Divergence of a vector field in a spherical polar coordinate system

[Robin04]

Homework Statement

I have to calculate the partial derivative of an arctan function. I have started to calculate it but I wonder if there is any simpler form, because if the simplest solution is this complex then it would make my further calculation pretty painful...

Homework Equations

$$\beta = \arctan{(\frac{x^2 + y^2}{z^2})}$$

The Attempt at a Solution

$$\frac{\partial \beta}{\partial x} = \frac{\partial}{\partial x} (\arctan{(\frac{x^2 + y^2}{z^2})}) = \frac{1}{1+ (\frac{x^2 + y^2}{z^2})^2} \frac{\partial}{\partial x}(\frac{x^2 + y^2}{z^2}) = \frac{\frac{2x}{z^2}}{1+(\frac{x^2 + y^2}{z^2})^2} = \frac{2x}{z^2 + \frac{(x^2 + y^2)^2}{z^2}}$$

 

[Orodruin]

Please do not just post part of a problem. Give us the entire problem.

Your form is correct assuming that this is indeed what you want to compute.

 

[Robin04]

Please do not just post part of a problem. Give us the entire problem.

Your form is correct assuming that this is indeed what you want to compute.

Well, I'm trying to calculate the divergence of a vector field in a spherical polar coordinate system. This is quite a long calculation so that's why I didn't mention any details of it. I'm about to arrive at the final equation of the divergence which is full of complex derivatives so that's why I am looking for a simpler form of this arctan derivative.

So you're saying this is the simplest form I can get?

 

[Mark44]

$$\frac{\partial \beta}{\partial x} = \frac{\partial}{\partial x} (\arctan{(\frac{x^2 + y^2}{z^2})}) = \frac{1}{1+ (\frac{x^2 + y^2}{z^2})^2} \frac{\partial}{\partial x}(\frac{x^2 + y^2}{z^2}) = \frac{\frac{2}{z^2}}{1+(\frac{x^2 + y^2}{z^2})^2} = \frac{2x}{z^2 + \frac{(x^2 + y^2)^2}{z^2}}$$

You have a mistake in the next-to-last expression. The numerator should be $\frac {2x}{z^2}$. It looks like the following expression is correct though.
Also, it would be better to simplify your final expression so that you don't have a fraction in the denominator.

 

[Robin04]

You have a mistake in the next-to-last expression. The numerator should be $\frac {2x}{z^2}$. It looks like the following expression is correct though.
Also, it would be better to simplify your final expression so that you don't have a fraction in the denominator.

Ah, thank you. That's what I missed. :)

 

[Ray Vickson]

Homework Statement

I have to calculate the partial derivative of an arctan function. I have started to calculate it but I wonder if there is any simpler form, because if the simplest solution is this complex then it would make my further calculation pretty painful...

Homework Equations

$$\beta = \arctan{(\frac{x^2 + y^2}{z^2})}$$

The Attempt at a Solution

$$\frac{\partial \beta}{\partial x} = \frac{\partial}{\partial x} (\arctan{(\frac{x^2 + y^2}{z^2})}) = \frac{1}{1+ (\frac{x^2 + y^2}{z^2})^2} \frac{\partial}{\partial x}(\frac{x^2 + y^2}{z^2}) = \frac{\frac{2x}{z^2}}{1+(\frac{x^2 + y^2}{z^2})^2} = \frac{2x}{z^2 + \frac{(x^2 + y^2)^2}{z^2}}$$

Depending on what you want to do later, it may be better to re-write the result as
$$\partial \beta / \partial x = \frac{2 x z^2}{(x^2+y^2)^2 + z^4}$$

 

[Orodruin]

Well, I'm trying to calculate the divergence of a vector field in a spherical polar coordinate system. This is quite a long calculation so that's why I didn't mention any details of it.

This is exactly why you should at least give the full problem! How else are we going to know if your calculations are correct up to this point?

 

[Ray Vickson]

Well, I'm trying to calculate the divergence of a vector field in a spherical polar coordinate system. This is quite a long calculation so that's why I didn't mention any details of it. I'm about to arrive at the final equation of the divergence which is full of complex derivatives so that's why I am looking for a simpler form of this arctan derivative.

So you're saying this is the simplest form I can get?

Why don't you use the formulas for the divergence in spherical coordinates?

 

[Robin04]

This is exactly why you should at least give the full problem! How else are we going to know if your calculations are correct up to this point?

I'm sorry, I was misunderstandable. What I wanted to say is that the reason why I started this thread is only because of this little problem of simplifying the expression (which is already solved, thank you :) ). I don't want to bring my whole project here because I'm still working on it and I haven't encountered any other problems yet. I'm following a method my teacher told me and I want to continue and see where I can get with it. If I get stuck with it then I'll start a new thread where I'll describe the whole problem precisely. Thank you for your help, I really appreciate it! :)

 

[Robin04]

Why don't you use the formulas for the divergence in spherical coordinates?

Because I want to find it on my own. We're learning orthogonal curvilinear coordinate systems at uni and I'm practicing the method of finding these equations. I have the formula in my handbook so I can check in the end if I got it right. :)

 

[Orodruin]

I'm sorry, I was misunderstandable. What I wanted to say is that the reason why I started this thread is only because of this little problem of simplifying the expression

I still think you should post the entire problem. Nobody here will solve it for you, only give you hints and guidance where you need it. The reason I am telling you this is that your expression looks strange. I would expect such an expression if you had gone wrong at an earlier stage, but you are giving me and everyone else zero chance of checking this. Otherwise your original vector field would have to be rather funny and I think less likely to be given as an exercise. Again, you give us zero opportunity to check whether this is the case or not. To be honest, I find it quite annoying.

 

[Orodruin]

Also note this passage from the homework guidelines:

Reproduce the problem statement accurately.
Type the problem statement exactly as worded. If you're only asking about one part of a long problem it may not be necessary to type up the entire problem, but you need to ensure you've provided the proper context for the sub-problem. If you paraphrase or summarize, make sure you're not changing the meaning or omitting important information. It's very frustrating trying to help with a problem only to discover that critical information is missing.

As it is, you are not providing the proper context.

 

[Robin04]

I still think you should post the entire problem. Nobody here will solve it for you, only give you hints and guidance where you need it. The reason I am telling you this is that your expression looks strange. I would expect such an expression if you had gone wrong at an earlier stage, but you are giving me and everyone else zero chance of checking this. Otherwise your original vector field would have to be rather funny and I think less likely to be given as an exercise. Again, you give us zero opportunity to check whether this is the case or not. To be honest, I find it quite annoying.

Okay, if you insist, then I would appreciate your help. :) I'll try to explain what I'm doing.

I'm starting my physics BSc degree next september (I'm studying computer science at the moment) but I decided to attend a course which is for physics students in the second semester. I'm lacking quite some knowledge that they already learned in the first semester but I'm trying to understand as much as I can. The goal of the course is to give an introduction to advanced concepts of mathematics used in theoretical physics. The first lecture was about orthogonal curvilinear coordinate systems and we calculated the divergence of a vector field in a cylindrical polar coordinate system. I decided to practice it a bit in order to understand the concept better so I started to do the same in spherical polar coordinate system. The method we learned is not the simplest one, the lecturer said the next week we're going to learn a better method.

I uploaded the pictures of what I've done so far. I hope it's readable.

Little explanation:
Page 1:
- I definine the relationship between the cartesian and the spherical polar coordinates.
- I choose the bases (to represent vectors) $\vec{t_r}, \vec{t_\alpha}, \vec{t_\beta}$ and then I set their magnitude to $1$ and name them $\vec{e_r}, \vec{e_\alpha}, \vec{e_\beta}$
- I check if they're perpendicular to each other by their scalar product.
- I define my vector field $\vec{v}(\vec{r})$ with the components $v_x(x, y, z), v_y(x, y, z), v_z(x, y, z)$ and write the equation for the divergence.

Page 2:
- I define new unity vectors $\vec{E_x}, \vec{E_y}, \vec{E_z}$ that are aligned with the x,y,z axis.
- I represent a vector with both bases and then express the relationship of the components.
- The next problem is that I can't differentiate with respect to x, y, z as written in the divergence formula, I need to express them with the relationship between the cartesian and the spherical polar coordinates and that part with the f function is like a skeleton for how I want to do it and it gives me which derivatives to calculate. This is where I encountered with the arctan function.

Page 3:
- I continue calculating these derivatives.

Page 4:
I'm substituting everything in the divergence formula and it's waiting to be solved :D

 

[Orodruin]

I am sorry, but it is impossible to see what you write in those pages. Anyway, as I suspected, you have gone wrong long before the issue you posted in this thread. Based on what you are looking to do, the angle $\beta$ is the polar angle. It is not given by
$$
\tan(\beta) = \frac{x^2+y^2}{z^2}.
$$
Instead, I would suggest that you look back at your derivation for the coordinate expressions.

 

[Robin04]

I am sorry, but it is impossible to see what you write in those pages.

Is it because of the quality of the images or my handwriting style?

Anyway, as I suspected, you have gone wrong long before the issue you posted in this thread. Based on what you are looking to do, the angle β\beta is the polar angle. It is not given by

tan(β)=x2+y2z2.​

Oh, yes, you're right. Thank you! :)

 

[Orodruin]

Is it because of the quality of the images or my handwriting style?

Image quality. Because of it I really cannot say anything about your handwriting.

Oh, yes, you're right. Thank you! :)

So what do you get instead?

 

[Robin04]

Image quality. Because of it I really cannot say anything about your handwriting.

So what do you get instead?

I missed a square root, so $\beta = \arctan{\sqrt{\frac{x^2+y^2}{z^2}}}$
I haven't recalculated the derivatives yet, so in the images I still have the wrong formulas for $\beta$

 

[Orodruin]

I missed a square root, so $\beta = \arctan{\sqrt{\frac{x^2+y^2}{z^2}}}$

Indeed, or written a bit differently, $\beta = \arctan\left(\rho/z\right)$, where $\rho^2 = x^2 + y^2$. This should help you significantly in determining the derivative.