Relativistic Momentum-Energy Relation

Problem

Begin with the relativistic momentum and energy:

$p= \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$

$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$

Derive the relativistic energy-momentum relation:

$E^2 = (pc)^2 + (mc^2)^2$.

Solution

With a little algebra we discover that

$\frac{p}{E}=\frac{v}{c^2}$.

Square the equation for relativistic energy

$E^2=\frac{(mc^2)^2}{1-\frac{v^2}{c^2}}$

And rearrange to arrive at

$E^2-\frac{E^2 v^2}{c^2} =(mc^2)^2$

$E^2=\frac{E^2 v^2}{c^2} +(mc^2)^2$.

From the relation

$\frac{p}{E}=\frac{v}{c^2}$

we find

$pc=\frac{Ev}{c}$

and

$(pc)^2=\frac{E^2 v^2}{c^2}$.

Substitute this result into

$E^2=\frac{E^2 v^2}{c^2} +(mc^2)^2$

to get

$E^2 = (pc)^2 + (mc^2)^2$.