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Noether's Theorem and Conservation Laws

Attempt these from memory before reading.

  • State Noether’s theorem in one sentence.
  • Which symmetry gives conservation of momentum? Of angular momentum? Of energy?
  • What does “up to a boundary term” mean in the statement?
  • Is Noether’s theorem about symmetries of the equations, or of the Lagrangian?

Motivation: the dictionary between symmetry and conservation

Section titled “Motivation: the dictionary between symmetry and conservation”

Before Noether (1918), conservation laws were empirical facts, each proved separately: momentum conservation from Newton’s third law, energy conservation from the work-energy theorem, angular momentum from central forces. Nothing connected them.

Noether’s theorem says they are all the same phenomenon. A continuous symmetry of the action produces a conserved quantity, and each familiar conservation law is the conserved quantity for one particular symmetry:

Symmetry of the LagrangianConserved quantity
translation in xxlinear momentum pxp_x
translation in ttenergy HH
rotation about zzangular momentum LzL_z
Galilean boostcentre-of-mass motion

Reading the table backwards is the useful direction. If you can guess a symmetry, you get a conservation law without solving anything. That is how conserved quantities are found in particle physics, where the Lagrangian is written down from symmetry requirements rather than from forces — and it is why Emmy Noether’s theorem is arguably the most important result in theoretical physics.

Theorem 5.1 (Noether’s Theorem). For every continuous symmetry of the action, there is a Corresponding conserved quantity.

More precisely: if the action is invariant (up to a boundary term) under the infinitesimal transformation qj→qj+ϵfj(q,q˙,t)q_j \to q_j + \epsilon f_j(q, \dot{q}, t) Then

Q=∑j∂L∂q˙jfjQ = \sum_j \frac{\partial L}{\partial \dot{q}_j} f_j

Is a constant of motion.

Theorem 5.2 (Noether’s Theorem --- Full Proof). Suppose the Lagrangian transforms under an infinitesimal transformation qj→qj+ϵδqjq_j \to q_j + \epsilon \delta q_j as:

L→L+ϵdFdtL \to L + \epsilon \frac{dF}{dt}

For some function F(q,t)F(q, t). Then the quantity

Q=∑jpj δqj−FQ = \sum_j p_j\, \delta q_j - F

Is conserved.

Proof. The variation of the action is:

δS=∫t1t2[∑j(∂L∂qjδqj+∂L∂q˙jδq˙j)]dt=∫t1t2dFdt dt\delta S = \int_{t_1}^{t_2} \left[\sum_j \left(\frac{\partial L}{\partial q_j}\delta q_j + \frac{\partial L}{\partial \dot{q}_j}\delta\dot{q}_j\right)\right] dt = \int_{t_1}^{t_2} \frac{dF}{dt}\, dt

Where the second equality uses the assumption that the action changes by at most a boundary term. Using the Euler-Lagrange equations ∂L∂qj=ddt∂L∂q˙j\frac{\partial L}{\partial q_j} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}:

δS=∫t1t2∑j[ddt(∂L∂q˙j)δqj+∂L∂q˙jddtδqj]dt\delta S = \int_{t_1}^{t_2} \sum_j \left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)\delta q_j + \frac{\partial L}{\partial \dot{q}_j}\frac{d}{dt}\delta q_j\right] dt =∫t1t2ddt(∑jpj δqj)dt= \int_{t_1}^{t_2} \frac{d}{dt}\left(\sum_j p_j\, \delta q_j\right) dt

Setting this equal to ∫t1t2dFdt dt\int_{t_1}^{t_2} \frac{dF}{dt}\, dt:

ddt(∑jpj δqj−F)=0\frac{d}{dt}\left(\sum_j p_j\, \delta q_j - F\right) = 0

Therefore Q=∑jpj δqj−FQ = \sum_j p_j\, \delta q_j - F is constant. ■\blacksquare

5.3 Worked Example: Spatial Translation and Linear Momentum

Section titled “5.3 Worked Example: Spatial Translation and Linear Momentum”

Problem. Show that spatial translation invariance implies conservation of linear momentum.

Solution

Consider an infinitesimal translation x→x+ϵx \to x + \epsilon I.e., δx=1\delta x = 1, δy=0\delta y = 0, δz=0\delta z = 0.

For a free particle, L=12m(x˙2+y˙2+z˙2)L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) Which is invariant (δL=0\delta L = 0 So F=0F = 0).

By Noether’s theorem:

Q=px⋅1+py⋅0+pz⋅0−0=px=constQ = p_x \cdot 1 + p_y \cdot 0 + p_z \cdot 0 - 0 = p_x = \mathrm{const}

This is conservation of the xx-component of linear momentum. Translation invariance in all three directions gives conservation of the full momentum vector p\mathbf{p}. ■\blacksquare

5.4 Worked Example: Rotation and Angular Momentum

Section titled “5.4 Worked Example: Rotation and Angular Momentum”

Problem. Show that rotational invariance implies conservation of angular momentum.

Solution

Consider an infinitesimal rotation by angle ϵ\epsilon about the zz-axis:

δx=−ϵy,δy=ϵx,δz=0\delta x = -\epsilon y, \quad \delta y = \epsilon x, \quad \delta z = 0

For a free particle, δL=m(x˙ δx˙+y˙ δy˙)=m(x˙(−ϵy˙)+y˙(ϵx˙))=0\delta L = m(\dot{x}\,\delta\dot{x} + \dot{y}\,\delta\dot{y}) = m(\dot{x}(-\epsilon\dot{y}) + \dot{y}(\epsilon\dot{x})) = 0 So F=0F = 0.

By Noether’s theorem:

Q=px(−y)+py(x)−0=x py−y px=LzQ = p_x(-y) + p_y(x) - 0 = x\, p_y - y\, p_x = L_z

This is the zz-component of angular momentum. Full rotational invariance gives conservation of the entire angular momentum vector L=r×p\mathbf{L} = \mathbf{r} \times \mathbf{p}. ■\blacksquare

5.5 Worked Example: Time Translation and Energy

Section titled “5.5 Worked Example: Time Translation and Energy”

Problem. Show that time translation invariance implies conservation of energy.

Solution

Consider an infinitesimal time translation t→t+ϵt \to t + \epsilon. The coordinates transform as qj(t)→qj(t+ϵ)≈qj(t)+ϵq˙j(t)q_j(t) \to q_j(t + \epsilon) \approx q_j(t) + \epsilon \dot{q}_j(t) So δqj=q˙j\delta q_j = \dot{q}_j.

If LL does not depend explicitly on time, then:

δL=∑j(∂L∂qjq˙j+∂L∂q˙jq¨j)ϵ=dLdtϵ=ddt(ϵL)\delta L = \sum_j \left(\frac{\partial L}{\partial q_j}\dot{q}_j + \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j\right)\epsilon = \frac{dL}{dt}\epsilon = \frac{d}{dt}\left(\epsilon L\right)

So F=ϵLF = \epsilon L Giving F=LF = L (per unit ϵ\epsilon).

By Noether’s theorem:

Q=∑jpjq˙j−L=hQ = \sum_j p_j \dot{q}_j - L = h

This is the energy function, which equals T+VT + V for natural systems. ■\blacksquare

5.6 Summary: Symmetry-Conservation Correspondence

Section titled “5.6 Summary: Symmetry-Conservation Correspondence”
SymmetryTransformationConserved Quantity
Time translationt→t+ϵt \to t + \epsilonEnergy HH
Spatial translationx→x+ϵx \to x + \epsilonLinear momentum pxp_x
Rotation about zzϕ→ϕ+ϵ\phi \to \phi + \epsilonAngular momentum LzL_z
Galilean boostx→x+ϵtx \to x + \epsilon tCentre-of-mass motion

Noether’s theorem is the universe’s most elegant bookkeeping principle. Every symmetry you can identify is a guarantee that something is conserved. If the laws of physics are the same here as they are on the other side of the room, then linear momentum is conserved. If the laws are the same now as they were yesterday, then energy is conserved. If the laws do not care which way you point your coordinate axes, then angular momentum is conserved. Think of it like a conservation bank: every symmetry deposits a conserved quantity that you can withdraw later to solve problems. The deep insight is that conservation laws are not separate accidents of nature but consequences of the underlying symmetry structure. When you discover a new symmetry, you automatically gain a new conservation law.

Problem. A particle moves in a central potential V(r)V(r). Show that angular momentum is conserved.

Solution. In spherical coordinates (r,θ,ϕ)(r, \theta, \phi) with V=V(r)V = V(r):

L=12m(r˙2+r2θ˙2+r2sin⁡2θ ϕ˙2)−V(r)L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\,\dot{\phi}^2) - V(r)

Since LL does not depend on ϕ\phi (rotational symmetry about the zz-axis):

pϕ=∂L∂ϕ˙=mr2sin⁡2θ ϕ˙=constp_\phi = \frac{\partial L}{\partial \dot{\phi}} = mr^2\sin^2\theta\,\dot{\phi} = \mathrm{const}

This is the zz-component of angular momentum. By Noether’s theorem, the full angular momentum vector Is conserved for any central potential. ■\blacksquare

Mistake 1: Assuming every symmetry implies a conservation law without checking continuity Noether’s theorem applies only to continuous symmetries. Discrete symmetries like parity or time reversal do not yield conserved quantities via Noether’s theorem. For example, a crystal lattice has discrete translational symmetry but does not conserve crystal momentum in the same sense as continuous translation invariance conserves linear momentum.

Mistake 2: Confusing the conserved quantity with the symmetry generator The conserved quantity Q=∑jpjδqj−FQ = \sum_j p_j \delta q_j - F is not the same as the infinitesimal generator of the transformation. The generator is the vector field δqj\delta q_j, while QQ is the corresponding momentum map. For time translation, the generator is q˙j\dot{q}_j but the conserved quantity is the Hamiltonian H=∑jpjq˙j−LH = \sum_j p_j \dot{q}_j - L.

Mistake 3: Applying Noether’s theorem to systems with explicit time dependence in the Lagrangian When the Lagrangian depends explicitly on time, the action is not invariant under time translations, so energy is not conserved. The quantity h=∑jpjq˙j−Lh = \sum_j p_j \dot{q}_j - L is still well-defined but is not conserved. Students often assume hh is always the energy, but it only equals the conserved energy when ∂L/∂t=0\partial L/\partial t = 0.

The Galilean boost: a conservation law that surprises people

Section titled “The Galilean boost: a conservation law that surprises people”

Boost symmetry is the case where the naive statement is wrong, and it is worth working through.

A free particle has L=12mx˙2L = \frac12 m \dot x^2, which is not invariant under a Galilean boost x→x+vtx \to x + vt: the kinetic energy picks up mvx˙+12mv2tm v \dot x + \frac12 m v^2 t. But those extra terms are a total time derivative, ddt(mvx+12mv2t)\frac{d}{dt}(m v x + \frac12 m v^2 t), and Noether’s theorem is stated “up to a boundary term” exactly for this reason.

The conserved quantity is then

Q=mx˙−mv,Q = m\dot x - m v,

which is the momentum in the boosted frame. The boost symmetry gives conservation of the velocity of the centre of mass, not a new independent quantity. Getting this wrong — expecting a new conserved quantity from every transformation — is the standard error.

ClaimTrue?Counterexample
Discrete symmetries give conserved quantitiesnoparity is a symmetry of electromagnetism but nothing is conserved by it
Every conserved quantity comes from a symmetry of qqnomomentum can be conserved with no obvious coordinate symmetry
Symmetry of the equations is enoughnothe theorem needs symmetry of the action
Noether’s theorem applies to dissipative systemsnofriction is not derivable from a Lagrangian

The first row is the one most often asserted wrongly. It is continuous symmetries only, because the proof differentiates with respect to the transformation parameter. Parity is a symmetry and charge conjugation is a symmetry, and neither produces a Noether current.

Where the theorem is used at research level

Section titled “Where the theorem is used at research level”
  • Particle physics. The Standard Model Lagrangian is written down by requiring invariance under SU(3)×SU(2)×U(1)SU(3) \times SU(2) \times U(1). Each gauge symmetry gives a conserved current, and the associated charges are electric charge, weak isospin and colour.
  • General relativity. Killing vectors of a spacetime metric play the role of symmetry generators; a Killing vector gives a conserved quantity along geodesics, which is how energy and angular momentum are defined for orbits around a black hole.
  • Integrable systems. The solitons of the Korteweg-de Vries equation carry infinitely many conserved quantities, each from a symmetry of a Lax pair — the reason that equation is exactly solvable.

Mix these with material from Lagrangian mechanics, Hamiltonian mechanics and the symplectic structure page.

  1. (Lagrangian.) Show that for a central potential V(r)V(r) the Lagrangian is invariant under rotations, and recover conservation of LzL_z from Noether’s theorem.
  2. (Hamiltonian.) Show that if GG generates a symmetry then {G,H}=0\{G, H\} = 0, connecting the Poisson-bracket statement of conservation to Noether’s.
  3. (Relativity.) Show that the Lagrangian of a free relativistic particle, L=−mc21−v2/c2L = -mc^2\sqrt{1 - v^2/c^2}, is invariant under Lorentz boosts up to a boundary term, and identify the conserved quantity.
  4. (Fields.) For a complex scalar field with L=∣∂ϕ∣2−m2∣ϕ∣2L = |\partial\phi|^2 - m^2|\phi|^2, show that invariance under ϕ→eiαϕ\phi \to e^{i\alpha}\phi gives a conserved current jμj^\mu, and compute the charge.
  5. (Symplectic structure.) Show that a symmetry generated by GG preserves the symplectic form, connecting Noether’s theorem to the geometry of phase space.