Generalized momentum

Also known as conjugate momentum.

Definition and motivation

Given a Lagrangian classical system $(Q, L)$ with coordinates $q_{i}$ , the generalized momentum or conjugate momentum is defined as

p_{i} := \frac{\partial L}{\partial {\dot{q}}_{i}}

Motivation: If we compare the Euler-Lagrange equations

\frac{\partial L}{\partial q} - \frac{d}{d t} \frac{\partial L}{\partial \dot{q}} = 0

with Newton equation

\frac{d}{d t} p = F

we get the feeling that

on the one hand, $\frac{\partial L}{\partial q} = \frac{\partial}{\partial q} (T - V) = - \frac{\partial V}{\partial q} = F$ .
on the other, $\frac{\partial L}{\partial \dot{q}}$ must play the role of $p$ .

Momentum as a covector

Informal approach

It turns out that generalized momentum is a covector, a 1-form. Roughly speaking, it can be understood in the following way. The Lagrangian $L$ is a modification of the kinetic energy $T$ , which is a "kind of" squared length of the velocity ${\dot{q}}_{i}$ . In a vector space with an inner product $g (-, -)$ , a length is computed in the following way:

we take the vector $v$
we obtain the dual vector $g (v, -)$ .
we compute $S q u a r e d L e n g t h = g (v, v)$
Therefore, loosely speaking

\frac{\partial S q u a r e d L e n g t h}{\partial v} = \frac{\partial g (v, v)}{\partial v} = 2 g (v, -)

which is a covector.

Coordinate transformation

To understand why $p_{i}$ is a covector, we need to examine how it transforms under a change of coordinates. Suppose we have two sets of coordinates on $Q$ : $q^{i}$ and $θ^{a}$ . The transformation between these coordinates is given by a diffeomorphism $φ : Q \to Q$ , such that $θ^{a} = φ^{a} (q^{i})$ .
The velocities transform as:

{\dot{θ}}^{a} = \frac{\partial θ^{a}}{\partial q^{i}} {\dot{q}}^{i} .

This is the transformation rule for the components of a vector in the tangent space $T Q$ . Now, consider the generalized momentum in the new coordinates:

p_{a} = \frac{\partial L}{\partial {\dot{θ}}^{a}} .

Using the chain rule, we can express this in terms of the original coordinates:

p_{a} = \frac{\partial L}{\partial {\dot{q}}^{i}} \frac{\partial {\dot{q}}^{i}}{\partial {\dot{θ}}^{a}} = p_{i} \frac{\partial q^{i}}{\partial θ^{a}} .

Here, we used the fact that $\frac{\partial {\dot{q}}^{i}}{\partial {\dot{θ}}^{a}} = \frac{\partial q^{i}}{\partial θ^{a}}$ because the transformation is linear in the velocities.
This is precisely the transformation rule for the components of a covector (a 1-form) on $Q$ . In contrast, vectors transform with the inverse Jacobian:

v^{a} = \frac{\partial θ^{a}}{\partial q^{i}} v^{i} .

Thus, the generalized momentum $p_{i}$ transforms as a covector, not as a vector.

Lagrangian as a Metric

In many physical systems, the Lagrangian includes a kinetic energy term that is a quadratic form in the velocities:

L = \frac{1}{2} g_{i j} (q) {\dot{q}}^{i} {\dot{q}}^{j} - U (q),

where $g_{i j} (q)$ is a symmetric, positive-definite matrix (the metric on $Q$ ). We can think of $L$ as a generalized metric. The generalized momentum is then:

p_{i} = \frac{\partial L}{\partial {\dot{q}}^{i}} = g_{i j} (q) {\dot{q}}^{j} .

The generalized momentum $p_{i}$ is the image of the velocity ${\dot{q}}^{i}$ under something similar to the musical isomorphism.

Why Generalized Momentum?

Let's take a moment to explore an intuitive approach to Hamiltonian mechanics. Traditionally, position, $q$ , and velocity, $q_{t}$ , have been considered the fundamental quantities that define the state of a system. If the system's dynamics are known, then from any given $q$ and $\dot{q}$ , we can determine all future states $(q, q_{t})$ .
On the other hand, we have the potential energy $U$ , which typically depends only on $q$ , and the kinetic energy $K$ , which usually depends only on $q_{t}$ . The total energy, given by $H = K + U$ , has been observed time and again as a conserved quantity, making it natural to assign it special importance.

Wouldn't it be elegant if $U$ and $K$ exhibited a symmetric relationship with respect to $q$ and $q_{t}$ ? The one summarizes all the dependence on $q$ , and the other summarizes all dependence on $q_{t}$ .
Differentiating $U$ (or equivalently $H$ ) with respect to $q$ gives us $- F$ , but this can be interpreted more fundamentally as the rate of change of momentum, $p = m q_{t}$ , that is,

\frac{\partial U}{\partial q} = \frac{\partial H}{\partial q} = - \dot{p} .

On the other hand, differentiating $K$ with respect to $q_{t}$ yields

\frac{\partial K}{\partial q_{t}} = \frac{\partial H}{\partial q_{t}} = m q_{t},

which represents the rate of change of $q$ up to a factor of $m$ .
If mass were constant, we could write $\dot{p} = m \dot{q_{t}}$ , leading to the symmetric relationships:

\frac{\partial H}{\partial q_{t}} = m \dot{q},

\frac{\partial H}{\partial q} = - m \dot{q_{t}} .

This is indeed a beautiful way of viewing mechanics. However, mass is not always constant, and, more importantly, when using generalized coordinates, the kinetic energy expression often contains additional terms, disrupting this symmetry.

Hamilton’s key insight (I guess) was to resolve this issue by taking $p$ (instead of $q_{t}$ ) as a fundamental quantity in describing the system’s state. This leads to the expression for kinetic energy:

K = \frac{1}{2} m q_{t}^{2} = \frac{p^{2}}{2 m},

and thus,

\frac{\partial H}{\partial p} = \frac{p}{m} = \dot{q} .

Combining this with our earlier result,

\frac{\partial H}{\partial q} = - \dot{p},

we arrive at a wonderfully symmetric formulation of mechanics: Hamiltonian equations.