Galilean Invariance of Newton's Laws

Derivation

Galilean Invariance of Newton's Laws

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Statement

Under a Galilean transformation between two frames — a constant spatial translation \(\mathbf{d}\), a fixed rotation \(R\in SO(3)\), and a constant-velocity boost \(\mathbf{V}\), with absolute time \(t'=t\) — Newton's second law \(\mathbf{F}=m\mathbf{a}\) keeps exactly the same algebraic form, provided the force transforms as a vector built from relative positions and relative velocities. The transformations that leave the law form-invariant are precisely those with zero relative acceleration; these define the inertial class of frames.

Why it matters

Galilean invariance is the physical statement that no experiment in classical mechanics can detect uniform motion or absolute position: a laboratory drifting at constant velocity obeys the same dynamical equations as one at rest. This is why we speak of the inertial frames rather than one privileged frame, and it is the classical ancestor of the relativity principle later sharpened by Einstein.

The requirement singles out inertial frames operationally. Any frame accelerating or rotating relative to an inertial one acquires extra terms (fictitious forces) that spoil the form \(\mathbf{F}=m\mathbf{a}\). The symmetry group of the free equation of motion is therefore the diagnostic that separates inertial frames from non-inertial ones, and by Noether's theorem it is the source of the conservation laws of mechanics.

Assumptions

Absolute, universal time.If time were frame-dependent (as in special relativity), \(t'=t\) fails, boosts mix space and time, and the invariance group becomes the Lorentz group instead of the Galilean group.

The boost velocity \(\mathbf{V}\) and rotation \(R\) are constant in time.If \(\mathbf{V}\) or \(R\) depended on \(t\), differentiating twice would produce inertial terms (\(-m\dot{\mathbf V}\), Coriolis, centrifugal, Euler) and the law would no longer be form-invariant.

Mass is an invariant scalar.If \(m\) transformed between frames, the two sides of the law would scale differently and the equality could not be preserved.

Forces depend only on relative positions and relative velocities of the interacting bodies, and transform as vectors under rotations.If a force referred to an absolute position or absolute velocity, a translation or boost would change it, the third-law pair would no longer be invariant, and the interaction would secretly select a preferred frame.

Derivation

\[ \mathbf{r}' = R\,\mathbf{r} + \mathbf{V}\,t + \mathbf{d}, \qquad t' = t \]

Write the most general Galilean map: fixed rotation \(R\) (orthogonal, \(R^{\mathsf T}R=\mathbb{1}\)), constant boost \(\mathbf{V}\), constant translation \(\mathbf{d}\), and absolute time. Stated once in full so every later operation is explicit. A

\[ \frac{d\mathbf{r}'}{dt'} = \frac{d\mathbf{r}'}{dt} = R\,\frac{d\mathbf{r}}{dt} + \mathbf{V} \]

Differentiate once. Because \(t'=t\) we have \(d/dt'=d/dt\), and \(R,\mathbf{V},\mathbf{d}\) are constants. This is the Newtonian velocity-addition rule: \(\mathbf{v}'=R\mathbf{v}+\mathbf{V}\). A

\[ \frac{d^2\mathbf{r}'}{dt'^2} = R\,\frac{d^2\mathbf{r}}{dt^2} \]

Differentiate again; the constant \(\mathbf{V}\) drops out. Accelerations are related by the rotation alone — the boost is invisible at second order. A

\[ \mathbf{a}' = R\,\mathbf{a}, \qquad \mathbf{a}\equiv\frac{d^2\mathbf{r}}{dt^2} \]

Rename the second derivatives as accelerations. This is the key intermediate result: acceleration is a Galilean vector, insensitive to boost and translation. A

\[ \mathbf{r}_i' - \mathbf{r}_j' = R\big(\mathbf{r}_i - \mathbf{r}_j\big), \qquad \mathbf{v}_i' - \mathbf{v}_j' = R\big(\mathbf{v}_i - \mathbf{v}_j\big) \]

Subtract the map (step 1) and its derivative (step 2) for two bodies: the \(\mathbf{V}t\) and \(\mathbf{d}\) terms cancel. Relative separations and relative velocities rotate but are unshifted by boost or translation. B

\[ \mathbf{F} = \mathbf{F}\big(\mathbf{r}_i-\mathbf{r}_j,\ \mathbf{v}_i-\mathbf{v}_j\big) \;\Longrightarrow\; \mathbf{F}' = R\,\mathbf{F} \]

By assumption the force is a vector function of relative positions and relative velocities, both of which transform by \(R\) only (step 5). A vector built from rotated inputs is itself rotated. C

\[ \mathbf{F} = m\,\mathbf{a} \;\Longrightarrow\; R\,\mathbf{F} = m\,(R\,\mathbf{a}) = R\,(m\,\mathbf{a}) \]

Apply \(R\) to the original law and use linearity of \(R\) and the invariance of the scalar \(m\). Both sides carry the same rotation. B

\[ R^{-1}(R\,\mathbf{F}) = R^{-1}(R\,m\,\mathbf{a}) \;\Longrightarrow\; \mathbf{F}' = m\,\mathbf{a}' \]

Since \(R\) is invertible (\(R^{\mathsf T}R=\mathbb{1}\)), and identifying \(\mathbf{F}'=R\mathbf{F}\) (step 6) and \(\mathbf{a}'=R\mathbf{a}\) (step 4), the primed quantities satisfy the identical equation. The law reproduces itself verbatim. C

Result

\[ \mathbf{F}' = m\,\mathbf{a}' \quad\Longleftrightarrow\quad \mathbf{F} = m\,\mathbf{a} \]

Reading. Newton's second law written in the primed frame is algebraically indistinguishable from the law in the unprimed frame. The boost \(\mathbf{V}\) and translation \(\mathbf{d}\) leave no residue at all; the rotation \(R\) acts identically on both sides and cancels. No local mechanical measurement can reveal which inertial frame you occupy — only relative motions and relative separations enter, and those are exactly what the force law depends on.

Units check. Every term is a force: \([m\mathbf{a}]=\mathrm{kg}\cdot\mathrm{m\,s^{-2}}=\mathrm{N}\). The rotation \(R\) is dimensionless, so \(R\mathbf{F}\) has units of \(\mathrm{N}\); the boost contributes \(\mathbf{V}t\) with units \((\mathrm{m\,s^{-1}})(\mathrm{s})=\mathrm{m}\) to position, consistent with \(\mathbf{d}\) in metres, and disappears after two time derivatives. Dimensions balance on both sides.

Limiting cases

Pure boost (\(R=\mathbb{1},\ \mathbf{d}=\mathbf{0}\)): \(\mathbf{r}'=\mathbf{r}+\mathbf{V}t\); velocities shift by \(\mathbf{V}\), accelerations and forces unchanged — the classic "moving train" invariance.
Pure translation (\(R=\mathbb{1},\ \mathbf{V}=\mathbf{0}\)): homogeneity of space; shifting the origin changes no force, giving momentum conservation via Noether.
Pure rotation (\(\mathbf{V}=\mathbf{d}=\mathbf{0}\)): isotropy of space; both \(\mathbf{F}\) and \(\mathbf{a}\) rotate together, giving angular-momentum conservation.
Identity map (\(R=\mathbb{1},\ \mathbf{V}=\mathbf{d}=\mathbf{0}\)): trivially \(\mathbf{F}'=\mathbf{F}\); the group's identity element.
Non-relativistic limit of Lorentz (\(|\mathbf{V}|\ll c\)): the Lorentz boost reduces to the Galilean boost and this invariance is recovered as the leading approximation, correct to order \((V/c)^2\).

Breaks when

The frame accelerates or rotates in time. If \(\mathbf{V}=\mathbf{V}(t)\) or \(R=R(t)\), step 3 no longer kills the boost: extra terms \(-m\dot{\mathbf V}\) (linear), \(-2m\,\boldsymbol\omega\times\mathbf{u}\) (Coriolis), \(-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf{r})\) (centrifugal), and \(-m\,\dot{\boldsymbol\omega}\times\mathbf{r}\) (Euler) appear. The law is not form-invariant; these fictitious forces are the signature of a non-inertial frame.
Speeds approach \(c\). Absolute time \(t'=t\) fails; velocities do not add linearly (the rule becomes \((u+V)/(1+uV/c^2)\)), and the correct invariance group is the Lorentz/Poincaré group. Newton's second law must be replaced by \(\mathbf{F}=d\mathbf{p}/dt\) with the relativistic momentum \(\mathbf{p}=\gamma m\mathbf{v}\).
Velocity-dependent forces that reference an absolute frame. A drag law \(\mathbf{F}=-b\,\mathbf{v}\) written with velocity relative to a fixed medium is not Galilean-invariant: it secretly picks out the medium's rest frame. Only forces built from relative velocities preserve the form.

Failure modes

Boosting the acceleration. Writing \(\mathbf{a}'=\mathbf{a}+\mathbf{V}\) — confusing the velocity transformation (step 2) with the acceleration transformation (step 3). The boost enters velocity but not acceleration.
Forgetting the rotation acts on both sides. Concluding invariance holds only for \(R=\mathbb{1}\), because a student rotates \(\mathbf{a}\) but treats \(\mathbf{F}\) as fixed. A genuine force is a vector and rotates identically.
Treating fictitious forces as real interactions. Insisting centrifugal force is a Galilean-invariant force; it is a frame artifact that vanishes in any inertial frame and has no third-law partner.
Using absolute velocity in the force. Applying air resistance \(\mathbf{F}=-b\mathbf{v}\) with \(\mathbf{v}\) measured in the lab and then claiming the setup is Galilean-invariant.
Assuming momentum is invariant rather than covariant. Momentum \(\mathbf{p}=m\mathbf{v}\) changes under a boost (\(\mathbf{p}'=\mathbf{p}+m\mathbf{V}\)); it is the law's form, not each quantity, that is preserved.

Discussion

The derivation shows a precise division of labour among the three transformation types. Translation and boost drop out entirely at the level of acceleration — the boost survives one differentiation (shifting velocity) but not two — while rotation is the only piece that acts non-trivially on the law, and it acts identically on force and on acceleration so it cancels. This is why acceleration, not velocity or position, is the quantity Newton's law is built from: it is the lowest time-derivative that is a genuine Galilean vector insensitive to boost and translation.

The full Galilean group is ten-dimensional: three translations, three rotations, three boosts, and one time translation. By Noether's theorem each continuous symmetry corresponds to a conserved quantity — spatial translations give linear momentum, rotations give angular momentum, time translation gives energy, and the boosts give the uniform motion of the centre of mass, \(\mathbf{X}_{\text{cm}}(t)=\mathbf{X}_0+\mathbf{V}_{\text{cm}}t\). The invariance of the dynamics and the conservation laws of mechanics are two faces of the same structure.

Operationally, this invariance is what "inertial frame" means. Rather than defining inertial frames by appeal to absolute space (Newton's bucket) and then asserting the laws hold in them, one can invert the logic: the inertial frames are exactly the set within which \(\mathbf{F}=m\mathbf{a}\) holds in identical form, and they are related to one another by Galilean transformations. Any frame outside this set betrays itself through fictitious forces proportional to mass — which is precisely why fictitious forces produce a mass-independent acceleration, the observation that led Einstein to the equivalence principle.

The deeper modern statement is that the Galilean group is a contraction (the \(c\to\infty\) limit, in the sense of Inönü–Wigner) of the Poincaré group. In the Poincaré group boosts and rotations are intertwined because time and space mix; in the Galilean limit time becomes absolute and the boosts commute among themselves, forming an abelian subgroup. The projective (rather than true) representations of the Galilean group carry a central extension whose central charge is the mass — the quantum-mechanical reason mass appears as a superselection label (Bargmann's theorem) and that a non-relativistic wavefunction acquires the phase \(\exp\!\big[\tfrac{i}{\hbar}\big(m\mathbf{V}\cdot\mathbf{r}-\tfrac12 mV^2 t\big)\big]\) under a boost.

Common misconceptions. Galilean invariance does not say every quantity is the same in all frames — velocities, kinetic energies, and momenta all change. It says the equation of motion has the same form. Nor does it privilege a rest frame: there is no experiment distinguishing rest from uniform motion. And it is not exact in nature — it is the low-speed approximation to Lorentz invariance.

Worked examples

Example 1 — A collision looks the same from a moving platform.

\[ m_1 = 2.0\ \mathrm{kg},\ u_1 = +3.0\ \mathrm{m\,s^{-1}};\qquad m_2 = 1.0\ \mathrm{kg},\ u_2 = -1.0\ \mathrm{m\,s^{-1}} \]

Set up a one-dimensional collision in the lab frame; symbols and initial data. A

\[ p_{\text{lab}} = m_1 u_1 + m_2 u_2 = (2.0)(3.0) + (1.0)(-1.0) = 5.0\ \mathrm{kg\,m\,s^{-1}} \]

Total momentum in the lab frame. A

\[ V = +2.0\ \mathrm{m\,s^{-1}}:\quad u' = u - V \;\Rightarrow\; u_1' = +1.0,\ \ u_2' = -3.0\ \mathrm{m\,s^{-1}} \]

Apply the Galilean velocity transformation (step 2 of the derivation, with \(R=\mathbb{1}\)). A

\[ p' = m_1 u_1' + m_2 u_2' = (2.0)(1.0) + (1.0)(-3.0) = -1.0\ \mathrm{kg\,m\,s^{-1}} \neq p_{\text{lab}} \]

The total momentum itself differs between frames — as expected, it is covariant, not invariant. B

\[ p' = p_{\text{lab}} - (m_1+m_2)V = 5.0 - (3.0)(2.0) = -1.0\ \mathrm{kg\,m\,s^{-1}}\ \checkmark \]

The two frames' momenta differ by exactly the total mass times the boost, a constant. Hence if momentum is conserved in one frame it is conserved in the other. B

\[ \Delta p = 0 \;\Longleftrightarrow\; \Delta p' = 0 \]

Reading. Although the numerical momentum (\(5.0\) vs \(-1.0\ \mathrm{kg\,m\,s^{-1}}\)) is frame-dependent, the law — momentum before equals momentum after — is form-invariant, because the constant offset \((m_1+m_2)V\) is unchanged by the collision. Units: \(\mathrm{kg\,m\,s^{-1}}\) throughout.

Example 2 — A fictitious force appears the moment the frame accelerates.

\[ A = 2.5\ \mathrm{m\,s^{-2}},\qquad m = 0.20\ \mathrm{kg} \]

A bus accelerates at \(A\); a strap of mass \(m\) hangs from the ceiling. Find the hang angle \(\theta\) from the vertical. A

\[ \text{Ground (inertial): } \quad T\sin\theta = mA, \qquad T\cos\theta = mg \]

In an inertial frame only real forces (tension, gravity) appear; the strap genuinely accelerates horizontally with the bus. A

\[ \tan\theta = \frac{A}{g} = \frac{2.5}{9.81} = 0.2549 \;\Rightarrow\; \theta = 14.3^\circ \]

Divide the two equations to eliminate \(T\), then substitute numbers. A

\[ \text{Bus (non-inertial): add } \mathbf{F}_{\text{fict}} = -m\mathbf{A}, \quad |\mathbf{F}_{\text{fict}}| = (0.20)(2.5) = 0.50\ \mathrm{N} \]

Because this frame accelerates, the derivation's step 3 fails and a mass-proportional pseudo-force (pointing backward) must be inserted to keep \(\mathbf{F}=m\mathbf{a}\) usable. C

\[ \text{Strap static in bus frame: } T\sin\theta = mA,\ \ T\cos\theta = mg \;\Rightarrow\; \tan\theta = \frac{A}{g} \]

Same angle, different bookkeeping. The fictitious force is the price of using Newton's law in a frame where Galilean invariance is broken. C

\[ \theta = \arctan\!\frac{A}{g} = 14.3^\circ, \qquad F_{\text{fict}} = 0.50\ \mathrm{N} \]

Reading. The two frames agree on the observable (the hang angle) but only the accelerating frame requires an extra force with no source and no third-law partner. Its appearance, proportional to \(mA\), is exactly the diagnostic that the bus frame is not inertial. Units check. \(\tan\theta = (\mathrm{m\,s^{-2}})/(\mathrm{m\,s^{-2}})\) is dimensionless; \(F_{\text{fict}} = \mathrm{kg\,m\,s^{-2}} = \mathrm{N}\).

Problems

A puck slides frictionlessly at \(4.0\ \mathrm{m\,s^{-1}}\) east in the lab. A camera drifts at \(4.0\ \mathrm{m\,s^{-1}}\) east. What are the puck's velocity and acceleration in the camera frame, and does \(\mathbf{F}=m\mathbf{a}\) hold there?
Solution
Velocity: \(u' = u - V = 4.0 - 4.0 = 0\ \mathrm{m\,s^{-1}}\) (puck at rest in the camera frame). Acceleration is unchanged by a boost, \(a' = a = 0\). Force: \(F = 0\) in both frames. So \(F' = ma'\) reads \(0 = 0\) — the law holds identically. The camera frame moves at constant velocity, hence is inertial, and Galilean invariance guarantees this.
Two blocks, \(3.0\ \mathrm{kg}\) at \(+2.0\ \mathrm{m\,s^{-1}}\) and \(2.0\ \mathrm{kg}\) at \(-4.0\ \mathrm{m\,s^{-1}}\), are viewed from a frame boosted at \(V = -1.5\ \mathrm{m\,s^{-1}}\). Compute the total momentum in both frames and verify the offset relation.
Solution
Lab: \(p = (3.0)(2.0) + (2.0)(-4.0) = 6.0 - 8.0 = -2.0\ \mathrm{kg\,m\,s^{-1}}\). Boosted velocities \(u' = u - V = u + 1.5\), giving \(+3.5\) and \(-2.5\ \mathrm{m\,s^{-1}}\). Then \(p' = (3.0)(3.5) + (2.0)(-2.5) = 10.5 - 5.0 = 5.5\ \mathrm{kg\,m\,s^{-1}}\). Offset check: \(p' = p - M_{\text{tot}}V = -2.0 - (5.0)(-1.5) = -2.0 + 7.5 = 5.5\ \mathrm{kg\,m\,s^{-1}}\ \checkmark\). Momentum differs but transforms by the constant \(M_{\text{tot}}V\).
A force law is proposed as \(\mathbf{F} = -k\,\mathbf{x}_{\text{abs}}\), where \(\mathbf{x}_{\text{abs}}\) is position measured from a fixed point in absolute space. Show it is not Galilean-invariant, and state the modification that restores invariance.
Solution
Under translation \(\mathbf{x}\to\mathbf{x}'=\mathbf{x}+\mathbf{d}\), the force becomes \(\mathbf{F}'=-k(\mathbf{x}+\mathbf{d})=\mathbf{F}-k\mathbf{d}\neq\mathbf{F}\). It changes by the shift, so the equation of motion is not the same in the translated frame — the law selects a preferred origin (absolute space). Restoring invariance requires a relative coordinate, e.g. \(\mathbf{F}=-k(\mathbf{x}_1-\mathbf{x}_2-\boldsymbol\ell)\) between two interacting bodies (a spring); the difference \(\mathbf{x}_1-\mathbf{x}_2\) is translation-invariant.
A frame rotates at constant angular velocity \(\omega = 0.50\ \mathrm{rad\,s^{-1}}\). A \(0.10\ \mathrm{kg}\) bead sits \(0.40\ \mathrm{m}\) from the axis, at rest in the rotating frame. Compute the centrifugal force seen in the rotating frame and explain why this frame is not admitted by Galilean invariance.
Solution
Centrifugal magnitude: \(F = m\omega^2 r = (0.10)(0.50)^2(0.40) = (0.10)(0.25)(0.40) = 0.010\ \mathrm{N}\), directed outward. This term is proportional to mass and has no physical source or third-law partner. The derivation assumed \(R\) is time-independent (step 1); a rotating frame has \(R=R(t)\), so differentiating twice produces centrifugal (and Coriolis) terms that break the form \(\mathbf{F}=m\mathbf{a}\). Hence rotating frames are non-inertial and lie outside the Galilean class.
In the lab a projectile has acceleration \(\mathbf{a} = (0,\,-9.81)\ \mathrm{m\,s^{-2}}\). A second inertial frame is rotated by \(\varphi = 30^\circ\) in that vertical plane. Using \(\mathbf{a}'=R\mathbf{a}\), find \(\mathbf{a}'\) and confirm its magnitude is unchanged.
Solution
Take \(R=\begin{pmatrix}\cos30^\circ & \sin30^\circ\\ -\sin30^\circ & \cos30^\circ\end{pmatrix}=\begin{pmatrix}0.866 & 0.500\\ -0.500 & 0.866\end{pmatrix}\). Then \(a_x' = (0.866)(0)+(0.500)(-9.81) = -4.905\); \(a_y' = (-0.500)(0)+(0.866)(-9.81) = -8.496\). So \(\mathbf{a}' = (-4.91,\,-8.50)\ \mathrm{m\,s^{-2}}\). Magnitude: \(\sqrt{4.905^2+8.496^2} = \sqrt{24.06+72.18} = \sqrt{96.24} = 9.81\ \mathrm{m\,s^{-2}}\ \checkmark\). Rotation preserves magnitude (\(R^{\mathsf T}R=\mathbb{1}\)), and because \(\mathbf{F}=m\mathbf{a}\) rotates identically on both sides, the law is unchanged: \(\mathbf{F}'=m\mathbf{a}'\) with \(|\mathbf{F}'|=|\mathbf{F}|\).