A little math never hurt anybody

A quick snippet of the derivation of Einstein’s Mass-energy equivalence equation. This is always good for some good late night reading.


In the particle’s rest frame, the momentum is (mc,0) and so for the force four-vector to be orthogonal, its time component must be zero in the rest frame as well, so F = (0,F). Applying a Lorentz transformation to an arbitrary frame, we find

F=\left(\frac{\gamma}{c}(\mathbf{F}\cdot\mathbf{v}),\mathbf{F} + \frac{\gamma^2}{\gamma + 1}(\mathbf{F}\cdot\mathbf{v})\right)^T.

Thus the time component of the relativistic version of Newton’s second law is

\frac{\gamma}{c}(\mathbf{F}\cdot\mathbf{v})=\frac{d(m\gamma c)}{d\tau}.

Recalling the definition of work done by the applied force as

W=\int \mathbf{F}\cdot\,d\mathbf{r} = \int \mathbf{F}\cdot\mathbf{v}\,dt,

and since the change in energy is given by the work done, we have

\frac{dE}{d\tau} = \gamma\mathbf{F}\cdot\mathbf{v},

and so finally we see that, up to an additive constant,

E=m\gamma c^2 \,

from Wikipedia

This entry was posted in Interesting. Bookmark the permalink.

Leave a Reply