Well. I’m finally getting around to talking about the Hamiltonian formalism of classical mechanics. Of cover if you’ve been keeping up with the good populate over there undergo discussed recently (with some of the Lagrangian formalism thrown in) but I conclude 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 measure will still be modeled by a 1-dimensional Euclidean space but we ordain allow space to be an arbitrary manifold
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 change state together) but an added bonus of having a metric is it induces a bundle isomorphism between the tangent bundle and cotangent bundle of
denominate the lay of physical paths a particle can take. We will anticipate that a path in
called the symplectic gradient. An obvious challenge that can be asked at this inform is if every infinitesimal symmetry is the symplectic gradient of some function in this manner. The short answer is no in command. Remember that
is claim. Also we see that not every answer corresponds to a different symplectic gradient we can dress the function by a constant value on each connected component of
is connected. In classical mechanics we have a distinguished 1-parameter group of measure translations. We assume that
the corresponding infinitesimal generator is an infinitesimal symmetry and that
is exact. (Back in the Newtonian picture this would correspond to our assumption that
was time-independent.) We now undergo a choice of corresponding observables (up to constant). We pick one call it
- a Hamiltonian system is just a symplectic copy with a distinguished observable.
We define an infinitesimal symmetry of a Hamiltonian system to be an infinitesimal symmetry
is an observable corresponding to an infinitesimal symmetry of the Hamiltonian system then
We label such an observable a conserved rush. Why do we call it conserved? come up bequeath how the Poisson bracket was defined:
and so energy is conserved! (Big broach this follows straight from the definition. We have yet to show that this
corresponds to what we usually evaluate of as energy.) In general for any observable
That’s plenty of this abstraction; let’s take a end and see that in the familiar inspect of a particle moving through Euclidean space.
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 understand the
as coordinates of momentum. But why should we consider these to be momentum coordinates. Remember that we are on a Riemannian copy. An element of the tangent pack is interpreted as a velocity vector
Therefore momentum is a tangent vector but since we have a canonical way of identifying the tangent and cotangent bundles we can (and often like to) evaluate of momentum as a covector.
In our case the metric (and so the identification of bundles) is trivial thus the second equation becomes
Plugging this formula into the first equation we get Newton’s Second Law:
So we’ve recovered Newtonian mechanics from the Hamiltonian formalism and the Hamiltonian did take the expected create 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 bring about 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.
that is in a free particle system momentum is conserved. (Actually the conserved charges we get are
but if one is conserved then the other is. Also say that the like the Hamiltonian the momentum charges are only determined up to a constant. Fixing a constant is akin to fixing a close in of reference to make your measurements.)
The free particle system also has rotational symmetry so we would expect the vector field
to be an infinitesimal symmetry of our Hamiltonian system. However this is not a symmetry: it is easy to analyse that
is not closed. This is because when we rotate our arrange system we also have to turn the corresponding coordinates in the cotangent space. Thus we should define:
It is easy to analyse that this is in fact an infinitesimal symmetry and the corresponding rush is
These are precisely the components of the angular momentum tensor. Thus we undergo derived the law of conservation of angular momentum. It could be noted that in fact we did not be a constant potential to find this conservation law. It is enough to undergo a potential that depends only on
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.
Instead of using the canonical inner product on R^n you should really think of crowd 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 crowd (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.
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 crowd (inertia) is also nice in rigid be 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 lay particularly alter.
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 compose for this?
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We’re a assort of graduate students in mathematics at several universities just starting out. We’ve all open some interesting things floating out there that no one seems to experience about or just that we’d desire a good displace to post a rant about. So we’ve decided to start this blog.
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Related article:
http://rigtriv.wordpress.com/2007/10/24/hamiltonian-mechanics/
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