# Del operator with coordinate transformations

• Curl
In summary, the del operator after a change of variables can be expressed using the chain rule for partial derivatives. In cylindrical coordinates, the del operator has components in the "r" and "\theta" directions, which can be derived using the chain rule and the relationships between the Cartesian and cylindrical unit vectors.
Curl
How can you express the del operator after a change of variables? For example, if I want to use cylindrical coordinates for a fluids problem, what is the del operator in terms of the new coordinates?

And how do you derive it for any other arbitrary coordinate transforms?

1. A (3-D) set of coordinates can be said to specify three types of basic surfaces, each associated with holding one of the variables constant.
For example, for cylindrical coordinates, z=constant are planes parallell to the xy-plane, theta=constant are half-planes (ending in the line x=y=0) , whereas r=constant specifies a cylindrical shell.

2. To each of these surfaces, there will be, at every point, an associated normal vector, and the proper length element associated with a(n infinitesemal) change in a particular variable.
For the cylindrical coordinates, those length elements are:
$$dz, rd\theta, dr$$

dz is the length element by changing from one z=constant plane to the "next", rdtheta the arc length you traverse between two half-planes each specified by a constant theta, and dr moving from one cylindrical shell to the next.

3. For an arbitrary coordinate system, our del-operator will look like:
$$\nabla=\vec{i}_{1}\frac{\partial}{\partial{l}_{1}}+\vec{i}_{2}\frac{\partial}{\partial{l}_{2}}+ \vec{i}_{3}\frac{\partial}{\partial{l}_{3}}$$
where the vectors are the normal vectors, whereas $\partial{l}_{1}$ denotes the associated length element.
Thus, for cylindrical coordinates, we have:
$$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+ \vec{i}_{z}\frac{\partial}{\partial{z}}$$

4. When using the del operator, it will often be necessary to remember that in non-Cartesian coordinate systems, the normal vectors will be non-constant functions of the variables themselves.

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Curl said:
How can you express the del operator after a change of variables? For example, if I want to use cylindrical coordinates for a fluids problem, what is the del operator in terms of the new coordinates?

And how do you derive it for any other arbitrary coordinate transforms?

Use the chain rule. Given a function f(x,y,z), the "del" or "grad" is
$\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$
in Cartesian coordinates.

In cylindrical coordinates,
$$\frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}$$

$r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2}$ so $\frac{\partial r}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)$$= \frac{x}{\sqrt{x^2+ y^2}}= \frac{r cos(\theta)}{r}= cos(\theta)$
$\theta= arctan(\frac{y}{x})$ so $\frac{\partial \theta}{\partial x}= \frac{1}{1+ \frac{y^2}{x^2}}\frac{-y}{x^2}$$= \frac{-y}{x^2+ y^2}= \frac{-r sin(\theta)}{r^2}= -\frac{1}{r}sin(\theta)$.
Of course, since x, y, and z are independent variables, of course $\frac{\partial z}{\partial x}= 0$.

So $\frac{\partial f}{\partial x}= cos(\theta)\frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}$
and similarly for $\frac{\partial f}{\partial y}$. $\frac{\partial f}{\partial z}$ is the same in cylidrical coordinates.

Here, I have left the vector in the $\vec{i}$, $\vec{j}$, and $\vec{k}$ basis. Arildno put them in terms of vectors in the "r" and "$\theta$" directions.

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HallsofIvy has given you a detailed derivation of a specific case of transformation.

In order to prove the general result (i.e, what I gave you without any proof), it would be necessary to make a generalization out of HallsofIvy's proof for the special case.

Okay, so you transform from F(x,y,z) to f(r,θ,z), and grad f is simply i ∂f/∂x + j ∂f/∂y + k ∂f/∂z, using chain rule to compute the partial derivatives?

Here, I have left the vector in the $\vec{i}$, $\vec{j}$, and $\vec{k}$ basis. Arildno put them in terms of vectors in the "r" and "$\theta$" directions.

I'm not following this part. How do you switch between the two bases?

Curl said:
Okay, so you transform from F(x,y,z) to f(r,θ,z), and grad f is simply i ∂f/∂x + j ∂f/∂y + k ∂f/∂z, using chain rule to compute the partial derivatives?
Correct.
I'm not following this part. How do you switch between the two bases?
We have:
$$\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$

Inverting this, we get the useful relations:
$$\vec{i}=\cos\theta\vec{i}_{r}-\sin\theta\vec{i}_{\theta}$$
$$\vec{j}=\sin\theta\vec{i}_{r}+\cos\theta\vec{i}_{\theta}$$

arildno said:
Correct.

We have:
$$\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$

Inverting this, we get the useful relations:
$$\vec{i}=\cos\theta\vec{i}_{r}-\sin\theta\vec{i}_{\theta}$$
$$\vec{j}=\sin\theta\vec{i}_{r}+\cos\theta\vec{i}_{\theta}$$

In accordance with HallsofIvy's post, we therefore get:
$$\frac{\partial{F}}{\partial{x}}\vec{i}+\frac{\partial{F}}{\partial{y}}\vec{j}=(\frac{\partial{f}}{\partial{r}}\frac{\partial{r}}{\partial{x}}+\frac{\partial{f}}{\partial\theta}\frac{\partial\theta}{\partial{x}})(\cos\theta\vec{i}_{r}-\sin\theta\vec{i}_{\theta})+(\frac{\partial{f}}{\partial{r}}\frac{\partial{r}}{\partial{y}}+\frac{\partial{f}}{\partial\theta}\frac{\partial\theta}{\partial{y}})(\sin\theta\vec{i}_{r}+\cos\theta\vec{i}_{\theta})$$
Substituting, we get:
$$=(\frac{\partial{f}}{\partial{r}}\cos\theta-\frac{\partial{f}}{\partial\theta}\frac{\sin\theta}{r})(\cos\theta\vec{i}_{r}-\sin\theta\vec{i}_{\theta})+(\frac{\partial{f}}{\partial{r}}\sin\theta+\frac{\partial{f}}{\partial\theta}\frac{\cos\theta}{r})(\sin\theta\vec{i}_{r}+\cos\theta\vec{i}_{\theta})$$
Simplify and rearrange that expression to get the planar components of the gradient in cylindrical co-ordinates (the vertical component is, of course, unchanged).

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Thanks a lot to both of you.

What I was confused about is switching bases in the general case. In cylindrical coordinates its easier to see it geometrically, but in the general case its not always obvious; so I was wondering if there is a technique for solving the problem with other transforms. Maybe its too difficult and I should stick to simpler problems, I don't know.

## 1. What is the "Del operator" in coordinate transformations?

The "Del operator" is a mathematical symbol, also known as the nabla symbol, that represents a vector differential operator used in vector calculus. It is commonly used to express the gradient, divergence, and curl of a vector field in three-dimensional Cartesian coordinates.

## 2. How is the Del operator used in coordinate transformations?

In coordinate transformations, the Del operator is used to express the partial derivatives of a vector field with respect to each of the coordinate variables. It allows for the transformation of vector fields from one coordinate system to another, making it easier to solve problems in different coordinate systems.

## 3. What is the relationship between the Del operator and the gradient?

The gradient is a vector that points in the direction of the steepest increase of a scalar field. The Del operator is used to represent the gradient, and its magnitude is equal to the rate of change of the scalar field in the direction of the gradient.

## 4. Can the Del operator be used in non-Cartesian coordinate systems?

Yes, the Del operator can be used in non-Cartesian coordinate systems, such as cylindrical and spherical coordinates. In these systems, the Del operator takes on a slightly different form, but it still represents the gradient, divergence, and curl of a vector field in that coordinate system.

## 5. Are there any important properties of the Del operator in coordinate transformations?

Yes, there are several important properties of the Del operator in coordinate transformations. These include the chain rule, product rule, and quotient rule, which allow for the calculation of derivatives of vector fields in different coordinate systems. Additionally, the Del operator follows the rules of vector calculus, such as the commutative and associative properties.

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