Coriolis force

Every weather forecast you have ever seen is a Coriolis calculation. Global forecast models integrate the rotating-frame momentum equation — precisely equation (3) plus pressure and friction — on a spinning sphere, and the leading horizontal balance almost everywhere outside the tropics is geostrophic: pressure gradient against Coriolis, $f\,\hat z\times\mathbf{v} = -\nabla p/\rho$. That single balance explains why winds run parallel to isobars, why Northern-Hemisphere lows spin counterclockwise, and why the jet stream exists at all. Oceanographers live one derivative deeper in the same equation: Ekman spirals, western boundary currents like the Gulf Stream, and the great subtropical gyres are all shaped by the latitude dependence of $f = 2\omega\sin\lambda$. The sharpest modern twist is in your pocket. The gyroscope chip in a phone is a MEMS vibrating structure gyroscope: a micron-scale mass driven to oscillate along one axis. Rotate the phone and the Coriolis force $-2m\,\boldsymbol{\omega}\times\mathbf{v}$ shakes the mass along the perpendicular axis; capacitive electrodes read that picometre-scale quiver and report $\boldsymbol{\omega}$. The same trick, scaled up in precision, guides aircraft and spacecraft (hemispherical resonator gyros), and Coriolis mass-flow meters — which infer flow rate from the Coriolis twist of a vibrating pipe — are the precision standard of the process industry. Foucault's 1851 theatre piece is now a sensor shipped by the billion.

Slide a hockey puck straight across a spinning merry-go-round. A friend standing on the ground sees the most boring physics imaginable: no horizontal force, a dead-straight line at constant speed. But you, riding the platform, see the puck swerve sideways in a graceful arc, as though some invisible hand pushed it. Nothing pushed it. Your coordinates turned underneath it. The Coriolis force is the bookkeeping entry that lets you keep writing $m\mathbf{a} = \mathbf{F}$ on the platform anyway — geometry masquerading as a force.

Three features of the formula deserve to be internalized before any algebra. First, the velocity dependence: a passenger standing still on the merry-go-round sees no Coriolis force at all (only the outward centrifugal pull); the force switches on only when you move across the rotating frame. Second, the perpendicularity: because $\boldsymbol{\omega}\times\mathbf{v}$ is always at right angles to $\mathbf{v}$, the Coriolis force does zero work — like a magnetic force on a charge, it steers but never accelerates along the path, so kinetic energy is untouched. Third, the factor of 2, which puzzles everyone once: it is two equal effects stacked. Half comes from the rotating frame's basis vectors turning under a moving object (a constant inertial velocity acquires an apparent rotation), and half from the rotation acting on the $\boldsymbol{\omega}\times\mathbf{r}$ sweep velocity of the point the object currently occupies. Each contributes $\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rel}}$; together they make $2\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rel}}$. On Earth the acceleration $2\omega v$ is minute — for a car on a highway it is about a ten-thousandth of gravity — but it never lets up and it never changes sign. Over the hours and the thousand kilometres of a weather system, the accumulated deflection becomes the dominant horizontal force balance of the atmosphere. Which way it deflects depends on where you stand:

At latitude $\lambda$, the planet's angular velocity vector splits into a local vertical part $\omega\sin\lambda$ and a local horizontal part $\omega\cos\lambda$. For horizontal motion it is the vertical component that matters, and its sign flips across the equator: moving bodies veer to the right of their velocity in the Northern Hemisphere, to the left in the Southern, and not at all on the equator itself. Every cyclone photo from orbit is this one cross product made visible.

Everything flows from a single operator identity. Fix an inertial frame $S$ and a frame $S'$ rotating with angular velocity $\boldsymbol{\omega}$ about a common origin. Any vector $\mathbf{A}$ changes for two reasons: because its components change within the rotating frame, and because the frame's own basis vectors sweep around. Writing that sentence in symbols gives the master key of this whole subject:

where: $\mathbf{A}$ — any vector whatsoever; the subscripts mark in which frame the time derivative is evaluated. The $\boldsymbol{\omega}\times\mathbf{A}$ term is the velocity of the tip of a frozen vector carried around by the rotation. Applied to $\boldsymbol{\omega}$ itself it gives $\dot{\boldsymbol{\omega}}_{\mathrm{in}} = \dot{\boldsymbol{\omega}}_{\mathrm{rot}}$ — the angular velocity is the one vector both frames agree about.

Which formula answers which question

For geophysics one number rules them all. Set up local axes at latitude $\lambda$ ($\hat{x}$ east, $\hat{y}$ north, $\hat{z}$ up) and ask what the Coriolis force does to horizontal motion. Only the vertical component of $\boldsymbol{\omega}$ produces horizontal deflection of horizontal velocities, and meteorologists give twice that component its own name:

where: $f$ — the Coriolis parameter (also called planetary vorticity); $\omega = \omega_\oplus \approx 7.29\times10^{-5}\ \mathrm{rad/s}$ — Earth's sidereal rotation rate. At the poles $f = \pm 2\omega$ takes its extreme values; at the equator $f = 0$ and horizontal motion feels no horizontal deflection at all — the fundamental reason hurricanes cannot form within about five degrees of the equator. The most beautiful consequence is Foucault's. A pendulum is a velocity in perpetual oscillation, so the Coriolis force perpetually nudges its swing plane sideways. The plane precesses at

where: $\Omega_F$ — the precession rate of the swing plane, clockwise seen from above in the Northern Hemisphere. Note it is $\omega\sin\lambda$, not $2\omega\sin\lambda$ — the pendulum's plane turns at the local vertical rotation rate itself, completing a full circle in $23.93\ \mathrm{h}/\sin\lambda$: one sidereal day at the pole, 31.8 hours at Paris, never at the equator.

One more Earth effect closes the set: drop a stone down a deep shaft and it lands slightly to the east. Falling from height $h$, the stone starts with the eastward sweep velocity of its high starting point, which exceeds that of the ground below — equivalently, the Coriolis force $-2m\,\boldsymbol{\omega}\times\mathbf{v}$ acting on the downward velocity points east. Integrating that acceleration twice over the fall time $t = \sqrt{2h/g}$ gives

where: $\Delta x$ — eastward landing displacement; the $\cos\lambda$ appears because it is the horizontal component of $\boldsymbol{\omega}$ that couples to vertical motion — the deflection is largest at the equator and vanishes at the poles, the mirror image of the Foucault effect. The $t^3$ growth marks a doubly integrated, velocity-proportional force.

1 · The sideways push on a high-speed train (confidence builder)

A 400-tonne train runs due north at $v = 200\ \mathrm{km/h}$ at latitude $\lambda = 45°\mathrm{N}$. Find the Coriolis acceleration, the sideways force on the rails, and compare with the train's weight.

Step 1. Convert: $v = 200/3.6 = 55.6\ \mathrm{m/s}$; $m = 4.0\times10^5\ \mathrm{kg}$; $\omega = 7.29\times10^{-5}\ \mathrm{rad/s}$.

Step 2. For horizontal motion, the horizontal deflecting acceleration uses the vertical component of $\boldsymbol{\omega}$: $a_{\mathrm{Cor}} = 2\omega v\sin\lambda = 2 \times 7.29\times10^{-5} \times 55.6 \times 0.707 = \boxed{5.7\times10^{-3}\ \mathrm{m/s^2}}$, directed east — to the train's right, as always in the Northern Hemisphere.

Step 3. Force on the rails: $F = ma_{\mathrm{Cor}} = 4.0\times10^5 \times 5.7\times10^{-3} = \boxed{2.3\ \mathrm{kN}}$ pressing on the eastern rail. Over years of one-directional traffic this asymmetric wear is real railway-engineering lore.

Step 4. Compare with weight: $a_{\mathrm{Cor}}/g = 5.7\times10^{-3}/9.81 = 5.8\times10^{-4}$ — about six parts in ten thousand. Imperceptible to passengers, yet a hundred times larger per kilogram than what organizes a hurricane; the atmosphere just has far more time and room to accumulate it.

2 · Dropping a stone down a mineshaft, then checking with Foucault (exam level)

A stone is dropped from rest into a $h = 100\ \mathrm{m}$ shaft at latitude $\lambda = 40°\mathrm{N}$. Find where it lands. Then find the Foucault precession period a museum at the same latitude should observe, and explain why one effect uses $\cos\lambda$ and the other $\sin\lambda$.

Step 1. Fall time: $t = \sqrt{2h/g} = \sqrt{200/9.81} = 4.52\ \mathrm{s}$.

Step 2. Eastward deflection from (8): $\Delta x = \tfrac13\,\omega g t^3\cos\lambda = \tfrac13 \times 7.29\times10^{-5} \times 9.81 \times (4.52)^3 \times 0.766$. With $t^3 = 92.1\ \mathrm{s^3}$: $\Delta x = \tfrac13 \times 7.29\times10^{-5} \times 9.81 \times 92.1 \times 0.766 = \boxed{1.7\ \mathrm{cm}\ \text{east}}$. (At the equator the same drop gives $2.2\ \mathrm{cm}$ — measurable, and measured: such experiments in the 19th century confirmed both sense and size.)

Step 3. North–south deflection: none to this order — the Coriolis force from the vertical velocity is purely east–west, and the induced eastward velocity ($\sim$ mm/s) is too small to be re-deflected appreciably during a 4.5-second fall.

Step 4. Foucault period at $\lambda = 40°$: $T = \dfrac{23.93\ \mathrm{h}}{\sin 40°} = \dfrac{23.93}{0.643} = \boxed{37.2\ \mathrm{h}}$, clockwise seen from above.

Step 5. Why the complementary trig factors: falling motion is vertical, and vertical velocity couples to the horizontal component of $\boldsymbol{\omega}$ (magnitude $\omega\cos\lambda$) to give a horizontal force. Pendulum motion is horizontal, and horizontal velocity is turned by the vertical component (magnitude $\omega\sin\lambda$). The two experiments together measure both components of the Earth's angular velocity vector — a complete vector determination without ever looking at the sky.

Texts. Morin, Introduction to Classical Mechanics, ch. 10 (Accelerating frames of reference) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §4.10 · Landau & Lifshitz, Mechanics, §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, ch. 9 (Mechanics in non-inertial frames).

Provenance. Equations and derivations cross-checked against Morin ch. 10, Goldstein §4.10, Taylor ch. 9, and Landau-Lifshitz §39. A verifier should reproduce: (1) the four-term acceleration transformation via the double operator identity, including the factor of 2; (2) the Foucault precession rate $\Omega_F=\omega\sin\lambda$ and period $24\,\mathrm{h}/\sin\lambda$; (3) the eastward-deflection formula $\tfrac13\omega g t^3\cos\lambda$; (4) Earth's rotation rate $\omega_\oplus\approx7.29\times10^{-5}\,\mathrm{rad/s}$ (one sidereal day) and Coriolis parameter $f=2\omega\sin\lambda$. Worked-example numbers on this page ($a_{\mathrm{Cor}} = 5.7\times10^{-3}\ \mathrm{m/s^2}$ and $F = 2.3$ kN for the train; $\Delta x = 1.7$ cm and $T_{\mathrm{prec}} = 37.2$ h at $\lambda = 40°$) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Time derivatives in rotating frames · Transformation of acceleration in rotating frames · Angular velocity vector of a rotating frame · Fictitious (inertial) forces · Non-inertial reference frames. Leads to: Effective gravity on the rotating Earth · Centrifugal force · Euler force · Equatorial bulge and the figure of the Earth.