Momentum, symmetry, and conservation

1. What a conserved quantity tells us

An equation of motion predicts acceleration from the present state. A conserved quantity supplies a different kind of information: a combination of positions and velocities keeps the same value along the entire motion. It can reduce the order of a calculation and reveal a symmetry that was hidden in the original force components.

In Cartesian coordinates we already recognize linear momentum and angular momentum. Generalized coordinates place them in one scheme. The momentum belonging to a coordinate is determined by how the Lagrangian depends on that coordinate's velocity.

2. Generalized momentum and cyclic coordinates

Definition. The momentum conjugate to $q_i$ is

Its dimensions are energy times time divided by the dimensions of $q_i$. For a Cartesian distance, this is linear momentum. For an angle, it is angular momentum. For more general coordinates, $p_i$ need not equal a particle mass times $\dot q_i$.

A coordinate is cyclic, or ignorable, when $L$ has no explicit dependence on that coordinate: $\partial L/\partial q_i=0$. Its velocity may still occur. Lagrange's equation reads

Therefore, for a cyclic coordinate with $Q_i=0$,

This is the complete proof. Its assumptions are indispensable: cyclicity of $L$ alone does not imply conservation in a driven or damped system. A cyclic coordinate is not necessarily constant either. A free particle's position changes even while its momentum stays fixed.

3. Translation and rotation

Suppose translating an entire isolated system by the same small distance $\epsilon$ along $x$ leaves its Lagrangian unchanged. Then $\delta x_a=\epsilon$, $\delta\dot x_a=0$. To first order,

For unforced Euler–Lagrange motion, $\partial L/\partial x_a=d(\partial L/\partial\dot x_a)/dt$. Thus

For the ordinary Cartesian kinetic energy and a velocity-independent potential, $p_{ax}=m_a\dot x_a$, so the conserved quantity is total linear momentum in the $x$ direction. Translation symmetry in all three directions gives conservation of the full total momentum vector. Individual particle momenta may still change through internal forces.

For planar rotations about a fixed origin, a small rotation $\epsilon$ gives $\delta x_a=-y_a\epsilon$, $\delta y_a=x_a\epsilon$, and the same relations for velocities. Rotational invariance implies

Replace the coordinate derivatives by $\dot p_{ax},\dot p_{ay}$:

The bracket is the $z$ component of total angular momentum. In a central potential it is usually shorter to choose the polar angle itself as a coordinate: its absence from $L$ makes the same conservation law immediate. The Cartesian proof shows why that conserved conjugate momentum is angular momentum.

These are specific symmetry proofs. The general Noether theorem extends the idea to much broader transformations; its full machinery is not needed here.

4. Time independence and the energy function

Time is the evolution parameter, not an extra generalized coordinate in these equations. To obtain the relevant conserved quantity, define the Lagrangian energy function

Differentiate the product and expand the total derivative of $L$:

Since $p_i=\partial L/\partial\dot q_i$, the acceleration terms cancel in pairs. The remaining terms are

Use Lagrange's equation to obtain the energy balance:

If $Q_i=0$ and $\partial L/\partial t=0$, then $\mathcal E$ is constant. This proof does not assume that $\mathcal E=T+V$. That further identification requires a velocity-independent potential and kinetic energy homogeneous of degree two in generalized velocities, as happens for time-independent coordinate transformations. The next chapter proves that condition carefully.

The partial derivative $\partial L/\partial t$ means time dependence with $q,\dot q$ held fixed. An autonomous pendulum has $\partial L/\partial t=0$ even though $L(\theta(t),\dot\theta(t))$ usually changes during its motion.

5. Applications

Example 1 · Horizontal momentum during a fall

For $L=m(\dot x^2+\dot y^2)/2-mgy$, $x$ is cyclic and $Q_x=0$. Therefore $p_x=m\dot x$ is conserved. The vertical coordinate is not cyclic: $\dot p_y=\partial L/\partial y=-mg$.

The autonomous energy function is $\mathcal E=m(\dot x^2+\dot y^2)/2+mgy$. Its constancy allows loss of height to become kinetic energy. Horizontal translation is a symmetry; vertical translation changes this chosen gravitational potential by a constant, a case where a broader boundary-term symmetry argument is possible. The direct force equation nevertheless shows that vertical momentum is not conserved.

Example 2 · A particle in a radial potential

Choose plane polar coordinates. The energies are $T=m(\dot r^2+r^2\dot\theta^2)/2$, $V=V(r)$, and $L=T-V$. Since $\theta$ is absent,

The constant $\ell$ is the perpendicular angular momentum. At $r=2\,\mathrm m$, a particle of mass $0.5\,\mathrm{kg}$ with $\dot\theta=3\,\mathrm{s^{-1}}$ has $\ell=6\,\mathrm{kg\,m^2\,s^{-1}}$. At $r=3\,\mathrm m$, the same orbit has $\dot\theta=\ell/(mr^2)=4/3\,\mathrm{s^{-1}}$.

Angular speed decreases as $r^{-2}$, whereas tangential speed $r\dot\theta$ decreases as $r^{-1}$. Confusing the two produces an incorrect area law.

Example 3 · Internal forces and total momentum

Two particles on a line interact through $V=k(x_2-x_1-a)^2/2$. Choose center position $X=(m_1x_1+m_2x_2)/(m_1+m_2)$ and separation $s=x_2-x_1$. Put $M=m_1+m_2$ and $\mu=m_1m_2/M$. Then

Expanding the two squared velocities, the cross coefficient is $-m_1m_2/M+m_2m_1/M=0$. The kinetic energy becomes $T=M\dot X^2/2+\mu\dot s^2/2$, while $V=k(s-a)^2/2$. Thus $L$ is independent of $X$, and $P_X=M\dot X=m_1\dot x_1+m_2\dot x_2$ is constant.

The separation equation is $\mu\ddot s+k(s-a)=0$. The particles exchange momentum through the spring while their center of mass moves uniformly. This same center/relative decomposition will return in the two-body central-force problem.

Example 4 · A cyclic coordinate with drag

A free particle on a horizontal line has $L=m\dot x^2/2$. Although $x$ is cyclic, a drag force $Q_x=-c\dot x$ gives $\dot p=-cp/m$. Separate variables: $dp/p=-(c/m)dt$, so

The energy function is $p^2/(2m)$ and decreases at rate $-c\dot x^2$. Neither cyclicity nor absence of explicit time in $L$ overrides the right-hand-side force. As $c\to0$, expand $1-e^{-ct/m}\simeq ct/m$ to recover $x=x_0+p_0t/m$.

6. Conservation arguments in written solutions

A complete cyclic-coordinate argument identifies the missing coordinate, calculates its conjugate momentum, states that the corresponding remaining force vanishes, and then sets its time derivative to zero. For energy, write the energy function before identifying it with mechanical energy. For translation or rotation, identify which transformation leaves the physical system unchanged; a fixed external support or field can remove a symmetry even when the particles' mutual interaction has it.

The useful conclusion is structural: conservation laws follow from the dependence of the equations on position, orientation, and time. They are not separate assumptions to add after solving a problem.

Exercises

Hints and solutions are collected separately.

1. 9.1 Can a cyclic coordinate change with time?

2. 9.2 Prove the cyclic-momentum result including its force qualification.

3. 9.3 Derive the energy balance for $\mathcal E=\sum_i\dot q_ip_i-L$.

4. 9.4 For $L=m(\dot x^2+\dot y^2)/2-V(y)$, identify a conserved momentum.

5. 9.5 A central orbit changes radius from $r$ to $2r$. By what factors do $\dot\theta$ and tangential speed change?

6. 9.6 For $L=\tfrac12m\dot x^2-\tfrac12k(t)x^2$, find $\dot{\mathcal E}$.

7. 9.7 Prove conservation of total $x$ momentum from simultaneous translation invariance.

8. 9.8 What angular momentum is conserved for the spherical pendulum?

9. 9.9 Two particles interacting only through $V(x_2-x_1)$ have masses $m,2m$ and velocities $4v,-v$. Find their center velocity.

10. 9.10 Explain why $dL/dt=0$ and $\partial L/\partial t=0$ are different statements.

11. 9.11 If a potential is unchanged by rotation about $z$, derive the conserved Cartesian quantity.

12. 9.12 Does conservation of energy imply conservation of every momentum?