Del operator with coordinate transformations

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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?
 

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  • #2
arildno
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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:
[tex]dz, rd\theta, dr[/tex]

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:
[tex]\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}}[/tex]
where the vectors are the normal vectors, whereas [itex]\partial{l}_{1}[/itex] denotes the associated length element.
Thus, for cylindrical coordinates, we have:
[tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+ \vec{i}_{z}\frac{\partial}{\partial{z}}[/tex]

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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  • #3
HallsofIvy
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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
[itex]\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}[/itex]
in Cartesian coordinates.

In cylindrical coordinates,
[tex]\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}[/tex]

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

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

Here, I have left the vector in the [itex]\vec{i}[/itex], [itex]\vec{j}[/itex], and [itex]\vec{k}[/itex] basis. Arildno put them in terms of vectors in the "r" and "[itex]\theta[/itex]" directions.
 
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  • #4
arildno
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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.
 
  • #5
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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 [itex]\vec{i}[/itex], [itex]\vec{j}[/itex], and [itex]\vec{k}[/itex] basis. Arildno put them in terms of vectors in the "r" and "[itex]\theta[/itex]" directions.

I'm not following this part. How do you switch between the two bases?
 
  • #6
arildno
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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:
[tex]\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]

Inverting this, we get the useful relations:
[tex]\vec{i}=\cos\theta\vec{i}_{r}-\sin\theta\vec{i}_{\theta}[/tex]
[tex]\vec{j}=\sin\theta\vec{i}_{r}+\cos\theta\vec{i}_{\theta}[/tex]
 
  • #7
arildno
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Correct.

We have:
[tex]\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]

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

In accordance with HallsofIvy's post, we therefore get:
[tex]\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})[/tex]
Substituting, we get:
[tex]=(\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})[/tex]
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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  • #8
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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.
 

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