Why Is Momentum Conserved?

This article explores the conservation of momentum using Lagrangian mechanics.

Momentum in physics

Momentum is defined by \(\mathbf{p}=m\mathbf{v}\). It is a vector quantity, so its direction matters, and its rate of change is the force, \(\dot{\mathbf{p}} = \mathbf{F}\).

The total momentum of an isolated system, one with no external forces acting on it, is conserved. This is not an accident: it follows from translational symmetry, the fact that the laws of physics are the same everywhere in space.

Why should a symmetry guarantee a conservation law? Lagrangian mechanics, developed by Joseph-Louis Lagrange in 1788, gives a clean explanation. In the Lagrangian formulation, the equations of motion are

\[\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_j}\right) = \frac{\partial \mathcal{L}}{\partial q_j},\]

where \(\mathcal{L}=T-V\) and the \(q_j\) are generalised coordinates.

Suppose \(\mathcal{L}\) does not depend on some coordinate \(q_j\) (such a coordinate is called cyclic or ignorable), so that \(\frac{\partial \mathcal{L}}{\partial q_j} = 0\). Then

\[\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_j}\right) = 0,\]

and the generalised momentum \(p_j = \frac{\partial \mathcal{L}}{\partial \dot{q}_j}\) is a constant of the motion.

For a free particle in one dimension, \(T=\frac{1}{2}m\dot{x}^2\) and \(V=0\), so \(\mathcal{L}=\frac{1}{2}m\dot{x}^2\). This does not depend on \(x\), and the conserved quantity is \(\partial\mathcal{L}/\partial\dot x = m \dot{x}\): the familiar momentum.

Concluding remarks

This is a special case of Noether’s theorem, which states that every continuous symmetry of a system leads to a conserved quantity.

Here the Lagrangian does not depend on the position \(x\), only on the velocity \(\dot x\), and that absence produces a conserved (canonical) momentum. If the Lagrangian does depend on \(x\), for example for a particle in a position-dependent potential \(V(x)\), the corresponding momentum is no longer conserved.

A second conserved quantity, related to symmetry under translations in time, is the energy function

\[h=\sum_i\frac{\partial \mathcal{L}}{\partial \dot{q}_i}\, \dot{q}_i - \mathcal{L},\]

which is conserved whenever \(\mathcal{L}\) does not depend explicitly on time. For the free particle above, \(h = T\); more generally, when \(T\) is quadratic in the velocities and \(V\) does not depend on them, \(h = T + V\) is the total energy. The energy function leads naturally to Hamiltonian mechanics, a reformulation of Lagrangian mechanics.

Exercises

• What else is conserved for the free particle above? (Secondary school)
• Why is the potential taken to be zero? What would change if it were a non-zero constant? (Secondary school)
• Write down the Lagrangian for a particle moving in a plane under a central force, and show that angular momentum is conserved. (Undergraduate, years 1–2)
• () The Lagrangian of a charged particle in an electromagnetic field is \(\mathcal{L}(\mathbf{r}, \dot{\mathbf{r}}, t)=\frac{1}{2} m\lvert\dot{\mathbf{r}}\rvert^2-e\left[\phi(\mathbf{r}, t)-\dot{\mathbf{r}} \cdot \mathbf{A}(\mathbf{r}, t)\right]\). Show that the canonical momentum is \(\mathbf{p}=m\mathbf{v}+e\mathbf{A}\). (Undergraduate, years 2–3*)

Changes on 25 October 2026

What was wrong. The post said the energy function \(h\) equals the kinetic energy \(T\). That is only true for a free particle. When the kinetic energy is quadratic in the velocities and the potential does not depend on them, \(h = T + V\), the total energy.

What changed. This is corrected and the conditions for \(h\) to be conserved are given. The electromagnetic example is tidied.