Here we’re going to solve a few problems using what we’ve learned. If you notice any errors, send me an email!
Problem 1
(a) Show that the Euler-Lagrange equation for the functional
\[ S[y] = \int_0^X ((y’)^2 – y^2) dx, y(0) = 0, y(X) = 1 \]
where \(0 \lt X \lt \pi\), is
\[ y” + y = 0 \]
(b) Show that the stationary function is \(y = sin x / sin X\)
a) In this case \(F(x, y, y’) = (y’)^2 – y^2\), so:
\[ \frac{\partial F}{\partial y’} = 2y’ \;\;\;\text{and}\;\;\; \frac{\partial F}{\partial y} = -2y \]
Which gives us the Euler-Lagrange Equation \(2y” – (-2y) = 0\), or
\[ y” + y = 0 \]
b)We know that solutions of the Euler-Lagrange function are stationary paths of our functional. From here we can look at the equation \(y” + y = 0\) and realize that this is a simple harmonic oscillator, which have the general solution
\[ y = A \text{cos} x + B \text{sin} x \]
Where \(A\) and \(B\) are constants determined by the boundary conditions. Since \(y(0) = 0\), we see that \(A = 0\). Similarly, we can see that \(B \text{sin} X = 1\), so \(B = 1 / \text{sin} X\). Therefore, the stationary function is:
\[ y = 0 \text{cos} x + \frac{1}{\text{sin}X} \text{sin} x = \frac{\text{sin} x}{\text{sin}X} \]
Problem 2
Given a sphere with radius \(R\), what is the shortest distance between two points \(a\) and \(b\) that both lie upon the surface?
With our constraint that both points are on the surface of the sphere, we can define any point with 2 angles, \(\theta\) and \(\phi\).
Now in Cartesian coordinates we can define the distance between two points, \(s\), to be \(s = \sqrt{x^2 + y^2 + z^2} \) and similarly \(ds = \sqrt{dx^2 + dy^2 + dz^2}\). Converting this to spherical coordinates, we see that
\[ ds = \sqrt{R^2 \text{sin}\phi^2d\theta^2 + R^2 d\phi^2} = R\sqrt{1 + \text{sin}^2\phi \left(\frac{d \theta}{d \phi}\right)^2} d\phi \]
Which we can use to make the functional
\[ S[\theta] = int_a^b ds = R int_a^b \sqrt{1 + \text{sin}^2 \phi \left( \frac{d \theta}{d \phi}\right)^2} d \phi \]
Now what is our functional exactly? Usually we see it in the form \(S[f]\) or \(S[y]\). In this case, similarly to \(y(x)\) in Cartesian coordinates, we define our curve along the sphere’s surface as the function \(\theta(\phi)\). So similar to our work in part 2, we want to find the equation \(\theta(\phi)\) that minimizes \(S[\theta]\). Now if we redefine the Euler-Lagrange equation as
\[ \frac{d}{d\phi} \frac{\partial F}{\partial \theta’} – \frac{\partial F}{\partial \theta} = 0 \]
No since \(F = R \sqrt{1 + \text{sin}^2\phi \theta ‘ ^2}\) and \( \theta’ = \frac{d \theta}{d \phi}\), we can realize that \(F\) does not explicitly depend on \(\theta\) and so \(\partial F / \partial \theta = 0\). This allows us to reduce our equation to:
\[ \frac{d}{d\phi} \frac{\partial F}{\partial \theta’} = 0 \]
Since we know that the derivative of \(\frac{\partial F}{\partial \theta ‘}\) is zero, that means that \(\frac{\partial F}{\partial \theta ‘}\) itself is a constant, \(C\)
\[ \frac{\partial F}{\partial \theta ‘} = C = \frac{\partial}{\partial \theta ‘} \left( R\sqrt{1 + \text{sin}^2 \phi \theta’ ^2} \right) \]
which gives
\[ C = \frac{R\text{sin}^2 \phi \theta ‘}{\sqrt{1 + \text{sin}^2 \phi \theta’ ^2}} \]
or by multiplying through a constant factor,
\[ C = \frac{R\text{sin}^2 \phi \theta ‘}{1 + \text{sin}^2 \phi \theta’ ^2} \]
Next we’re going to do a clever bit of trickery. Spherical coordinates allow us to always orientate ourselves so that the z axis passes through point \(a\). That means that our expression is always equal to
\[ \frac{R\text{sin}^2 (0) \theta ‘}{1 + \text{sin}^2 (0) \theta’ ^2} = 0 \]
Therefore \(C = 0\) for all values of \(\phi\). But the only way for our definition of \(C\) to be true for any \(\phi\) and \(\theta\) is that we have \(\theta ‘ = 0\), which means that \(\theta\) is a constant. One a sphere, this equates to \(\phi\) being the only degree of freedom, so the shortest possible path must be a great circle.
Problem 3
In your previous physics classes you almost certainly covered Snell’s Law, which gives a relationship between the angle of incidence and the angle of refraction for light traveling between two different media. This law follows from Fermat’s principle, which states that the path given by a light ray between two points follows the path that can be traveled in the least time.
A caveat that may or may not have been brought up, is that Snell’s Law only holds true for isotropic media, or material where the index of refraction is constant throughout. By what about in an anisotropic material, where the refractive index could be modeled by a function of position?
3) Derive the path a light ray takes between points \(P_a\) and \(P_b\), while traveling through a 2-D medium who’s index of refraction is defined as
\[ n(x,y) = \frac{L}{y} \]
where \(L\) is the length of the material.
Answer) If we break up our path into infinitesimally small pieces of length \(ds = \sqrt{dx^2 + dy^2}\), then we realize it takes time \(dt = ds / v\) to travel each piece, where \(v\) is the speed of light during the traversal of that piece. We can then define the total time traveling from \(P_a\) to \(P_b\) to be:
\[ T = \int_{P_a}^{P_b} dt \]
\[ = \int_{P_a}^{P_b} \frac{ds}{v} \]
\[ = \int_{P_a}^{P_b} \frac{\sqrt{dx^2 + dy^2}}{v}\]
\[ = \int_{P_a}^{P_b} \frac{\sqrt{1 + y’^2}}{v} dx \]
\[ = \int_{P_a}^{P_b} \frac{n(y)}{c}\sqrt{1 + y’^2} dx \]
\[ T = \int_{P_a}^{P_b} \frac{L}{cy}\sqrt{1 + y’^2} dx \]
Now since Fermat’s principle tells us that light travels in the path that minimizes time, we can see that we are trying to minimize the functional
\[ S[y] = \int_{P_a}^{P_b} \frac{L}{cy}\sqrt{1 + y’^2} dx \]
and using the Beltrami identity, we can formulate this as:
\[ \frac{\partial F}{\partial y’} y’ – F = C \]
or
\[ \frac{L}{cy} \frac{y’^2}{\sqrt{1 + y’^2}} – \frac{L}{cy} \sqrt{1 + y’^2} = C \]
from which we get
\[ y’ = \sqrt{\frac{L^2}{y^2c^2C^2} – 1} \]
If we separate these so that everything involving \(y\) is on the left, and everything else is on the right, we get
\[ \frac{y}{\sqrt{1 – \frac{c^2C^2}{L}y^2}}y’ = \frac{L}{cC} \]
which if we express \(y’\) as \(dy/dx\), lets us construct the integrals
\[ \int \frac{y}{\sqrt{1 – \frac{c^2C^2}{L}y^2}} dy = \int \frac{L}{cC} dx \]
which gives
\[ \int \frac{y}{\sqrt{1 – \frac{c^2C^2}{L}y^2}} dy = \frac{L}{cC}x + A \]
where \(A\) is a constant of integration. With u-substitution, we can see that the this becomes
\[ \frac{L}{c^2C^2} \sqrt{1 – \frac{c^2C^2}{L}y^2} = \frac{L}{cC}x + A \]
Which can then be rearranged to
\[ y^2 + \left( x + \frac{c^3C^3A}{L^3} \right)^2 = \frac{L^2}{c^2C^2} \]
Fascinatingly, this is the equation for a circle of radius \(\frac{L^2}{c^2C^2} \) centered at \(x = \frac{c^3C^3A}{L^3} \). This means that in a material with a refractive index where \(n \propto \frac{1}{y}\), light rays travel along circular trajectories.