Non-inertial reference frames

Every navigation system that cannot see the sky runs on this article's equations. An inertial navigation unit — in an airliner, a submarine, a launch vehicle — is a box of accelerometers and gyroscopes that integrates equation (5) in reverse: it measures the frame's $\mathbf{A}$ and $\boldsymbol{\omega}$ thousands of times per second and reconstructs position without any external signal. The exquisite modern versions are ring-laser gyroscopes: the "G" ring laser at Wettzell in Germany senses Earth's rotation optically to parts in $10^{8}$, precise enough to watch the planet's spin axis wobble daily and to feel the Moon tugging the length of the day. Meanwhile every phone contains the budget version — MEMS gyros reading fictitious forces on a vibrating speck of silicon. The frontier application inverts the logic: instead of correcting for a non-inertial frame, build an inertial one. Drag-free spacecraft enclose a free-floating test mass and steer the surrounding satellite so the mass never feels the hull — LISA Pathfinder demonstrated residual accelerations below $10^{-14}\ \mathrm{m/s^2}$, and the full LISA gravitational-wave observatory now under construction for the 2030s will fly three such laboratories, the most inertial frames ever engineered. At that fidelity the only "force" left is the one Einstein taught us is also fictitious: gravity itself, encoded in the curvature that LISA is being built to hear.

Start with the test. Take a puck, remove every force you can — flat frictionless ice, no strings, no fields — and let it go. In an inertial frame the answer is Newton's first law verbatim: a dead-straight line at constant speed. Now watch the same puck from a braking car: it appears to accelerate forward, though nothing touches it. Watch it from a turntable: it sweeps out a curve, as if steered by an invisible hand. The puck did nothing new. Your coordinates did. A frame is non-inertial precisely when free particles refuse to move uniformly in it — that is the operational definition, no philosophy required.

Where does that leave the observers in the failing frames? They have a choice. They can transform everything back to an inertial frame, do honest Newtonian mechanics there, and transform the answer forward — always correct, often miserable. Or they can stay put and repair the law: add correction forces, invented precisely so that $\mathbf{F}=m\mathbf{a}$ works again with locally measured accelerations. Those corrections are the fictitious forces (their full taxonomy and physical status get their own article), and this article's job is to derive where they come from. Two prototypes generate everything. A frame that only translates with acceleration $\mathbf{A}(t)$ needs a single uniform correction $-m\mathbf{A}$ — the forward lurch in the braking car, explored in depth in the linearly accelerating frames article. A frame that rotates at $\boldsymbol{\omega}$ needs three more, of which the velocity-dependent one — the Coriolis force — is subtle enough to command its own companion deep-dive. The geometry behind the repair kit is worth seeing before the algebra. An arbitrary frame $S'$ differs from an inertial frame $S$ in two independent ways: its origin moves, and its axes turn.

Each sin will cost us. The moving origin donates one correction term; the turning axes donate three, because rotation contaminates every time derivative taken in the primed frame. And here is the deepest single fact in the subject: every one of those corrections is proportional to the mass of the body it acts on. Heavy toolbox, light feather — same extra acceleration. Only one other force in nature behaves that way: gravity. That coincidence, promoted by Einstein to a principle, is why a sealed elevator cannot mechanically distinguish acceleration from a gravitational field, and it is the conceptual on-ramp from this article to general relativity. Newton's bucket completes the picture, because it shows that non-inertial effects are facts, not artifacts. The concave water surface is there for all observers — inertial, rotating, or upside-down. What changes between frames is only the explanation: the inertial observer says each water parcel needs a centripetal force so the surface tilts to supply it; the co-rotating observer says an outward centrifugal force pushes the water up the walls. Same parabola either way.

How non-inertial is your frame? A laboratory bolted to Earth rides three accelerations at once: the daily spin contributes up to $\omega_\oplus^2 R_\oplus \approx 0.034\ \mathrm{m/s^2}$ at the equator, the orbit around the Sun about $0.006\ \mathrm{m/s^2}$, and the Sun's orbit around the Galaxy a serene $2\times10^{-10}\ \mathrm{m/s^2}$. There is no perfectly inertial frame in the building, the Solar System, or the observable universe — only a hierarchy of better and better approximations, and the professional skill is knowing which rung of the ladder your problem requires.

The whole subject follows from differentiating one picture twice. Let a particle sit at $\mathbf{r}_{\mathrm{in}}$ in the inertial frame $S$, while the non-inertial frame $S'$ has its origin at $\mathbf{R}(t)$ and its axes rotating at $\boldsymbol{\omega}(t)$. The particle's position bookkeeping is pure vector addition — the diagram above written as an equation:

where: $\mathbf{R}(t)$ — the inertial-frame position of $S'$'s origin; $\mathbf{r}_{\mathrm{rel}}$ — the particle's position relative to that origin. Nothing has happened yet: (1) is true for any two frames whatsoever. The physics enters when we differentiate, because the components of $\mathbf{r}_{\mathrm{rel}}$ are kept along axes that may themselves be turning. That turning is handled once and for all by the master identity of rotating-frame mechanics. If the primed axes rotate at $\boldsymbol{\omega}$, then any vector $\mathbf{Q}$ — position, velocity, angular momentum, anything — has two notions of time derivative, and they differ by a single cross product:

where: the subscripts mark the frame in which the rate of change is evaluated. The extra term $\boldsymbol{\omega}\times\mathbf{Q}$ is the velocity of the tip of a vector frozen in the rotating frame and carried around by it. Applied to $\boldsymbol{\omega}$ itself the identity gives $\dot{\boldsymbol{\omega}}_{\mathrm{in}} = \dot{\boldsymbol{\omega}}_{\mathrm{rot}}$ — both frames agree on the angular acceleration, a small mercy we will use freely.

With the identity in hand, differentiating (1) once relates the velocities. The origin contributes its own velocity $\dot{\mathbf{R}}$, and the relative position, differentiated through the identity, contributes both the velocity the rotating observer measures and the sweep of the axes:

where: $\mathbf{v}_{\mathrm{rot}} = (\mathrm{d}\mathbf{r}_{\mathrm{rel}}/\mathrm{d}t)_{\mathrm{rot}}$ — the velocity measured with the primed rulers and clock. Read it as three stacked motions: ride the origin, move across the platform, get carried around by the platform's turn. For a child walking on a moving, spinning carousel, the three terms are literally her stroll, the carousel's travel, and its spin. Differentiating once more — carefully, with the identity acting on each rotating-frame quantity — produces the acceleration transformation, and with it the entire non-inertial repair kit.

Which formula answers which question

Two of the four fictitious forces deserve their sizes on record here; their full personalities live in the sibling articles. The centrifugal force always points radially away from the rotation axis, with the double cross product collapsing to a clean magnitude:

where: $\boldsymbol{\rho}$ — the perpendicular vector from the rotation axis to the particle. It grows linearly with distance from the axis and quadratically with spin rate, acts on moving and stationary bodies alike, and is the term that shapes Newton's parabolic water surface. And the honest answer to "is my Earth-fixed lab inertial?" is a number. For a body at rest on the equator, the centrifugal correction to gravity is

where: $\omega_\oplus \approx 7.29\times10^{-5}\ \mathrm{rad/s}$ — Earth's sidereal rotation rate; $R_\oplus \approx 6.37\times10^{6}\ \mathrm{m}$. A third of a percent: invisible to a bathroom scale's user, glaring to a geodesist, and — accumulated over continental scales and days — the organizing principle of every weather map via the Coriolis term.

1 · The toolbox on the carousel (confidence builder)

A playground carousel turns once every $6.0\ \mathrm{s}$. A $2.0\ \mathrm{kg}$ toolbox rests on its deck at $r = 1.5\ \mathrm{m}$ from the axis. In the carousel frame, find the fictitious force on the toolbox and the minimum coefficient of static friction that keeps it in place.

Step 1. The spin rate: $\omega = 2\pi/T = 2\pi/6.0 = 1.047\ \mathrm{rad/s}$, constant, so the Euler force is zero; the toolbox is at rest in the rotating frame, so the Coriolis force is zero too. Equation (5) leaves exactly one fictitious force: centrifugal.

Step 2. Its magnitude from (6): $F_{\mathrm{cf}} = m\omega^{2}r = 2.0 \times (1.047)^{2} \times 1.5 = 2.0 \times 1.096 \times 1.5 = \boxed{3.3\ \mathrm{N}}$, pointing radially outward.

Step 3. In the carousel frame the toolbox is in equilibrium, so static friction must supply $3.3\ \mathrm{N}$ inward. The requirement $\mu_s m g \geq m\omega^{2}r$ gives $\mu_s \geq \omega^{2}r/g = 1.645/9.81 = \boxed{0.17}$ — easily met by rubber on steel, marginal for a smooth plastic case on wet paint.

Step 4. Translate to the inertial ledger as a cross-check: there, no centrifugal force exists; instead the toolbox travels a circle and needs a centripetal force $m\omega^{2}r = 3.3\ \mathrm{N}$, which friction provides, pointing inward. Same friction, same number, opposite narrative — the two frames must and do agree on every measurable.

2 · Designing a spinning space habitat (exam level)

A ring-shaped station of radius $R = 200\ \mathrm{m}$ spins to give its crew Earth weight at the rim. Find the required spin rate and period, the head-to-foot gravity difference for a $1.8\ \mathrm{m}$ astronaut, and the apparent weight change for a crew member running at $5.0\ \mathrm{m/s}$ along the rim, with and against the spin.

Step 1. Standing "on" the rim means the centrifugal force plays gravity: $\omega^{2}R = g$, so $\omega = \sqrt{9.81/200} = \sqrt{0.04905} = \boxed{0.221\ \mathrm{rad/s}}$ — a period $T = 2\pi/\omega = 28.4\ \mathrm{s}$, about $2.1$ revolutions per minute, comfortably below the $\sim 3$ rpm threshold where inner-ear disorientation sets in.

Step 2. Head-to-foot gradient: centrifugal "gravity" scales as $\omega^{2}\rho$, linear in distance from the axis, so $\Delta g/g = 1.8/200 = \boxed{0.9\%}$ — imperceptible. In a small $R = 20\ \mathrm{m}$ station the same astronaut would lose 9% of her weight between feet and head, one reason habitat designs favour large rings.

Step 3. Now the run. In the rotating frame a runner has $\mathbf{v}_{\mathrm{rot}} \neq 0$, so the Coriolis term in (5) wakes up, pointing radially outward or inward depending on direction. Cleanest is the inertial cross-check: the rim moves at $\omega R = 44.3\ \mathrm{m/s}$, so running with the spin the runner's true speed is $49.3\ \mathrm{m/s}$ and her required centripetal acceleration is $(49.3)^{2}/200 = 12.15\ \mathrm{m/s^2}$; against the spin, $(39.3)^{2}/200 = 7.72\ \mathrm{m/s^2}$.

Step 4. The rotating-frame bookkeeping reproduces this exactly: $g_{\mathrm{app}} = \omega^{2}R \pm 2\omega v + v^{2}/R = 9.81 \pm 2.21 + 0.13$, giving $\boxed{12.15\ \mathrm{m/s^2}}$ with the spin (24% heavier) and $\boxed{7.72\ \mathrm{m/s^2}}$ against it (21% lighter) — the middle term is the Coriolis force, the small last term the centrifugal effect of her own circling. Sprinting against the spin is the budget version of anti-gravity; the full story of that velocity-dependent term is the companion Coriolis force article. Answer summary: $\omega = 0.221\ \mathrm{rad/s}$ ($T = 28.4\ \mathrm{s}$, 2.1 rpm); head-to-foot difference 0.9%; running changes apparent weight by roughly $\pm 22\%$ at jogging speed.

Texts. Morin, Introduction to Classical Mechanics, ch. 10 (Accelerating frames of reference) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §4.9–4.10 · Landau & Lifshitz, Mechanics, §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, ch. 9 (Mechanics in non-inertial frames). The master law follows from applying the operator identity $(\mathrm{d}/\mathrm{d}t)_{\text{in}}=(\mathrm{d}/\mathrm{d}t)_{\text{rot}}+\boldsymbol{\omega}\times$ twice and collecting terms; the factor-2 Coriolis coefficient emerges from that expansion. Numerical constants used throughout: $\omega_\oplus=7.29\times10^{-5}\,\mathrm{rad\,s^{-1}}$ (one rotation per sidereal day, $86164\,\mathrm{s}$), $R_\oplus=6.37\times10^{6}\,\mathrm{m}$, $g=9.81\,\mathrm{m\,s^{-2}}$, giving $\omega^2 R/g\approx 3.4\times10^{-3}$. Worked-example results ($F_{\mathrm{cf}} = 3.3\ \mathrm{N}$ and $\mu_s \geq 0.17$ for the carousel; $\omega = 0.221\ \mathrm{rad/s}$, $T = 28.4\ \mathrm{s}$, and $12.15/7.72\ \mathrm{m/s^2}$ for the habitat) follow directly from the boxed formulas.

Builds on: Newton's laws and inertial frames · Vector calculus and cross products · Angular velocity vector of a rotating frame. Leads to: Fictitious (inertial) forces · Coriolis force · Effective gravity on the rotating Earth · Equivalence of uniform acceleration and gravity.