Question on Divergence in Cylindrical Coords

In summary, the textbook equation states that the vector product between two vectors is the sum of their components multiplied by the angle between them. When working with Cartesian coordinates, the dot product of two vectors is simply the product of their components.
  • #1
Spectre5
182
0
Hey, I have a question on a derivation

The following is in my textbook (V = vector):

[tex] \nabla \cdot V = \frac {1}{r} \frac {\partial{(rV_{r}})}{\partial{r}} + \frac {1}{r} \frac {\partial{V_{\theta}}}{\partial{\theta}} + \frac {\partial{V_{z}}}{\partial{z}}[/tex]

where:

[tex] \nabla = \hat {r} \frac {\partial}{\partial {r}} + \hat {\theta} \frac {1}{r} \frac {\partial}{\partial {\theta}} + \hat {k} \frac {\partial}{\partial {z}} [/tex]

and

[tex] V = \hat {r}V_{r} + \hat {\theta}V_{\theta} + \hat {k}V_{z} [/tex]

This is, obviously, in cylindrical coordinates.

However, I would expect the result to be as follows:

[tex]\nabla \cdot V = \frac {\partial{V_{r}}}{\partial{r}} + \frac {1}{r} \frac {\partial{V_{\theta}}}{\partial{\theta}} + \frac {\partial{V_{z}}}{\partial{z}}[/tex]

Where did I go wrong? The second two terms I get the same thing, but I am confused on where that first term comes from in the given formula. Thanks in advance.
 
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  • #2
Remember that the radial and angular unit VECTORS are also functions of the angular variable. That's where your mistake lies..
 
  • #3
Don't we just multiply the components in the r, theta, and z direction and add the results just like we do in the x, y, z space?
 
  • #4
I do know that

[tex] \frac {\partial{\hat {r}}}{\partial{\theta}} = \hat {\theta} [/tex]

and

[tex] \frac {\partial{\hat {\theta}}}{\partial{\theta}} = - \hat {r} [/tex]

I would assume I need these, but I have no idea where to use them or exactly why.
 
  • #5
Nope, here's one way of doing this:
In a slightly different notation than yours, 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}},\vec{V}=v_{r}\vec{i}_{r}+v_{\theta}\vec{i}_{\theta}+v_{z}\vec{i}_{z}[/tex]
Furthermore, we have the orthogonality relations:
[tex]\vec{i}_{r}\cdot\vec{i}_{\theta}=\vec{i}_{r}\cdot\vec{i}_{z}=\vec{i}_{\theta}\cdot\vec{i}_{z}=0[/tex]
Along with the differential relations:
[tex]\frac{\partial\vec{i}_{r}}{\partial{r}}=\frac{\partial\vec{i}_{\theta}}{\partial{r}}=\frac{\partial\vec{i}_{z}}{\partial{r}}=\frac{\partial\vec{i}_{r}}{\partial{z}}=\frac{\partial\vec{i}_{\theta}}{\partial{z}}=\frac{\partial\vec{i}_{z}}{\partial{z}}=\vec{0}[/tex]
[tex]\frac{\partial\vec{i}_{r}}{\partial\theta}=\vec{i}_{\theta},\frac{\partial\vec{i}_{\theta}}{\partial\theta}=-\vec{i}_{r},\frac{\partial\vec{i}_{z}}{\partial\theta}=\vec{0}[/tex]
Thus:
[tex]\nabla\cdot\vec{V}=\vec{i}_{r}\cdot\frac{\partial\vec{V}}{\partial{r}}+\vec{i}_{\theta}\cdot\frac{\partial\vec{V}}{r\partial\theta}+\vec{i}_{z}\cdot\frac{\partial\vec{V}}{\partial{z}}[/tex]
All that remains, is now to differentiate properly, calculate the various dot products appearing in your expression, and simplify...
 
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  • #6
Why do those two have that special relation but all of the other ones are zero?

Also...why are you including the unit vectors in the dot product? I thought you just multiply the parts in the same direction and add them?


I thought the dot product was defined as:

a (dot) b = (a_x)(b_x) + (a_y)(b_y) + (a_z)(b_z) (no unit vectors here, so why are there unit vectors with cylindrical)
 
  • #7
No,no,the dot product is what it is:a contracted tensor product between the 2 vectors.Vectors should be expressed in a (preferably orthonormal) basis,so that the computations should be easier...

[tex] \vec{a}\cdot\vec{b}=\left(a_{x}\vec{i}+...+a_{z}\vec{k}\right)\cdot\left(b_{x}\vec{i}+...+b_{z}\vec{k}\right) [/tex]

Daniel.
 
  • #8
"Why do those two have that special relation but all of the other ones are zero?"
Because none of the unit vectors depend on "r" or "z", whereas only the radial and angular unit vectors depend on the angle.


Well, look at the Cartesian case (i.e, with [tex]\vec{V}=\vec{V}(x,y,z)[/tex]):
You can perfectly well write that in the same manner:
[tex]\nabla\cdot\vec{V}=\vec{i}_{x}\cdot\frac{\partial{\vec{V}}}{\partial{x}}+\vec{i}_{y}\cdot\frac{\partial\vec{V}}{\partial{y}}+\vec{i}_{z}\cdot\frac{\partial\vec{V}}{\partial{z}}[/tex]
We now have:
[tex]\frac{\partial\vec{V}}{\partial{x}}=\frac{\partial{v}_{x}\vec{i}_{x}}{\partial{x}}+\frac{\partial{v}_{y}\vec{i}_{y}}{\partial{x}}+\frac{\partial{v}_{z}\vec{i}_{z}}{\partial{x}}[/tex]
Now, observe the following in the Cartesian case:
1) None of the unit vectors depends on "x", that is, we have for example:
[tex]\frac{\partial{v}_{x}\vec{i}_{x}}{\partial{x}}=\frac{\partial{v}_{x}}{\partial{x}}\vec{i}_{x}[/tex]
2) [tex]\vec{i}_{x}\cdot\vec{i}_{y}=\vec{i}_{x}\cdot\vec{i}_{z}=0,\vec{i}_{x}\cdot\vec{i}_{x}=1[/tex]
Thus, we have the following simplification:
[tex]\vec{i}_{x}\cdot\frac{\partial\vec{V}}{\partial{x}}=\frac{\partial{v}_{x}}{\partial{x}}[/tex]
Similar relations hold for the y-and z derivations; thus, we regain the familiar expression:
[tex]\nabla\cdot\vec{V}=\frac{\partial{v}_{x}}{\partial{x}}+\frac{\partial{v}_{y}}{\partial{y}}+\frac{\partial{v}_{z}}{\partial{z}}[/tex]
 
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  • #9
ok, ok...so I'm an idiot...I think I got it now, thanks for all the help!

One last thing though, why do the r and theta vectors depend on theta whereas all the others don't?

Obviously z would not depend on theta, and obviously theta would depend on the angle theta, but why does r depend on theta. I wouldn't think that the distance you move outwards would depend on the angle you are at?
 
  • #10
"r" does not depend on the angle, but THE RADIAL UNIT VECTOR does!
 
  • #11
The unit vector along "r" does,since it's a vector.Its modulus doesn't change as a function of angle,but its direction does...

EDIT:Mine,too.Arildno,take it easy.The shock wave that came to Belgium with the sound of your scream almost broke my window...

Daniel.
 
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  • #12
Sure, but the "outward" direction at one angle value is not PARALLELL to the "outward" direction at another angle value..

EDIT: I see a post got deleted here.
My post is an answer to that deleted post.
 
  • #13
Isn't the unit r vector just in the outward direction from the z axis though? I realize that in x-y-z coordinates that the r vector would change with changing theta, but cylindrical coords are different

EDIT: No, you just looked like a mind reader because it was a response to this post (which was deleted then reposted with different wording) :smile:
 
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  • #14
errr...um ok...I guess i'll just have to try thinking about it some more...anyways thanks for the help guys!
 

Related to Question on Divergence in Cylindrical Coords

1. What is divergence in cylindrical coordinates?

Divergence is a mathematical operation that measures the flow of a vector field through a given point. In cylindrical coordinates, it represents the amount of fluid flowing in or out of a point in a cylindrical coordinate system.

2. How is divergence calculated in cylindrical coordinates?

In cylindrical coordinates, the divergence can be calculated using the following formula:

∇ · F = (1/ρ) ∂(ρFρ)/∂ρ + (1/ρ) ∂Fφ/∂φ + ∂Fz/∂z

where ∇ · F represents the divergence, ρ represents the radial distance, φ represents the angle, and z represents the height coordinate.

3. What is the physical interpretation of divergence in cylindrical coordinates?

The physical interpretation of divergence in cylindrical coordinates is the rate at which a fluid is either expanding or contracting at a given point. A positive divergence indicates fluid is flowing away from a point, while a negative divergence indicates fluid is flowing towards a point.

4. Can the divergence be negative in cylindrical coordinates?

Yes, the divergence can be negative in cylindrical coordinates. This indicates that the fluid is flowing towards a point, causing it to contract.

5. What are some applications of divergence in cylindrical coordinates?

Divergence in cylindrical coordinates has various applications in physics and engineering. It is commonly used in fluid mechanics to model fluid flow in cylindrical systems, such as pipes or turbines. It is also used in electromagnetic theory to understand the behavior of electric and magnetic fields in cylindrical systems. Additionally, divergence is used in computer graphics to simulate smoke or fluid movements in cylindrical shapes.

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