Fermat's Principle Of Least Time: Feynman's Intro & A Proof From Classical Mechanics

Fermat's principle is an intriguing one in physics, first pointed out by Feynman in his lectures.  Therein he noted Fermat discovered this principle in 1650, basically given all possible paths that light might take to get from one point to another, the one it actually takes requires the shortest time.  

In Sec. 26-3 of his Feynman Lectures, Richard Feynman uses a plane mirror and states Fermat's law includes both 'the law of straight line propagation' and 'the law for the mirror'.  He uses the diagram shown below for reference with MM' the mirror surface and asks for the way to get from A to B  - in the shortest time - but the ray striking the mirror first.

The evident point needed for the shortest time is C, but how to get it? Feynman proposes a 'geometrical trick' - to wit, constructing an 'artificial point' B' on the other side of MM' which is the same distance below MM' as point B is above it. Since BFM is a right angle and BF = FB', then EB is equal to EB'.

Then the sum of the two distances: AE + EB is proportional to the time it will take, assuming a constant value c (speed of light). Which is also the sum of the two lengths AE + EB'.  The problem is then easily re-phrased as: When is the sum of the two lengths the least? The answer, by inspection, is when the reflection line goes through point C as a straight line from A to B'.  C works because it is the best choice to get to the artificial point B' in the most direct manner from A.  To break it down: if ACB' is a straight line then the angle BCF is equal to the angle B'CF and thence to angle ACM. So saying the angle of incidence is equal to the angle of reflection is equivalent to saying the light travels to the mirror in such a way that it comes back to point B' in the least possible time.

 

In a more modern context, given Fermat's least time principle also applies in classical mechanics, it can be proven using a more quantitative method from calculus. We use the diagram below:

                                  Sketch for mathematical analysis of situation

The purpose here is to demonstrate that for a ray going from a to b  the angle of incidence will equal the angle of reflection and hence be in accordance with the principle of least time.  Consider then, from the diagram, a path such that the time from a to b is a minimum (i.e. x1,a to xo,o back to x2,b), then we have;

t₁ =  ò _{x1, a} ^xo,0     ds/  C     (1st integral for path 1) 

where:   ds   =   Ö (dx² + dy ² )  =    dx Ö [ 1  +  (dy/dx) ² ]

Let:   f  =   1  +  (dy/dx) ²     

Þ

f  -  y'   ¶ f / ¶ y'    =  const.

t₁ =   1/ c ò _{x1, a} ^xo,0     dx f

Therefore, 

Ö  1  -  y ²    =   y'   (½ )   [y' / Ö (1  +  y'  ²  ) ]  =  const.  = A

Þ

1  +  y'  ²    - 2 y' ²  =   A ((1  +  y'  ²  )  (½ )

1  +  y' ²    =   1/  A²  =  B

But:   y'   =  dy/dx  =   Ö B  -1 

 Þ

y   =   (Ö B  -1 )  x  + C      (straight line)

Then:   

 For x 1,a  to   x 0,0  to  x1, b  :

t₁ +   t₂    =   

1/ C  [(x 1   - x 0 )²  +   a ²] ^1/2  +  [(x 0   - x 2 )²  +   b ²] - ^1/2  (2 (x 0   - x 2 )

For minimum:  

¶ ( t₁ + t₂ ) / ¶ x 0    =  0

Þ

 ½ [(x 1   - x 0 )²  +   a ²] ^-1/2   2 (x 1   - x 0) (-1) 

+ ½ [(x 0   - x 2 )²  +   b ²] ^{- 1/2}  (2 (x 0   - x 2 )   =  0   

Þ

(x 1   - x 0 )²/ (x 1   - x 0 )²  +   a ²   =  (x 0   - x 2 )² / (x 0   - x 2 )²  +   b ²

Therefore:   cos² q 1   =    cos² q 2 

q 1   =    q 2      I.e.

q i   =    q r 

The statement of the law of reflection that the angle of incidence ( q i) is equal to the angle of reflection (q r ) is equivalent to the statement that light travels (via the mirror)  from point  x 1,a  to  x1, b    in  the least possible time.   Hence, the principle of least time or Fermat's principle.