Direct Energy, 2018a

11 CALCULUS OF VARIATIONS 265
The distance between the points can be described by the action
√
ˆ x1
( ) 2 dy
S = 1+ dx.
dx
x 0
To nd the path y(x) that minimizes
√
the action, we can solve the
Euler-Lagrange equation, with L = 1+ ( dy 2
dx)
as the Lagrangian,
for this shortest path y(x). This approach can be used because we
want to minimize the integral of some functional L even though this
functional does not represent an energy dierence [163, p. 33].
Set up the Euler-Lagrange equation, and solve it for the shortest
path, y(x).
Hint 1: The answer to this problem is that the shortest path between
two points is a straight line. Here, you will derive this result.
Hint 2: In the examples of this chapter, the Lagrangian had the form
L ( )
t, y, dy
dt with independent variable t and path y(t). Here, the Lagrangian
has the form L ( x, y,
dx) dy where the independent variable is
position x, and the path is y(x).
d
Hint 3: If (something) =0, then you know that (something) is
dx
constant.
(x 1 ,y 1 )
d −→ l
(x 0 ,y 0 )
11.7. Light travels along the quickest path between two points. This idea
is known as Fermat's principle. In a material with relative permittivity
ɛ r and permeability μ 0 , light travels at the constant speed √ c
ɛr
where c is the speed of light in free space. In Prob. 11.6, we showed
that the shortest path between two points is a straight line, so in a
uniform material, light will travel along a straight line between two
points. However, what if light travels across a junction between two
materials? In this problem, we will answer this question and derive