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Cleaning up after Matt Parker

September 16, 2026 at 11:36 PM by Dr. Drang

This video Matt Parker posted a couple of days ago really pissed me off:

It’s cheating to say that a result arrived at through classical mechanics—in this case, that $\frac{1}{2}m{v}^2$ is the kinetic energy of a particle—is wrong because it’s only an approximation to relativistic physics. Of course that’s the case, and I swore at my phone when the “reveal” came.

By the end of the video, though, I had calmed down because I knew I could put together a quick post of my own by rewriting the infinite series equation he derived in a better form. Also, I could provide a general expression for the individual terms.

Let’s start with this result: that the kinetic energy of a particle (including relativistic effects) is

$$m{c}^2(\gamma -1)$$

where m is the mass of the particle, c is the speed of light, and

$$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

is the Lorentz factor, with v as the velocity of the particle. The video gets to this result at about the 8-minute mark.

Matt then does a series expansion of the $\gamma$ term to get

$$\gamma =1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\frac{35}{128}\frac{v^8}{c^8}+\dots$$

The leading 1 of this series gets canceled by the 1 that’s subtracted from γ, giving

$$\frac{1}{2}m{v}^2+\frac{3}{8}m\frac{v^4}{c^2}+\frac{5}{16}m\frac{v^6}{c^4}+\frac{35}{128}m\frac{v^8}{c^6}+\dots$$

for the kinetic energy. The leading term is the one we get from classical mechanics and the others are essentially zero unless you’re in a particle accelerator (which is discussed later in the video).

I don’t like this form for the equation. It’s cleaner if you factor out all the terms that give the expression the units of energy and then have a nondimensional expression afterward. Like this:

$$\frac{1}{2}m{v}^2(1+\frac{3}{4}\frac{v^2}{c^2}+\frac{5}{8}\frac{v^4}{c^4}+\frac{35}{64}\frac{v^6}{c^6}+\dots )$$

Isn’t this nicer? The expression in the parentheses is a function of the ratio of the velocity of the particle to the speed of light. If we call that

$$\varphi =\frac{v}{c}$$

then the kinetic energy is

$$\frac{1}{2}m{v}^2(1+\frac{3}{4}{\varphi }^2+\frac{5}{8}{\varphi }^4+\frac{35}{64}{\varphi }^6+\dots )$$

and it’s much easier to see why the terms after the 1 are vanishingly small for most situations—all the situations for which classical mechanics applies.

One last thing. Matt sort of explained how to calculate the terms of the expansion of γ in the companion video, but there was a lot of handwaving and he bailed out after the second term. It doesn’t take too much effort to show that the series expansion of γ can be written like this:

$$\gamma =\sum _{n=0,2,4,\dots }^{\infty }\frac{n!}{2^n{\left[\left(n/2\right)!\right]}^2}{\varphi }^n$$

That means the kinetic energy, $m{c}^2(\gamma -1)$, is

$$\frac{1}{2}m{v}^2\sum _{n=2,4,6,\dots }^{\infty }\frac{n!}{2^{n-1}{\left[\left(n/2\right)!\right]}^2}{\varphi }^{n-2}$$

You can confirm that these terms match the equation we saw earlier by plugging values of n from 2 through 8 into this expression. And now we can extend the series as far as we like, even though the additional terms add essentially nothing.

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