# A Roller Coaster Ride through Relativity/Appendix G

## The relation between Energy and Momentum

The total relativistic energy E and the relativistic momentum p of a body are given by the following expressions:

$E=\gamma M_{0}c^{2}={M_{0}c^{2} \over {\sqrt {1-v^{2}/c^{2}}}}$
$p=\gamma M_{0}v={M_{0}v \over {\sqrt {1-v^{2}/c^{2}}}}$

We wish to eliminate v from these equations. First square and multiply across:

$E^{2}(1-v^{2}/c^{2})=M_{0}^{2}c^{4}$
$p^{2}(1-v^{2}/c^{2})=M_{0}^{2}v^{2}$

Now for a diabolically cunning move, multiply the second equation by c2 and subtract!

$(E^{2}-p^{2}c^{2})(1-v^{2}/c^{2})=m_{0}^{2}c^{4}(1-v^{2}/c^{2})$

from which we obtain:

$E^{2}-p^{2}c^{2}=M_{0}^{2}c^{4}$

An alternative (and in my opinion better) way of writing this equation is:

$E^{2}-E_{0}^{2}=p^{2}c^{2}$

where E0 is the rest-mass energy of the body.

It is instructive to compare this expression with the non-relativistic relation between energy and momentum which is calculated as follows

$KE={\tfrac {1}{2}}Mv^{2}~~~{\text{and}}~~~p=Mv$

so

$KE={p^{2} \over 2M}$

It is not easy to see, at first, how the relativistic expression will reduce (as it must) to the non-relativistic one when v is small, but it does. Watch!

Since

$E^{2}-E_{0}^{2}=p^{2}c^{2}$

we can write

$(E-E_{0})(E+E_{0})=p^{2}c^{2}$

Now (E - E0) is just the relativistic kinetic energy KEr which, at low speeds approximates to the ordinary kinetic energy KE.

At low speeds, the total relativistic energy E and the rest-mass energy E0 are virtually equal and equal to Mc2 so:

$KE~.~2Mc^{2}=p^{2}c^{2}$

from which it is easy to see that

$KE={p^{2} \over 2M}$

as expected.