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Hamilton-Jacobi equation
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Definition/Summary
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The Hamilton-Jacobi equation is an additional reformulation of classical mechanics, one that uses a function S variously called the action, the generating function, and Hamilton's principal function.
It is very convenient for problems where one can separate the variables, and it is also convenient for expressing constants of the motion. |
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Equations
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The generating function S is a function of canonical coordinates [itex]q_a(t)[/itex] and constants of the motion [itex]\alpha_a[/itex]. The canonical momenta [itex]p_a(t)[/itex] are given in terms of it by
[itex]p_a = \frac{\partial S}{\partial q_a}[/itex]
and some additional constants of the motion [itex]\beta_a[/itex] by:
[itex]\beta_a = \frac{\partial S}{\partial \alpha_a}[/itex]
Plugging these canonical-momentum values into the Hamiltonian, we find the Hamilton-Jacobi equation:
[itex]\frac{\partial S}{\partial t} + H = 0[/itex] |
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Recent forum threads on Hamilton-Jacobi equation
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Breakdown
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Physics
> Classical Mechanics
>> Lagrangian/Hamiltonian
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Extended explanation
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To demonstrate what the Hamilton-Jacobi equation can be used for, here is a solution of the central-force problem with potential V(r) using it, complete with separation of variables. We start out with its Lagrangian
[itex]L = \frac12 m \left[ \left( \frac{dr}{dt} \right)^2 + r^2 \left( \frac{d\theta}{dt} \right)^2 \right] - V[/itex]
Its canonical momenta are
[itex]p_r = m \frac{dr}{dt}[/itex]
and
[itex]p_\theta = m r^2 \frac{d\theta}{dt}[/itex]
and thus its Hamiltonian is
[itex]H = \frac{(p_r)^2}{2m} + \frac{(p_\theta)^2}{2mr^2} + V[/itex]
The Hamilton-Jacobi equation becomes
[itex]\frac{\partial S}{\partial t} + \frac{1}{2m}\left( \frac{\partial S}{\partial r} \right)^2 + \frac{1}{2mr^2}\left( \frac{\partial S}{\partial \theta} \right)^2 + V = 0[/itex]
Aside from the partial derivatives, this equation has explicit dependence only on r. This suggests a decomposition of S using constants of the motion E (energy) and L (angular momentum):
[itex]S = - Et + L\theta + S'(E,L,r)[/itex]
We easily find S' from it:
[itex]S' = \int \left[ 2m(E - V) - \left( \frac{L}{r} \right)^2 \right]^{1/2} dr[/itex]
and the remaining constants of the motion, which are zero points:
[itex]-t_0 = - t + \frac{\partial S'}{\partial E}[/itex]
[itex]\theta_0 = \theta + \frac{\partial S'}{\partial L}[/itex] |
Commentary
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