# Solving the Laplacian in Spherical Coordinates (#1)

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.

$x = rsin\theta cos\phi$
$y = rsin\theta cos\phi$
$z = rcos\theta$

where

$0 \leq r < \infty$
$0 \leq \theta \leq \pi$
$0 \leq \phi < 2\pi$

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

$\nabla ^{2}f ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \phi ^{2}}} = 0$

To solve it we use the method of separation of variables.

$f = R(r)\Theta(\theta)\Phi(\phi)$

Plugging in the value of $f$ into the Laplacian, we get that :

$\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0$

Dividing throughout by $R\Theta\Phi$ and multiplying throughout by $r^2$, further simplifies into:

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0$

It can be observed that the first expression in the differential equation is merely a function of $r$ and the remaining a function of $\theta$ and $\phi$ only. Therefore, we equate the first expression to be $\lambda = l(l+1)$ and the second to be $-\lambda = -l(l+1)$. The reason for choosing the peculiar value of $l(l+1)$ is explained in another post.

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0$ (1)

The first expression in (1) the Euler-Cauchy equation in $r$.

$\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R$

The general solution of this has been in discussed in a previous post and it can be written as:

$R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}$

The second expression in (1) takes the form as follows:

$\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta + \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = 0$

The following observation can be made similar to the previous analysis

$\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0$ (2)

The first expression in the above equation (2) is the Associated Legendre Differential equation.

$\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta = m^2$

$sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0$

The general solution to this differential equation can be given as:
$\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)$

The solution to the second term in the equation (2) is a trivial one:

$\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi$
$\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}$

Therefore the general solution to the Laplacian in Spherical coordinates is given by:

$R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)$

# An infinite joke

$\frac{1}{0} = \infty$

Now flip this over by 90 degree counter clockwise :

$- 10 = 8$

$- 18 = 0$

Now flip this over again by 90 degree clockwise :

$\frac{1}{\infty} = 0$

# A strange operator

In a previous post on using the Feynman’s trick for Discrete calculus, I used a very strange operator ( $\triangledown$ ). And whose function is the following :

$\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}$

What is this operator? Well, to be quite frank I am not sure of the name, but I used it as an analogy to Integration. i.e

$\int x^{n} = \frac{x^{n+1}}{n+1} + C$

What are the properties of this operator ? Let’s use the known fact that $n^{\underline{k+1}} = (n-k) n^{\underline{k}}$

$\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}$

$\triangledown n^{\underline{k}} = \frac{(n-k) n^{\underline{k}}}{k+1}$

And applying the operator twice yields:

$\triangledown^2 n^{\underline{k}} = \frac{n^{\underline{k+2}}}{(k+1)(k+2)}$

$\triangledown^2 n^{\underline{k}} = \frac{(n-k-1) n^{\underline{k+1}}}{(k+1)(k+2)}$

$\triangledown^2 n^{\underline{k}} = \frac{(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)}$

We can clearly see a pattern emerging from this already, applying the operator once more :

$\triangledown^3 n^{\underline{k}} = \frac{(n-k-2)(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)(k+3)}$

$\vdots$

Or in general, the operator that has the characteristic prescribed in the previous post is the following:

$\triangledown^m n^{\underline{k}} = \frac{n^{\underline{k+m}}}{(k+m)^{\underline{m}}} n^{\underline{k}}$

If you guys are aware of the name of this operator, do ping me !

# Matrix Multiplication and Heisenberg Uncertainty Principle

We now understand that Matrix multiplication is not commutative (Why?). What has this have to do anything with Quantum Mechanics ?

Behold the commutator operator:
$[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

where $\hat{A},\hat{B}$ are operators that are acting on the wavefunction $\psi$. This is equal to 0 if they commute and something else if they don’t.

One of the most important formulations in Quantum mechanics is the Heisenberg’s Uncertainty principle and it can be written as the commutation of the momentum operator (p) and the position operator (x):

$[\hat{p}, \hat{x}] = \hat{p}\hat{x} - \hat{x}\hat{p} = i\hbar$

If you think of p and x as some Linear transformations. (just for the sake of simplicity).

This means that measuring distance and then momentum is not the same thing as measuring momentum and then distance. Those two operators do not commute! You can sort of visualize them in the same way as in the post.

But in Quantum Mechanics, the matrices that are associated with $\hat{p}$ and $\hat{x}$ are infinite dimensional. ( The harmonic oscillator being the simple example to this )

$\hat{x} = \sqrt{\frac{\hbar}{2m \omega}} \begin{bmatrix} 0 & \sqrt{1} & 0 & 0 & \hdots \\ \sqrt{1} & 0 &\sqrt{2} & 0 & \hdots \\ 0 & \sqrt{2} & 0 &\sqrt{3} & \hdots \\ 0 & 0 & \sqrt{3} & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}$

$\hat{p} = \sqrt{\frac{\hbar m \omega}{2}} \begin{bmatrix} 0 & -i & 0 & 0 & \hdots \\ i & 0 & -i \sqrt{2} & 0 & \hdots \\ 0 & i\sqrt{2} & 0 &\-i \sqrt{3} & \hdots \\ 0 & 0 & i\sqrt{3} & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}$

# Beautiful proofs(#2): Euler’s Sum

$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}$

Say what? This one blew my mind when I first encountered it. But it turns out Euler was the one who came up with it and it’s proof is just beautiful!

Prerequisite
Say you have a quadratic equation $f(x)$ whose roots are $r_1,r_2$, then you can write $f(x)$ as follows:

$f(x) = (x-r_1)(x-r_2) = 0$  (or)

$f(x) = (r_1-x)(r_2-x) = 0$  (or)

$f(x) = (1- \frac{x}{r_1})(1- \frac{x}{r_2}) = 0$

$f(x) = 1 - (\frac{1}{r_1} + \frac{1}{r_2}) + \frac{x^2}{r_1 r_2} = 0$

As for as this proof is concerned we are only worried about the coefficient of $x$, which you can prove that for a n-degree polynomial is:

$a_1 = - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})$

where $r_1,r_2 \hdots r_n$ are the n-roots of the polynomial.

Now begins the proof

It was known to Euler that

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots$

But this could also be written in terms of the roots of the equation as:

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})y + \hdots$

Now what are the roots of $f(y)$ ?. Well, $f(y) = 0$ when $\sqrt{y} = n \pi$ i.e $y = n^2 \pi^2$ *

The roots of the equation are $y = \pi^2, 4 \pi^2, 9 \pi^2, \hdots$

Therefore,

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots = 1 -( \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \hdots )y + \hdots$

Comparing the coefficient of y on both sides of the equation we get that:

$\frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{ 9 \pi^2} + \hdots$

$\zeta(2) = \frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots$

Q.E.D

* n=0 is not a root since
$\frac{sin(\sqrt{y})}{\sqrt{y}} = 1$ at y = 0

** Now if all that made sense but you are still thinking : Why on earth did Euler use this particular form of the polynomial for this problem, read the first three pages of this article. (It has to do with convergence)

# Why on earth is matrix multiplication NOT commutative ? – Intuition

One is commonly asked to prove in college as part of a linear algebra problem set that matrix multiplication is not commutative. i.e If A and B are two matrices then :

$AB \neq BA$

But without getting into the Algebra part of it, why should this even be true ? Let’s use linear transformations to get a feel for it.

If A and B are two Linear Transformations namely Rotation and Shear. Then it means that.

$(Rotation)(Shearing) \neq (Shearing)(Rotation)$

Is that true? Well, lets perform these linear operations on a unit square and find out:

(Rotation)(Shearing)

(Shearing)(Rotation)

You can clearly see that the resultant shape is not the same upon the two transformations. This means that the order of matrix multiplication matters a lot ! ( or matrix multiplication is not commutative.)

# Basis Vectors are instructions !

Basis vectors are best thought of in the context of roads.

Imagine you are in a city – X which has only roads that are perpendicular to one another.

You can reach any part of the city but the only constraint is that you need to move along these perpendicular roads to get there.

Now lets say you go to another city-Y which has a different structure of roads.

In this case as well you can get from one part of the city to any other, but you have to travel these ‘Sheared cubic’ pathways to get there.

Just like these roads determine how you move about in the city, Basis Vectors encode information on how you move about on a plane. What do I mean by that ?

The basis vector of City-X is given as:

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

This to be read as – ” If you would like to move in City-X you can only do so by taking 1 step in the x-direction or 1 step in the y-direction ”

The basis vector of City-Y is given as:

$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$

This to be read as – ” If you would like to move in City-Y you can only do so by taking 1 step in the x-direction or  1 step along the diagonal OB ”

Conclusion:

By having the knowledge about the Basis Vectors of any city, you can travel to any destination by merely scaling these basis vectors.

As an example, lets say need to get to the point (3,2), then in City-X,  you would take 2 steps in the x-direction and 3 steps in the y-direction

$\begin{bmatrix} 3 \\ 2 \end{bmatrix} = 3* \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 * \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

And similarly in City-Y, you would take 1 step along the x -direction and 2 steps along the diagonal OB.

$\begin{bmatrix} 3 \\ 2 \end{bmatrix} = 1* \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 * \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Destination Arrived 😀

# On the origins of Taylor/Maclaurin Series

Many a times it is not discussed as to How the Taylor/Maclaurin series came to be in its current form. This short snippet is all about it.

Let us assume that some function $f(x)$ can be written as a power series expansion. i.e

$f(x) = a_0 + a_1 x + a_2 x^2 + \hdots$.

We are left with the task of finding out the coefficients of the power series expansion.

Substitution x = 0, we obtain the value of $a_0$.

$a_0 = f(0)$.

Lets differentiate $f(x)$ wrt x.

$\frac{d}{dx} f(x) = a_1 + 2a_2 x + \hdots$

Evaluating at x =0 , we get

$\frac{d}{dx} f(0) = a_1$

And likewise:

$\frac{d^2}{dx^2} f(0) = 2.1.a_2 = 2! \space a_2$

$\frac{d^3}{dx^3} f(0) = 3.2.1.a_3 = 3! \space a_3$

$\vdots$

$\frac{d^n}{dx^n} f(0) = n.n-1...3.2.1.a_n = n! a_n$

That’s it we have found all the coefficient values, the only thing left to do is to plug it back into the power series expression:

$f(x) = f(0) + \frac{d}{dx}f(0) \frac{x}{1!} + \frac{d^2}{dx^2}f(0) \frac{x^2}{2!} + \frac{d^3}{dx^3} f(0) \frac{x^3}{3!} \hdots$.

The above series expanded about the point x = 0 is called as the ‘Maclaurin Series’. The same underlying principle can be extended for expanding about any other point as well i.e ‘Taylor Series’.

# The generalized product rule ( Leibniz Formula )

If f and g are n-times differentiable functions, then :

$(fg)^{'} = fg^{'} + gf^{'}$

Now, we would like to find out a generalized expression for the n-th derivative of fg. In order to arrive at that formulation lets calculate a few derivatives to see whether we can find a pattern:

$(fg)^{'} = fg^{'} + gf^{'}$

$(fg)^{''} = \left(fg^{'} + gf^{'}\right)^{'} = fg^{''} + 2 f^{'}g^{'} + gf^{''}$

$(fg)^{'''} = \left(fg^{''} + 2 f^{'}g^{'} + gf^{''} \right)^{'} = fg^{'''} + 3 f^{''}g^{'} + 3 f^{'}g^{''} + gf^{'''}$

$(fg)^{''''} = fg^{''''} + 4 f^{'''}g^{'} + 6f^{''}g^{''} + 4 f^{'}g^{'''} + gf^{''''}$

$\vdots$

You must have noticed a pattern in the above expressions. The coefficients seem are the one in the binomial expansion of $(x+y)^n$

Therefore we can write the expression for the n-derivative of fg as the following expression:

$(fg)^n = \sum\limits_{i=0}^{n} \binom{n}{i} f^{(i)}g^{(n-i)}$
where (i) means to differentiate i-times.

This is also known as Leibniz Formula.

** This plays an important role when we start discussing about the Associated Legendre Differential Equation.

# On the travelling wave: An intuition

The aim of this post is to understand the travelling wave solution. It is sometimes not explained in textbook as to why the solution “travels”.

We all know about our friend – ‘The sinusoid’.

$y = sin(x)$

y becomes 0 whenever sin(x) = 0 i.e $x = n \pi$

Now the form of the travelling sine wave is as follows:

$y= sin(x - \omega t)$

When does the value for y become 0 ? Well, it is when

$x - \omega t = n \pi$

$x = n \pi + \omega t$

As you can see this value of x is dependent on the value of time ‘t’, which means as time ticks, the value of x is pushed forward/backward by a $\omega$.

When the value of $\omega > 0$, the wave moves forward and when $\omega < 0$, the wave moves backward.

Here is a slowly moving forward sine wave.