Hamilton-Jacobi Equation

1. Aug 18, 2010

Piano man

I'm having a bit of difficulty understanding part of this problem:

Using the Hamilton-Jacobi equation ﬁnd the trajectory and the motion of a particle in the
potential $$U(r)=-Fx$$

The Hamilton-Jacobi Equation: $$\frac{\partial S}{\partial t}+H(q_1,...,q_s;\frac{\partial S}{\partial q_1},...,\frac{\partial S}{\partial q_s};t)=0$$

Starting off with the Hamiltonian:
$$H(p_x,p_y,p_z,x,y,z)=\frac{p_x^2}{2m}-Fx+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}$$

From HJE, since y and z are cyclic,
$$S(x,y,z;p_x,p_y,p_z;t)=-Et+p_yy+p_zz+S(x,p_x)$$

All this is grand, but the next step in the solutions I have say that we can now say that $$E=p_x+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}$$

I don't see where this comes from.

Any ideas?
Thanks

2. Aug 18, 2010

Dickfore

The action is not a function of the momenta explicitly. The conserved momenta (those that are conjugate to the cyclic coordinates, i.e. the coordinates that do not enter in the Hamiltonian explicitly) take the role of the arbitrary constants in finding the complete integral of the HJ eqn by the method of separation of variables.

3. Aug 18, 2010

Piano man

Ok, but how does that explain where E comes from?

4. Aug 18, 2010

Dickfore

If the Hamiltonian is time-indpendent, then energy is a sonserved quantity and $-E$ is the corresponding "conjugate variable". By writing the action as:

$$S(q, t) = S_{0}(q) - E \, t$$

the time-dependent HJ eqn:

$$\frac{\partial S}{\partial t} + H(q, \frac{\partial S}{\partial q}) = 0$$

becomes:

$$\frac{\partial S}{\partial t} = -E, \; \frac{\partial S}{\partial q_{j}} = \frac{\partial S_{0}}{\partial q_{j}}$$

$$H(q, \frac{\partial S_{0}}{\partial q}) = E$$

This is the time-independent HJ eqn.

5. Aug 18, 2010

Piano man

Right, but surely then, in the example above, E should equal H, since the Hamiltonian is time independent anyway.

But the Hamiltonian is $$H(p_x,p_y,p_z,x,y,z)=\frac{p_x^2}{2m}-Fx+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}$$

and the required value for E is
$$E=p_x+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}$$

Why is that?

6. Aug 18, 2010

Dickfore

I don't know what you are talking about. After you had substituted:

$$S_{0}(x, y, z) = f(x) + p_{y} \, y + p_{z} \, z$$

into the time independent HJ eqn, you should get:

$$f'(x) = \left(2 m \, E - p^{2}_{y} - p^{2}_{z} + 2 m \, F \, x\right)^{\frac{1}{2}}$$

Then, you should integrate this and you will get the complete integral. The arbitrary constants are $p_{y}, p_{z} and E$.