How does the theory of relativity explain changes in mass and energy?

[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?

 

[bassplayer142]

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

 

[Crazy Tosser]

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

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?

 

[DopplerDog]

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?

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).

 

[WarPhalange]

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

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

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

 

[jablonsky27]

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

$$E=m_{0}c^2$$ 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.

 

[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).