×

# Relativistic Energy-momentum Relation

Begin with the relativistic momentum and energy: $p = \frac{mv}{\sqrt{1-{v}^{2}/{c}^{2}}}$

$E = \frac{m{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}.$ Derive the relativistic energy-momentum relation: ${E}^{2} = {(pc)}^{2} + {(m{c}^{2})}^{2}.$

Solution

There are some technical details we must note:

1) The momentum $$p$$ and velocity $$v$$ are vectors.

2) The gamma factor is usually written as $\gamma = \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}.$

However, it is more accurate to write $\gamma = \frac{1}{\sqrt{1-\frac{v\cdot v}{{c}^{2}}}}$ or $\gamma = \frac{1}{\sqrt{1-\frac{{\left|v\right|}^{2}}{{c}^{2}}}}.$

Part 1

We begin with the relativistic momentum and energy: $p = \frac{mv}{\sqrt{1-{v}^{2}/{c}^{2}}}$

$E = \frac{m{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}.$

With a little algebra we discover that

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

Part 2

Now we square the relativistic energy and rearrange the equation

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

Using the relation derived in Part 1, we find that

${E}^{2} = {(pc)}^{2} + {(m{c}^{2})}^{2}.$

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
2 years, 11 months ago

Sort by:

it can also be derived from Rayleigh jean's experimental equation · 2 years, 11 months ago

Really? I can't see how the Rayleigh-Jean experiment on radiation has anything to do with the gamma factor. · 2 years, 11 months ago

What is the simplest set of assumptions you need to arrive at your first equation? Staff · 2 years, 11 months ago

Good point. I was debating on whether I should include the relativistic momentum and energy. I will add them. · 2 years, 11 months ago

I was thinking of how to arrive at those expressions in the first place. Staff · 2 years, 11 months ago