

### xArcherx

That's why you should go with the full equation...

E^2 = (mc^2)^2 + (pc)^2

Since m is the rest mass, you have to add the energy from the momentum, p, which (when regarding mass bearing objects) is...

p = ɣmv
ɣ = (1-v^2/c^2)^(-1/2)

So there you get a change in momentum with a change in velocity, changing the total energy of the object and giving you the energy of the same object at rest (E = mc^2) when not at rest (E^2 = (mc^2)^2 + (((1-v^2/c^2)^(-1/2) * m * v) * c)^2).