## Hamiltonian Mechanics

Hi everybody,

Well, I’m finally getting around to talking about the Hamiltonian formalism of classical mechanics. Of course, if you’ve been keeping up with The Everything Seminar, the good people over there have discussed this very topic recently (with some of the Lagrangian formalism thrown in), but I feel it never hurts to have someone go over it again. Besides, I promised in an earlier post that I would talk about this, so here I go:

In this discussion, time will still be modeled by a 1-dimensional Euclidean space, but we will allow space to be an arbitrary manifold $X$ with a Riemannian metric. The Riemannian metric is an obvious requirement since we want to be able to measure distances between points (at least points that are close together), but an added bonus of having a metric is it induces a bundle isomorphism between the tangent bundle and cotangent bundle of $X$. As in this post, we will let $M$ denote the space of physical paths a particle can take. We will assume that a path in $M$ is determined by its initial position and velocity so that $M \cong TX \cong T^*X$. Thus, we can pull the canonical symplectic structure of $T^*X$ back onto $M$. The goal of Hamiltonian mechanics is to encode the symmetries of $M$ in the Lie algebra $(C^\infty(M), \{\cdot, \cdot\}_\omega)$.

By a symmetry of $M$, I mean a symplectomorphism, but we will only be concerned with infinitesimal symmetries (symplectomorphisms): $\xi\in \Gamma(TM)$ such that $Lie_\xi(\omega)=0$. Let $\Gamma_\omega(TM)$ be the set of all infinitesimal symplectomorphisms. We call elements of $C^\infty(M)$ observables. We saw last time that we can associate to every $f\in C^\infty(M)$ a vector field $\xi_f\in \Gamma_\omega(TM)$ called the symplectic gradient. An obvious question that can be asked at this point is if every infinitesimal symmetry is the symplectic gradient of some function in this manner. The short answer is no in general. Remember that $-df = \omega(\xi_f, \cdot_)$, so $\xi\in \Gamma_\omega(TM)$ to corresponds to a function if and only if $\iota_{\xi_f}\omega$ is exact. Also, we see that not every function corresponds to a different symplectic gradient, we can change the function by a constant value on each connected component of $M$. To summarize, we have the following exact sequence:

$0 \to H^0_{dR}(M) \to C^\infty(M) \to \Gamma_\omega(TM) \to H^1_{dR}(M) \to 0$

The first map is injection of the constant functions into $C^\infty(M)$; the second assigns to a function its symplectic gradient; the last is given by $\xi\mapsto [\iota_\xi\omega]$. To show that the last map is well defined, we need to show that indeed $\iota_\xi\omega$ is closed. This follows immediately from the fact that $\omega$ is closed and Cartan’s formula: $d(\iota_\xi\omega) = Lie_\xi\omega = 0$.

From this point on, we will assume $M$ is connected. In classical mechanics, we have a distinguished 1-parameter group of time translations. We assume that $\xi_t$, the corresponding infinitesimal generator, is an infinitesimal symmetry, and that $\iota_\xi\omega$ is exact. (Back in the Newtonian picture, this would correspond to our assumption that $V(x)$ was time-independent.) We now have a choice of corresponding observables (up to constant). We pick one, call it $-H$, where $H$, the Hamiltonian (or energy). Thus, $\iota_{\xi_t}\omega = dH$. We now have what is known as a Hamiltonian system $(M, H)$ – a Hamiltonian system is just a symplectic manifold with a distinguished observable.

We define an infinitesimal symmetry of a Hamiltonian system to be an infinitesimal symmetry $\xi$ of our symplectic manifold with the added condition that $\xi(H) = 0$. Note that if $Q$ is an observable corresponding to an infinitesimal symmetry of the Hamiltonian system, then $\{Q,H\}_\omega = 0$. We call such an observable a conserved charge. Why do we call it conserved? Well, remember how the Poisson bracket was defined: $0 = \{Q,H\} = \xi_t Q$, and so $Q$ is independent of time. A trivial example of this is $\xi_t H = \{H,H\}_\omega = 0$, and so energy is conserved! (Big deal, this follows straight from the definition. We have yet to show that this $H$ corresponds to what we usually think of as energy.) In general, for any observable $Q \in C^\infty(M)$, we can define its time derivative by $\dot{Q} = \{Q, H\}$.

That’s plenty of this abstraction; let’s take a break and see that in the familiar case of a particle moving through Euclidean space, $H$ takes the form we expect, namely that of energy. Here, our space is $\mathbb{R}^n$ with the usual Euclidean metric, and say our particle has mass $m$. If $(x_1, \ldots, x_n)$ are the usual global coordinate functions on $\mathbb{R}^n$, then we can write my canonical symplectic form as $\omega = \sum_i dx^i\wedge dp_i$, as we did at the end of this post. Now we can compute $dH = \iota_{\xi_t}\omega$ in these coordinates. $dH = \sum_i dx^i(\xi_t)dp_i - dp_i(\xi_t)dx^i = \xi_t(x^i)dp_i - \xi_t(p_i)dx^i$ $= \dot{x}^idp^i - \dot{p}_idx^i$

Thus, we get two sets of equations:

$\frac{\partial H}{\partial x}^i = -\dot{p}_i \quad \frac{\partial H}{\partial p_i} = \dot{x}^i$

These are called Hamilton’s equations. You probably recognize them as the equations of motion from when you originally learned Hamiltonian mechanics, provided we the interpret the $p_i$ as coordinates of momentum. But why should we consider these to be momentum coordinates. Remember that we are on a Riemannian manifold. An element of the tangent bundle is interpreted as a velocity vector $\dot{x}(t)$. The classical mechanics definition of momentum of a particle at time $t$ is simply $p(t) = m\dot{x}(t)$. Therefore, momentum is a tangent vector, but since we have a canonical way of identifying the tangent and cotangent bundles, we can (and often prefer to) think of momentum as a covector.

In our case, the metric (and so the identification of bundles) is trivial, thus, the second equation becomes $\frac{\partial H}{\partial p_i} = \frac{p_i}{m}$. Therefore, $H = \frac{p^2}{2m} + V(x)$ for some function $V$. Plugging this formula into the first equation, we get Newton’s Second Law: $V'(x) = -\dot{p}_i = -m\ddot{x}^i$. So we’ve recovered Newtonian mechanics from the Hamiltonian formalism, and the Hamiltonian did take the expected form of total energy (kinetic + potential).

As we stated before, the main lesson to take away from Hamiltonian mechanics is that symmetries of the Hamiltonian system lead to conserved charges. Before saying good-bye, let’s get some practice computing some of these charges in a situation with a lot of symmetry: a free particle moving in Euclidean space. In this case, $H = \frac{p^2}{2m}+C$ for some constant $C$. It is easy to check that the vector fields on $T^*\mathbb{R}^n$ given by $\frac{\partial}{\partial x^i}$ are infinitesimal symmetries of the Hamiltonian system. Indeed, $\frac{\partial H}{\partial x^i} = 0$ and $\iota_{\frac{\partial}{\partial x^i}}\omega = dp_i$. Hence, we get conserved charges $p_i$, that is, in a free particle system momentum is conserved. (Actually, the conserved charges we get are $-p_i$, but if one is conserved, then the other is. Also, note that the like the Hamiltonian, the momentum charges are only determined up to a constant. Fixing a constant is akin to fixing a frame of reference to make your measurements.)

The free particle system also has rotational symmetry, so we would expect the vector field $\xi_{ij} = x_i\frac{\partial}{\partial x^j} - x^j\frac{\partial}{\partial x_i}$ for $i\neq j$ to be an infinitesimal symmetry of our Hamiltonian system. However, this is not a symmetry: it is easy to check that $\iota_{\xi_{ij}}\omega$ is not closed. This is because when we rotate our coordinate system, we also have to rotate the corresponding coordinates in the cotangent space. Thus, we should define:

$\xi_{ij} := x^i\frac{\partial}{\partial x^j} - x^j\frac{\partial}{\partial x^i} + p_i\frac{\partial}{\partial p_j} - p_j\frac{\partial}{\partial p_i}$

It is easy to check that this is in fact an infinitesimal symmetry, and the corresponding charge is $L_{ij} = x^ip_j - x^jp_i$. These are precisely the components of the angular momentum tensor. Thus, we have derived the law of conservation of angular momentum. It could be noted that in fact we did not need a constant potential to find this conservation law. It is enough to have a potential that depends only on $|x|$. One final thought: in the Hamiltonian formalism, there is an ambiguity in finding the conserved charges. Each charge is only defined up to a constant. In the Lagrangian formalism, we will see a method of computing these charges in which there is no ambiguity.

That’s all for now. ‘Till next time.

-Joe

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### 6 Responses to Hamiltonian Mechanics

1. Eitan says:

Instead of using the canonical inner product on R^n, you should really think of mass as your inner product. <>=m where is the canonical inner product. Then kinetic energy is proportional to the squared norm of a velocity K=(1/2)<>. And this gives a natural reason to think of momentum as a covector, if the particle has velocity v the momentum is given by p(w)=<> (so K=(1/2)p(v)). Thinking of the inner product as mass (inertia) is also nice in rigid body mechanics where now the inner product should be a Riemannian metric on SO(3), given by the inertia tensor making the definition of angular momentums as the cotangent space particularly cool.

2. Eitan says:

oops…stuff between angle brackets went away.

Instead of using the canonical inner product on R^n, you should really think of mass as your inner product. ((v,w))=m(v,w) where (v,w) is the canonical inner product. Then kinetic energy is proportional to the squared norm of a velocity K=(1/2)((v,v)). And this gives a natural reason to think of momentum as a covector, if the particle has velocity v the momentum is given by p(w)=((v,w)) (so K=(1/2)p(v)). Thinking of the inner product as mass (inertia) is also nice in rigid body mechanics where now the inner product should be a Riemannian metric on SO(3), given by the inertia tensor making the definition of angular momentums as the cotangent space particularly cool.

3. jwwalsh says:

That is indeed a very useful interpretation of mass that I haven’t thought of before. And it sounds like it would generalize very naturally to the configuration space of a many-particle system. Do you have a reference for this?

• mlbaker says:

Folland’s book “Quantum Field Theory: A Tourist Guide for Mathematicians”, page 16.

4. Eitan says:

nope.