I’m reading The Feynman Physics Lectures vol I for the zillionth time. Excellent stuff if you’re into that.
Fermat’s Principle basically says that light will take the path that requires the shortest time. Which explains why it reflects at the angle it does on a mirror or why it refracts when going through a second medium like glass or water.
Feynman gives an explanation of why this is so, and it’s mostly lost on me. (Which is certainly not Feynmans fault; he’s damn good at this physics for the common man talk.) If you’ve got the book, I’m referring to sections 26-5 and 26-6. But even if you don’t have the book, I ask:
I might make a fool out of myself, but I’ll try to answer your question.
The way I imagine it ought to be proved is the following:
There is a function f§ which for each possible path gives the time it would take for the light to traverse it.
We also need to assume that the paths p, somehow can be assigned real continous numbers. (This is where my mathematics get a bit shaky, but I hope you follow the drift) We now want to ‘differentiaete’ F§ with respect to p, and then do something lke a Taylor expansion.
Here comes the trick: For each path where the time has a local minima (or maxima!) ‘surrounding’ paths will have approximatelly the same time. (as the first order in the Taylor expansion is zero!) And thus cause constructive interference!
However for every other path, the surrounding paths will have a very different time, and the total sum of ‘neighbouring’ paths will add up to a complete random phase-shift, and destructive interference.
It is interesting to note that this shows more than Fermat said, as it will also allow for the path with the longest path.
Formal mathematics was never my strongest subject, and I’m aware of having messed up above.
I hope someone will understand what I have written…
hopefully someone will soon come along and do a better job at explaining.
tc, thanks, but I still don’t get it. I’m with you about surrounding paths having minimal difference. And I see how those paths have constructive interference. But then… Why do surrounding paths of incorrect paths have destructive interference? And how does the destructive interference cause this effect?
Fynman says that quantum mechanics ties into this and says that if you add up all the possible probabilities, you find that incorrect paths tend to cancel themselves out; I assume this is where the destructive interference comes in.
Hmm, I have always seen it as a purely classical effect. Maybe I ought to try to find Feynmanns books. (I remember looking at the set at a bookstore at university, but they were outrageously overpriced…)
Anyway. Todays lesson. Why paths with non-extreme pathlengths give rise to destructive interference (classical version):
Look at the light as a wave, here expressed as the intensity after travelling length l:
F(l)=Acos(c (c*t-l)/lambda) (or sommething like that)
If you now look at light arriving somewhere where l has a maxima (or minima), neighbouring waves, will have more or less the same phase, and interfere constructively, whereas if l varies, the phase will be all over the place, with a flat distribution.
And of course the integral of cos(wt+b), wrt b is zero!
What you might do to make it all ‘modern’ is to put in Psi instead, as the wave function, to see that the |Psi|[sup]2[/sup] == 0 !
I’ve just re-read this, and I can’t say that I make much sense. Sorry. We might have to wait till Chronos comes in and puts everything in place.