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Conserved quantities and Noether's theorem

Ignorable coordinates and generalized momenta, momentum and angular momentum from symmetry, the energy function $\sum p\dot{q} - \mathcal{L}$, and when it is not the total energy.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to find a system's conserved quantities from the symmetries of its Lagrangian, construct its energy function, and use conservation laws to reduce a problem.

2. What you already have

You can write a Lagrangian in generalized coordinates and derive Lagrange's equations. You know from earlier lessons that momentum is conserved without external force, angular momentum under central forces, and energy for conservative forces. You also met first integrals in the calculus of variations. This lesson shows that every one of those laws comes from a symmetry of the Lagrangian, and gives a way to find conserved quantities by looking.

3. Words for this lesson

4. Symmetries give conservation laws

Lagrange's equation for a coordinate $q_i$ is $\frac{d}{dt}p_i = \partial\mathcal{L}/\partial q_i$, with the generalized momentum $p_i = \partial\mathcal{L}/\partial\dot{q}_i$. If $q_i$ does not appear in $\mathcal{L}$, the right side is zero and

$$p_i = \frac{\partial\mathcal{L}}{\partial\dot{q}_i} = \text{constant}.$$

Such a coordinate is ignorable. It is missing because the system looks the same whatever its value: a symmetry. Choosing coordinates in which the symmetry shows as a missing coordinate turns it into a conservation law with no work at all.

Time works the same way. If $\mathcal{L}$ does not depend explicitly on $t$, then the energy function

$$\mathcal{H} = \sum_ip_i\dot{q}_i - \mathcal{L}$$

is conserved. When the constraints do not move and $T$ is quadratic in the velocities, $\mathcal{H} = T + U$, the total energy. The general statement, that each continuous symmetry gives a conserved quantity, is Noether's theorem, proved by Emmy Noether in 1915.

5. Momentum from translation

For a system of particles interacting through forces that depend on their separations, the Lagrangian is unchanged if every particle is moved by the same distance $\epsilon$ along $x$. Using the center of mass $X$ as one coordinate and the separations as the others, $X$ does not appear in $\mathcal{L}$, and its generalized momentum, $M\dot{X}$, the total momentum, is conserved.

A uniform external field breaks the symmetry in its own direction only. For a projectile, $\mathcal{L} = \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mgy$ contains $y$ but not $x$: horizontal momentum is conserved and vertical momentum is not. Reading which coordinates are absent tells you which components of momentum survive, before any equation is solved.

6. Angular momentum from rotation

If the Lagrangian is unchanged when the whole system is rotated about an axis, choose the rotation angle $\phi$ about that axis as a coordinate. It is ignorable, and $p_\phi$ is the angular momentum about the axis. For a particle in a central potential, $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2) - U(r)$ gives $p_\phi = mr^2\dot{\phi}$ at once.

A symmetry about only one axis gives only one component. A top spinning on a table is symmetric about the vertical, so the vertical component of its angular momentum is conserved, but gravity's torque changes the horizontal components, which is why the top precesses. A later lesson solves that motion using exactly these conserved quantities.

7. The energy function and when it is the energy

Differentiate $\mathcal{L}(q, \dot{q}, t)$ along a motion with the chain rule and use Lagrange's equations. The result can be arranged as $\frac{d}{dt}\left(\sum p_i\dot{q}_i - \mathcal{L}\right) = -\partial\mathcal{L}/\partial t$. So if $t$ does not appear explicitly in $\mathcal{L}$, the energy function $\mathcal{H}$ is conserved.

When is $\mathcal{H}$ the total energy? If the kinetic energy is a quadratic form in the $\dot{q}$'s, which is true whenever the relation between Cartesian and generalized coordinates does not involve time, then $\sum p_i\dot{q}_i = 2T$, a result of Euler's theorem on homogeneous functions, and $\mathcal{H} = 2T - (T - U) = T + U$. With a moving constraint the kinetic energy has terms without a $\dot{q}$, and $\mathcal{H}$ is something else.

8. A conserved quantity that is not the energy

The bead on a rod spinning at a steady $\omega$ has $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + \omega^2r^2)$, with no $t$. Its energy function is $\mathcal{H} = \tfrac{1}{2}m\dot{r}^2 - \tfrac{1}{2}m\omega^2r^2$, which is conserved. But the bead's kinetic energy, $\tfrac{1}{2}m(\dot{r}^2 + \omega^2r^2)$, grows as it slides out: the motor turning the rod does work on it through the rod's sideways push.

So $\mathcal{H}$, called the Jacobi integral in such problems, is conserved while $T + U$ is not. It is still useful: from rest at $r_0$, it gives $\dot{r}^2 = \omega^2(r^2 - r_0^2)$ directly, and integrating gives $r = r_0\cosh\omega t$. The same integral organizes the motion of a spacecraft near the Earth and Moon, in a frame rotating with them.

9. Canonical momentum is not always mass times velocity

For a charge $q$ in a magnetic field with vector potential $\vec{A}$, the Lagrangian is $\tfrac{1}{2}mv^2 - q\Phi + q\vec{v} \cdot \vec{A}$. Its generalized momentum is $\vec{p} = m\vec{v} + q\vec{A}$, the canonical momentum, which differs from the mechanical momentum $m\vec{v}$.

If the fields are symmetric about an axis, the canonical angular momentum about it is conserved, including the field's part. This is how magnetic mirrors trap charged particles in the Earth's Van Allen belts and in fusion experiments: a symmetry of the field produces a conserved quantity that stops the particles from reaching the regions of strongest field. It is also the momentum that becomes the operator $-i\hbar\nabla$ in quantum mechanics.

10. The method, step by step, and how to check it

1. Look for the symmetry first: a direction the forces ignore, an axis about which the system looks the same, or no clock in the problem.
2. Choose coordinates that make the symmetry a missing coordinate.
3. Write the conserved quantities: each $p_i$ for a missing $q_i$, and $\mathcal{H}$ if $t$ is missing.
4. Use them to replace second-order equations by first-order ones.

Checking an answer. Differentiate each conserved quantity along your solution and confirm it is constant. A conserved generalized momentum for a Cartesian coordinate must be the ordinary momentum. And if you find $T + U$ conserved for a system with a driven constraint, look again: energy is flowing in through the constraint.

11. Why Noether's theorem matters beyond mechanics

Noether's theorem turned conservation laws from lucky facts into consequences of symmetry, and physicists now use it in reverse: to find what is conserved, look for what does not change. Electric charge is conserved because of a symmetry in the phase of quantum fields. In particle physics, symmetries of the equations predicted conserved quantities, such as baryon number, and particles, before experiments found them.

It also explains when conservation fails. In an expanding universe the equations do depend on time, and the energy of light traveling across it is not conserved: it is redshifted. The theorem tells us exactly which symmetry is broken, and so which law no longer holds.

12. Using conserved quantities to reduce a problem

Each conserved quantity removes one integration. A particle in a central potential has two coordinates, $r$ and $\phi$, and two second-order equations. The ignorable $\phi$ gives $mr^2\dot{\phi} = \ell$, so $\dot{\phi} = \ell/mr^2$ can be substituted wherever it appears. The energy function then gives $\tfrac{1}{2}m\dot{r}^2 + \ell^2/2mr^2 + U(r) = E$, a single first-order equation in $r$ alone, which is a one-dimensional problem of exactly the kind the energy lesson solved with a potential curve.

The term $\ell^2/2mr^2$ acts like an extra repulsive potential, keeping the particle away from the origin unless $\ell = 0$. The next unit is built on this reduction: the orbit of a planet or a comet is found by reading turning points off the curve of $U(r) + \ell^2/2mr^2$, with no need to solve the full two-dimensional motion. The rule to take away is to find every conserved quantity before writing any equation of motion; each one you find is an equation you do not have to solve.

13. In the world: Halley's comet

Halley's comet swings within $0.59$ AU of the Sun, inside Venus's orbit, and out to $35$ AU, beyond Neptune. The Sun's gravity is central, so the angular momentum $mr^2\dot{\phi}$ is conserved. At the ends of the orbit, where the velocity is perpendicular to the radius, this gives $r_pv_p = r_av_a$.

At perihelion the comet moves at about $54.6$ km/s; at aphelion, sixty times farther out, at only about $0.91$ km/s, slower than a rifle bullet. Because it crawls so slowly in the outer part of its orbit, it spends most of its $76$-year period far from the Sun and only a few months near enough to grow a visible tail. Its last visit was in 1986, and it will be back in 2061, when observers across the United States will be able to see it again.

14. In the world: divers and figure skaters

A diver leaving the board has a fixed angular momentum about her center of mass, since gravity exerts no torque about it. By tucking, she cuts her moment of inertia to about a third of its stretched value, and her rotation rate triples: enough to complete two and a half somersaults in the second or so before she reaches the water from a $10$ m platform.

Figure skaters do the same about a vertical axis. Pulling in her arms, a skater at the U.S. Championships can spin up from two revolutions a second to six. Her kinetic energy, $L^2/2I$, rises threefold as $I$ falls; the extra energy comes from the work her muscles do pulling her arms inward against their tendency to fly outward. The conserved quantity is angular momentum, not energy, which is exactly the lesson of the puck on the string.

15. The conserved energy function is not always the energy

Because $\mathcal{H} = T + U$ in most problems, it is easy to assume it always is. With a moving constraint, such as a spinning rod or hoop, $\mathcal{H}$ is conserved but $T + U$ changes, fed by the motor that drives the constraint. Conversely, a Lagrangian that depends explicitly on time, such as a pendulum whose length is being changed, conserves neither.

A second error is to think a conservation law needs the whole potential to vanish. Only the particular coordinate must be absent: a central potential depends on $r$ but still conserves angular momentum.

16. Angular momentum in a central force

1. Write the Lagrangian of a particle in a central potential in polar coordinates.

   $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2) - U(r)$

   The potential depends only on distance.

2. Look for a missing coordinate.

   $\dfrac{\partial\mathcal{L}}{\partial\phi} = 0$

   Rotational symmetry.

3. Write its generalized momentum.

   $p_\phi = \dfrac{\partial\mathcal{L}}{\partial\dot{\phi}} = mr^2\dot{\phi}$

   The angular momentum about the origin.

4. Relate it to the area swept.

   $\dfrac{dA}{dt} = \tfrac{1}{2}r^2\dot{\phi} = \dfrac{p_\phi}{2m}$

   A thin triangle of base $r$ and height $r\,d\phi$.

5. State the consequence.

   $\dfrac{dA}{dt} = \text{constant}$

   Kepler's second law, for any central force.

17. The energy function is conserved

1. Differentiate $\mathcal{L}(q, \dot{q}, t)$ along a motion.

   $\dfrac{d\mathcal{L}}{dt} = \sum_i\left(\dfrac{\partial\mathcal{L}}{\partial q_i}\dot{q}_i + \dfrac{\partial\mathcal{L}}{\partial\dot{q}_i}\ddot{q}_i\right) + \dfrac{\partial\mathcal{L}}{\partial t}$

   The chain rule.

2. Replace the generalized force using Lagrange's equation.

   $\dfrac{\partial\mathcal{L}}{\partial q_i} = \dot{p}_i$

   Each coordinate's equation of motion.

3. Recognize a product rule.

   $\dot{p}_i\dot{q}_i + p_i\ddot{q}_i = \dfrac{d}{dt}(p_i\dot{q}_i)$

   With $p_i = \partial\mathcal{L}/\partial\dot{q}_i$.

4. Collect the terms.

   $\dfrac{d}{dt}\left(\sum_ip_i\dot{q}_i - \mathcal{L}\right) = -\dfrac{\partial\mathcal{L}}{\partial t}$

   Move $d\mathcal{L}/dt$ to the other side.

5. State the conservation law.

   $\dfrac{\partial\mathcal{L}}{\partial t} = 0 \Rightarrow \mathcal{H} = \text{constant}$

   Symmetry under shifts of time.

18. The bead on a spinning rod

1. Write the Lagrangian of a bead on a rod spinning at $\omega$.

   $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + \omega^2r^2)$

   Horizontal, so no potential.

2. Form the energy function.

   $\mathcal{H} = m\dot{r}^2 - \mathcal{L} = \tfrac{1}{2}m\dot{r}^2 - \tfrac{1}{2}m\omega^2r^2$

   $p_r = m\dot{r}$.

3. Compare it with the total energy.

   $T + U = \tfrac{1}{2}m\dot{r}^2 + \tfrac{1}{2}m\omega^2r^2 \ne \mathcal{H}$

   The constraint moves, so $T$ is not quadratic in $\dot{r}$ alone.

4. Use conservation from rest at $r_0$.

   $\tfrac{1}{2}m\dot{r}^2 - \tfrac{1}{2}m\omega^2r^2 = -\tfrac{1}{2}m\omega^2r_0^2$

   $\mathcal{H}$ has its starting value throughout.

5. Solve for the radial speed.

   $\dot{r} = \omega\sqrt{r^2 - r_0^2}$

   Outward, since the bead is flung out.

6. Integrate for the motion.

   $r = r_0\cosh\omega t$

   Separating and using $\int dr/\sqrt{r^2 - r_0^2} = \cosh^{-1}(r/r_0)$.

19. Your turn: a particle has $\mathcal{L} = \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - kz$. Which momenta are conserved?

1. List the coordinates in the Lagrangian.

   $\mathcal{L} \text{ contains } z \text{ but not } x \text{ or } y$

   Only the potential depends on position.

2. Write the momenta of the missing coordinates.

   $p_x = m\dot{x}, \quad p_y = m\dot{y}$

   $\partial\mathcal{L}/\partial\dot{x}$ and $\partial\mathcal{L}/\partial\dot{y}$.

3. Your turn: work this step out. Its working is at the end of the packet.

   State the conservation laws.

20. Guided practice

A particle moves in a plane with $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2) - 5r^2$. Which quantity is conserved because a coordinate is missing from $\mathcal{L}$?

• the angular momentum $mr^2\dot{\phi}$
• the radial momentum $m\dot{r}$
• the transverse momentum $mr\dot{\phi}$
• nothing, because the potential depends on $r$

21. Guided practice

Complete the worked solution: a $2$ kg ball flies with $\dot{x} = 5$ m/s and $\dot{y} = 2$ m/s at height $y = 2$ m, with $g = 10$ m/s². Find its conserved horizontal momentum in kg m/s, its kinetic energy in J, and its conserved energy function in J.

1. Multiply the mass by the horizontal velocity.

   $p_x = m\dot{x} =$ _p

   Conserved, because $\mathcal{L}$ does not contain $x$.

2. Take half the mass times the speed squared.

   $T = \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2) =$ _t

   The kinetic energy at this instant.

3. Add the gravitational potential energy.

   $\mathcal{H} = T + mgy =$ _h

   For fixed constraints and a velocity-independent potential, $\mathcal{H}$ is the total energy.

4. Check which quantities keep their values in flight.

   $\text{horizontal momentum and energy function fixed; kinetic energy not}$

   $x$ and $t$ are absent from $\mathcal{L}$, but $y$ is present.

22. Guided practice

Match each symmetry of the Lagrangian to the quantity it conserves.

23. Practice

A projectile of mass $8$ kg has $\mathcal{L} = \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mgy$ with $g = 10$ m/s². At one instant $\dot{x} = 1$ m/s, $\dot{y} = 5$ m/s and $y = 5$ m. Fill in $p_x$ in kg m/s, $p_y$ in kg m/s, and the energy function $\mathcal{H}$ in J.

24. Practice

A bead on a rod spinning at $\omega = 6$ rad/s has $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + \omega^2r^2)$. It starts at rest (along the rod) at $r = 2$ m. Use the conserved energy function to write $\dot{r}^2$, in m²/s², as a formula in $r$.

Answer:

25. Practice

A puck on a frictionless table circles a hole at $30$ cm with $\dot{\phi} = 1$ rad/s, held by a string through the hole. The string is pulled until the puck circles at $10$ cm. What is its new angular velocity, in rad/s?

Answer: rad/s after pulling

26. Somewhere new

Comet Tempel–Tuttle passes perihelion $0.976$ AU from the Sun at $41.62$ km/s, and its aphelion is $19.7$ AU out. How fast is it moving at aphelion, in km/s?

Answer: km/s at aphelion

27. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

28. Test question

A bead on a rod spinning at $\omega = 5$ rad/s has $\mathcal{L} = \tfrac{1}{2}m(\dot{r}^2 + \omega^2r^2)$. It starts at rest (along the rod) at $r = 2$ m. Use the conserved energy function to write $\dot{r}^2$, in m²/s², as a formula in $r$.

Answer:

29. What you can do now

You can connect symmetries to conservation laws. Explain to someone why a figure skater's energy rises when she pulls in her arms while her angular momentum does not.

Working for the steps left to you

19. Your turn: a particle has $\mathcal{L} = \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - kz$. Which momenta are conserved?, step 3