Lesson 1.3 · Foundations

Symmetries & Conservation

This lesson contains one of the most beautiful results in all of physics: Noether's theorem. It reveals a deep connection between symmetry and conservation laws — a connection that runs through all of modern physics.

The Big Idea

You already know some conservation laws:

Energy is conserved
Momentum is conserved
Angular momentum is conserved

But why? Where do these laws come from? Are they separate facts, or connected?

Emmy Noether answered this in 1918:

Noether's Theorem

Every continuous symmetry of the action corresponds to a conserved quantity.

The conservation laws aren't accidents. They're consequences of symmetries in the laws of physics:

Each symmetry of physics generates a conservation law.

What is a Symmetry?

A symmetry is a transformation that leaves something unchanged.

In physics, we care about symmetries of the action. If we transform the coordinates in some way and the action doesn't change, we have a symmetry.

Time Translation Symmetry

If I do an experiment today or tomorrow, I get the same results. The laws of physics don't depend on when you are.

Mathematically: if we shift time by a constant $\epsilon$

t \to t + \epsilon

and the Lagrangian doesn't explicitly depend on $t$, then the action is unchanged.

Space Translation Symmetry

If I do an experiment here or 10 meters to the left, I get the same results. The laws don't depend on where you are.

x \to x + \epsilon

Rotation Symmetry

If I rotate my apparatus, physics works the same way. The laws don't depend on which direction you face.

\theta \to \theta + \epsilon

Deriving Noether's Theorem

Let's prove the connection. Consider a continuous transformation of coordinates:

q \to q + \epsilon \, \delta q

where $\epsilon$ is infinitesimal and $\delta q$ describes the type of transformation.

Step 1: Change in the Lagrangian

Under this transformation:

$$ \delta L = \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} $$

Step 2: Use the Euler-Lagrange equation

We know that $\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$, so:

$$ \delta L = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \cdot \delta q + \frac{\partial L}{\partial \dot{q}} \cdot \delta \dot{q} $$

Step 3: Recognize a total derivative

This is just the product rule:

$$ \delta L = \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \delta q \right) $$

Step 4: If the action is symmetric

If the transformation is a symmetry, then $\delta L = 0$ (or $\delta L = \frac{d}{dt}(\text{something})$, which doesn't affect equations of motion). Therefore:

$$ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \delta q \right) = 0 $$

Step 5: The conserved quantity

If the time derivative is zero, the quantity is constant:

$$ Q = \frac{\partial L}{\partial \dot{q}} \delta q = \text{constant} $$

This $Q$ is the conserved charge associated with the symmetry.

The Three Great Conservation Laws

Energy from Time Symmetry

If $L$ doesn't depend explicitly on time, consider the quantity:

H = \dot{q} \frac{\partial L}{\partial \dot{q}} - L

Taking its time derivative and using Euler-Lagrange:

\frac{dH}{dt} = \dot{q} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} + \ddot{q}\frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q}\dot{q} - \frac{\partial L}{\partial \dot{q}}\ddot{q} - \frac{\partial L}{\partial t}

The middle terms cancel. Using Euler-Lagrange on the first and third terms:

\frac{dH}{dt} = -\frac{\partial L}{\partial t}

If $\frac{\partial L}{\partial t} = 0$ (time-translation symmetry), then $H$ is conserved.

For $L = T - V = \frac{1}{2}m\dot{q}^2 - V(q)$:

H = \dot{q}(m\dot{q}) - \left(\frac{1}{2}m\dot{q}^2 - V\right) = \frac{1}{2}m\dot{q}^2 + V = T + V = E

Energy Conservation

If physics is the same today as tomorrow (time-translation symmetry), then energy is conserved.

Momentum from Space Symmetry

If $L$ doesn't depend on position $x$ (only on $\dot{x}$), then:

\frac{\partial L}{\partial x} = 0

The Euler-Lagrange equation gives:

\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 0

So $p = \frac{\partial L}{\partial \dot{x}} = m\dot{x}$ is conserved.

Momentum Conservation

If physics is the same here as there (space-translation symmetry), then momentum is conserved.

Angular Momentum from Rotation Symmetry

For a system with rotational symmetry (central force), the Lagrangian in polar coordinates is:

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

Note: $L$ doesn't depend on $\theta$, only on $r$. This is rotational symmetry — the physics doesn't change if we rotate the system.

Since $\frac{\partial L}{\partial \theta} = 0$:

\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} = 0

So $L_z = \frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta}$ is conserved.

Angular Momentum Conservation

If physics is the same in all directions (rotational symmetry), then angular momentum is conserved.

The Deep Meaning

Noether's theorem says conservation laws aren't separate facts to memorize. They're all manifestations of symmetry.

This goes far beyond classical mechanics:

Electric charge conservation comes from a symmetry called $U(1)$ gauge invariance
Color charge conservation (quarks) comes from $SU(3)$ gauge invariance
Lepton number, baryon number — more symmetries, more conservation laws

The entire Standard Model of particle physics is built on symmetry principles.

Connection: Broken Symmetries

What if a symmetry is "broken"? Then the conservation law fails. The universe's time-translation symmetry is broken by cosmic expansion — and indeed, energy is not conserved on cosmological scales. This is why the CMB has cooled from 3000K to 2.7K since the Big Bang.

Cyclic Coordinates

A coordinate $q_i$ that doesn't appear in the Lagrangian (only $\dot{q}_i$ appears) is called cyclic or ignorable.

For a cyclic coordinate, the Euler-Lagrange equation immediately gives:

\frac{\partial L}{\partial q_i} = 0 \quad \Rightarrow \quad \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} = 0 \quad \Rightarrow \quad p_i = \frac{\partial L}{\partial \dot{q}_i} = \text{const}

The conjugate momentum to a cyclic coordinate is conserved. This is Noether's theorem in its simplest form.

Example: Kepler Problem

A planet orbiting the sun. In polar coordinates:

L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + \frac{GMm}{r}

(We use $+V$ because $V = -GMm/r$ is negative.)

Symmetries:

$\theta$ is cyclic → angular momentum $L_z = mr^2\dot{\theta}$ is conserved
No explicit time dependence → energy $E = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - \frac{GMm}{r}$ is conserved

These two conservation laws are enough to solve the entire problem — no need to solve differential equations directly.

Kepler orbits: angular momentum conservation keeps the planet in a plane.

Exercises

Verify energy conservation: For $L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2$ (harmonic oscillator), compute $H = \dot{x}\frac{\partial L}{\partial \dot{x}} - L$ and verify it equals the total energy.
Find the conserved quantity: A bead slides on a rotating hoop. The Lagrangian is $L = \frac{1}{2}mR^2\dot{\theta}^2 + \frac{1}{2}mR^2\omega^2\sin^2\theta - mgR\cos\theta$. Is energy conserved? (Hint: does $L$ depend explicitly on $t$?)
Boost symmetry: For a free particle, the action is invariant under Galilean boosts $x \to x + vt$. What is the associated conserved quantity? (This one is tricky — the conserved quantity involves both position and momentum.)

Key Takeaways

Every continuous symmetry has a corresponding conservation law (Noether's theorem)
Time symmetry → energy conservation
Space symmetry → momentum conservation
Rotation symmetry → angular momentum conservation
Cyclic coordinates give conserved conjugate momenta
This principle extends to all of physics, including gauge symmetries in QFT

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