July 15th: Spherical Harmonics (Continuation)

Spherical harmonics corresponding to different eigenvalues are orthogonal

Let $f_n, f_m$ be two spherical functions such that:

$\tilde{\Delta} f_n = n \cdot f_n$

$\tilde{\Delta} f_m = m \cdot f_m$ , where $n \neq m$ .

We want to show $\int_{S^2} f_n \cdot f_m d{\sigma} = 0$ .

From the two equations above, we obtain:

$\int_{S^2} f_m \tilde{\Delta} f_n - f_n \tilde{\Delta} f_m d{\sigma} = (n - m) \int_{S^2} f_m \cdot f_n d\sigma$ (*)

Key: We show that $\int_{S^2} f_m \tilde{\Delta} f_n d{\sigma} = \int_{S^2} f_n \tilde{\Delta} f_m d{\sigma}$ .

The $\tilde{\Delta}$ operator takes homogeneous functions, restricts them to the unit sphere, applies the Laplacian, and then extend the resulting function to $\mathbb{R}^3$ by multiplying $r^2$ (where $r = \sqrt{x^2 + y^2 + z^2}$ ).

Since the multiple of $r^2$ does not change the equality, it suffices to show that

$\int_{S^2} f_m \Delta|_{S^2} f_n d{\sigma} = \int_{S^2} f_n \Delta|_{S^2} f_m d{\sigma}$ (**)

where $\Delta|_{S^2}$ is the operator that restricts a homogeneous function to the sphere and then applies the Laplacian.

$\Delta|_{S^2} = \frac{1}{sin(\theta)} \frac{\partial}{\partial \theta}(sin(\theta) \frac{\partial}{\partial \theta}) + \frac{1}{sin^2(\theta)}\frac{\partial^2}{\partial \phi^2}$ .

Hence,

$\int_{S^2} f_m \Delta|_{S^2} f_n d{\sigma}$

$= \int_{0}^{2\pi} \int_{0}^{\pi} f_n (\frac{1}{sin(\theta)} \frac{\partial}{\partial \theta}(sin(\theta) \frac{\partial f_m}{\partial \theta}) + \frac{1}{sin^2(\theta)}\frac{\partial^2 f_m}{\partial \phi^2}) sin(\theta) d{\theta} d{\phi}$ .

We apply product rule twice (and cancellation from $sin(0) = sin(\pi) = 0$ and $\frac{\partial f_n(0, \phi)}{\partial \phi} = \frac{\partial f_n(2\pi, \phi)}{\partial \phi}$ ) to obtain:

$\int_{0}^{2\pi} \int_{0}^{\pi} (\frac{1}{sin(\theta)} \frac{\partial}{\partial \theta}(sin(\theta) \frac{\partial f_n}{\partial \theta}) + \frac{1}{sin^2(\theta)}\frac{\partial^2 f_n}{\partial \phi^2}) sin(\theta) f_m sin(\theta) d{\theta} d{\phi}$ .

This proves (**). Thus, the LHS of (*) is 0, and since $n \neq m$ , we have:

$\int_{S^2} f_n \cdot f_m d{\sigma} = 0$ .

Question: Does a harmonic function, f(x, y, z), that is harmonic and bounded in the complement of a line segment in $\mathbb{R}^3$ have a removable singularity?

Answer: Yes

Our proof will use the barrier method.

Suppose the line segment to be on the x-axis and is given by the interval [-a, b].

We take a sphere centered at the origin that contains this line segment. Let such sphere be denoted by S and it has radius R.

Key: since the restriction of the f to the sphere is continuous, we can find a harmonic function, u(x, y, z), defined inside and on S such that it coincides with f on S.

Define $u_0 = f - u$ . Then, $u_0$ is 0 on the surface of S. We wish to show that $u_0 = 0$ inside S. This is where we have to choose a sequence of harmonic functions that can ‘squeeze’ $u_0$ .

The function that we pick is the potential of this line segment with unit charge. It is given by:

$V(x, y, z) = \int_{\text{line segment}} \frac{dq}{r}$

$V(x, y, z) = \int_{-a}^{b} \frac{1}{\sqrt{{\xi}^2 + d^2}} d{\xi}$ (where $d$ is the distance between (x, y, z) and the line segment)

$V(x, y, z) = ln(\frac{b + \sqrt{b^2 + d^2}}{-a + \sqrt{a^2 + d^2}})$ .

Since $u_0$ is bounded inside S, then let $|u_0| \leq M$ for some $M \geq 0$ .

By multiplying V by a constant C, we can allow V to be greater than $u_0$ on the surface $\Sigma_{\epsilon}$ (the set of points a distance $\epsilon$ away from the line segment) contained inside S.

Such C is given by $C = \frac{M}{ln(\frac{b + \sqrt{b^2 + \epsilon^2}}{-a + \sqrt{a^2 + \epsilon^2}})}$ .

Notice that as $\epsilon \rightarrow 0$ , $C \rightarrow 0$ . Hence, for $(x, y, z)$ in the region between $\Sigma_{\epsilon}$ and S, V(x, y, z) tends to 0. Since V and – V bounds $u_0$ on both the surfaces and V and -V are both monotone, it follows that $u_0$ is bounded by both V and -V in the region between these two surfaces as well. Hence, $u_0 = 0$ , which is what we want.

Follow up from last post concerning the proof of Liouville’s theorem in Euclidean space: given two balls with fixed centers, when the radius is large enough, the two balls coincide except for a small proportion of their volumes.

Key: finding a formula for the intersection volume of the two spheres.

We can let one of the sphere to be centered at (0, 0, 0) and the other to be centered at (d, 0, 0). In the previous proof, we assumed the radii of the two spheres to be the same. Thus, the intersection of the 2 spheres is a curve (in fact, it is a circle) lying in the plane $x = \frac{d}{2}$ .

The volume of the intersection can be found by adding the two congruent spherical caps.

The distances from the spheres’ centers to the base of the caps are both $\frac{d}{2}$ . So the heights of the caps are both $h = R - \frac{d}{2}$ .

The volume of a spherical cap of height h for a sphere of radius R is given by:

$V(h, R) = \frac{1}{3} \pi h^2 (3R - h)$

Source: https://mathworld.wolfram.com/SphericalCap.html

Hence, the volume of the intersection after some simplification is given by:

$V = \frac{1}{12} \pi (4R + d)(2R - d)^2$ .

The ratio of V to the volume of one sphere is given by:

$\frac{\frac{1}{12} \pi (4R + d)(2R - d)^2}{\frac{4}{3}\pi R^3} = \frac{\frac{1}{12} \pi (4 + \frac{d}{R})(2 - \frac{d}{R})^2}{\frac{4}{3} \pi} \rightarrow 1$ as $R \rightarrow \infty$ .

Hence, the proportion of the sphere that is not in the intersection goes to 0.

July 15th: Spherical Harmonics (Continuation)

Share this:

Leave a Reply Cancel reply