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Derivation of Eikonal equations from Fermat's Principle

  1. Jan 11, 2012 #1


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    So, I'm reading Holm's Geometric Mechanics Part I, and in it he wants to derive the Eikonal equations from Fermat's principle.

    There's one part of the derivation that I don't understand. He gives the "Optical path" as:

    [tex]A=\int_a^b n(\vec{r}(s))ds[/tex]

    Where ds is the infinitessimal arc length.

    And then proceeds to take the variation of:

    [tex]\delta \int_a^b n(\vec{r}(s))\sqrt{\frac{d\vec{r}}{ds}\cdot\frac{d\vec{r}}{ds}}ds[/tex]

    What's the deal with adding the square root term in there? It seems that it's legitimate to add this term since it's actually 1, but it appears that without adding this term, you don't get the Eikonal equations back by taking that variation to be 0, so what rational is there that "mandates" the adding of this term?
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  3. Jan 11, 2012 #2


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    I think the issue is that you need to express the integral in a form where the dependence on the path length is made explicit because you need to perform variations in which the length of the path varies also. While ds is path length in the first equation, in the second equation it's just any parameter that varies along the curve, and in fact he should have called it something else.
  4. Jan 11, 2012 #3


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    Hmmm, so, would it be fair to restate Fermat's principle as:

    [tex]0=\delta \int_C n(\vec{r}) d\vec{r}[/tex]

    And then parameterize the curve (arbitrarily) to obtain:

    [tex]0=\delta \int_a^b n(\vec{r}(s))|\vec{r}'(s)|ds[/tex]

    And then of course from the last expression, if we take s to be the arc-length so that [itex]|\vec{r}'(s)|=1[/itex], then we can recover the original form of the principle. Is this a valid method?

    From this view, then, we are performing variations on the parameterization of the curves as well then?
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