Is Huygens Principle just a fundamental way to understand light? It always seemed to me that it was somehow "derived" or that it should be-but is it simply a well-founded theory?
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$\begingroup$ It's not a quantum theory for one, so it can never be fundamental in the literal sense... $\endgroup$– DanuCommented Dec 11, 2013 at 8:01
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2$\begingroup$ Good question. I'd say it's a model that happens to fit rather than derived from anything more fundamental. There's an interesting article discussing some of these issues at mathpages.com/home/kmath242/kmath242.htm $\endgroup$– John RennieCommented Dec 11, 2013 at 8:48
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$\begingroup$ @JohnRennie: What do you mean by "rather than derived from anything more fundamental"? If it's more coarse than anything presumably fundamental, then I guess it, as a model, is derivable. At Danu: So quantum theory is necessarily fundamental? At Anonymous: Once one settles on fields and writes them down as propagators, I'd intuitively assume that the principle is equivalent to the requirement of some linearity of amplitude composition. $\endgroup$– Nikolaj-KCommented Dec 11, 2013 at 8:55
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$\begingroup$ Isn't it a statement about being able to decompose any wave on a basis of spherical harmonics? $\endgroup$– gatsuCommented Dec 11, 2013 at 10:14
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$\begingroup$ @gatsu Yes it is. And a rather useful statement as well! $\endgroup$– Carl WitthoftCommented Dec 11, 2013 at 12:44
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Actually, it can be theoretically derived from D'Alembert equation (that is satisfied by each component of ${\bf E}$ and ${\bf B}$ in absence of sources in view of free Maxwell equations). The idea is to compute the field (any component of ${\bf E}$ or ${\bf B}$) in $p$, when it is generated by a spherical point source localized in $q$ emitting a spherical monochromatic field with fixed (scalar) wavenumber $k$, and between $q$ and $p$ there is a screen with an aperture of known area. The mathematical tool is an integral formula, due to Kirchoff, which produces the solution in $p$ when it is known the value of the field and its normal derivative on a surface surrounding $p$. The surface is chosen to have a part adapted to the screen, including the aperture, and the remaining part is taken far away from $p$. Here, i.e., to fix the value of the field and its normal derivative on the surface, some approximations enter the computation and they usually have physical sense for $k >> d$, where $d$ is the "diameter" of the aperture. This situation is discussed in details in Jackson's textbook. The final formula obtained this way can be shown to be equivalent to directly apply Huygen's principle from scratch.
answered Dec 11, 2013 at 10:48
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2$\begingroup$ Isn't this simply a Green's function method to find the effect of the source? $\endgroup$– AbhinavCommented Dec 11, 2013 at 13:04
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$\begingroup$ I am not an expert on these issues, but I think that certainly the Green's function approach is the starting point. However you next have to perform some appropriate approximation. Birchoff's name here is, in fact, related with the appropriate choice of approximated boundary conditions... $\endgroup$ Commented Dec 11, 2013 at 13:46
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$\begingroup$ @Abhinav not quite: it's more centred on deriving a field from its values on a boundary. The whole topic is thoroughly treated (if a bit archaically) in Chapter 8 of Born and Wolf "Principles of Optics"; archaic notation and treatment notwithstanding, it's still a great deal more lucid than Jackson IMO. $\endgroup$ Commented Dec 11, 2013 at 13:49
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$\begingroup$ Yes sure, for "boundary conditions" I actually intended "values on a boundary" (the values of the field and its normal derivative on the surface where one integrates and where ranges one argument of the Green function). Thanks for the reference! $\endgroup$ Commented Dec 11, 2013 at 13:52
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1$\begingroup$ Actually, at least in classical physics, Maxwell's equations could be considered fundamental equations, they summarize all properties of EM fields, at least in classical (not quantum) physics. They were, more or less, derived by experiments by physicists (not only Maxwell). $\endgroup$ Commented Dec 11, 2013 at 18:20