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]\displaystyle{ S=vt }[/math]

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

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

The index of refraction is denoted as

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

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

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

[math]\displaystyle{ = \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]\displaystyle{ x }[/math] to find a minimum for the time of flight

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

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

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

using chain rule

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

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

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

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

the path integral can now be written in terms of dx such that

[math]\displaystyle{ S= \int_{(x_1)}^{(x_2)} f(y,y^{\prime},x) dx }[/math]

To consider deviation away from [math]\displaystyle{ y(x) }[/math] the function [math]\displaystyle{ \eta(x) }[/math] is introduced to denote deviations away from the shortest line and the parameter [math]\displaystyle{ \alpha }[/math] is introduced to weight that deviation

[math]\displaystyle{ \eta(x) }[/math] = the difference between the current curve and the shortest path.

let

[math]\displaystyle{ Y(x) = y(x) + \alpha \eta(x) }[/math] = A path that is not the shortest path between two points.

let

[math]\displaystyle{ S(\alpha)= \int_{(x_1)}^{(x_2)} f(Y,Y^{\prime},x) dx }[/math]

[math]\displaystyle{ = \int_{(x_1)}^{(x_2)} f(y+\alpha \eta,y^{\prime}+ \alpha \eta^{\prime},x) dx }[/math]

the deviations in the various paths can be expressed in terms of a differentiate of the above integral with respect to the parameter [math]\displaystyle{ \alpha }[/math] as this parameter changes the deviation

[math]\displaystyle{ \frac{\partial}{\partial \alpha} f(y+\alpha \eta,y^{\prime}+ \alpha \eta^{\prime},x) = \frac{\partial y}{\partial \alpha}\frac{\partial f}{\partial y}+\frac{\partial y^{\prime}}{\partial \alpha} \frac{\partial f}{\partial y^{\prime}}= \eta \frac{\partial f}{\partial y}+\eta^{\prime} \frac{\partial f}{\partial y^{\prime}} }[/math]

[math]\displaystyle{ \frac{dS(\alpha)}{d \alpha}=\frac{d}{d \alpha} \int_{(x_1)}^{(x_2)} f(Y,Y^{\prime},x) dx }[/math]

[math]\displaystyle{ = \int_{(x_1)}^{(x_2)} \frac{d}{d \alpha}f(y+\alpha \eta,y^{\prime}+ \alpha \eta^{\prime},x) dx }[/math]

[math]\displaystyle{ = \int_{(x_1)}^{(x_2)} \left ( \eta \frac{\partial f}{\partial y}+\eta^{\prime} \frac{\partial f}{\partial y^{\prime}} \right )dx }[/math]

[math]\displaystyle{ = \int_{(x_1)}^{(x_2)} \left ( \eta \frac{\partial f}{\partial y} \right ) + \int_{(x_1)}^{(x_2)} \left ( \eta^{\prime} \frac{\partial f}{\partial y^{\prime}} \right )dx }[/math]

The second integral above can by evaluated using integration by parts as

let

[math]\displaystyle{ u = \frac{\partial f}{\partial y^{\prime}} \;\;\;\;\;\;\;\;\;\;\;\; dv = \eta^{\prime} dx }[/math]

[math]\displaystyle{ du = \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right ) \;\;\;\;\;\;\;\;\;\;\;\; v=\eta }[/math]

[math]\displaystyle{ \int_{(x_1)}^{(x_2)} \left ( \eta^{\prime} \frac{\partial f}{\partial y^{\prime}} \right )dx = \left [ \eta \frac{\partial f}{\partial y^{\prime}} \right ]_{x_1}^{x_2} - \int_{(x_1)}^{(x_2)} \eta \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right )dx }[/math]

[math]\displaystyle{ \left [ \eta \frac{\partial f}{\partial y^{\prime}} \right ]_{x_1}^{x_2} }[/math]  : the difference between the end points should be zero for a minimum

[math]\displaystyle{ \int_{(x_1)}^{(x_2)} \left ( \eta^{\prime} \frac{\partial f}{\partial y^{\prime}} \right )dx = 0 - \int_{(x_1)}^{(x_2)} \eta \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right )dx }[/math]

[math]\displaystyle{ \frac{dS(\alpha)}{d \alpha}=\frac{d}{d \alpha} \int_{(x_1)}^{(x_2)} f(Y,Y^{\prime},x) dx }[/math]

[math]\displaystyle{ = \int_{(x_1)}^{(x_2)} \left ( \eta \frac{\partial f}{\partial y} \right ) - \int_{(x_1)}^{(x_2)} \eta \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right )dx }[/math]

[math]\displaystyle{ = \int_{(x_1)}^{(x_2)} \eta \left [ \left ( \frac{\partial f}{\partial y} \right ) - \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right ) \right ] dx }[/math]

to force the deviations in the path length to be zero the first derivative of the integral should be zero which is possible if

[math]\displaystyle{ \left [ \left ( \frac{\partial f}{\partial y} \right ) - \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right ) \right ] =0 }[/math]

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

[math]\displaystyle{ \frac{\partial f}{\partial y} = \frac{\partial }{\partial y} \sqrt{1+y^{\prime}(x)^2} = 0 }[/math]

[math]\displaystyle{ \frac{d}{dx} \left ( \frac{\partial f}{\partial y^{\prime}} \right ) = \frac{d}{dx} \left ( \frac{\partial ( \sqrt{1+y^{\prime}(x)^2})}{\partial y^{\prime}} \right ) = \frac{\partial f}{\partial y} = 0 }[/math]

[math]\displaystyle{ \Rightarrow \left ( \frac{\partial ( \sqrt{1+y^{\prime}(x)^2})}{\partial y^{\prime}} \right ) }[/math]= Constant

[math]\displaystyle{ \frac{y^{\prime(x)}}{ \sqrt{1+y^{\prime}(x)^2}} = C }[/math]

[math]\displaystyle{ \left ( y^{\prime}(x)\right )^2=C^2 \left ( 1+y^{\prime}(x)^2\right ) }[/math]

[math]\displaystyle{ \left (1 - C^2 \right ) \left ( y^{\prime(x) }\right )^2=C^2 }[/math]

[math]\displaystyle{ \left ( y^{\prime}(x)\right )^2=\frac{C^2}{1 - C^2) } }[/math]

[math]\displaystyle{ y^{\prime}(x)=\sqrt{\frac{C^2}{1 - C^2) } } \equiv m }[/math]

integrating

[math]\displaystyle{ y =mx+b }[/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