The big ideas of physics, with the fine print

Statement

With $H(q,p)$, Hamilton's equations $\dot q=\partial_pH$, $\dot p=-\partial_qH$ generate a flow $\Phi_t$ on phase space that is symplectic: it preserves $dq\wedge dp$, and so in particular phase-space volume (Liouville). For the pendulum $H=\tfrac12p^2-\cos q$, the orbits are level sets of $H$, and the separatrix $H=1$ divides librations from rotations.

Why it's true

The Hamiltonian vector field $(\partial_pH,-\partial_qH)$ has zero divergence, because $\partial_q\partial_pH-\partial_p\partial_qH=0$. So the flow is volume-preserving. Explicit Euler, $(q,p)\mapsto(q+hp,\;p-h\sin q)$, has Jacobian determinant $1+h^2\cos q\ne1$. It expands phase space, and energy grows steadily. Symplectic Euler updates $p$ first and then $q$ with the new $p$. Its Jacobian determinant is exactly 1, and backward error analysis shows it exactly conserves a nearby "shadow" Hamiltonian $\tilde H=H+O(h)$, so the energy error stays bounded for exponentially long times.

Statement

For a ball thrown upward, $L=\tfrac12m\dot y^2-mgy$, and the classical path is the parabola. If $L$ is unchanged by a continuous symmetry $q\to q+\epsilon K(q)$, then $\frac{\partial L}{\partial\dot q}K$ is conserved (Noether, 1918).

Why it's true

Vary the path, $q\to q+\epsilon\eta$ with $\eta(t_0)=\eta(t_1)=0$, and integrate by parts. Then $\delta S=\epsilon\int\eta\,\big[\partial_qL-\tfrac{d}{dt}\partial_{\dot q}L\big]dt$, which vanishes for all $\eta$ exactly when the Euler–Lagrange equation holds. In quantum mechanics, Feynman's path integral weights every path by $e^{iS/\hbar}$. Paths near a stationary point add in phase, and the rest cancel. That is why the classical path dominates when $S\gg\hbar$.

Statement

Near a stable equilibrium, $L=\tfrac12\dot q^\top M\dot q-\tfrac12q^\top Kq$. The generalized eigenproblem $K\mathbf a=\omega^2M\mathbf a$ gives normal modes that oscillate independently. For two identical pendulums ($\omega_0^2=g/\ell$) joined by a spring of strength $\kappa$:

Why it's true

$M$ is positive definite and $K$ symmetric, so they can be simultaneously diagonalized. In normal coordinates the Lagrangian becomes a sum of independent oscillators. Starting with $x_1=A$, $x_2=0$ excites both modes equally. Then $x_1=A\cos\bar\omega t\cos\tfrac{\Delta\omega}2t$ and $x_2=A\sin\bar\omega t\sin\tfrac{\Delta\omega}2t$: a fast oscillation inside a slow envelope, with energy passing completely between the pendulums.

Statement

A system is chaotic when nearby trajectories separate as $|\delta(t)|\approx|\delta(0)|e^{\lambda t}$ with maximal Lyapunov exponent $\lambda>0$, on a bounded invariant set. The predictability horizon then grows only logarithmically with precision:

Why it's true

A system with two degrees of freedom that is integrable has two conserved quantities, and its motion lies on invariant tori. The double pendulum conserves only energy. At low energy most tori survive (KAM theorem) and motion is quasi-periodic. At high energy they break up, and the tangent dynamics stretch and fold phase space, which produces positive Lyapunov exponents. Improving the initial measurement by a factor of $10^6$ buys only about $\ln10^6/\lambda$ extra seconds.

Statement

A packet $\psi(x,t)=\int A(k)e^{i(kx-\omega(k)t)}dk$ centred on $k_0$ moves at $v_g=\omega'(k_0)$, and its width grows as

For deep water waves $\omega=\sqrt{gk}$, so $v_g=\tfrac12v_p$. For a free quantum particle $\omega=k^2/2m$, so $v_g=2v_p$. For light in vacuum $\omega=ck$, so there is no dispersion at all.

Why it's true

Expand $\omega(k)=\omega_0+v_g(k-k_0)+\tfrac12\omega''(k-k_0)^2$. The linear term translates the envelope rigidly. The quadratic term multiplies each component by a chirped phase, which for a Gaussian packet gives the formula above. The method of stationary phase gives the same result for any smooth $A(k)$.

Statement

For any normalized $\psi\in L^2(\mathbb R)$ with transform $\tilde\psi(k)$, the spreads satisfy $\sigma_x\sigma_k\ge\tfrac12$, with equality exactly for Gaussians. More generally, for observables $A,B$ (Robertson, 1929):

Why it's true

Let $f=(A-\langle A\rangle)\psi$ and $g=(B-\langle B\rangle)\psi$. Cauchy–Schwarz gives $\|f\|^2\|g\|^2\ge|\langle f,g\rangle|^2\ge(\operatorname{Im}\langle f,g\rangle)^2=\tfrac14|\langle[A,B]\rangle|^2$. Equality needs $f\propto g$ with an imaginary constant, which for $x$ and $p$ is the first-order ODE whose solution is the Gaussian.

Statement

For $H=-J\sum_{\langle ij\rangle}s_is_j$ with $s_i=\pm1$ on the square lattice, the spontaneous magnetization is (Onsager 1944; Yang 1952)

Near $T_c$, $m\sim(T_c-T)^\beta$ with $\beta=1/8$, and the correlation length diverges with $\nu=1$. These exponents are universal across the 2D Ising class.

Why it's true

Peierls' argument (1936) shows order at low $T$: flipping a domain costs energy proportional to its boundary, and the entropy of boundaries of length $\ell$ grows only like $3^\ell$, so small domains win at low temperature. Onsager's exact solution diagonalizes a transfer matrix. The lab samples the Boltzmann distribution with the Metropolis algorithm: flip a spin with probability $\min(1,e^{-\Delta E/T})$. That satisfies detailed balance, so the stationary distribution is the Gibbs measure.

Statement

For a particle with mobility $\mu$ (drag coefficient $\gamma=1/\mu$) in a fluid at temperature $T$, the mean squared displacement in $d$ dimensions grows linearly:

Why it's true

In the Langevin equation $m\dot v=-\gamma v+\xi(t)$ with white noise $\langle\xi(t)\xi(t')\rangle=2\Gamma\delta(t-t')$, equipartition $\tfrac12m\langle v^2\rangle=\tfrac12k_BT$ forces $\Gamma=\gamma k_BT$: the noise strength is fixed by the friction. Integrating twice, for $t\gg m/\gamma$, gives $\langle x^2\rangle=2(k_BT/\gamma)t$. The central limit theorem explains why the displacement distribution is Gaussian whatever the details of the kicks.

Statement

For a rectangular barrier of height $V_0$ and width $a$, with $\hbar^2/2m=1$, $k=\sqrt E$ and $\kappa=\sqrt{V_0-E}$:

with $q=\sqrt{E-V_0}$. For thick barriers $T\approx16\frac{E}{V_0}\big(1-\frac{E}{V_0}\big)e^{-2\kappa a}$.

Why it's true

Inside the barrier the solution of Schrödinger's equation is a combination of $e^{\pm\kappa x}$, decaying rather than oscillating. Matching $\psi$ and $\psi'$ at both edges gives four linear equations for the amplitudes, which solve to the formula. Above the barrier, $T=1$ exactly when $qa=n\pi$: the barrier region holds a whole number of half-wavelengths, and reflections from the two edges cancel (a Fabry–Pérot resonance).

Statement

Alice measures with setting $a$ or $a'$ and Bob with $b$ or $b'$. The outcomes are $\pm1$, and $E(a,b)$ is their average product. Then

For the singlet state, $E(a,b)=-\cos(a-b)$. With $a=0$, $a'=90°$, $b=45°$, $b'=135°$ this gives $|S|=2\sqrt2$.

Why it's true

With hidden variable $\lambda$ and deterministic local outcomes $A(a,\lambda),B(b,\lambda)\in\{\pm1\}$: $A(a)[B(b)-B(b')]+A(a')[B(b)+B(b')]$. One bracket is 0 and the other is $\pm2$, so each run contributes $\pm2$, and averaging over $\lambda$ gives $|S|\le2$. The quantum bound comes from $\|A\otimes(B-B')+A'\otimes(B+B')\|\le2\sqrt2$ for operators with eigenvalues $\pm1$.

Statement

Writing $v=\tanh\phi$ with rapidity $\phi$ makes the boost a hyperbolic rotation, $\begin{pmatrix}\cosh\phi&-\sinh\phi\\-\sinh\phi&\cosh\phi\end{pmatrix}$. Rapidities add: velocities compose as $\tanh(\phi_1+\phi_2)$.

Why it's true

Linearity follows from homogeneity of space and time. Invariance of the light cone $t=\pm x$ forces the transformation to preserve $t^2-x^2$ up to a factor, and isotropy plus the group property fixes that factor to 1. The $t'$ axis is the worldline $x=vt$. The $x'$ axis (where $t'=0$) is the line $t=vx$, tilted toward the light cone by the same angle. Events on a horizontal line in one frame aren't simultaneous in the other.

Statement

For timelike geodesics in the Schwarzschild metric, with $u=1/r$ and angular momentum per unit mass $h$,

Why it's true

Without the $u^2$ term the solution is the conic $u=\frac{GM}{h^2}(1+e\cos\phi)$. Treat $\varepsilon=3GM/c^2$ as a perturbation. The resonant term generates $u\propto\cos\big((1-\varepsilon GM/h^2)\phi\big)$ to first order, so perihelion recurs after $\Delta\phi=2\pi(1+3G^2M^2/h^2c^2)$. Substituting $h^2=GMa(1-e^2)$ gives the formula. For Mercury that's $5.0\times10^{-7}$ rad per orbit, and 415 orbits per century give 43″.