Conservation Laws & Noether's Theorem
Conservation Laws & Noether's Theorem
One of the deepest results in theoretical physics connects symmetries of the Lagrangian to conserved quantities. Noether's theorem (Emmy Noether, 1915) states: for every continuous symmetry of the action, there exists a corresponding conserved quantity. Time-translation symmetry gives conservation of energy; spatial-translation symmetry gives conservation of linear momentum; rotational symmetry gives conservation of angular momentum. These are not coincidences but mathematical necessities, and they hold in every domain of physics from particle mechanics to field theory.
Formally, a continuous symmetry is a one-parameter family of transformations $q_i\to q_i+\epsilon\delta q_i(q,\dot{q},t)$ under which the Lagrangian changes by at most a total time derivative: $\delta L=\epsilon\,d\Lambda/dt$. The associated Noether charge (conserved quantity) is $$I=\sum_i p_i\,\delta q_i - \Lambda = \text{const along any solution}.$$ The proof follows directly from the Euler-Lagrange equations and is one of the most elegant arguments in all of mathematics.
Beyond Noether's theorem, there is the energy function (Jacobi integral): $h=\sum_i p_i\dot{q}_i-L$. When $L$ has no explicit time dependence, $h$ is conserved and equals the total energy $T+V$ (provided $T$ is a homogeneous quadratic in $\dot{q}$). When $t$ appears explicitly in $L$ — as in a time-varying potential — $h$ is not conserved and $dh/dt=-\partial L/\partial t$.
Theorem — Noether's Theorem
If the action $S[q]=\int L\,dt$ is invariant under a continuous one-parameter transformation $q_i\to q_i+\epsilon\delta q_i$ (up to a boundary term $d\Lambda/dt$), then $$I=\sum_i\frac{\partial L}{\partial\dot{q}_i}\delta q_i-\Lambda$$ is conserved along every solution of the equations of motion.
Definition — Energy Function (Jacobi Integral)
The energy function is $h(q,\dot{q},t)=\sum_i p_i\dot{q}_i-L$. If $\partial L/\partial t=0$, then $dh/dt=0$ and $h$ equals the total mechanical energy $T+V$ when $T$ is quadratic in the velocities.
Example 1: Linear Momentum from Translation Symmetry
For a system of particles with $L=\sum_i\frac{m_i}{2}|\dot{\mathbf{r}}_i|^2-V(|\mathbf{r}_i-\mathbf{r}_j|)$, a simultaneous shift $\mathbf{r}_i\to\mathbf{r}_i+\epsilon\hat{\mathbf{n}}$ leaves $L$ unchanged (potential depends only on differences). Noether's theorem gives the conserved quantity $I=\hat{\mathbf{n}}\cdot\sum_i m_i\dot{\mathbf{r}}_i=\hat{\mathbf{n}}\cdot\mathbf{P}$, i.e., the component of total linear momentum along $\hat{\mathbf{n}}$ is conserved.
Example 2: Angular Momentum from Rotation Symmetry
Under an infinitesimal rotation about the $z$-axis, $\delta\mathbf{r}_i=\hat{z}\times\mathbf{r}_i$. For a rotationally symmetric potential (central force), $\delta L=0$. The Noether charge is $I=\sum_i m_i(\mathbf{r}_i\times\dot{\mathbf{r}}_i)\cdot\hat{z}=L_z$, the $z$-component of total angular momentum. Full rotational symmetry in 3D gives conservation of all three components $L_x,L_y,L_z$.
Example 3: Runge-Lenz Vector
The Kepler problem ($V=-k/r$) has a hidden symmetry beyond rotation: the orbit does not precess. This extra symmetry corresponds to the conserved Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times\mathbf{L}-mk\hat{r}$. Its conservation implies that the orbit is a closed ellipse (Bertrand's theorem for $1/r^2$ and $r$ forces). The symmetry group is $SO(4)$ (not just $SO(3)$), a result only transparent in the Hamiltonian framework.
Example 4: Scale Invariance & Virial Theorem
For a homogeneous potential $V(\lambda\mathbf{r})=\lambda^n V(\mathbf{r})$, the virial theorem states that for bounded orbits $\langle T\rangle=\frac{n}{2}\langle V\rangle$. For the harmonic oscillator ($n=2$): $\langle T\rangle=\langle V\rangle=E/2$. For gravity ($n=-1$): $\langle T\rangle=-\frac{1}{2}\langle V\rangle$, so the total energy $E=\langle V\rangle/2<0$ (bound state). The virial theorem follows from a quasi-symmetry (scaling) of the action.
Practice Problems
- Verify Noether's theorem for the Lagrangian $L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-V(x^2+y^2)$: identify the symmetry and the conserved quantity.
- A particle moves in the potential $V=V(x-y)$ depending only on the difference. What quantity is conserved? Prove it using Noether's theorem.
- For $L=\frac{1}{2}m\dot{q}^2 e^{2\alpha t}$, compute $\partial L/\partial t$ and show that the energy function $h$ is not conserved. Find $dh/dt$ explicitly.
- The Lagrangian for a particle in a magnetic field is $L=\frac{1}{2}m\dot{r}^2+q\dot{\mathbf{r}}\cdot\mathbf{A}$. Show that canonical momentum $p_i=m\dot{r}_i+qA_i$ is conserved when $\mathbf{A}$ is uniform.
- Prove the virial theorem: for a system in a periodic orbit with homogeneous potential $V\propto r^n$, show $2\langle T\rangle=n\langle V\rangle$ by differentiating $G=\sum_i \mathbf{p}_i\cdot\mathbf{r}_i$ with respect to time and time-averaging.
Show Answer Key
1. Symmetry: rotation $\delta\phi$ about $z$-axis: $\delta x = -y\delta\phi$, $\delta y = x\delta\phi$. Noether charge: $Q = \frac{\partial L}{\partial \dot{x}}(-y) + \frac{\partial L}{\partial \dot{y}}(x) = m(-y\dot{x}+x\dot{y}) = -L_z$. So angular momentum $L_z = m(x\dot{y}-y\dot{x})$ is conserved. ✓
2. Translation symmetry $x \to x+\epsilon$, $y \to y+\epsilon$ (both shift equally, preserving $x-y$). Noether charge: $Q = p_x + p_y = m(\dot{x}+\dot{y})$ — total momentum is conserved.
3. $L = \frac{1}{2}m\dot{q}^2 e^{2\alpha t}$. $p = m\dot{q}e^{2\alpha t}$. $h = p\dot{q}-L = \frac{1}{2}m\dot{q}^2 e^{2\alpha t}$. $\frac{\partial L}{\partial t} = m\alpha\dot{q}^2 e^{2\alpha t} = 2\alpha h$. So $dh/dt = -\partial L/\partial t = -2\alpha h$, giving $h(t) = h_0 e^{-2\alpha t}$. Energy dissipates exponentially.
4. Canonical momentum: $p_i = \frac{\partial L}{\partial \dot{r}_i} = m\dot{r}_i + qA_i$. $\dot{p}_i = \frac{\partial L}{\partial r_i}$. If $\mathbf{A}$ is uniform (constant), $\frac{\partial}{\partial r_i}(q\dot{\mathbf{r}}\cdot\mathbf{A}) = q\dot{A}_i/... $ Actually: $\partial L/\partial r_i = q\dot{r}_j \partial A_j/\partial r_i$. If $A$ is uniform, $\partial A_j/\partial r_i = 0$, so $\dot{p}_i = 0$. ✓
5. $G = \sum_i \mathbf{p}_i\cdot\mathbf{r}_i$. $\dot{G} = \sum(\dot{\mathbf{p}}_i\cdot\mathbf{r}_i + \mathbf{p}_i\cdot\dot{\mathbf{r}}_i) = \sum(\mathbf{F}_i\cdot\mathbf{r}_i + 2T_i)$. Time-average over period $\tau$: $\langle\dot{G}\rangle = 0$ (periodic). So $2\langle T\rangle = -\langle\sum\mathbf{F}_i\cdot\mathbf{r}_i\rangle$. For $V \propto r^n$: $\sum\mathbf{F}_i\cdot\mathbf{r}_i = -nV$ (Euler's theorem). Hence $2\langle T\rangle = n\langle V\rangle$.