Download Symmetries and Observables-Classical and Relativistic Mechanics-Lecture Handout and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity! Sketch of Proof: We will define α but not show α([x, y]) = [α(x), α(y)] or linearity of α. Given x ∈ g we form exp(tx) ∈ G, which gives a flow on X : φ:R×X → X (t, x) 7→ A(exp(tv))x Why is this a flow? Check: 1. A(exp(0v))x = A(1)x = x; 2. A(exp(t+ s)v)x = A(exp(tv) exp(sv))x = A(exp(tv))A(exp(sv))x. Then to get α(v) ∈ Vect(X) we just differentiate this flow: α(v)(x) = d dt A(exp(tv))x |t=0 ∈ TxX, ∀x ∈ X Let’s look at an example: Example: SO(3) acts on R3 in an obvious way, so we get α: so(3)→ Vect(R3). For example, take ez = 0 −1 0 1 0 0 0 0 0 This is called the “generator of rotations around the z-axis” since exp(tez) = cos t − sin t 0 sin t cos t 0 0 0 1 which describes rotation around z axis, which as t varies gives a flow φ:R× R3 → R3 with φt x y z = cos t − sin t 0 sin t cos t 0 0 0 1 x y z Differentiating this we get a vector field: picture of flow around z-axis In equations this vector field is d dt φt x y z ∣∣∣∣∣∣ t=0 = −y x 0 Now, suppose X is a Poisson manifold and we have an action A:G×X → X . Then we get two Lie algebra homomorphisms: g α // γ ""E E E E E Vect(X) C∞(X) β 99ssssssssss 2 docsity.com We have Lie algebra homomorphisms α and β, where β(f) := vf := {f, ·} and we say the action A is Hamiltonian if we can find a Lie algebra homomorphism γ such that α = βγ. Such a γ gives an observable γ(v) ∈ C∞(X) for any infinitesimal symmetry v ∈ g, such that α(v) = β(γ(v)) i.e. d dt A(exp(tv))x|t=0 = {γ(v), ·} i.e. the observable γ(v) generates the flow (t, x) 7→ A(exp(tv))x. In this case we have a nice map from (infinitesimal) symmetries to observables! Example: G = SO(3) acts on R3, the configuration space of a particle in R3, and thus it acts on the phase space X = T ∗R3 ∼= R3 × R3 3 (q, p) In detail: we have A:G×X → X (g, q, p) 7→ (gq, gp) Is this action Hamiltonian? Yes. What is γ: so(3)→ C∞(X)? In 3-dimensions, we have so(3) ∼= R3 where the standard basis of R3 corresponds to ex = 0 0 0 0 0 −1 0 1 0 ey = 0 0 1 0 0 0 −1 0 0 ez = 0 −1 0 1 0 0 0 0 0 Check: [ex, ey] = ez and cyclic permutations - so [·, ·] in so(3) corresponds to × in R3. Identify so(3) with R3 using this isomorphism. Then γ(v) = v · J where J = q × p ∈ R3 is the angular momentum. Let’s check that this works: α = βγ 3 docsity.com
