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:

$E=\gamma M_0{c}^2=\frac{M_0{c}^2}{\sqrt{1-v^2/c^2}}$
$p=\gamma M_{0}v=\frac{M_{0}v}{\sqrt{1-v^2/c^2}}$

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

$KE=\frac{1}{2}M{v}^2\text{and}p=Mv$

so

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

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 KE_r 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:

$KE.2M{c}^2=p^2{c}^2$

from which it is easy to see that

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

as expected.

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