Cosmology That ClicksEpisode 3 · the math-friendly edition

Last time: 1/137 (= v/c) inside the atom → This time: what happens if you set that v/c to zero?

Schrödinger Is the World
with Light Set to Infinity The most famous equation in quantum mechanics is really "relativity switched off."
Why does setting the speed of light c to infinity make that equation appear — traced through the ratio v/c.

Tools you'll need: fractions, limits (the feel of "make it big and it vanishes"), an energy formula \(c\to\infty\ \Rightarrow\ v/c\to 0\)

Last time we found the hydrogen electron orbits at \(1/137\) the speed of light, i.e. \(v/c=\alpha\). That \(v/c\) was also the gauge of "how much relativity matters." This time we step further and ask: what if the speed of light \(c\) were infinite? Then \(v/c\) goes to zero and all relativistic effects vanish. What's left is the Schrödinger equation — the most basic equation in quantum mechanics, a name you meet even in high school. "The Schrödinger equation is the rulebook of a world with the speed of light set to infinity" — we'll verify this, as much as possible, in equations.

STEP 01Relativity always arrives through "v/c"

In Einstein's relativity, the energy of a fast-moving object can be written like this (the famous \(E=mc^2\), extended to the moving case):

Energy of a moving object in relativity
$$E = mc^2\sqrt{1+\left(\frac{p}{mc}\right)^2}$$

Here \(m\) is mass, \(p\) is momentum (roughly "oomph," \(p\approx mv\)), and \(c\) is the speed of light. Notice where \(c\) enters: in the form \(\dfrac{p}{mc}\approx\dfrac{v}{c}\). In other words, relativistic effects always enter through "the ratio of speed \(v\) to the speed of light \(c\)." Last time's \(v/c\) is the star again here.

If \(c\) is very large, \(v/c\) is very small. The square of a small number is smaller still. So \(\left(\dfrac{p}{mc}\right)^2\) approaches zero as \(c\) grows — and this act of "ignoring the small thing" is the key to switching relativity off.

STEP 02Try it — open up the square root and drop terms as c→∞

Let's expand the energy formula into its close approximation for small \(v/c\). There's a handy fact: for small \(x\), \(\sqrt{1+x}\) is very close to \(1+\dfrac{x}{2}\) (the first step of a Taylor expansion). We use it.

Try it — unpack the energy into a visible form

Apply √(1+x) ≈ 1 + x/2, with x = (p/mc)²

$$E = mc^2\sqrt{1+\left(\frac{p}{mc}\right)^2} \approx mc^2\left(1+\frac{1}{2}\left(\frac{p}{mc}\right)^2\right)$$

Expand the inside

$$E \approx mc^2 + \frac{1}{2}mc^2\cdot\frac{p^2}{m^2c^2} = \underbrace{mc^2}_{\text{rest energy}} + \underbrace{\frac{p^2}{2m}}_{\text{the familiar kinetic energy}}$$

It splits cleanly into two. The first half, \(mc^2\), is the energy something has even at rest (yes, that \(E=mc^2\)). The second half, \(\dfrac{p^2}{2m}\) — put in \(p\approx mv\) and it becomes \(\dfrac{1}{2}mv^2\), the very kinetic energy you know from school. Out of the relativistic energy formula, the familiar kinetic energy shows its face.

Now the meaning of \(c\to\infty\) becomes clear. Finer relativistic effects continue as terms divided by big powers of \(c\): \(\dfrac{1}{c^2},\dfrac{1}{c^4},\dots\). Set \(c\) to infinity and they all drop to zero, leaving only the rest energy \(mc^2\) and the kinetic energy \(\dfrac{p^2}{2m}\). That \(\dfrac{p^2}{2m}\) is the heart of the Schrödinger equation.

STEP 03The surviving p²/2m becomes the Schrödinger equation

In quantum mechanics, rewriting an "energy formula" in the language of waves gives you an equation. We won't go into the detailed rewriting rules here, but the point is just this: what the energy is made of decides which equation you get.

What the energy is made of → which equation

If the energy is \(mc^2 + \dfrac{p^2}{2m}\) (relativity switched off) → the Schrödinger equation.
If the energy is the original \(mc^2\sqrt{1+(p/mc)^2}\) (relativity kept) → a relativistic equation (the Dirac equation, etc.).

So the Schrödinger equation is quantum mechanics with "\(c\to\infty\), relativity switched off" — the rulebook when you view the world through the flattened approximation "light is infinitely fast." Conversely, admit that light's speed is finite and look finely, and it gets upgraded to a relativistic equation. As teased last time, \(c\) is the dial for "how finely you view the world."

The reveal — the dropped term IS "fine structure" The next term we threw away in STEP 02, of order \(\left(\dfrac{p}{mc}\right)^4\), matters at the size of last time's \(\alpha^2\approx 1/18800\). This is "fine structure," the tiny splitting of the atom's colors. What Schrödinger dropped comes back when you look precisely — which is why \(\alpha\) is the gauge of "at what resolution relativity shows up." This is where it links to Episode 2.
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STEP 04"Coarse view = simple" is everywhere in physics

Our star this time was "relativity vanishes as \(c\to\infty\)," but this idea — push something to an extreme and the fine effects vanish, leaving a simple world — appears all over physics. A few familiar faces:

Set the speed of light to infinity and relativity vanishes, leaving Newtonian mechanics. Set Planck's constant \(\hbar\) to zero and quantum fluctuations vanish, again leaving classical mechanics. Move slowly enough and air resistance vanishes, leaving parabolic motion. Each has the same structure: turn some dial to its extreme, the fine terms drop out, and a familiar simple law appears. The complicated theory is the real one; the simple theory is its "coarse-grained view." Knowing this order of things makes physics far clearer.

The honest line (just one, kept light this time)

"\(c\to\infty\)" is a limit you take in your head — the speed of light doesn't actually become infinite. The real \(c\) is finite, so the Schrödinger equation is always an approximation: it works beautifully for slow electrons, but for electrons near light-speed (like the inner electrons of heavy atoms) the discrepancy becomes non-negligible and a relativistic equation is needed. It's enough to remember: "the simple formula is the coarse-grained view of the exact one."

Practice problems (solvable with just this and the last episode)
  1. Using \(\sqrt{1+x}\approx 1+\dfrac{x}{2}\), estimate \(\sqrt{1.01}\) as a decimal (\(x=0.01\)).
    Show answer
    \(1+0.01/2=1.005\). A calculator gives \(\sqrt{1.01}=1.00499\ldots\) — nearly identical. For small \(x\), this first-step approximation is plenty accurate.
  2. Substitute \(p=mv\) into the second half of the expanded energy \(E\approx mc^2+\dfrac{p^2}{2m}\) to reach the familiar form.
    Show answer
    \(\dfrac{p^2}{2m}=\dfrac{(mv)^2}{2m}=\dfrac{1}{2}mv^2\). The familiar kinetic energy.
  3. For the hydrogen electron (last time, \(v/c=1/137\)), find the scale of the first dropped relativistic term \((v/c)^2\), and say in one line how good the Schrödinger approximation is.
    Show answer
    \((1/137)^2\approx 5.3\times10^{-5}\). Only about one part in twenty thousand off, so for hydrogen the Schrödinger equation works extremely well. (For heavy atoms whose inner electrons approach light-speed, this approximation gets worse.)

Wrap-upThe simple formula is the "coarse-grained view" of the exact one

Expanding the relativistic energy \(mc^2\sqrt{1+(p/mc)^2}\) for small \(v/c\), it split into rest energy \(mc^2\) and the familiar kinetic energy \(\dfrac{p^2}{2m}\). That \(\dfrac{p^2}{2m}\) is the heart of the Schrödinger equation, and the form with every fine term dropped as \(c\to\infty\) is the most basic equation in quantum mechanics.

And the dropped terms didn't vanish for good — they come back as "fine structure" at the size of \(\alpha^2\). \(c\) is the dial for how finely you view the world; \(\alpha\) is the gauge of at what resolution relativity appears — Episodes 1, 2, and 3 all tie together through this one idea.

This is Episode 3 of "Cosmology That Clicks," a reading piece for curious high-schoolers. The expansion of the relativistic energy \(mc^2\sqrt{1+(p/mc)^2}\approx mc^2+p^2/2m\), and obtaining the Schrödinger equation in the \(c\to\infty\) limit, are correct relations; numbers are approximate. Strictly, the equation is derived via replacing momentum with an operator, but here we focus on the point that "what the energy is made of decides the equation." The Schrödinger equation is a non-relativistic approximation; the precise case needs a relativistic equation (the Dirac equation, etc.). — To print, use your browser's Print → Save as PDF (in the printed version, answers are hidden automatically).

Print / save as PDF: ⌘+P (Ctrl+P on Windows). Printing hides the answers, turning it into a problem set.