The Full Equation for Mass-energy Equivalence

[Jason Kim]

Hi.

I've seen a video by MinutePhysics that talked about the mass-energy equivalence equation,
usually known as E=mc^2.

It said that there is an extra part to it, and I didn't really understand what it meant.

(E^2)=((mc^2)^2)+((pc)^2) seems to be the full one (p being momentum)

So, any ideas?

By the way:

 

[ImaLooser]

Hi.

I've seen a video by MinutePhysics that talked about the mass-energy equivalence equation,
usually known as E=mc^2.

It said that there is an extra part to it, and I didn't really understand what it meant.

(E^2)=((mc^2)^2)+((pc)^2) seems to be the full one (p being momentum)

So, any ideas?

By the way:

Each equation uses a different definition of mass. e=mc^2 uses m="relativistic mass," which increases the faster the mass is moving relative to the observer. The second equation uses m="rest mass." The first equation was Einstein's, the second is used more these days because it is usually more convenient and less confusing.

Each equation is correct, given its definition of m.

 

[grzz]

If one starts from E = mc$^{2}$ and replaces m by $\frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ where m$_{o}$ is the mass at rest, one gets the other version of the equivalence equation.

 

[bcrowell]

E=mc² is just the special case where p=0.

Each equation uses a different definition of mass. e=mc^2 uses m="relativistic mass," which increases the faster the mass is moving relative to the observer. The second equation uses m="rest mass." The first equation was Einstein's, the second is used more these days because it is usually more convenient and less confusing.

I don't think this interpretation works, since nobody today uses relativistic mass, but everyone uses E=mc².