Deriving Divergence in Cylindrical and Spherical coordinate systems.

Divergence of the vector field is an important operation in the study of Electromagnetics and we are well aware with its formulas in all the coordinate systems. Generally, we are familiar with the derivation of the Divergence formula in Cartesian coordinate system and remember its Cylindrical and Spherical versions intuitively. This article explains the step by step procedure for deriving the Deriving Divergence in Cylindrical and Spherical coordinate systems.

What is Divergence of vector field?

Divergence of a vector field is a measure of the “outgoingness” of the field at that point. If more and more field lines are sourcing out, coming out of the point then we say that there is a positive divergence. While if the field lines are sourcing in or contracting at a point then there is a negative divergence. The uniform vector field posses a zero divergence. Go through the following article for complete discussion of the Divergence of the vector field.

What is the Divergence of vector field?

Divergence formulas

Divergence formula in Cartesian

Divergence formula in Cylindrical

Divergence formula in Spherical

Derivation for the Divergence in Cartesian

The divergence formula in cartesian coordinate system can be derived from the basic definition of the divergence. Go through the following article for intuitive derivation.

The formulas of the Divergence with intuitive explanation!

Deriving Divergence in Cylindrical and Spherical

Let’s talk about getting the divergence formula in cylindrical first. Later by analogy you can work for the spherical coordinate system.

As read from above we can easily derive the divergence formula in Cartesian which is as below.

$\nabla\cdot\overrightarrow A=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}$

Now let me present the same in Cylindrical coordinates.

$\nabla\cdot\overrightarrow A=\frac1\rho\frac\partial{\partial\rho}\left(\rho A_\rho\right)+\frac1\rho\frac{\partial A_\phi}{\partial\phi}+\frac{\partial A_z}{\partial z}$

It is quite obvious to think that why some extra terms like (1/ρ) and ρ are present in first and second terms. Where do they come from? What is the logic behind them. Because thinking intuitively, one might expect the formula similar to cartesian one. We know, Cartesian is characterized by x, y and z while Cylindrical is defined by ρ, φ and z. So one can think of getting partial derivatives w.r.t. ρ, φ and z instead of x, y and z and A_ρ, A_φ and A_z instead of A_x, A_y and A_z. Right?

But it is not like that. The answer for this can be found in the steps for deriving the divergence in cylindrical system. Actually we can reach up to the result by two approaches.

Approach 1 for deriving the Divergence in Cylindrical

In this approach, you start with the divergence formula in Cartesian then convert each of its element into the cylindrical using proper conversion formulas. The partial derivatives with respect to x, y and z are converted into the ones with respect to ρ, φ and z. The x, y and z components of the vector are equivalently written in terms of ρ, φ and z components. But, for deriving Divergence in Cylindrical and Spherical, I am going to explain with another approach discussed below.

Approach 2 for deriving the Divergence in Cylindrical

We know that the divergence of the vector field is given as

$\nabla\cdot\overrightarrow A$

Here ∇ is the del operator and A is the vector field. If I take the del operator in cylindrical and dotted with A written in cylindrical then I would get the divergence formula in cylindrical coordinate system.

In cylindrical coordinates, any vector field is represented as follows:

$\overrightarrow A=A_\rho{\overrightarrow a}_\rho+A_\phi{\overrightarrow a}_\phi+A_z{\overrightarrow a}_z$

The Cylindrical del operator is as follows

$\nabla={\overrightarrow a}_\rho\frac\partial{\partial\rho}+{\overrightarrow a}_\phi\frac1\rho\frac\partial{\partial\phi}+{\overrightarrow a}_z\frac\partial{\partial z}$

Read here: – How to convert the Del operator from Cartesian to Cylindrical?

So let’s take the dot product.

$\nabla\cdot\overrightarrow A=\left({\overrightarrow a}_\rho\frac\partial{\partial\rho}+{\overrightarrow a}_\phi\frac1\rho\frac\partial{\partial\phi}+{\overrightarrow a}_z\frac\partial{\partial z}\right)\cdot\left(A_\rho{\overrightarrow a}_\rho+A_\phi{\overrightarrow a}_\phi+A_z{\overrightarrow a}_z\right)$

Be careful. This is not the simple dot product consisting of the multiplication of the respective components. You have to take an account of the derivative. For example let us consider the first term,

$\nabla\cdot\overrightarrow A=\left({\overrightarrow a}_\rho\frac\partial{\partial\rho}\right)\cdot\left(A_\rho{\overrightarrow a}_\rho+A_\phi{\overrightarrow a}_\phi+A_z{\overrightarrow a}_z\right)$

Let us put dot product outside and consider the derivative first.

$\nabla\cdot\overrightarrow A={\overrightarrow a}_\rho\cdot\left[\frac\partial{\partial\rho}\left(A_\rho{\overrightarrow a}_\rho+A_\phi{\overrightarrow a}_\phi+A_z{\overrightarrow a}_z\right)\right]$

Now consider the first derivative,

$\frac\partial{\partial\rho}\left(A_\rho{\overrightarrow a}_\rho\right)$

It is simplified using the product rule of the derivative.

$\frac\partial{\partial\rho}\left(A_\rho{\overrightarrow a}_\rho\right)\;=\frac{\partial A_\rho}{\partial\rho}{\overrightarrow a}_\rho+A_\rho\frac{\partial{\overrightarrow a}_\rho}{\partial\rho}$

Don’t assume that derivatives of the unit vectors are equal to zero or keep out of the derivation as a constant. No. Because cylindrical and spherical unit vectors are not universally constant. Though their magnitude is always 1, they can have different directions at different points of consideration. So unlike the cartesian these unit vectors are not global constants. I am going to cover the derivatives of the unit vectors in the independent article. But as if now, let us just use their results as follows:

$\frac\partial{\partial\rho}\left({\overrightarrow a}_\rho\right)\;=0;\;\frac\partial{\partial\rho}\left({\overrightarrow a}_\phi\right)=0;\;\frac\partial{\partial\rho}\left({\overrightarrow a}_z\right)=0$

$\frac\partial{\partial\phi}\left({\overrightarrow a}_\rho\right)\;={\overrightarrow a}_\phi;\;\frac\partial{\partial\phi}\left({\overrightarrow a}_\phi\right)=-{\overrightarrow a}_\rho;\;\frac\partial{\partial\phi}\left({\overrightarrow a}_z\right)=0$

$\frac\partial{\partial z}\left({\overrightarrow a}_\rho\right)\;=0;\;\frac\partial{\partial z}\left({\overrightarrow a}_\phi\right)=0;\;\frac\partial{\partial z}\left({\overrightarrow a}_z\right)=0$

So let us consider the complete terms together,

$={\overrightarrow a}_\rho\cdot\left[\left(A_\rho\frac{\partial{\overrightarrow a}_\rho}{\partial\rho}+{\overrightarrow a}_\rho\frac{\partial A_\rho}{\partial\rho}\right)+\left(A_\phi\frac{\partial{\overrightarrow a}_\phi}{\partial\rho}+{\overrightarrow a}_\phi\frac{\partial A_\phi}{\partial\rho}\right)+\left(A_z\frac{\partial{\overrightarrow a}_z}{\partial\rho}+{\overrightarrow a}_z\frac{\partial A_z}{\partial\rho}\right)\right]$

$+\left(\frac1\rho\right){\overrightarrow a}_\phi\cdot\left[\left(A_\rho\frac{\partial{\overrightarrow a}_\rho}{\partial\phi}+{\overrightarrow a}_\rho\frac{\partial A_\rho}{\partial\phi}\right)+\left(A_\phi\frac{\partial{\overrightarrow a}_\phi}{\partial\phi}+{\overrightarrow a}_\phi\frac{\partial A_\phi}{\partial\phi}\right)+\left(A_z\frac{\partial{\overrightarrow a}_z}{\partial\phi}+{\overrightarrow a}_z\frac{\partial A_z}{\partial\phi}\right)\right]$

$+{\overrightarrow a}_z\cdot\left[\left(A_\rho\frac{\partial{\overrightarrow a}_\rho}{\partial z}+{\overrightarrow a}_\rho\frac{\partial A_\rho}{\partial z}\right)+\left(A_\phi\frac{\partial{\overrightarrow a}_\phi}{\partial z}+{\overrightarrow a}_\phi\frac{\partial A_\phi}{\partial z}\right)+\left(A_z\frac{\partial{\overrightarrow a}_z}{\partial z}+{\overrightarrow a}_z\frac{\partial A_z}{\partial z}\right)\right]$

$={\overrightarrow a}_\rho\cdot\left[\left(A_\rho(0)+{\overrightarrow a}_\rho\frac{\partial A_\rho}{\partial\rho}\right)+\left(A_\phi(0)+{\overrightarrow a}_\phi\frac{\partial A_\phi}{\partial\rho}\right)+\left(A_z(0)+{\overrightarrow a}_z\frac{\partial A_z}{\partial\rho}\right)\right]$

$+\left(\frac1\rho\right){\overrightarrow a}_\phi\cdot\left[\left(A_\rho({\overrightarrow a}_\phi)+{\overrightarrow a}_\rho\frac{\partial A_\rho}{\partial\phi}\right)+\left(A_\phi(-{\overrightarrow a}_\rho)+{\overrightarrow a}_\phi\frac{\partial A_\phi}{\partial\phi}\right)+\left(A_z(0)+{\overrightarrow a}_z\frac{\partial A_z}{\partial\phi}\right)\right]$

$+{\overrightarrow a}_z\cdot\left[\left(A_\rho(0)+{\overrightarrow a}_\rho\frac{\partial A_\rho}{\partial z}\right)+\left(A_\phi(0)+{\overrightarrow a}_\phi\frac{\partial A_\phi}{\partial z}\right)+\left(A_z(0)+{\overrightarrow a}_z\frac{\partial A_z}{\partial z}\right)\right]$

Now, we know that,

${\overrightarrow a}_\rho\cdot{\overrightarrow a}_\rho={\overrightarrow a}_\phi\cdot{\overrightarrow a}_\phi={\overrightarrow a}_z\cdot{\overrightarrow a}_z=1$

${\overrightarrow a}_\rho\cdot{\overrightarrow a}_\phi={\overrightarrow a}_\phi\cdot{\overrightarrow a}_z={\overrightarrow a}_z\cdot{\overrightarrow a}_\rho=0$

$\therefore\nabla\cdot\overrightarrow A=\boxed{\frac{\partial A_\rho}{\partial\rho}+\frac{A_\rho}\rho}+\frac1\rho\frac{\partial A_\phi}{\partial\phi}+\frac{\partial A_z}{\partial z}$

The highlighted terms can be rewritten for consistency as follows

$\nabla\cdot\overrightarrow A=\frac1\rho\frac\partial{\partial\rho}\left(\rho A_\rho\right)+\frac1\rho\frac{\partial A_\phi}{\partial\phi}+\frac{\partial A_z}{\partial z}$

Which is our required divergence operator in cylindrical. Similar steps can be followed for deriving the Divergence in Spherical.

Suggested Community: Electromagnetics for GATE & ESE

44 Shares

Share42

Pin2