Why is Momentum Conserved?

In this article we explore the conservation of momentum using Lagrangian Dynamics.

Momentum in Physics

Momentum is defined by \(p=mv\). It is a vector quantity, meaning the direction matters. The derivative of momentum is force.

In an inertial frame, momentum is a conserved quantity, meaning that it is not affected by external forces in a closed system. This is an example of one of the most important concepts in physics - translational symmetry.

The astute reader can notice that this is highly unobvious — how can a system allow and ensure a symmetry? Below, I aim to provide a intuitive explanation using more advanced classical mechanics techniques, namely the Lagragian Dynamics Formulation, developed in 1788 by Joseph-Louis Lagrange.

The equation of motion of an object in Lagrangian Formulation is given by:

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

where \(\mathcal{L}=T-V\).

Assuming we can equate the expression on RHS to \(0\), i.e. \(\frac{\partial \mathcal{L}}{\partial q_j} = 0\), we have \(\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_j}\right) = 0\) , which implies \(\frac{\partial \mathcal{L}}{\partial \dot{q}_j}\) is a constant. There appears to be a “freedom” in the system!

We can verify this is true for a 1 dimensional object. Set \(T=\frac{1}{2}m\dot{x}^2\) and \(V=0\), we have \(\mathcal{L}=\frac{1}{2}m\dot{x}^2\). Indeed, \(m \dot{x}\) is the constant obtained, which is the standard result in classical mechanics.

Concluding Remarks

This is expected to be true due to Noether’s Theorem, which states that every symmetry will lead to a conservation, which is explained here.

Therefore, the energy of a system being independent of position (lack of \(x\) in expression, while \(\dot{x}\) is present) causes a constant in the system, known as the (canonical) momentum. In other words, momentum is a term of freedom caused by the lack of a particular spatial coordinate (\(x\) in our case). Momentum is not conserved if the energy term involves spatial coordinates, say in a particle is in the presence of electromagnetic force.

The other constant is known as the energy function and is given by \(h=\sum_i\frac{\partial L}{\partial \dot{q}_i} \dot{q}_i - \mathcal{L}\). The reader can show that this corresponds to the energy of the system, which is \(T\). More of these is related to Hamiltonian Dynamics, which is a reformulation of Lagrangian Dynamics (very closely related to the energy function).

Exercises for the reader

• What is also conserved in the system mentioned above? (Difficulty: F4 HKDSE)

• What are the restrictions of the form taken by the potential, if so, why? Why is the potential taken as \(0\)? (Difficulty: F6 HKDSE)

• By writing down the Lagrangian for a ball undergoing circular motion, show that the angular momentum is conserved. (Difficulty: Year 1-2 University)

• (*) Given that \(\mathcal{L}(\underline{r}, \underline{\dot{r}}, t)=T-\tilde{V}=\frac{1}{2} m|\underline{\dot{r}}|^2-e[\phi(\underline{r}, t)-\underline{\dot{r}} \cdot \underline{A}(\underline{r}, t)]\) is the Lagrangian of a particle in a EM field, show that \(p=mv+eA\). (Difficulty: Year 2-3 University)

Last Updated - 3/6/2024