Spherical Del Operator | The Conversion from Cartesian to Spherical

# How to convert the Del operator from Cartesian to Spherical?

The Del Operator is useful in vector differentiation particularly for finding Gradient, Divergence, Curl etc. Let us obtain the expression for the Spherical Del Operator starting from Cartesian.

#### What is Del Operator?

It is significant in vector differentiation for finding Gradient, Divergence, Curl, Laplacian etc. In a cartesian coordinate system it is expressed as follows:-

You can notice that, it is a vector differential operator. It can be operated on a scalar or a vector field and depending on the operation the outcome can be a scalar or vector. For example, if it is operated on a scalar field, the operation is known as Gradient whose answer is a vector. For the vector fields, its application yields two operations namely Divergence and Curl. The outcome of the Divergence of a vector field is a scalar while that of Curl is a vector.

Now, Electromagnetics generally deals with three types of coordinate systems viz. Cartesian, Cylindrical and Spherical. So it is obvious to convert the above expression in Cartesian into Cylindrical and Spherical.

## Proof to get the Spherical Del Operator

Let me present the formula for the del operator in Cartesian Coordinate System which we are going to convert into another system.

And the same del operator in Spherical Coordinate System is as follows: –

$\nabla={\overrightarrow a}_{r}\frac\partial{\partial r}+{\overrightarrow a}_{\theta}\frac1r\frac\partial{\partial\theta}+{\overrightarrow a}_{\phi}\frac1{r\sin\theta}\frac\partial{\partial\phi}$

Let us work mathematically to prove that both are the same.

First of all, as we are trying to convert the formula from Cartesian to Spherical, let us recall the transformation formulas between these coordinate systems. We know that the Cartesian coordinate System is characterized by x, y and z while the Spherical Coordinate System is characterized by r, θ and φ. The conversion formulas are as follows:-

Have a look at the Cartesian Del Operator. To convert it into the spherical coordinates, we have to convert the variables of the partial derivatives. In other words, the Cartesian Del operator consists of the derivatives are with respect to x, y and z. But Spherical Del operator must consist of the derivatives with respect to r, θ and φ.

$\frac\partial{\partial x};\frac\partial{\partial y}\;and\;\frac\partial{\partial z}\Rightarrow in\;terms\;of\;\frac{\displaystyle\partial}{\displaystyle\partial r};\frac{\displaystyle\partial}{\displaystyle\partial\theta}\;and\;\frac{\displaystyle\partial}{\displaystyle\partial\phi}$

For that let us apply the basic rule of the differentiation called the chain rule.

$\frac\partial{\partial x}=\frac\partial{\partial r}\frac{\partial r}{\partial x}+\frac\partial{\partial\theta}\frac{\partial\theta}{\partial x}+\frac\partial{\partial\phi}\frac{\partial\phi}{\partial x}$

$\frac\partial{\partial y}=\frac\partial{\partial r}\frac{\partial r}{\partial y}+\frac\partial{\partial\theta}\frac{\partial\theta}{\partial y}+\frac\partial{\partial\phi}\frac{\partial\phi}{\partial y}$

$\frac\partial{\partial z}=\frac\partial{\partial r}\frac{\partial r}{\partial z}+\frac\partial{\partial\theta}\frac{\partial\theta}{\partial z}+\frac\partial{\partial\phi}\frac{\partial\phi}{\partial z}$

So let us calculate the required derivatives.

$\frac{\partial r}{\partial x}=\frac\partial{\partial x}\sqrt{x^2+y^2+z^2}=\frac{x}{\sqrt{x^2+y^2+z^2}}$

$\therefore\frac{\partial r}{\partial x}=\frac{r\sin\theta\cos\phi}r=\sin\theta\cos\phi$

Note the simplification in the above step. As we are going to convert into the Spherical coordinates from the Cartesian ones, we must simplify to the extent so that to get spherical variables. Similarly,

$\frac{\partial\theta}{\partial x}=\frac\partial{\partial x}\tan^{-1}\left(\frac{\sqrt{x^2+y^2}}z\right)=\frac1{1+{\displaystyle\frac{x^2+y^2}{z^2}}}\frac\partial{\partial x}\left(\frac{\displaystyle\sqrt{x^2+y^2}}{\displaystyle z}\right)$

$=\frac{z^2}{x^2+y^2}\frac1z\frac x{\sqrt{x^2+y^2}}=\frac{xz}{\left(x^2+y^2+z^2\right){\displaystyle\sqrt{x^2+y^2}}}$

$\therefore\frac{\partial\theta}{\partial x}=\frac{r\;\sin\theta\;\cos\phi\;r\;\cos\theta}{r\sqrt{r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi}}=\frac{r\sin\theta\cos\theta\cos\phi}{r\sin\theta}$

$\therefore\frac{\partial\theta}{\partial x} =\cos\theta\cos\phi$

$\frac{\partial\phi}{\partial x}=\frac\partial{\partial x}\left(\tan^{-1}\frac yx\right)=\frac1{1+\left({\displaystyle\frac yx}\right)^2}y\left(\frac{-1}{x^2}\right)=\frac{-y}{x^2+y^2}$

$\therefore\frac{\partial\phi}{\partial x}=-\frac{r\sin\theta\sin\phi}{r^2\sin^2\theta}=-\frac{\sin\phi}{r\sin\theta}$

Working on the similar lines, we can get following derivatives,

$\frac{\partial r}{\partial y}=\sin\theta\sin\phi;\;\frac{\partial\theta}{\partial y}=\cos\theta\sin\phi;\;\frac{\partial\phi}{\partial y}=\frac{\cos\phi}{r\sin\theta}$

$\frac{\partial r}{\partial z}=\cos\theta;\;\frac{\partial\theta}{\partial z}=\frac{-1}{r\sin\theta};\;\frac{\partial\phi}{\partial z}=0$

Now let us put everything together, i.e. unit vectors and derivatives in spherical coordinates as follows: –

$\nabla={\overrightarrow a}_x\frac\partial{\partial x}+{\overrightarrow a}_y\frac\partial{\partial y}+{\overrightarrow a}_z\frac\partial{\partial z}$

$=(\sin\theta\cos\phi{\overrightarrow a}_r+\cos\theta\cos\phi{\overrightarrow a}_\theta-\sin\phi{\overrightarrow a}_\phi)\left[\left(\frac\partial{\partial r}\right)\sin\theta\cos\phi+\left(\frac\partial{\partial\theta}\right)\cos\theta\cos\phi+\left(\frac\partial{\partial\phi}\right)\left(\frac{-\sin\phi}{r\sin\theta}\right)\right]$

$+(\sin\theta\sin\phi{\overrightarrow a}_r+\cos\theta\sin\phi{\overrightarrow a}_\theta+\cos\phi{\overrightarrow a}_\phi)\left[\left(\frac\partial{\partial r}\right)\sin\theta\sin\phi+\left(\frac\partial{\partial\theta}\right)\cos\theta\sin\phi+\left(\frac\partial{\partial\phi}\right)\left(\frac{\cos\phi}{r\sin\theta}\right)\right]$

$+(\cos\theta{\overrightarrow a}_r-\sin\theta{\overrightarrow a}_\theta)\left[\left(\frac\partial{\partial r}\right)\cos\theta+\left(\frac\partial{\partial\theta}\right)\left(\frac{-1}{r\sin\theta}\right)+\left(\frac\partial{\partial\phi}\right)0\right]$

For simplicity, initially collect the r derivative terms together as:

$\left(\frac\partial{\partial r}\right)\lbrack\left(\sin\theta\cos\phi\right)\left(\sin\theta\cos\phi{\overrightarrow a}_r+\cos\theta\cos\phi{\overrightarrow a}_\theta-\sin\phi{\overrightarrow a}_\phi\right)$

$+\left(\sin\theta\sin\phi\right)\left(\sin\theta\sin\phi{\overrightarrow a}_r+\cos\theta\sin\phi{\overrightarrow a}_\theta+\cos\phi{\overrightarrow a}_\phi\right)$

$+\left(\cos\theta\right)\left(\cos\theta{\overrightarrow a}_r-\sin\theta{\overrightarrow a}_\theta\right)\rbrack$

$=\lbrack\left(\sin^2\theta\cos^2\phi+\sin^2\theta\sin^2\phi+\cos^2\theta\right){\overrightarrow a}_r$

$+\left(\sin\theta\cos\theta\cos^2\phi+\sin\theta\cos\theta\sin^2\phi-\sin\theta\cos\theta\right){\overrightarrow a}_\theta$

$+\left(-\sin\theta\sin\phi\cos\phi+\sin\theta\sin\phi\cos\phi\right){\overrightarrow a}_\phi\rbrack\left(\frac\partial{\partial r}\right)$

$={\overrightarrow a}_r\frac\partial{\partial r}$

Similarly collecting the theta and phi derivative terms and working similarly we get,

$\nabla={\overrightarrow a}_{r}\frac\partial{\partial r}+{\overrightarrow a}_{\theta}\frac1r\frac\partial{\partial\theta}+{\overrightarrow a}_{\phi}\frac1{r\sin\theta}\frac\partial{\partial\phi}$

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