Variation of eikonal (phase) set to zero

  • Thread starter neelakash
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We know, in optics Fermat's principle is written in analogy with principle of Maupertuis in Classical mechanics (given by [tex]\delta S=\delta\int\vec{p}\cdot d\vec{l}=0[/tex]). In terms of the wave vector it is written as [tex]\delta\psi=\delta\int\vec{k}\cdot d\vec{l}=0[/tex]. Here [tex]\psi[/tex] is known as eikonal (or, phase).

For a monochromatic wave of fixed frequency,[tex]|\vec{k}|=\frac{\omega}{v}=\frac{\omega}{c}\frac{c}{v}=|\vec{k_0}|\frac{c}{v}=|\vec{k_0}|n[/tex] ([tex]n[/tex] is refractive index). Using that the above reduces to
[tex]|\vec{k_0}|\delta\int\ n\ dl=0[/tex] or [tex]\delta\int\ n\ dl=0[/tex] which can be identified as more familiar form of Fermat's principle.[we have calculated the line integral along the direction of [tex]\vec{k}[/tex]]

Notice that it is reached by setting the variation of [tex]\psi[/tex] to zero like in mechanics we have principle of Maupertuis by setting variation of action [tex]S[/tex] equal to zero. It is known that action is minimized in mechanics to get the classical path traced by the system. But is the case same with the phase eikonal also? Landau (section 53-54) identifies [tex]\vec{k}=\nabla\psi[/tex] and thus [tex]\psi[/tex] is a scalar potential of a conservative field. That means whatever path you choose from A to B,the integral [tex]\int_{A}^{B}\vec{k}\cdot\ dl=\int_{A}^{B}|\vec{k}|\ dl[/tex] will remain the same. Anyway, the minimization of optical path is evident from the familiar form of Fermat's principle: [tex]\delta\int\ n\ dl=0[/tex] what can be proved using geometry.

The only difference is that action [tex]S[/tex] is not constrained by any partial differential equation like [tex]\psi[/tex] is. [tex]\psi[/tex] is constrained the eikonal equation given by [tex](\nabla\psi)^2=\ n^2[/tex]-here [tex]n[/tex] is refractive index...Anyway, what I wonder is that whether it is possible to call the problem of optics a variational problem: we are not actually minimizing [tex]\psi[/tex] for [tex]\psi(B)-\psi(A)[/tex] is the same for all the paths...(unless you constrain it via eikonal equation).

Do people know the explanation for it? May be I am missing something...
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Maupertuis in Classical mechanics





No scalar product!

Eikonal is solution of equation

[tex](\frac{\partial F}{\partial x})^2+(\frac{\partial F}{\partial y})^2+(\frac{\partial F}{\partial z})^2=\frac{\omega^2}{c^2}n^2(x,y,z)[/tex]

So I think is good to go with Hamilton - Jacobi equation

[tex](\frac{\partial W}{\partial x})^2+(\frac{\partial W}{\partial y})^2+(\frac{\partial W}{\partial z})^2=2m[E-U(\vec{r})]=p^2[/tex]
Yes...but I did not ask that...what I asked is following:-

In mechanics, the problem is posed like this: you are to find the desired path by minimizing action...not all paths will lead to the minimum value of action.

But in optics, the variational principle [tex]\delta\int\vec{k}\cdot\ d\vec{l}=0[/tex] does NOT minimize the eikonal (for the integral is the same along all the paths).Yet when you reduce this to [tex]\delta\int\ n\ dl=0[/tex],the minimization is apparent...

Why is this difference?
I do not know why you say there would not be any scalar product...In many texts including Landau the equation is written in terms of scalar product. Only when you choose your contour along the direction of motion,you get cosine term to be unity.

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