Download Poisson Manifolds Part 1-Classical and Relativistic Mechanics-Lecture Handout and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity!
1 Poisson Manifolds
We have described Poisson brackets for functions on R^2 n^ - the phase space for a system whose config- uration space is Rn. Now let’s generalize this to systems whose configuration space is any manifold, M. Here we will see that the phase space is the “cotangent bundle” T ∗M and this is a “Poisson manifold” - a manifold such that the commutative algebra of smooth real-valued functions on it, C∞(T ∗M ) is equipped with Poisson bracket {·, ·} making it into a Poisson algebra - the Poisson algebra of “observables” for our system.
Example: a particle on a sphere S^2. (picture of a sphere M = S^2 with point q ∈ M ) The position and momentum of this particle give a point in T ∗S^2 : q ∈ S^2 , p ∈ T (^) q∗ M , so (q, p) ∈ T ∗S^2. Recall that a manifold M is a topological space such that every point q ∈ M has a “neighborhood that looks like Rn.” In other words, there is an open set U ⊂ M with q ∈ U and a bijection
φ: U → Rn.
Indeed we have a collection of these (Ui, φi: Ui → Rn) and they are compatible: (picture of charts overlapping) that is, φj ◦ φ− i 1 is smooth (infinitely differentiable) where defined. A collection of this sort is an atlas, and the functions φi: Ui → Rn^ are called charts. We will usually use a maximal atlas, i.e. one containing all charts that are compatible with all charts in the atlas. So - a manifold is a topological space with a maximal atlas.
If the manifold M is the configuration space of some physical system, the a point q ∈ M de- scribes the position of the system and a “tangent vector” v ∈ Tq M describes its velocity, where Tq M is the tangent space of M at q: (picture of tangent space to S^2 at q) which can be defined in various ways:
- A tangent vector v at the point q is an equivalence class of (smooth) curves
γ: R → M
such that γ(0) = q, where γ 1 ∼ γ 2 if and only if for every smooth function f ∈ C∞(M ) (smooth real-valued functions on M ) we have d dt
f (γ 1 (t))|t=0 =
d dt
f (γ 2 (t))|t=0.
You can show the set of such equivalence classes is an n-dimensional vector space, the tangent space Tq M. Given v = [γ] ∈ Tq M , we can define the derivative vf ∈ R for any f ∈ C∞(M ) by vf =
d dt
f (γ(t))|t=
- A tangent vector v at q ∈ M is a derivation
v: C∞(M ) → R
i.e., a map that is:
docsity.com
- v(αf + βg) = αv(f ) + βv(g) - linearity
- v(f g) = v(f )g(q) + f (q)v(g) - product rule These clearly form a vector space, the tangent space Tq M.
The set of all position-velocity pairs is a manifold, the tangent bundle of M :
T M = {(q, v) : q ∈ M, v ∈ Tq M }
Naively, we might define momentum by p = mv, in which case it would be a tangent vector. It is better to think of it is a “cotangent vector”. Every vector space V has a dual V ∗:
V ∗^ = {l: V → R : l linear}
The cotangent space of M at q ∈ M is:
T (^) q∗ M = (Tq M )∗
and the cotangent bundle of M is:
T ∗M = {(q, p) : q ∈ M, p ∈ T (^) q∗ M }
So T ∗M will be the system’s “phase space” - space of position-momentum pairs.
What good are cotangent vectors, though? The “gradient” or “differential” of a function f ∈ C∞(M ) at q ∈ M is a cotangent vector, (df )q ∈ T (^) q∗ M : (df )q (v) = v(f ), v ∈ Tq M
(or in low-brow notation: (∇f )(q) · v = vf ). The potential energy for our system (e.g. a particle on M ) is some fucntion of its position: V ∈ C∞(M ). We have seen already that “∇V = −F ” - but this really means
(dV )q = −F (q)
where F (q), the force at q ∈ M , is a cotangent vector: F (q) ∈ T (^) q∗ M. A tangent vector looks like a little arrow: (picture of a tangent vector) A cotangent vector looks like a “stack of hyperplanes” - its level surfaces: (picture of level surfaces) Together they give a number l(v) ∈ R: (picture of tangent vector crossing level curves)
In physics, the velocity v ∈ TpM is a tangent vector and the force F ∈ T (^) q∗ M is a cotangent vector, so F (v) ∈ R. We have seen this before in the formula: ∫ (^) t 2
t 1
F · q˙(t)dt = work
Now we would say, given a particle’s path q: R → M , that work from t 1 to t 2 is equal to ∫ (^) t 2
t 1
F (q(t))( ˙q(t))dt
Since force is a cotangent vector, so is momentum, since:
dp dt
= F
So: a position-momentum pair (q, p) is really a point in T ∗M :
q ∈ M, p ∈ T (^) q∗ M.
docsity.com
