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.

Solution

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, $E = \sqrt{{(pc)}^{2} + {(m{c}^{2})}^{2}}$

hence,

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

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

#Mechanics #Relativity #SpecialRelativity #Force #RelativisticMechanics

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Comments

Elegant!! Good job

Racchit Jain - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$