What is the divergence of a unit vector not in the r direction?


by SiggyYo
Tags: direction, divergence, unit, vector
SiggyYo
SiggyYo is offline
#1
Feb1-13, 01:44 PM
P: 5
Hi guys,

I've run across a problem. In finding the potential energy between two electrical quadrupoles, I've come across the expression for the energy as follows:

[itex]U_{Q}=\frac{3Q_{0}}{4r^{4}}\left[(\hat{k}\cdot \nabla)(5(\hat{k}\cdot \hat{r})^3-2(\hat{k}\cdot \hat{r})^2-(\hat{k}\cdot \hat{r}))\right],[/itex]

where [itex]\hat{k}[/itex] is the orientation of the quadrupoles, and [itex]\hat{r}[/itex] is the direction between the quadrupoles.

If I let [itex]\hat{r}[/itex] be in the [itex]\hat{z}[/itex]-direction, I get

[itex]U_{Q}=\frac{3Q_{0}}{4r^{4}}\left[(\hat{k}\cdot \nabla)(5(\cos{\theta})^3-2(\cos{\theta})^2-(\cos{\theta}))\right].[/itex]

My problem now is, that I don't know what to do about the divergence of the [itex]\hat{k}[/itex]-vector. I would like to do the differentiation in cartesian coordinates, but have them translated into spherical polar coordinates. I know, that the result should probably involve a [itex]\frac{1}{r}[/itex]-factor, but I can't seem to do it right. I've tried to rewrite [itex]\hat{k}[/itex] in polar coordinates and tried using the chain rule on the derivative, but I get 3 as an answer. So I don't know if the initial expression is wrong, or I just dont know how to take the derivative. Can anyone please help?

Thanks,
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
chiro
chiro is offline
#2
Feb1-13, 10:30 PM
P: 4,570
Hey SiggyYo.

Have you tried representing a transformation between spherical and cartesian?

(Example: for (r,theta) -> (x,y) we have y/x = arctan(theta) and x^2 + y^2 = r^2 which can be used to get (x,y)).
SiggyYo
SiggyYo is offline
#3
Feb2-13, 10:47 AM
P: 5
Thank you chiro for the quick response.

I am afraid I don't know what you mean. Wouldn't I just obtain the usual
[itex]x=r\sin{\theta}\cos{\phi}[/itex]
[itex]y=r\sin{\theta}\sin{\phi}[/itex]
[itex]z=r\cos{\theta}[/itex]?

Also, I want [itex]\hat{k}[/itex] to be a unit vector, which gives me [itex]r=1[/itex]. How do I take this into account, when trying to get a result with a factor of [itex]\frac{1}{r}[/itex]? I am really lost on this one :P

chiro
chiro is offline
#4
Feb2-13, 08:26 PM
P: 4,570

What is the divergence of a unit vector not in the r direction?


If k is a unit vector, then I don't think you will have any extra terms.

I'm not really sure what you are doing or trying to say: you have a conversion from polar to R^3 and provided the formula is correct, you should be able to plug these definitions in.

Also is the r term in your equation related to some vector in polar or is it some other variable?


Register to reply

Related Discussions
Is it valid to subtract a position vector of direction E with one of direction W? General Physics 6
unit vector in direction of max increase of f(x,y,z) Calculus & Beyond Homework 4
Vector Calc Homework Help! Divergence Free Vector Fields Calculus & Beyond Homework 0
Divergence and Curl of Unit Vectors? Calculus & Beyond Homework 2
Check Divergence Theorem on Unit Cube Calculus & Beyond Homework 2