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The Calculus Behind
The World's Quickest Derivation of E = mc²

John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh

Consider a body of mass $m$ that moves at a speed $v$ very close to the speed of light. A force $F$ acts on it and, as a result, the energy $E$ of the body increases. The speed of the body $v$ cannot exceed c and, as the force continues to act, the speed $v$ approaches c asymptotically. We have:

$${\displaystyle \frac{dE}{dt}}=Fv={\displaystyle \frac{d(mv)}{dt}}v=v^2{\displaystyle \frac{dm}{dt}}+mv{\displaystyle \frac{dv}{dt}}$$

In this last equation, t is the independent variable. So the equation tracks how energy $E(t)$, mass $m(t)$ and speed $v(t)$ grown as a function of time t.

Our concern, however, is to see how mass grows as we increase the energy of the body. To track that, we need to make energy $E$ the independent variable; and to take time $t(E)$, mass $m(E)$ and speed $v(E)$ all to be functions of $E$. Multiplying the last equation by ${\displaystyle \frac{dt(E)}{dE}}$and using the chain rule, we recover:

$$1=v^2{\displaystyle \frac{dm}{dt}}{\displaystyle \frac{dt}{dE}}+mv{\displaystyle \frac{dv}{dt}}{\displaystyle \frac{dt}{dE}}=v^2{\displaystyle \frac{dm}{dE}}+mv{\displaystyle \frac{dv}{dE}}$$ Rearranging we recover $${\displaystyle \frac{dm}{dE}}={\displaystyle \frac{1}{v^2}}-{\displaystyle \frac{m}{v}}{\displaystyle \frac{dv}{dE}}$$

We now take the limit in which the energy $E$ grows large. In that limit, $v$ approaches c asymptotically and ${\displaystyle \frac{dv}{dE}}$ approaches zero. In this limit we have: $$\underset{E\to \mathrm{\infty }}{lim}{\displaystyle \frac{dm}{dE}}={\displaystyle \frac{1}{c^2}}$$

This equation is the result. It tells us that mass $m$ grows incrementally by ${\displaystyle \frac{1}{c^2}}$ for each unit of energy $E$ added. While this result holds generally, this simple demonstration returns the result only in the special case in which the energy $E$ of the body has grown so large that its speed is close to c.

Thanks to Christian Seberino for suggesting this calculation.

Copyright John D. Norton. May 8, 2015.