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# Hamilton's Equations - CBU

By Tony Cole,2014-04-02 04:44
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Hamilton's Equations - CBUS,s,#,CBU,# 39

Hamilton’s Equations

Review: generalized momentum: p = ;T/;q jj

generalized force: Q = Σ[F(;x/;q) + F(;y/;q) + F(;z/;q)] kiixikiyikizik

Lagrange's Equations: d/dt[;T/;q’] - ;T/;q = Q kkkIF all the forces are conservative so that Q = -;V/;q, AND IF V is not a function of velocities, kk

only positions, so ;V/;q’ = 0, then we define what is called the: k

LAGRANGIAN: L = T - V .

and Lagrange's equation of motion is: d/dt[;L/;q’] - ;L/;q = 0 . kk

Basic Idea: Recall that L is a function of the q, the q’ and t: L(q,q’,t). In principle, from kkkkp = ;L/;q’, we could find p(q’,q), then find q’(p,q), and then get a new function, H, where kkkkkkkk

H(q,p,t). kk

To do this, consider: dL = Σ{(;L/;q’)dq’ + (;L/;q)dq} + (;L/;t)dt. kkkkk

From the definition of generalized momentum: p = ;T/;q and with V not a function of the q’, jjjwe have p = ;T/;q = ;L/;q’ = p , and from Lagrange's Eq: d/dt[;L/;q’] - ;L/;q = 0, or jjjjkk

;L/;q = d/dt[p] = p , kkk

so we can write: dL = Σ{pdq’ + p’dq} + (;L/;t)dt. kkkkk

Now we define the Hamiltonian, H, as: H = Σpq- L kkk

so that

dH = Σ(dpq’ + pdq’) - dL = Σ(dpq’ + pdq’) - Σ{pdq’ + p’dq} - (;L/;t)dt kkkkkkkkkkkkkkk

and canceling the second and third terms gives: dH = Σ{dpq’ – p’dq} - (;L/;t)dt . kkkkk

But since H = H(q,p,t), we have: dH = Σ{(;H/;q)dq + (;H/;p)dp} + (;H/;t)dt. kkkkkkk

From the above two expressions for dH we can identify the following equations, called

Hamiltonian Equations of Motion:

;H/;p = q , ;H/;q = -p , and ;H/;t = -;L/;t . kkkk

The first equation in integral form gives:

q(t) - q(0) = (;H/;p)dt . kkk

In the special case of an ignorable coordinate, (q does not appear), the second equation gives: k

;H/;q = -p’ , so p = constant if q does not appear in H. kkkk

Further, let's consider dH/dt:

dH = Σ{(;H/;q)dq + (;H/;p)dp} + (;H/;t)dt . kkkkk

so

dH/dt = Σ{(;H/;q)q’ + (;H/;p)p’} + (;H/;t) , kkkkk

and using the Hamiltonian equations, ;H/;p = q, ;H/;q = -p, and ;H/;t = -;L/;t we have: kkkk

dH/dt = Σ{-p’q’ + q’p’} + ;H/;t = ;H/;t = -;L/;t = dH/dt . kkkkk

As mentioned above, the function, H, is called the Hamiltonian function, and it is a function of the

q's, the p's and time. kk

Important Note: IF V is NOT a function of the q' (normal situation), AND IF the q are NOT kka function of time explicitly [often normal situation: q = q(x, x, …, x) but not q = q(x, x, …,x,t) ] , kk123Nkk123NTHEN, since

a) T = ΣΣ{?Aq’q’} + Σ{Bq’} + Tkkkkkko , ;;;

but here B = 0 and T = 0 (since any partial with respect to time is zero), so we have ko

T = ΣΣ{?Aq’q’} ; kkk;;;

b) p = ;L/;q’ , kk

but here V is not a function of the q’, so k

p = ;L/;q’ = ;T/;q’ = Σ{Aq’} kkkk;;;

since the B = 0; k

c) so pq = Σ{Aq’}q ; kkkk;;;

d) so Σpq = ΣΣ{Aq’}q = 2T ; kkkkkk;;;

e) so H = Σpq- L = 2*T (T-V) = T+V = E , which means kkk

H = E, the energy of the system!

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EXAMPLE:

Set up the Hamiltonian and write out the Hamiltonian Equations of Motion for the simple Harmonic

Oscillator in 1-D.

2Since V is not a function of x’ (V = ?kx in this case), and x is not an explicit function of time (normal case since we are not in a rotating or other such moving frame), then H = E, so

22 H = T + V = p/2m + ?kx

;H/;p = q , ;H/;q = -p , and ;H/;t = -;L/;t . kkkk

;H/;p = p/m = x , ;H/;x = kx = -p’ , and ;H/;t = 0 , or

p = mx’ , p’ = dp/dt = F = -kx , H = constant = E.

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WRITTEN HOMEWORK PROBLEM #7: (9-28 in Symon)

Start with the Lagrangian for the spherical pendulum (R = constant): L = T V where 2222T = kinetic energy = ? mv + ? mv = ?m(Rθ’) + ?m(R sin(θ) φ’), and θφ

V = mgR cos(θ). Remember that the Hamiltonian is a function only of the p and q . ii

a) Write down the Hamiltonian function for the spherical pendulum. b) Write down the Hamiltonian equations of motion. 22222c) Derive from them the equation: E = ? mRθ’ + p/[2mRsin(θ)] + mgRcos(θ) . φ

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