Lagrangian mechanics primer.md

Lagrangian mechanics primer

Comments and corrections to J. M. F. Tsang ([email protected]).

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Lagrangian, momentum and energy

The Lagrangian $L$ is $$ L = \text{kinetic energy} - \text{potential energy} = T - V $$ For a particle in one dimension with position $x$, the Lagrangian is in general $L = L(x, \dot{x}, t)$.

Expressions for the kinetic energy in 3D coordinate systems:

Coordinates	KE
Cartesians	$\frac{1}{2} m \left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right)$
Cylindricals	$\frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 \right)$
Sphericals	$\frac{1}{2} m \left( \dot{r}^2 + r^2 (\dot{\theta}^2+ \dot{\phi}^2 \sin^2 \theta ) \right)$

The conjugate momentum of the coordinate $x$ is defined as $$ p_x = \frac{\partial L}{\partial \dot{x}} $$ Examples assuming the potential $V$ does not depend on velocities:

Coordinates	Coordinate	Conjugate momentum
Cartesians	$x$	$p_x = m\dot{x}$
Cylindricals	$r$	$p_r = m\dot{r}$
Cylindricals	$\theta$	$p_\theta = mr^2\dot{\theta}$
Note that angular momentum $p_\theta$ is not a constant multiple of $\dot\theta$.

In general if the coordinates are $(q_1, q_2, \dots)$ and the conjugate momenta are $(p_1, p_2, \dots)$ then the energy of the system is defined as $$ E = \sum_i q_i p_i - L $$ Usually, $T$ is a quadratic function of the velocities and $V$ does not depend on the velocities. In this case we recover $$ E = T + V $$

Principle of Least Action

For a system (e.g. particle moving in a potential) with the Lagrangian $L$, define the action $S$ as $$ S = \int_{t_1}^{t_2} L,\mathrm{d} t $$ for a given time interval $[t_1, t_2]$. The action has units $[S] = \mathrm{J}\mathrm{s} = \mathrm{kg}\mathrm{m}^2\mathrm{s}^{-1}$. It does not necessarily have a physical meaning (see here for a discussion of the metaphysics).

The principle of least action states that amongst all possible paths $(q_i(t))$, the actual path taken by the system is one that gives extremises $S$, giving $\delta S = 0$. Equivalently, $(q_i)$ must be a solution to the Euler-Lagrange equation $$ \frac{\mathrm{d}p_i}{\mathrm{d}t} = \frac{\partial L}{\partial q_i} $$ This is a generalised form of Newton's second law, in curvilinear coordinates. In Cartesian coordinates, $p_i$ are linear momenta and so the RHS has the interpretation of force. In cylindrical polars, $p_\theta$ is the angular momentum about the origin and the RHS is a torque.

First integrals and Noether's theorem

Noether's theorem states that

Any continuous symmetry is associated with a corresponding conserved quantity.

A symmetry is a transformation that leaves $L$ unchanged; a continuous symmetry is one that can be parameterised smoothly (contrast discrete symmetries such as reflections, which are either on or off). If a system has a continous symmetry then it is possible to change to a coordinate system where one of the coordinates parameterises that symmetry. Then $\partial L /\partial q = 0$ for that coordinate and so $p = \partial L / \partial\dot{q}$ is conserved.

Examples:

Coordinates	Symmetry	Conserved quantity
Cartesians	translations in $x$	linear momentum $p_x = m\dot{x}$
Cylindricals	rotations	angular momentum $p_\theta = r^2\dot\theta$
Cylindricals	translations in $z$	$p_z = m\dot{z}$
Sphericals	rotations in $\phi$	$p_\phi$ (exercise)
Any	time translations	energy $E = T + V$

For example, in cylindricals the following conditions are equivalent:

• The potential $V$ has no explicit $\theta$ dependence.
• The force $F = -\nabla V$ has no component in the $\theta$ direction.
• The force exerts no torque on the particle about the origin.
• The angular momentum is conserved.

Hamiltonian

The Hamiltonian $H = H(q, p, t)$ is the Legendre transform of the Lagrangian $L = L(q, \dot{q}, t)$, swapping out each of the velocities $q_i$ for the corresponding conjugate momenta $p_i$. So it is the same quantity as the energy. $$ H = \sum_i q_i p_i - L = E $$ However, the Hamiltonian is written in terms of momenta. When $p = m\dot{x}$ we have this form for the kinetic energy: $$ \frac{1}{2} m \dot{x}^2 = \frac{p^2}{2m} $$ In Cartesians the distinction may seem trivial since $p \propto \dot{x}$ by a constant factor $m$. The difference is more important in cylindricals, where $p_\theta = r^2\dot{\theta}$ depends on $r$ as well as $\dot{\theta}$. This is because the partial derivative $$ \frac{\partial H}{\partial r} $$ means taking a derivative keeping $p_\theta$ constant, not keeping $\dot{\theta}$ constant.

Hamilton's equations state that for each of the coordinates $q_i$ and its corresponding conjugate momentum $p_i$, $$ \frac{\mathrm{d}q_i}{\mathrm{d}t} = \frac{\partial H}{\partial p_i}, \qquad \frac{\mathrm{d}p_i}{\mathrm{d}t} = -\frac{\partial H}{\partial q_i}. $$ (*) The Hamiltonian itself is conserved if $\partial{H}/\partial{t} = 0$.