CYCLIC COORDINATES AND POISSON BRACKETS

Link to: physicspages home page. To leave a comment or report an error, please use the auxiliary blog and include the title or URL of this post in your comment. Post date: 18 Jan 2021. Hamilton’s canonical equations are:

∂H = q˙i (1) ∂pi ∂H − = p˙i (2) ∂qi

1. CYCLIC COORDINATES

If a coordinate qi is missing in the Hamiltonian (that is, H is independent of qi), then ∂H p˙i = − = 0 (3) ∂qi Thus the conjugate momentum pi is conserved. Such a missing coordinate qi is known as a cyclic coordinate. [I’m not sure of the origin of this term. Again Google doesn’t provide a definitive answer.]

2. POISSONBRACKETS There is a general method for calculating the rate of change of some func- tion ω (p,q) that depends on the momenta and coordinates, but not explicitly on the time (ω is allowed to depend implicitly on time since p and/or q can depend on time). The time derivative can then be written using the chain rule:

dω  ∂ω ∂ω  = ∑ q˙i + p˙i dt i ∂qi ∂pi  ∂ω ∂H ∂ω ∂H  = ∑ − i ∂qi ∂pi ∂pi ∂qi ≡ {ω,H} where in the second line we used Hamilton’s equations 1 and 2. The last line defines the Poisson bracket of the function ω with the Hamiltonian H. Poisson bracket 1 CYCLIC COORDINATES AND POISSON BRACKETS 2

 ∂ω ∂H ∂ω ∂H  {ω,H} ≡ ∑ − (4) i ∂qi ∂pi ∂pi ∂qi The rate of change of a quantity ω with respect to time is given by its Poisson bracket with the Hamiltonian, provided that H doesn’t depend ex- plicitly on time:

dω = ω˙ = {ω,H} (5) dt We can see that if {ω,H} = 0, the function ω is conserved. Since {H,H} = 0 automatically, the total energy (represented by the Hamiltonian) is conserved, provided there is no explicit time dependence. Such a time dependence can arise if the system is subject to some external force, for example. From the definition 4 we can derive a few fundamental properties of Pois- son brackets. We’ll consider a general Poisson bracket between two arbi- trary functions ω (p,q) and λ(p,q). Then

 ∂ω ∂λ ∂ω ∂λ  {ω,λ} = ∑ − (6) i ∂qi ∂pi ∂pi ∂qi  ∂ω ∂λ ∂ω ∂λ  = −∑ − (7) i ∂pi ∂qi ∂qi ∂pi  ∂λ ∂ω ∂λ ∂ω  = −∑ − (8) i ∂qi ∂pi ∂pi ∂qi = −{λ,ω} (9) A Poisson bracket is distributive, in the sense that

 ∂ω ∂ (λ + σ) ∂ω ∂ (λ + σ) {ω,λ + σ} = ∑ − (10) i ∂qi ∂pi ∂pi ∂qi  ∂ω  ∂λ ∂σ  ∂ω  ∂λ ∂σ  = ∑ + − + (11) i ∂qi ∂pi ∂pi ∂pi ∂qi ∂qi  ∂ω ∂λ ∂ω ∂λ   ∂ω ∂σ ∂ω ∂σ  = ∑ − + ∑ − (12) i ∂qi ∂pi ∂pi ∂qi i ∂qi ∂pi ∂pi ∂qi = {ω,λ} + {ω,σ} (13) One more identity is useful, which we can derive using the product rule: CYCLIC COORDINATES AND POISSON BRACKETS 3

 ∂ω ∂ (λσ) ∂ω ∂ (λσ) {ω,λσ} = ∑ − (14) i ∂qi ∂pi ∂pi ∂qi  ∂ω ∂λ ∂ω ∂λ   ∂ω ∂σ ∂ω ∂σ  = ∑σ − + ∑λ − (15) i ∂qi ∂pi ∂pi ∂qi i ∂qi ∂pi ∂pi ∂qi = {ω,λ}σ + {ω,σ}λ (16)

The Poisson brackets involving the coordinates qi and momenta pi turn up frequently, so it’s worth deriving them in detail. We have   ∂qi ∂qj ∂qi ∂qj {qi,qj} = ∑ − = 0 (17) k ∂qk ∂pk ∂pk ∂qk This follows because, in the Hamiltonian formalism, the qis and pis are independent variables, so ∂qj = ∂pj = 0 for all j and k. For the same reason, ∂pk ∂qk we have   ∂pi ∂pj ∂pi ∂pj {pi,pj} = ∑ − = 0 (18) k ∂qk ∂pk ∂pk ∂qk The mixed Poisson bracket is a different story, however:

  ∂qi ∂pj ∂qi ∂pj {qi,pj} = ∑ − (19) k ∂qk ∂pk ∂pk ∂qk = ∑δikδjk − 0 (20) k = δij (21) Hamilton’s equations 1 and 2 can be written using Poisson brackets by setting ω equal to qi and pi respectively in 5:

q˙i = {qi,H} (22) p˙i = {pi,H} (23)

3. ANGULAR MOMENTUM CONSERVATION In two dimensions, suppose we have a Hamiltonian:

2 2 2 2 H = px + py + ax + by (24) If a = b, then in polar coordinates, the only coordinate appearing in H is p the radial distance from the origin r = x2 + y2, which means that the polar CYCLIC COORDINATES AND POISSON BRACKETS 4 angle θ is a cyclic coordinate. This means that the conjugate momentum pθ must be conserved. That is,

p˙θ = {pθ,H} = 0 (25)

However, pθ is the angular momentum `z, so this just says that angular momentum is conserved. To see this explicitly, it’s easier to convert to polar coordinates. From Hamilton’s equations

∂H x˙ = = 2px (26) ∂px y˙ = 2py (27) 1 p2 + p2 = x˙2 + y˙2
(28) x y 4 v2 = (29) 4 1 = r˙2 + r2θ˙2
(30) 4 where in the fourth line, v is the linear velocity and in the fifth line we converted this to polar coordinates. Thus the Hamiltonian becomes, in the case where a = b:

1 H = r˙2 + r2θ˙2
+ ar2 (31) 4 To find the conjugate momenta in polar coordinates, we can write out the x˙2 y˙2 Lagrangian. We use pxx˙ = 2 and pyy˙ = 2 and get

L = ∑piq˙i − H (32) i 1 1 = x˙2 + y˙2
− r˙2 + r2θ˙2
− ar2 (33) 2 4 1 = r˙2 + r2θ˙2
− ar2 (34) 4 The conjugate momenta are thus

∂L 1 2 pθ = = r θ˙ (35) ∂θ˙ 2 ∂L r˙ p = = (36) r ∂r˙ 2 CYCLIC COORDINATES AND POISSON BRACKETS 5

From this we can see that pθ is indeed angular momentum as it’s pro- portional to the product of r and the tangential velocity vθ = rθ˙. (’Real’ momentum and angular momentum must, of course, also contain a factor of a mass, but from the definition of the Hamiltonian above, we see that the mass has been incorporated into the momentum parameters.) Plugging these back into 31 we get

2 2 2 H = pr + pθ + ar (37) We can now calculate the Poisson brackets:

  ∂pθ ∂H ∂pθ ∂H {pθ,H} = ∑ − (38) i ∂qi ∂pi ∂pi ∂qi ∂p ∂H = 0 − θ = 0 (39) ∂pθ ∂θ   ∂pr ∂H ∂pr ∂H {pr,H} = ∑ − (40) i ∂qi ∂pi ∂pi ∂qi ∂p ∂H = 0 − r (41) ∂pr ∂r = −2ar (42)

Thus pθ (the angular momentum) is conserved, whilep ˙r < 0, meaning that its radial velocity is decreasing.

PINGBACKS Pingback: Canonical transformations Pingback: Poisson brackets are invariant under a canonical transforma- tion Pingback: Passive, regular and active transformations.