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Topic 9 of 11

Symmetry & Conservation Laws

Emmy Noether's theorem is one of the deepest results in physics: every continuous symmetry of the action corresponds to a conserved quantity. Translational symmetry gives momentum conservation; rotational symmetry gives angular momentum conservation; time-translation symmetry gives energy conservation.

Key Concepts

Noether's Theorem

If the Lagrangian is invariant under a continuous one-parameter transformation

q\to q+\epsilon\delta q

, then

Q=\sum_i\frac{\partial L}{\partial\dot q_i}\delta q_i

is a constant of motion.

Cyclic Coordinate

A coordinate

q_i

that does not appear in

L

(only

\dot q_i

does). The conjugate momentum

p_i=\partial L/\partial\dot q_i

is then conserved. This is the simplest application of Noether's theorem.

Energy Conservation

L

has no explicit time dependence (

\partial L/\partial t=0

), the Hamiltonian

H=\sum_i p_i\dot q_i-L

is conserved. This is conservation of energy.

Discrete Symmetries

Parity (spatial reflection) and time reversal are discrete symmetries. Unlike continuous symmetries, they do not generate conserved quantities (Noether's theorem requires continuous transformations), but they impose constraints on allowed interactions.

Key Equations

Noether's conserved charge

Q = \sum_i \frac{\partial L}{\partial \dot{q}_i}\,\delta q_i

Conserved quantity associated with the symmetry transformation δqᵢ.

Energy (Jacobi integral)

E = \sum_i \dot{q}_i \frac{\partial L}{\partial \dot{q}_i} - L

Conserved when ∂L/∂t = 0; equals T+V for natural systems.

Angular momentum

L_z = m(x\dot{y} - y\dot{x}) = mr^2\dot{\phi}

Conserved due to rotational symmetry; conjugate to the cyclic angle φ.

Worked Example

Angular Momentum from Rotational Symmetry

Problem

A particle moves in a central potential $V(r)$ . Show that angular momentum is conserved using cyclic coordinates.

Solution

In polar coordinates: $T=\frac{1}{2}m(\dot r^2+r^2\dot\phi^2)$ , $V=V(r)$ .

L = \frac{1}{2}m(\dot r^2 + r^2\dot\phi^2) - V(r)

$\phi$ is cyclic ( $\partial L/\partial\phi=0$ ), so the conjugate momentum is conserved:

p_\phi = \frac{\partial L}{\partial\dot\phi} = mr^2\dot\phi = l = \text{const}

$p_\phi=l$ is the angular momentum — conserved because $L$ has rotational ( $\phi$ ) symmetry.

Answer

l=mr^2\dot\phi=

const; conservation follows from

\phi

being cyclic in

L

Practice

Exercises

7 problems

1 of 7

A particle in 2D has $L=\frac{1}{2}m(\dot x^2+\dot y^2)-V(x)$ (no $y$ in $V$ ). Which momentum is conserved?

Your answer

(p_y = const)

2 of 7

A particle ( $m=2.0$ kg) moves in a circle of radius $r=3.0$ m at $\dot\phi=4.0$ rad/s. Find the conserved angular momentum $l=mr^2\dot\phi$ (in kg·m²/s).

Your answer

kg·m²/s

3 of 7

A free particle with $L=\frac{1}{2}m\dot x^2$ has $p_x=5.0$ kg·m/s at $t=0$ . What is $p_x$ at $t=10$ s?

Unlock Exercise 3