It is possible to demonstrate, using the Lagrangian formalism, that fundamental conservation laws, such as the conservation of linear momentum or angular momentum, arise from underlying symmetries of the considered systems. We will begin by defining a cyclic variable and show how translational symmetry implies the conservation of the momentum component along the direction of translation. Subsequently, we will explore how rotational symmetry leads to the conservation of angular momentum projected onto the axis of rotation, and finally, how energy conservation is expressed using Lagrange’s equations.
For a physical system described by a Lagrangian \mathcal L, which is a function of generalized coordinates q_j and generalized velocities \dot q_j, the generalized momentum p_j associated with the coordinate q_j is defined as:
p_j = \frac{\partial \mathcal L}{\partial \dot q_j}
A coordinate q_j is said to be cyclic if the Lagrangian \mathcal L does not explicitly depend on q_j, meaning \partial \mathcal L/\partial q_j = 0. In such a case, Lagrange’s equation,
\frac{\mathrm d}{\mathrm dt}\biggl(\frac{\partial \mathcal L}{\partial \dot q_j}\biggr) - \frac{\partial \mathcal L}{\partial q_j} = 0
reduces to:
\frac{\mathrm d}{\mathrm dt}\left(p_j\right) = 0
implying that p_j is conserved.
Consider a system of N point particles with positions given by vectors \mathbf{x}_i (i = 1, \dots, N). The system is conservative, so all forces are derived from a scalar potential energy function V(\mathbf{x}_1,\dots,\mathbf{x}_N). Suppose there exists a direction defined by a unit vector \mathbf{n} such that translating the entire system by an amount s along \mathbf{n} leaves the potential unchanged:
V(\mathbf{x}_1 + s\mathbf{n},\,\dots,\,\mathbf{x}_N + s\mathbf{n}) = V(\mathbf{x}_1,\dots,\mathbf{x}_N)
Since the kinetic energy T depends only on the velocities \dot{\mathbf{x}}_i, the Lagrangian \mathcal L = T - V is invariant under this translation. Therefore,
\frac{\mathrm d \mathcal L}{\mathrm ds} = 0
Treating \mathcal L as a function of \mathbf{x}_i(s)=\mathbf{x}_i(0)+s\mathbf{n} and \dot{\mathbf{x}}_i, the chain rule gives:
\frac{\mathrm d \mathcal L}{\mathrm ds} =\sum_{i=1}^N\sum_{\alpha=1}^3 \frac{\partial \mathcal L}{\partial x_{\alpha i}}\,\frac{\mathrm dx_{\alpha i}}{\mathrm ds} =\sum_{i=1}^N\sum_{\alpha=1}^3 \frac{\partial \mathcal L}{\partial x_{\alpha i}}\,n_\alpha = 0
From Lagrange’s equations, \partial \mathcal L/\partial x_{\alpha i} = d(\partial \mathcal L/\partial\dot x_{\alpha i})/\mathrm dt. Substituting yields:
\sum_{i=1}^N\sum_{\alpha=1}^3 \frac{\mathrm d}{\mathrm dt}\!\left(\frac{\partial \mathcal L}{\partial\dot x_{\alpha i}}\right)\,n_\alpha =0
For a standard Lagrangian with T=\sum_j\frac12 m_j\dot{\mathbf{x}}_j^2, one has \partial \mathcal L/\partial\dot x_{\alpha i}=m_i\dot x_{\alpha i}=p_{\alpha i}. Moving \mathrm d/\mathrm dt outside the sum,
\frac{\mathrm d}{\mathrm dt} \left(\sum_{i=1}^N\sum_{\alpha=1}^3m_i\dot x_{\alpha i}\,n_\alpha\right) =\frac{\mathrm d}{\mathrm dt}\!\left(\mathbf{P} \cdot\mathbf{n}\right) =0
where \mathbf P=\sum_i\mathbf{p}_i. The component of total linear momentum along \mathbf{n} is conserved.
This is a specific instance of Noether’s theorem. For a conservative system of N point particles, translational symmetry of the potential along a direction \mathbf{n} implies conservation of the total linear momentum component along \mathbf{n}.
We now turn our attention to rotational symmetry. Consider the same conservative system of N point particles. We now assume that if the entire system is rotated by an angle \theta around an axis passing through the origin, defined by a fixed unit vector \mathbf{n} in the reference frame, the potential energy V remains unchanged. We will demonstrate that this symmetry leads to the conservation of the component of angular momentum along this axis.
To mathematically represent this rotation, we consider an infinitesimal rotation by a small angle \delta\theta. The change in the position vector \mathbf{x}_i of the i^{th} particle under such a rotation is given by:
\delta\mathbf{x}_i = \delta\theta\, (\mathbf{n} \times \mathbf{x}_i)
So, the transformed position vector \mathbf{x}_i' can be written as \mathbf{x}_i' = \mathbf{x}_i + \delta\theta (\mathbf{n} \times \mathbf{x}_i).
The invariance of the potential energy V under this rotation means that V(\mathbf{x}_1', \dots, \mathbf{x}_N') = V(\mathbf{x}_1, \dots, \mathbf{x}_N). The kinetic energy T, which depends on velocities, is also invariant under a rotation of the coordinate system (or an equivalent rotation of the physical system, as the velocities would rotate along with the positions). Therefore, the Lagrangian \mathcal L = T - V is invariant under this infinitesimal rotation.
This implies that the derivative of \mathcal L with respect to the rotation angle \theta (considering \delta\theta as an infinitesimal d\theta) is zero when evaluated at \theta = 0 (or for any \theta if the symmetry holds for finite rotations, but the infinitesimal argument is sufficient here):
\frac{\mathrm d \mathcal L}{\mathrm d\theta} = 0
Using the chain rule, and treating \mathbf{x}_i(\theta) = \mathbf{x}_i(0) + \theta (\mathbf{n} \times \mathbf{x}_i(0)) for small \theta, we have \frac{\mathrm d\mathbf{x}_i}{\mathrm d\theta} = \mathbf{n} \times \mathbf{x}_i. The derivative of the Lagrangian is:
\frac{\mathrm d \mathcal L}{\mathrm d\theta} = \sum_{i=1}^N \sum_{\alpha=1}^3 \frac{\partial \mathcal L}{\partial x_{\alpha i}} \frac{\mathrm dx_{\alpha i}}{\mathrm d\theta} = \sum_{i=1}^N (\nabla_{\mathbf{x}_i}\mathcal L) \cdot (\mathbf{n} \times \mathbf{x}_i) = 0
where x_{\alpha i} is the \alpha^{th} Cartesian component of \mathbf{x}_i, and (\mathbf{n} \times \mathbf{x}_i)_\alpha is the \alpha^{th} component of \mathbf{n} \times \mathbf{x}_i.
From Lagrange’s equations, \frac{\partial \mathcal L}{\partial x_{\alpha i}} = \frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal L}{\partial \dot{x}_{\alpha i}} \right). Substituting this, we get:
\sum_{i=1}^N \left( \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot{\mathbf{x}}_i} \right) \cdot (\mathbf{n} \times \mathbf{x}_i) = 0
Recognizing \frac{\partial \mathcal L}{\partial \dot{\mathbf{x}}_i} = \mathbf{p}_i (the linear momentum of the i^{th} particle, as V is velocity-independent and \mathcal L = T - V), the equation becomes:
\sum_{i=1}^N \left( \frac{\mathrm d\mathbf{p}_i}{\mathrm dt} \right) \cdot (\mathbf{n} \times \mathbf{x}_i) = 0
We use the property of the scalar triple product that allows cyclic permutation: \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot (\mathbf{C} \times \mathbf{A}).
Let \mathbf{A} = \frac{\mathrm d\mathbf{p}_i}{\mathrm dt}, \mathbf{B} = \mathbf{n}, and \mathbf{C} = \mathbf{x}_i. Then:
\left( \frac{\mathrm d\mathbf{p}_i}{\mathrm dt} \right) \cdot (\mathbf{n} \times \mathbf{x}_i) = \mathbf{n} \cdot \left( \mathbf{x}_i \times \frac{\mathrm d\mathbf{p}_i}{\mathrm dt} \right)
Substituting this into the sum:
\sum_{i=1}^N \mathbf{n} \cdot \left( \mathbf{x}_i \times \frac{\mathrm d\mathbf{p}_i}{\mathrm dt} \right) = 0
Since \mathbf{n} is a constant vector, we can factor it out of the sum:
\mathbf{n} \cdot \left( \sum_{i=1}^N \mathbf{x}_i \times \frac{\mathrm d\mathbf{p}_i}{\mathrm dt} \right) = 0
Now, consider the time derivative of the angular momentum \mathbf{L}_i = \mathbf{x}_i \times \mathbf{p}_i of the i-th particle:
\frac{\mathrm d\mathbf{L}_i}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} (\mathbf{x}_i \times \mathbf{p}_i) = \left(\frac{\mathrm d\mathbf{x}_i}{\mathrm dt} \times \mathbf{p}_i\right) + \left(\mathbf{x}_i \times \frac{\mathrm d\mathbf{p}_i}{\mathrm dt}\right)
Since \frac{\mathrm d\mathbf{x}_i}{\mathrm dt} = \mathbf{v}_i and \mathbf{p}_i = m_i \mathbf{v}_i, the first term \mathbf{v}_i \times m_i \mathbf{v}_i = 0 because the cross product of two parallel vectors is zero:
\frac{\mathrm d\mathbf{L}_i}{\mathrm dt} = \mathbf{x}_i \times \frac{\mathrm d\mathbf{p}_i}{\mathrm dt}
Substituting this back into our equation:
\mathbf{n} \cdot \left( \sum_{i=1}^N \frac{\mathrm d\mathbf{L}_i}{\mathrm dt} \right) = 0
This can be rewritten as:
\mathbf{n} \cdot \frac{\mathrm d}{\mathrm dt} \left( \sum_{i=1}^N \mathbf{L}_i \right) = 0
Let \mathbf{L}_{tot} = \sum_{i=1}^N \mathbf{L}_i be the total angular momentum of the system. Since \mathbf{n} is a constant vector, we can write:
\frac{\mathrm d}{\mathrm dt} (\mathbf{L}_{tot} \cdot \mathbf{n}) = 0
This equation states that the component of the total angular momentum of the system projected along the axis of rotational symmetry \mathbf{n} is a conserved quantity. If the system is symmetric with respect to rotation about an axis \mathbf{n} for any infinitesimal angle \delta\theta, this implies symmetry for finite angles as well (by integrating infinitesimal rotations).
For a system of point particles with conservative forces, if the system exhibits rotational symmetry about a fixed axis \mathbf{n} (i.e., the Lagrangian is unchanged by an infinitesimal rotation about this axis), then the component of the total angular momentum along that axis is conserved. This again demonstrates that a fundamental conservation law—the conservation of angular momentum—results from a fundamental symmetry principle of the system.
Finally, we address the question of energy conservation. We will first identify a conserved quantity that might initially seem somewhat unique. Consider a system where the Lagrangian \mathcal L does not explicitly depend on time, i.e., \frac{\partial \mathcal L}{\partial t} = 0. We define a quantity \mathcal H as:
\mathcal H = \sum_j p_j \dot{q}_j - \mathcal L
where p_j = \frac{\partial \mathcal L}{\partial \dot{q}_j} is the generalized momentum associated with the generalized coordinate q_j, and the sum is over all n degrees of freedom. We will show that this quantity \mathcal H is conserved under the condition that \mathcal L has no explicit time dependence.
To demonstrate that \mathcal H is a conserved quantity, we calculate its total time derivative, \frac{\mathrm d\mathcal H}{\mathrm dt}:
\frac{\mathrm d\mathcal H}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} \left( \sum_j p_j \dot{q}_j \right) - \frac{\mathrm d \mathcal L}{\mathrm dt}
\frac{\mathrm d\mathcal H}{\mathrm dt} = \sum_j \left( \frac{\mathrm dp_j}{\mathrm dt} \dot{q}_j + p_j \ddot{q}_j \right) - \frac{\mathrm d \mathcal L}{\mathrm dt}
From the definition of generalized momentum, p_j = \frac{\partial \mathcal L}{\partial \dot{q}_j}. Using Lagrange’s equations, \frac{\mathrm dp_j}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{q}_j}\right) = \frac{\partial \mathcal L}{\partial q_j}.
Substituting these into the expression for \frac{\mathrm dH}{\mathrm dt}:
\frac{\mathrm d\mathcal H}{\mathrm dt} = \sum_j \left( \frac{\partial \mathcal L}{\partial q_j} \dot{q}_j + \frac{\partial \mathcal L}{\partial \dot{q}_j} \ddot{q}_j \right) - \frac{\mathrm d \mathcal L}{\mathrm dt}
The total time derivative of the Lagrangian \mathcal L(q_j, \dot{q}_j, t) is given by:
\frac{\mathrm d \mathcal L}{\mathrm dt} = \sum_j \left( \frac{\partial \mathcal L}{\partial q_j} \frac{\mathrm dq_j}{\mathrm dt} + \frac{\partial \mathcal L}{\partial \dot{q}_j} \frac{\mathrm d\dot{q}_j}{\mathrm dt} \right) + \frac{\partial \mathcal L}{\partial t}
\frac{\mathrm d \mathcal L}{\mathrm dt} = \sum_j \left( \frac{\partial \mathcal L}{\partial q_j} \dot{q}_j + \frac{\partial \mathcal L}{\partial \dot{q}_j} \ddot{q}_j \right) + \frac{\partial \mathcal L}{\partial t}
Now, substituting this expression for \frac{\mathrm d \mathcal L}{\mathrm dt} back into the equation for \frac{\mathrm dH}{\mathrm dt}:
\frac{\mathrm d\mathcal H}{\mathrm dt} = \sum_j \left( \frac{\partial \mathcal L}{\partial q_j} \dot{q}_j + \frac{\partial \mathcal L}{\partial \dot{q}_j} \ddot{q}_j \right) - \left[ \sum_j \left( \frac{\partial \mathcal L}{\partial q_j} \dot{q}_j + \frac{\partial \mathcal L}{\partial \dot{q}_j} \ddot{q}_j \right) + \frac{\partial \mathcal L}{\partial t} \right]
\frac{\mathrm d\mathcal H}{\mathrm dt} = - \frac{\partial \mathcal L}{\partial t}
Thus, if the Lagrangian \mathcal L does not explicitly depend on time (i.e., \frac{\partial \mathcal L}{\partial t} = 0), then \frac{\mathrm dH}{\mathrm dt} = 0, which means \mathcal H is a conserved quantity. This quantity \mathcal H is known as the Jacobi integral or the Hamiltonian of the system.
If, in addition to \mathcal L not depending explicitly on time, the coordinate transformations defining the generalized coordinates q_j from Cartesian coordinates \mathbf{r}_\alpha also do not explicitly depend on time (i.e., \mathbf{r}_\alpha = \mathbf{r}_\alpha(q_1, \dots, q_n), which implies time-independent constraints), then \mathcal H is equal to the total mechanical energy E = T + V of the system.
Let’s prove this. The term \sum_j p_j \dot{q}_j in the expression for \mathcal H is \sum_j \frac{\partial \mathcal L}{\partial \dot{q}_j} \dot{q}_j. Since the potential energy V is generally assumed to be independent of generalized velocities (V = V(q_j, t)), we have \frac{\partial \mathcal L}{\partial \dot{q}_j} = \frac{\partial (T-V)}{\partial \dot{q}_j} = \frac{\partial T}{\partial \dot{q}_j}.
So:
\sum_j p_j \dot{q}_j = \sum_j \frac{\partial T}{\partial \dot{q}_j} \dot{q}_j
If the constraints are time-independent, the kinetic energy T is a homogeneous quadratic function of the generalized velocities \dot{q}_j. That is, T = \frac{1}{2} \sum_{k,l} M_{kl}(q) \dot{q}_k \dot{q}_l. By Euler’s theorem for homogeneous functions, if T is a homogeneous function of degree 2 in the variables \dot{q}_j, then \sum_j \frac{\partial T}{\partial \dot{q}_j} \dot{q}_j = 2T.
Therefore, under these conditions (\mathcal L has no explicit time dependence, and constraints are time-independent):
\mathcal H = \sum_j p_j \dot{q}_j - L = 2T - \mathcal L
Since \mathcal L = T - V:
\mathcal H = 2T - (T - V) = T + V
This is the total mechanical energy of the system. If the Lagrangian does not explicitly depend on time and the constraints are time-independent, the total mechanical energy of the system is conserved.