Forest UCM CoV

Calculus of Variations

Fermat's Principle

Fermats principle is that light takes a path between two points that requires the least amount of time.

If we let S represent the path of light between two points then

[math]S=vt[/math]

light takes the time [math]t[/math] to travel between two points can be expressed as

[math]t = \int_A^B dt =\int_A^B \frac{1}{v} ds [/math]

The index of refraction is denoted as

[math]n=\frac{c}{v}[/math]

[math]t = \int_A^B \frac{n}{c} ds [/math]

for light traversing an interface with an nindex of refraction $n_1$ on one side and $n_2$ on the other side we would hav e

[math]t = \int_A^I \frac{n_1}{c} ds+ \int_I^B \frac{n_2}{c} ds [/math]

[math]= \frac{n_1}{c}\int_A^I ds+ \frac{n_2}{c} \int_I^B ds [/math]

[math]= \frac{n_1}{c}\sqrt{h_1^2 + x^2}+ \frac{n_2}{c} \sqrt{h_2^2 + (\ell -x)^2} [/math]

take derivative of time with respect to [math]x[/math] to find a minimum for the time of flight

[math] \frac{d t}{dx} = 0[/math]

[math] \Rightarrow 0 = \frac{d}{dx} \left ( \frac{n_1}{c}\left ( h_1^2 + x^2 \right )^{\frac{1}{2}}+ \frac{n_2}{c} \left (h_2^2 + (\ell -x)^2 \right)^{\frac{1}{2}} \right )[/math]

[math] = \frac{n_1}{c}\left( h_1^2 + x^2 \right )^{\frac{-1}{2}} (2x) + \frac{n_2}{c} \left (h_2^2 + (\ell -x)^2 \right)^{\frac{-1}{2}} 2(\ell -x)(-1)[/math]

[math] = \frac{n_1}{c}\frac{x}{\sqrt{ h_1^2 + x^2 }} + \frac{n_2}{c} \frac{(\ell -x)(-1)}{\sqrt{h_2^2 + (\ell -x)^2}} [/math]

[math] = n_1\frac{x}{\sqrt{ h_1^2 + x^2 }} - n_2 \frac{\ell -x}{\sqrt{h_2^2 + (\ell -x)^2}} [/math]

[math] \Rightarrow n_1\frac{x}{\sqrt{ h_1^2 + x^2 }} = n_2 \frac{\ell -x}{\sqrt{h_2^2 + (\ell -x)^2}} [/math]

or

[math] \Rightarrow n_1\sin(\theta_1) = n_2 \sin(\theta_2) [/math]

Generalizing Fermat's principle to determining the shorest path

One can apply Fermat's principle to show that the shortest path between two points is a straight line.

In 2-D one can write the differential path length as

[math]ds=\sqrt{dx^2+dy^2}[/math]

using chain rule

[math]dy = \frac{dy}{dx} dx \equiv y^{\prime}(x) dx[/math]

the the path length between two points [math](x_1,y_1)[/math] and [math](x_2,y_2)[/math] is

[math]S = \int_{(x_1,y_1)}^{(x_2,y_2)} ds= \int_{(x_1,y_1)}^{(x_2,y_2)} \sqrt{dx^2+dy^2}[/math]

[math] = \int_{(x_1,y_1)}^{(x_2,y_2)} \sqrt{dx^2+\left ( y^{\prime}(x) dx\right)^2}[/math]

[math] = \int_{(x_1,y_1)}^{(x_2,y_2)} \sqrt{1+y^{\prime}(x)^2}dx[/math]

adding up the minimum of the integrand function is one way to minimize the integral ( or path length)

let

[math]f(y,y^{\prime},x) \equiv \sqrt{1+y^{\prime}(x)^2}[/math]

a critical point (where a min or max) exists if

[math]\frac{df}{dx} = 0[/math]

if the second derivative is zero there is a min, max, or neither

if the second derivative is negative there is a max

if the second derivative is positive there is a min

[math]\frac{df}{dx} = \frac{d}{xdx} \left ( y^{\prime} \frac{\partial f}{\partial y^{\prime} \right )[/math]

http://scipp.ucsc.edu/~haber/ph5B/fermat09.pdf

Euler-Lagrange Equation

https://www.fields.utoronto.ca/programs/scientific/12-13/Marsden/FieldsSS2-FinalSlidesJuly2012.pdf

Forest_Ugrad_ClassicalMechanics