Difference between revisions of "Forest UCM CoV"
Line 85: | Line 85: | ||
let | let | ||
− | :<math>S(\alpha)= \int_{(x_1)}^{(x_2)} f(y,y^{\prime},x) dx</math> | + | :<math>S(\alpha)= \int_{(x_1)}^{(x_2)} f(Y,Y^{\prime},x) dx</math> |
+ | :<math>= \int_{(x_1)}^{(x_2)} f(y+\alpha \eta,y^{\prime}+ \alpha \eta^{\prime},x) dx</math> | ||
+ | |||
+ | To find the critical point we need to differentiate that above integral with respect to the parameter <math>\alpha</math> as this parameter changes the deviation | ||
+ | :<math>\frac{\partial}{\partial \alpha} f(y+\alpha \eta,y^{\prime}+ \alpha \eta^{\prime},x) = \eta \frac{\partial f}{\partial y}+\eta^{\prime} \frac{\partial f}{\partial y^{\prime}}</math> | ||
+ | |||
a critical point (where a min or max) exists if | a critical point (where a min or max) exists if |
Revision as of 15:40, 13 October 2014
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
light takes the time
to travel between two points can be expressed as
The index of refraction is denoted as
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
take derivative of time with respect to
to find a minimum for the time of flightor
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
using chain rule
the the path length between two points
and isadding up the minimum of the integrand function is one way to minimize the integral ( or path length)
let
the path integral can now be written in terms of dx such that
To consider deviation away from the function is introduced to denote deviations away from the shortest line and the parameter is introduced to weight that deviation
- = the difference between the current curve and the shortest path.
let
- = A path that is not the shortest path between two points.
let
To find the critical point we need to differentiate that above integral with respect to the parameter
as this parameter changes the deviation
a critical point (where a min or max) exists if
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
http://scipp.ucsc.edu/~haber/ph5B/fermat09.pdf
Euler-Lagrange Equation
https://www.fields.utoronto.ca/programs/scientific/12-13/Marsden/FieldsSS2-FinalSlidesJuly2012.pdf