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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
3 years, 3 months ago

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it can also be derived from Rayleigh jean's experimental equation

Nibedan Mukherjee - 3 years, 3 months ago

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Really? I can't see how the Rayleigh-Jean experiment on radiation has anything to do with the gamma factor.

Steven Zheng - 3 years, 3 months ago

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What is the simplest set of assumptions you need to arrive at your first equation?

Josh Silverman Staff - 3 years, 3 months ago

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Good point. I was debating on whether I should include the relativistic momentum and energy. I will add them.

Steven Zheng - 3 years, 3 months ago

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I was thinking of how to arrive at those expressions in the first place.

Josh Silverman Staff - 3 years, 3 months ago

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@Josh Silverman Ah. That would be a different exercise. Consider the light-clock thought experiment.

Steven Zheng - 3 years, 3 months ago

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