Theorem: for every continuous symmetry of the action of a physical system, there exists a corresponding quantity that is conserved over time.
We consider a physical system described by a set of generalized coordinates q_i(t), where i = 1, \dots, n.
The dynamics of the system are determined by the principle of stationary action, which asserts that the trajectory of the system between two points in time, t_1 and t_2, is one that extremizes the action integral S:
S = \int_{t_1}^{t_2} \mathcal L(q_i, \dot{q}_i, t) \, \mathrm d t
where \mathcal L(q_i, \dot{q}_i, t) is the Lagrangian of the system, a function of the generalized coordinates, their time derivatives \dot{q}_i = \mathrm d q_i / \mathrm d t, and time t.
Now, let’s introduce a continuous one-parameter group of transformations for the generalized coordinates and time, characterized by an infinitesimal parameter \varepsilon:
\begin{aligned} t &\rightarrow t^\prime = t + \varepsilon \delta t \\ q_i(t) &\rightarrow q_i^\prime(t^\prime) = q_i(t) + \varepsilon \delta q_i \end{aligned}
This transformation is considered a symmetry of the system if the action remains invariant under it. More precisely, the Lagrangian is allowed to change by a total time derivative of some function F(q_i, t):
\mathcal L(q_i^\prime(t^\prime), \dot{q}_i^\prime(t^\prime), t^\prime) \frac{\mathrm d t^\prime}{\mathrm d t} = \mathcal L(q_i(t), \dot{q}_i(t), t) + \varepsilon \frac{\mathrm d F}{\mathrm d t}
If this condition holds for any path q_i(t), not just the one that satisfies the equations of motion, then Noether’s theorem states that the quantity J given by:
J = \sum_{i} \frac{\partial L}{\partial \dot{q}_i} (\delta q_i - \dot{q}_i \delta t) - F
is a conserved quantity, meaning its total time derivative is zero:
\frac{\mathrm d J}{\mathrm d t} = 0
Proof: the derivation of Noether’s theorem begins by examining the variation of the action under the infinitesimal transformation. The change in the action, \delta S, can be expressed as:
\delta S = \int_{t_1}^{t_2} \left[ \mathcal L(q_i^\prime(t^\prime), \dot{q}_i^\prime(t^\prime), t^\prime) \frac{\mathrm d t^\prime}{\mathrm d t} - \mathcal L(q_i(t), \dot{q}_i(t), t) \right] \mathrm d t
From the definition of a symmetry transformation, this becomes:
\delta S = \int_{t_1}^{t_2} \varepsilon \frac{\mathrm d F}{\mathrm d t} \mathrm d t = \varepsilon F\bigg|_{t_1}^{t_2}
Alternatively, we can compute the variation of the action by expanding the Lagrangian. The total variation of the Lagrangian, \Delta L, is given by:
\Delta L = \mathcal L(q_i^\prime(t), \dot{q}_i^\prime(t), t) - \mathcal L(q_i(t), \dot{q}_i(t), t)
The transformation of the coordinates induces a change in the functional form of the Lagrangian.
To first order in \varepsilon, the change in the coordinates and velocities at the same time t are:
\begin{aligned} \Delta q_i(t) &= q_i^\prime(t) - q_i(t) = \varepsilon (\delta q_i - \dot{q}_i \delta t) \\ \Delta \dot{q}_i(t) &= \dot{q}_i^\prime(t) - \dot{q}_i(t) = \frac{\mathrm d}{\mathrm d t} (\Delta q_i) \end{aligned}
The variation of the Lagrangian can then be written as:
\delta L = \sum_i \left( \frac{\partial L}{\partial q_i} \Delta q_i + \frac{\partial L}{\partial \dot{q}_i} \Delta \dot{q}_i \right)
The action is defined with respect to the transformed time t^\prime.
The differential element transforms as:
\mathrm d t^\prime = (1 + \varepsilon \frac{\mathrm d (\delta t)}{\mathrm d t}) \mathrm d t
The change in the action can be written as:
\delta S = \int_{t_1}^{t_2} \left( \delta L + L \varepsilon \frac{\mathrm d (\delta t)}{\mathrm d t} \right) \mathrm d t
Substituting the expression for \delta L:
\delta S = \int_{t_1}^{t_2} \left[ \sum_i \left( \frac{\partial L}{\partial q_i} \Delta q_i + \frac{\partial L}{\partial \dot{q}_i} \Delta \dot{q}_i \right) + \varepsilon L \frac{\mathrm d (\delta t)}{\mathrm d t} \right] \mathrm d t
For paths that satisfy the Euler-Lagrange equations:
\frac{\partial L}{\partial q_i} = \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial L}{\partial \dot{q}_i} \right)
we can manipulate the integrand:
\begin{aligned} \delta S &= \int_{t_1}^{t_2} \left[ \sum_i \left( \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial L}{\partial \dot{q}_i} \right) \Delta q_i + \frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm d}{\mathrm d t}(\Delta q_i) \right) + \varepsilon L \frac{\mathrm d (\delta t)}{\mathrm d t} \right] \mathrm d t \\ &= \int_{t_1}^{t_2} \frac{\mathrm d}{\mathrm d t} \left( \sum_i \frac{\partial L}{\partial \dot{q}_i} \Delta q_i + \varepsilon L \delta t \right) \mathrm d t \end{aligned}
This can be integrated to give:
\delta S = \sum_i \frac{\partial L}{\partial \dot{q}_i} \Delta q_i + \varepsilon L \delta t \bigg|_{t_1}^{t_2}
Equating the two expressions for \delta S:
\varepsilon F \bigg|_{t_1}^{t_2} = \sum_i \frac{\partial L}{\partial \dot{q}_i} \varepsilon (\delta q_i - \dot{q}_i \delta t) + \varepsilon L \delta t \bigg|_{t_1}^{t_2}
Since this must hold for any interval [t_1, t_2], the quantities inside the brackets must be equal up to a constant. This implies:
\frac{\mathrm d}{\mathrm d t} \left( \sum_i \frac{\partial L}{\partial \dot{q}_i} (\delta q_i - \dot{q}_i \delta t) - F + L \delta t \right) = 0
The conserved quantity is therefore:
J = \sum_{i} p_i (\delta q_i - \dot{q}_i \delta t) + L \delta t - F
where:
p_i = \frac{\partial L}{\partial \dot{q}_i}
is the canonical momentum. This completes the derivation.
The theorem provides a direct path from the symmetries of a system to its conservation laws, such as the conservation of energy, momentum, and angular momentum.