CLASSICAL-MECHANICS · Interactive Practice | Unit 1 · Video 7

The Universe Has Rules: Deriving Physics from Units Alone

IKey Formulas

Formula	Name	Description
$[L],\ [M],\ [T]$	The three base dimensions	Length, Mass, Time
$T = 2\pi\sqrt{\dfrac{l}{g}}$	Pendulum period	Independent of mass and amplitude
$\ell_p = \sqrt{\dfrac{G\hbar}{c^3}}$	Planck length	$\approx 1.6\times10^{-35}\,\text{m}$
$\alpha = \dfrac{e^2}{2\varepsilon_0 c h}$	Fine structure constant	Dimensionless, $\approx 1/137$
$R_y = \tfrac{1}{2} m_e c^2 \alpha^2$	Rydberg energy	$\approx 13.6\,\text{eV}$ (binds hydrogen)

The one rule that powers everything: both sides of any physical equation must have identical dimensions.

IIInteractive Visualizations

Visualization 1 — The pendulum's period

Dimensional analysis forces the form $T \propto \sqrt{l/g}$ — and forbids mass from entering at all.

Length l (m) 1.00 Gravity g (m/s²) 9.8 Bob mass m (kg) 1.0 Phase

💡 $[M]$ appears in no other candidate quantity, so nothing can cancel it — dimensional consistency keeps mass out of the period entirely.

Visualization 2 — The scales of the universe

Feeding $c$ , $\hbar$ , and $G$ into dimensional analysis yields the Planck length, $\ell_p = \sqrt{G\hbar/c^3} \approx 1.6\times10^{-35}\,\text{m}$ .

💡 No experiment sets the Planck length — it is built from pure constants, and dimensional analysis is the only tool that even names such a scale.

Visualization 3 — The fine structure constant sizes the atom

The dimensionless $\alpha \approx 1/137$ fixes hydrogen's binding energy through $R_y = \tfrac{1}{2} m_e c^2 \alpha^2$ .

Fine structure constant α 0.0073

IIIQuiz Questions

Question 1

Two identical pendulums have the same length but different bob masses — one is light, one is heavy. According to dimensional analysis, how do their periods compare?

True or False: The heavier bob has a longer period.

✅ Correct! Mass has no quantity to cancel its dimension, so it cannot appear — both periods are equal.

❌ Not quite. The period $T = 2\pi\sqrt{l/g}$ contains no mass at all — the bobs swing identically.

Show solution

Solution:

The candidate quantities are $l\,[L]$ , $m\,[M]$ , $g\,[LT^{-2}]$ , and $\theta_0$ (dimensionless). The dimension $[M]$ appears only in the mass — there is no other quantity to cancel it. For the equation to be dimensionally consistent, mass cannot appear at all.

$T = 2\pi\sqrt{\frac{l}{g}}$

The period is completely independent of mass, so both pendulums swing with the same period. The statement is False.

Question 2

A pendulum has length $l = 1\,\text{m}$ on Earth where $g = 9.8\,\text{m/s}^2$ . Using $T = 2\pi\sqrt{l/g}$ , what is its period (to one decimal place)?

✅ Correct! $2\pi\sqrt{1/9.8} \approx 2.0$ s.

❌ Not quite. Compute $\sqrt{1/9.8} \approx 0.32$ , then multiply by $2\pi$ .

Show solution

Solution:

$T = 2\pi\sqrt{\frac{l}{g}} = 2\pi\sqrt{\frac{1}{9.8}} = 2\pi \times 0.3197 \approx 2.0\,\text{s}$

This matches the well-known result that a 1-metre pendulum has a period of about two seconds. (The choice 6.3 s forgets the square root and just takes $2\pi$ ; 0.3 s is $\sqrt{l/g}$ without the $2\pi$ factor.)

Question 3

We build the Planck length from $c\,[LT^{-1}]$ , $\hbar\,[ML^2T^{-1}]$ , and $G\,[L^3M^{-1}T^{-2}]$ by writing $\ell_p = c^{a_1}\hbar^{a_2}G^{a_3}$ and demanding the result have dimension $[L]$ .

Matching the Mass dimension gives one equation. What is it?

✅ Correct! $\hbar$ contributes $M^{+1}$ and $G$ contributes $M^{-1}$ , so $a_2 - a_3 = 0$ .

❌ Not quite. Only $\hbar\,(M^{+1})$ and $G\,(M^{-1})$ carry mass; setting their total power to zero gives the equation.

Show solution

Solution:

Only $\hbar$ and $G$ carry the dimension of Mass: $\hbar$ has $M^{+1}$ and $G$ has $M^{-1}$ (and $c$ has $M^0$ ). The total power of $M$ is therefore $a_2 - a_3$ . Since the Planck length is pure $[L]$ , it contains no mass:

$a_2 - a_3 = 0$

Combined with the Time and Length equations this yields $a_2 = a_3 = \tfrac{1}{2}$ , $a_1 = -\tfrac{3}{2}$ , giving $\ell_p = \sqrt{G\hbar/c^3}$ .

Question 4

Why is the fine structure constant $\alpha \approx 1/137$ considered a more "universal" fingerprint of nature than the Planck length $\ell_p \approx 1.6\times10^{-35}\,\text{m}$ ?

✅ Correct! Being dimensionless, α has the same value in every unit system — independent of human choices.

❌ Not quite. The key is that α has no dimensions, so unit changes leave it unchanged — unlike a length.

Show solution

Solution:

A length like $1.6\times10^{-35}\,\text{m}$ becomes a different number if you measure in feet or miles — its value depends on a human-chosen unit. But $\alpha = e^2/(2\varepsilon_0 c h)$ is dimensionless: every unit cancels, so $\alpha \approx 1/137$ in SI units, in CGS units, in natural units, for any alien civilization. It depends on nothing humans chose, which is exactly why it is a genuine, universal fingerprint of nature.

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