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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:

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

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 \(\dfrac{dt(E)}{dE}\)and using the chain rule, we recover:

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

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

This equation is the result. It tells us that mass \(m\) grows incrementally by \(\dfrac{1}{c^2}\) 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.