## Clarification on curl and divergence in cylindrical and spherical coordinates.

Divergence and Curl in cylindrical and spherical co are:

$$\nabla \cdot \vec E \;=\; \frac 1 r \frac {\partial r E_r}{\partial r} + \frac 1 r \frac {\partial E_{\phi}}{\partial \phi} + \frac {\partial E_z}{\partial z} \;=\; \frac 1 {R^2} \frac {\partial R^2 E_R}{\partial R} + \frac 1 {R\;sin \;\theta} \frac {\partial sin\;\theta E_{\theta}}{\partial \theta} + \frac 1 {R\;sin \;\theta}\frac {\partial E_{\phi}}{\partial \phi}$$

AND

$$\nabla \times \vec A \;=\; \frac 1 r \left|\begin{array}{ccc}\hat r & \hat {\phi} r & \hat z \\\frac {\partial}{\partial r} & \frac {\partial}{\partial \phi} & \frac {\partial}{\partial z} \\ A_r & A_{\phi} & A_z \end{array}\right|\;=\;\frac 1 {R^2 sin \theta} \left | \begin {array}{ccc} \hat R & \hat {\theta} R & \hat {\theta} R\;sin\;\theta \\\frac {\partial}{\partial R} & \frac {\partial}{\partial \theta} & \frac {\partial}{\partial \phi} \\ A_R & R\;A_{\theta} & R\;sin\theta\;A_{\phi} \end{array}\right|$$

Both divergence and curl are spatial derivative of a scalar and vector function at a PARTICULAR point in space respectively. My questions are about r and R in the equation:

1) Is r and R the magnitude from origin to the particular point where the divergence and curl is calculate. eg. If I want to calculate the divergence at the point (1,2,3), then $r=R=\sqrt{1^2+2^2+3^2} = \sqrt{14}$?

2) Also if we perform div and curb of a vector field C at a point pointed by a position vector $$\vec P = \left\langle 1,2,3\right\rangle$$ then still $\;r=R=\sqrt{1^2+2^2+3^2} = \sqrt{14}\;$ in calculating $\nabla \cdot \vec C \;\hbox { and }\; \nabla \times \vec C\;$.

Alan

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Hi Alan!
 Quote by yungman 1) Is r and R the magnitude from origin to the particular point where the divergence and curl is calculate … 2) Also if we perform div and curb of a vector field …
r and R (and θ) are the coordinates of the point …
R happens to be the distance from the origin,

but r of course is only the "horizontal" distance.

 Recognitions: Science Advisor If by (1,2,3) you mean (x,y,z), then r^2=x^2+y^2 only.

## Clarification on curl and divergence in cylindrical and spherical coordinates.

 Quote by clem If by (1,2,3) you mean (x,y,z), then r^2=x^2+y^2 only.

So you mean for cylindrical coordinates, $r=\sqrt{x^2+y^2}$ when used in

$$\nabla \times \vec A \;=\; \frac 1 r \left|\begin{array}{ccc}\hat r & \hat {\phi} r & \hat z \\\frac {\partial}{\partial r} & \frac {\partial}{\partial \phi} & \frac {\partial}{\partial z} \\ A_r & A_{\phi} & A_z \end{array}\right| \;=\; \frac 1 {\sqrt{x^2+y^2}} \left|\begin{array}{ccc}\hat r & \hat {\phi} \sqrt{x^2+y^2} & \hat z \\\frac {\partial}{\partial r} & \frac {\partial}{\partial \phi} & \frac {\partial}{\partial z} \\ A_r & A_{\phi} & A_z \end{array}\right|$$

 Quote by tiny-tim Hi Alan! r and R (and θ) are the coordinates of the point … R happens to be the distance from the origin, but r of course is only the "horizontal" distance.
Hi Tiny-Tim

I think I understand the r as shown in the post reply to Clem that $r=\sqrt{x^2+y^2}$.

Is my question 2) correct? Of cause except $r=\sqrt{x^2+y^2}$?

 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi Alan! Yes, r is just the coordinate of the point at which you're calculating.

 Quote by tiny-tim Hi Alan! Yes, r is just the coordinate of the point at which you're calculating.
Thanks for your help. I also gone through one round of calculation. I think it is best to avoid doing derivative of vector field in Cylindrical and particular Spherical coordinate if at all possible. They are tedious to put it politely. I just did $\nabla \times \vec A$ in cylindrical co. Took me a while to calculate, check and double checking for mistakes to get the right answer. All I used was A=<x,y,z>. It was just so easy to make a small mistake.

 Recognitions: Homework Help I think you get it now. Just to clarify, the radius in cylindrical co-ords is the distance from the z axis. And the radius in spherical co-ords is the distance from the origin. It is unfortunate that they are both given the name 'radius' in literature, since they mean different things. Its as confusing as calling both your children the same name. Also, yeah, using the non-cartesian co-ord systems take aaages, unless some of the derivatives disappear.
 I have like 8 books in EM and ED, they really don't explain the coordinate system very well. Maybe the professor explain better in class, but I am a self studier and it has been very confusing. I have 3 books in Vector Calculus ( not the multi-variables of the third semester calculus) and they don't even talk about the Cylindrical and Spherical coordinates. The only book that touch lightly on this is in the PDE books.......just very lightly!!! Yes, I finally get $r=\sqrt{x^2 + y^2}\;,\; R =\sqrt{x^2+y^2+z^2}$ in cylindrical and spherical coordinates resp. Again thanks for all the help. Alan
 Recognitions: Homework Help I guess that a full explanation of different coordinate systems will be in your third semester calculus on multi-variables. A quick summary of non-cartesian coord systems: The curl, divergence and grad in other coord systems comes from transforming the cartesian partial differentials into a different coord system. The position vector can be transformed to a different coordinate system by making appropriate substitutions to get the position vector in terms of the new variables (and new unit vectors). The velocity in any coordinate system is defined as the rate of change of the position vector with time. In the cartesian system, this is simple, since the unit vectors don't depend on time. But in other coordinate systems, the unit vectors do depend on time, so the velocity is more complicated. Once you know all this stuff, then you've got a firm footing on the use of general coordinate systems.
 I never went to class on the multi-variables calculus. But I studied very thoroughly on this subject through 3 different books used in college and I communicated with some professors. They don't even cover coordinates systems, they really don't cover Cylindrical and Spherical coordinates in the same sense as $\hat R, \hat{\theta}, \hat {\phi}$. They don't even have materials of divergence, curl and gradient in those coordinates at all. As I said, I have 3 books in Vector Calculus with is an upper division class and non go into these coordinates. From the stuff I understand so far, it is from combination of reading a few of the EM books and on line research to learn. It has been like putting pieces together!!!! There is definitely a gap there in my opinion. Thanks Alan
 Recognitions: Homework Help That's strange. Maybe you should suggest to your professors or head of department that they cover non-cartesian coordinate systems more thoroughly for future students.
 Recognitions: Science Advisor Unfortunately my own manuscript on mathematical methods for physicists is written in German. There, all this is covered. You usually find this in any mathematics book on multi-dimensional calculus. The best quick introduction for physicists I know of is in vol. 2 of Sommerfeld's Lectures on Theoretical Physics (in my opinion the best physics textbook series about classical(=non-quantum) physics ever written). This is available in English: A. Sommerfeld, Lectures on Theoretical Physics, vol. 2 (Mechanics of deformable Bodies), Academic Press (1950)
 If you check all the calculus books used for the first three semesters calculus, you will not find equations like in post #1. They only talked about $r=\sqrt{x^2+y^2}\;,\; R=\sqrt{x^2+y^2+z^2}$ but still stay with rectangular coordinates. Nothing on gradient, divergence and curl in cylindrical and spherical coordinates. As I said I have books by Thomas and Finney, Anton, Stein, Warberg etc. Nothing like this. Even in Vector Calculus books by Barr, Lovric and even Advanced Calculus by R.Creighton Buck don't have anything like this. I have all these books in front of me!!! Even in engineering electromagnetics books don't get into this too deep. They really avoid using these or just keep it very simple. They get into phasor to avoid all the difficult calculus. Luckily I decided to study Griffiths over and that gets more into the calculus and coordinates. It is like continuation of the vector calculus books!!! And even at that, the explanation are very shot. I have to make up problems and experiment and then come here to verify.
 I remember liking Schey - Div, Grad, Curl and All That when I was first learning vector calculus. but i can't remember if it does a good job with the other coordinate systems. http://www.amazon.com/Div-Grad-Curl-.../dp/0393925161
 Recognitions: Science Advisor You could try Franklin "Classical Electromagnetism" which does coordinate systems in Chapter 4.

 Quote by qbert I remember liking Schey - Div, Grad, Curl and All That when I was first learning vector calculus. but i can't remember if it does a good job with the other coordinate systems. http://www.amazon.com/Div-Grad-Curl-.../dp/0393925161
I have this book, only the third edition. I just look through it quickly, IT ACTUALLY HAVE THESE!!! I have this book at these years, kept thinking because I got it really cheap used and never really gone through it.

Thanks for pointing this out. I don't understand all the popular books in calculus really don't get into these.

Alan