Conservation Theorems and Symmetry Properties
- Classical Mechanics 1.5
Canonical Momentum
The canonical momentum associated with a coordinate $q_i$ is defined as:
\[ p_i = \pdv{L}{\dot{q}_i} \]
$p_i$ does not necessarily have the dimensions of linear momentum. For example, in spherical coordinates, the canonical momentum associated with the angular coordinate $\phi$ is $p_\phi = m r^2 \sin^2\theta \dot{\phi}$, which has dimensions of angular momentum. For an example of particles in electromagnetic fields,
\[ \b{p}_i = m_i \dot{\b{r}}_i + q_i \b{A}(\b{r}_i, t) \]
where $q_i$ is the charge of the $i$-th particle and $\b{A}$ is the vector potential.
Let’s think of the Lagrange equation of motion:
\[ \odv{}{t} \left( \pdv{L}{\dot{q}_i} \right) - \pdv{L}{q_i} = 0 \]
This can be rewritten as:
\[ \dot{p}_i = \pdv{L}{q_i} \]
So that if the Lagrangian does not depend on a particular coordinate $q_i$, called cyclic or ignorable, then the corresponding canonical momentum $p_i$ is conserved. Note that the $q_i$s here should be independent coordinates.
Symmetry and Conservation Laws
Conservation of Linear Momentum
We first consider a generalized coordinated $q_j$, for which a change $\dd{q_j}$ represents a translation of the system in some direction. Then clearly $q_j$ cannot appear in $T$. Further, we assume that the potential energy $V$ is not a function of $\dot{q}_j$. Then the Lagrangian equation of motion reduces to:
\[ \dot{p}_j = -\pdv{V}{q_j} = Q_j \]
Denoting as $\b{n} = \pdv{\b{r}_i}{q_j}$ we have:
\[ Q_j = \sum_i \b{F}_i \cdot \b{n} = \b{n} \cdot \b{F} \]
where $\b{F}$ is the total force on the system. The conjugate momentum is:
\[ \begin{align*} p_j &= \pdv{T}{\dot{q}_j} = \sum_i m_i \dot{\b{r}}_i \cdot \pdv{\dot{\b{r}}_i}{\dot{q}_j} \nl &= \sum_i m_i \b{v}_i \cdot \pdv{\b{r}_i}{q_j} \nl &= \b{n} \cdot \b{P} \end{align*} \]
where $\b{P}$ is the total linear momentum of the system. Thus, if the potential does not depend on $q_j$, then the $Q_j = 0$ and the component of the total linear momentum in the direction of $\b{n}$ is conserved.
Conservation of Angular Momentum
In a similar fashion, it can be shown that if a cyclic coordinate $q_j$ is such that $\dd{q_j}$ corresponds to a rotation of the system about some axis, then the conserved conjugate momentum $p_j$ is the component of the total angular momentum $\b{L}$ along that axis. By the same argument as before, $T$ does not depend on $q_j$, and we assume that $V$ does not depend on $\dot{q}_j$. Then the Lagrange equation of motion reduces to:
\[ \dot{p}_j = -\pdv{V}{q_j} = Q_j \]
as before. Setting $\b{n}$ as the unit vector along the axis of rotation, we have:
\[ \pdv{\b{r}_i}{q_j} = \b{n} \times \b{r}_i \]
Thus,
\[ \begin{align*} Q_j &= \sum_i \b{F}_i \cdot (\b{n} \times \b{r}_i) \nl &= \b{n} \cdot \sum_i (\b{r}_i \times \b{F}_i) \nl &= \b{n} \cdot \b{N} \end{align*} \]
where $\b{N}$ is the total torque on the system. Similarly, the conjugate momentum is:
\[ \begin{align*} p_j &= \pdv{T}{\dot{q}_j} = \sum_i m_i \dot{\b{r}}_i \cdot \pdv{\dot{\b{r}}_i}{\dot{q}_j} \nl &= \sum_i m_i \b{v}_i \cdot (\b{n} \times \b{r}_i) \nl &= \b{n} \cdot \sum_i (\b{r}_i \times m_i \b{v}_i) \nl &= \b{n} \cdot \b{L} \end{align*} \]
where $\b{L}$ is the total angular momentum of the system. Thus, if the potential does not depend on $q_j$, then $Q_j = 0$ and the component of the total angular momentum along the axis of rotation is conserved.
