Problem: Deduce the Hamilton-Jacobi equation of classical mechanics.
Solution: Instead of viewing the action \(S=S[\mathbf x(t)]\) as a functional of the particle’s trajectory \(\mathbf x(t)\), it can be viewed more simply as a scalar field \(S(\mathbf x,t)\) in which the initial point in spacetime \((t_0,\mathbf x_0)\) is fixed and one simply takes the on-shell trajectory from \((t_0,\mathbf x_0)\) to \((t,\mathbf x)\). Then the total differential \(dS=\mathbf p\cdot d\mathbf x\) (follows from the usual Noetherian calculation) so in particular:
\[\mathbf p=\frac{\partial S}{\partial\mathbf x}\]
Intuitively, this is saying that the particle moves in a direction (the direction of the momentum \(\mathbf p\)) orthogonal to the contour surfaces of the action field \(S\), i.e. such isosurfaces can be viewed as “wavefronts”. Then the total time derivative is:
\[\dot S=L\]
But \(\frac{\partial S}{\partial t}+\frac{\partial S}{\partial\mathbf x}\cdot\dot{\mathbf x}=\frac{\partial S}{\partial t}+\mathbf p\cdot\dot{\mathbf x}\). Thus, isolating for \(H=\mathbf p\cdot\dot{\mathbf x}-L\) yields the Hamilton-Jacobi nonlinear \(1^{\text{st}}\)-order PDE for \(S(\mathbf x,t)\):
\[-\frac{\partial S}{\partial t}=H\left(\mathbf x,\frac{\partial S}{\partial\mathbf x},t\right)\]
Problem: When \(\partial H/\partial t=0\), the Hamiltonian is conserved with energy \(H=E\), so this motivates the additive separation of variables, \(S(\mathbf x,t):=S_0(\mathbf x)-Et\) for some constant \(E\). What does the Hamilton-Jacobi equation simplify to in this case? For a single non-relativistic particle of mass \(m\) moving in a potential \(V(\mathbf x)\), what does this look like? What about in \(1\) dimension?
Solution: \[H\left(\mathbf x,\frac{\partial S_0}{\partial\mathbf x}\right)=E\]
which for \(H(\mathbf x,\mathbf p)=|\mathbf p|^2/2m+V(\mathbf x)\) looks like:
\[\frac{1}{2m}\biggr|\frac{\partial S_0}{\partial\mathbf x}\biggr|^2+V(\mathbf x)=E\]
and in \(1\) dimension is integrable to the explicit solution:
\[S_0(x)=\pm\int ^xdx’\sqrt{2m(E-V(x’))}\]
In particular, the usual trajectory \(x(t)\) can be obtained by treating \(S_o=S_0(x,t;E)\) as a family of solutions parameterized by the energy \(E\).
\[\frac{\partial S_0}{\partial E}=-t_0\]
Problem: What does Fermat’s principle in ray optics assert? Hence, derive the ray equation.
Solution: The time functional \(T=T[\mathbf x(s)]\) of a ray trajectory \(\mathbf x(s)\) is stationary on-shell. That is:
\[cT[\mathbf x(s)]=\int ds n(\mathbf x(s))\]
This is reparameterization invariant, since one can arbitrarily parameterize \(\mathbf x=\mathbf x(t)\) and replace \(ds=dt|\dot{\mathbf x}|\). The corresponding Euler-Lagrange equations are:
\[\frac{d}{dt}\left(n(\mathbf x)\frac{\dot{\mathbf x}}{|\dot{\mathbf x}|}\right)=|\dot{\mathbf x}|\frac{\partial n}{\partial\mathbf x}\]
But by choosing the natural parameterization \(t:=s\) one has \(|d\mathbf x/ds|=1\), hence the ray equation:
\[\frac{d}{ds}\left(n\frac{d\mathbf x}{ds}\right)=\frac{\partial n}{\partial\mathbf x}\]
This can also be written in terms of the curvature vector \(\boldsymbol{\kappa}=d^2\mathbf x/ds^2\):
\[\boldsymbol{\kappa}=\left(\frac{\partial\ln n}{\partial\mathbf x}\right)_{\perp d\mathbf x}\]