I'm trying to deduce eikonal equation from Fermat's principle through variational calculus (namely, applying Euler-Lagrange equations)

$$\delta L=\delta\int_{t_{1}}^{t_{2}}n(x(s), y(s), z(s)) ds =0 \iff \dfrac{d}{ds}\left(\dfrac{\partial n}{\partial \dot{x} } \right)-\dfrac{\partial n}{\partial x}=0, \dot{x}=\dfrac{dx}{ds}.$$

And the same for $y(s)$ and $z(s)$. The point is that I don´t know how to prove, in order to be equivalent to the eikonal equation, that:

$$\dfrac{d}{ds}\left(n\dfrac{dx}{ds}\right)=\dfrac{d}{ds}\left(\dfrac{\partial n}{\partial \dot{x} } \right)$$

How can be proven? I'm not interested in other proofs of the eikonal equation with or without Fermat's principle, I'm just concerned with this path.

Eikonal equation:

$$\nabla n=\dfrac{d}{ds}\left(n\dfrac{d\vec{r}}{ds}\right)$$


I'm thinking that you have to get a lot clearer on what you mean.

Let me define an abstract coordinate $p \in (0, 1)$ that traces in some sense "how far" an object has gone from the starting point to the ending point, then the paths we're considering have the form $\vec r(p).$ We know that the medium has an index $n(\vec r) = c/v(\vec r)$ defining the speed that things actually go, and over an interval $dp$ we know that the path travels a distance $|\vec r'(p)|~dp$ for a total time,$$\int dt = c^{-1} \int_0^1 dp~|\vec r'(p)|~c/v(\vec r(p)) = c^{-1} \int_0^1 dp ~|\vec r'(p)| ~ n(\vec r(p)).$$I therefore gather you're using $ds = dp~|\vec r'(p)|$? That sounds like it could be useful but it probably is not because it makes you write things like $\partial n/\partial\dot x$ which we know must be $0$.

Instead it looks like your effective Lagrangian is (dots are derivatives with respect to $p$),$$ L = \sqrt{\dot x^2 + \dot y^2 + \dot z^2}~~n(x,~y,~z)$$ creating the Euler-Lagrange equation$$\frac{d}{dp}\left(\frac{\partial L}{\partial\dot x}\right) = \frac{\partial L}{\partial x},$$which expands to$$\frac{d}{dp}\left(\frac{\dot x}{\sqrt{\dot x^2 + \dot y^2 + \dot z^2}} ~ n\right ) = \sqrt{\dot x^2 + \dot y^2 + \dot z^2} ~~ n_x.$$

Following what you seem to be doing, $\dot s = \sqrt{\dot x^2 + \dot y^2 + \dot z^2}$ and for any symbol Q we define $dQ/ds = \dot Q / \dot s,$ we get straightforwardly $n_x = \frac{d}{ds} \left(n~\frac{dx}{ds}\right),$ which is just the $x$-component of what you're trying to prove and can be promoted directly into a vector notation.

The bottom line seems to be that you need to be very suspicious about $ds$ in that integral sign if $s$ must be recalculated based on your path; the Euler-Lagrange equations are derived by assuming that the path perturbation does not perturb this variable of integration $dt$.


Define $$L(\mathbf r, \dot{\mathbf r}) = n(\mathbf r(s))\sqrt{\dot{\mathbf r}(s)\cdot\dot{\mathbf r}(s)}$$ satisfying $\mathrm ds^2 = \mathrm d\mathbf r\cdot \mathrm d\mathbf r$

Now, define $$I = \int_{s_1}^{s_2}L ~\mathrm ds = \int_{s_1}^{s_2} n(\mathbf r(s))\sqrt{\dot{\mathbf r}(s)\cdot\dot{\mathbf r}(s)}~\mathrm ds$$

So, $$ \begin{align} \delta I &=\delta\int_{s_1}^{s_2}L ~\mathrm ds \\ &=\delta \int_{s_1}^{s_2} n(\mathbf r(s))\sqrt{\dot{\mathbf r}(s)\cdot\dot{\mathbf r}(s)}~\mathrm ds\\ &= \int _{s_1}^{s_2} \epsilon\left(\frac{\partial n}{\partial\mathbf r}|\dot{\mathbf r}| - \frac{\mathrm d}{\mathrm ds}\left(n\left(\frac{\dot{\mathbf r}}{|\dot{\mathbf r}|}\right)\right)\right)~\varphi^\prime~\mathbf ds\tag{*} \end{align} $$

Now, using proper boundary conditions, deduce the E-L equation for $\delta I/\epsilon = 0$ to be true.

$(*)$ The variation of $\mathbf r$ is defined as $$\delta \mathbf r = \bar{\mathbf r} - \mathbf r = \epsilon \varphi(s)$$ for any non-negative constant $\epsilon \to 0\,.$

$\delta \dot {\mathbf r} = \dot{(\delta \mathbf r)}\,.$


$\bullet$ Principles of Optics by M.Born, E.Wolf.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.