# Remembering the Laplacian in different coordinate systems

When one is learning about Laplace and Poisson equations it can be frustrating to remember its form in different coordinate systems.  But when one is introduced to four vectors, special relativity and so on, here is a simple way to remember the Laplacian in any coordinate system.

$\nabla^2 \phi = \frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^{j}} \left(\sqrt{|g|} g^{ij} \frac{\partial \phi}{\partial x^{j}} \right)$

where we $g^{ij}$ is the inverse of the metric $g_{ij}$, $|g|$ is the determinant of the metric . And one specifies the coordinate system by mentioning the form of the metric. Let’s look at how this works out:

Cartesian Coordinates

$x^{j} = (x,y,z)$

$g_{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$g^{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$|g_{ij}| = 1$

$\nabla^2 \psi = \frac{1}{1} \frac{\partial}{\partial x^{j}} \left(\sqrt{1} g^{ij} \frac{\partial \psi}{\partial x^{j}} \right) = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2}$

Cylindrical Coordinates

$x^{j} = (r,\phi,z)$

$g_{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$g^{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{r^2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$|g_{ij}| = r^2$

$\nabla^2 \psi = \frac{1}{r} \frac{\partial}{\partial x^{j}} \left(r g^{ij} \frac{\partial \phi}{\partial x^{j}} \right)$

$\nabla^2 \psi = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi}{\partial r} \right) + \frac{1}{r} \frac{\partial}{\partial \phi} \left( \frac{r}{r^2} \frac{\partial \psi}{\partial \phi} \right) + \frac{1}{r} \frac{\partial}{\partial z} \left( r \frac{\partial \psi}{\partial z} \right)$

Noting that $\frac{\partial r}{\partial \phi} = 0 = \frac{\partial r}{\partial z}$ because they are independent variables, we get

$\nabla^2 \psi = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \phi^2} + \frac{\partial^2 \psi}{\partial z^2}$

Spherical Coordinates

$x^{j} = (r,\phi,\theta)$

$g_{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2(\theta) \end{bmatrix}$

$g^{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{r^2} & 0 \\ 0 & 0 & \frac{1}{r^2 \sin^2(\theta)} \end{bmatrix}$

$|g_{ij}| = r^4 \sin^2(\theta)$

Following the same approach as the Cylindrical and Cartesian coordinates, we get the following form for the Laplacian in Spherical coordinates,

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

# Prof.Ghrist at his best!

To understand why this is true, we must start with the Fundamental Theorem of Vector calculus. If $F$ is a conservative field ( i.e $F = \nabla \phi$ ), then

$\int\limits_{A}^{B} F.dr = \int\limits_{A}^{B} \nabla\phi .dr = \phi_{A} - \phi_{B}$

What this means is that the value is dependent only on the initial and final positions. The path that you take to get from A to B is not important.

Now if the path of integration is a closed loop, then points A and B are the same, and therefore:

$\int\limits_{A}^{A} F.dr = \int\limits_{A}^{A} \nabla\phi .dr = \phi_{1} - \phi_{1} = 0$

Now that we are clear about this, according to Stokes theorem the same integral for a closed region can be represented in another form:

$\int_{C} F.dr = \int\int_{A} (\nabla X F) .\vec{n} dA = 0$

From this we get that Curl = $\nabla X F = 0$ for a conservative field (i.e $F = \nabla \phi$). Therefore when a conservative field is operated on by a curl operator ($\nabla X$), it yields 0.

Bravo Prof.Ghrist! Beautifully said 😀

# On the direction of the cross product of vectors

One of my math professors always told me:

Understand the concept and not the definition

A lot of times I have fallen into this pitfall where I seem to completely understand how to methodically do something without actually comprehending what it means. And only after several years after I first encountered the notion of cross products did I actually understand what they really meant. When I did, it was purely ecstatic!

## Why on earth is the direction of cross product orthogonal ? Like seriously…

I mean this is one of the burning questions regarding the cross product and yet for some reason, textbooks don’t get to the bottom of this. One way to think about this is :

It is modeling a real life scenario!!

The scenario being :

When you try to twist a screw (clockwise screws being the convention) inside a block in the clockwise direction like so, the nail moves down and vice versa.

i.e When you move from the screw from u to v, then the direction of the cross product denotes the direction the screw will move.

That’s why the direction of the cross product is orthogonal. It’s really that simple!

## Another perspective

Now that you get a physical feel for the direction of the cross product, there is another way of looking at the direction too:

Displacement is a vector. Velocity is a vector. Acceleration is a vector. As you might expect, angular displacement, angular velocity, and angular acceleration are all vectors, too.

But which way do they point ?

Let’s take a rolling tire. The velocity vector of every point in the tire is pointed in every other direction. BUT every point on a rolling tire has to have the same angular velocity – Magnitude and Direction.

How can we possibly assign a direction to the angular velocity ?

Well, the only way to ensure that the direction of the angular velocity is the same for every point is to make the direction of the angular velocity perpendicular to the plane of the tire.
Problem solved!