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Relativistic Force

Let there be a given mass \(m\), acted by a net force \(F\). If the mass travels a small distance \(dr\), prove that the infinitesimal change in energy \(dE\) equals the work done by \(F\).

Note that \(m, E\) are scalars while \(F, dr\) are vectors.


We begin by defining \(F = \frac{dp}{dt}\) and \(dr = v dt\), where \(v\) is the velocity-vector.

Thus, \[Fdr = \frac{dp}{dt}\cdot v dt.\]

In relativistic mechanics, \(E = \sqrt{{(pc)}^{2} + {(m{c}^{2})}^{2}}\)


\[\frac{dE}{dt} = \frac{p{c}^{2}}{\sqrt{{(pc)}^{2} + {(m{c}^{2})}^{2}}} \cdot \frac{dp}{dt}\]

This long expression can be reduced to \(\frac{p{c}^{2}}{E} \cdot \frac{dp}{dt}\), which can be further reduced to \(v\cdot \frac{dp}{dt}\).

Assembling the above results yield \(\frac{dE}{dt} = v\cdot\frac{dp}{dt}\).

Therefore, \[dE = F\cdot dr.\]

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

Note by Steven Zheng
2 years, 3 months ago

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Elegant!! Good job Racchit Jain · 9 months, 2 weeks ago

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