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In the geometrical optic limit, light rays can be seen as characteristics of the solutions of the eikonal equation:

$$|\nabla L|^2=n^2(x)$$

where $n(x)=c/v(x)$ is the refraction index.

On the other hand some texts (for instance Analytical mechanics, by Fasano & Marmi; and Methods of mathematical physics, II by Hilbert and Courant) say that eikonal equations are Hamilton-Jacobi equations. As such, I presume, there should be a Hamiltonian or Lagrangian to start with. Can the eikonal equation be derived as a Hamilton-Jacobi equation starting from the Lagrangian of the optical path length $$n(x) \mathrm{d}s~?$$

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ebenezer
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1 Answers1

2

  1. We start$^1$ by identifying the optical length with a square root action$^2$ $$\begin{align}S_0[{\bf r}]~=~&\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L_0, \qquad L_0~:=~n({\bf r})|\dot{\bf r}|, \cr |\dot{\bf r}|~:=~&\sqrt{\dot{\bf r}^2}, \qquad \dot{\bf r}~:=~\frac{d{\bf r}}{d\lambda}, \qquad {\bf r}~=~\begin{pmatrix}r^1\cr r^2\cr r^3\end{pmatrix}. \end{align}\tag{1} $$ The Euler-Lagrange (EL) equation is the ray equation. Note that the action (1) has a worldline (WL) reparametrizations gauge symmetry $$\lambda\quad\longrightarrow\quad \lambda^{\prime}~=~f(\lambda).\tag{2} $$

  2. To get rid of the square root, we introduce an auxiliary einbein field $e(\lambda)>0$: $$\begin{align} S[{\bf r},e]~=~&\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L, \cr L~:=~&\frac{\dot{\bf r}^2}{2e}+\frac{e}{2}n({\bf r})^2. \end{align}\tag{3}$$ The EL equation for the einbein field is$^3$ $$ e~\stackrel{(3)}{\approx}~\frac{|\dot{\bf r}|}{n({\bf r})}.\tag{4}$$ If we eliminate/integrate out the einbein field $e$, the action (3) reduces to the original action (1).

  3. The momentum is $$ {\bf p}~=~\frac{\partial L}{\partial \dot{\bf r}}~\stackrel{(3)}{=}~\frac{\dot{\bf r}}{e}.\tag{5}$$

  4. The Hamiltonian action becomes $$ \begin{align} S_H[{\bf r},{\bf p},e]~=~&\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L_H, \cr L_H~:=~&{\bf p}\cdot\dot{\bf r}-\underbrace{\frac{e}{2}({\bf p}^2-n({\bf r})^2)}_{\text{Hamiltonian}}.\end{align}\tag{6} $$ If we eliminate/integrate out the momentum ${\bf p}$, the Hamiltonian action (6) reduces to the Lagrangian action (3).

  5. The einbein $e$ in the Hamiltonian action (6) becomes a Lagrange multiplier that imposes the constraint $$ |{\bf p}|~\stackrel{(6)}{\approx}~ n({\bf r}).\tag{7}$$

  6. If we perform a canonical transformation $({\bf r},{\bf p})\to({\bf R},{\bf P})$ with a type 2 generator $W({\bf r},{\bf P})$ such that $$ {\bf p}\cdot\dot{\bf r}~=~-{\bf R}\cdot\dot{\bf P}+\frac{dW({\bf r},{\bf P})}{dt}, \tag{8}$$ then the constraint (7) becomes the Hamilton-Jacobi (HJ) equation $$ \left|\frac{\partial W}{\partial {\bf r}}\right|~\stackrel{(7)}{\approx}~ n({\bf r}), \tag{9}$$ which in turn is the eikonal equation, where the eikonal $W$ is Hamilton's characteristic function, which on-shell is the abbreviated action.

References:

  1. H. Goldstein, Classical Mechanics, 2nd ed; section 10-8.

  2. A. Fasano & S. Marmi, Analytic Mechanics, 2006; section 11.9.

  3. R. Hilbert & D. Courant, Methods of Math. Phys, Vol. 2; Chapter II $\S$9. NB: In Ref. 3 the Lagrangian is denoted $F$; the Hamiltonian is denoted $L$; the action and Hamilton's principal function are both denoted $J$. The discussion in Section II$\S$9.3 essentially amounts to choosing the gauge $e=1$.

$^1$ In Refs. 1 & 2 the eikonal equation is derived as some appropriate approximation to the wave equation and/or the Schrödinger equation. It is also noted there that the eikonal equation is structurally similar to Hamilton-Jacobi (HJ) equation in classical mechanics, where the eikonal and refractive index $n({\bf r})$ play the role of Hamilton's characteristic function $W$ and momentum, respectively.

$^2$ This is a modification of the action for a relativistic point particle, cf. e.g. this and this Phys.SE posts.

$^3$ The $\approx$ symbol means equality modulo eqs. of motion (EOM). The words on-shell and off-shell refer to whether the EOM are satisfied or not.

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