Partial derivative in Spherical Coordinates

Click For Summary

Homework Help Overview

The discussion revolves around the concept of partial derivatives in both Cartesian and spherical coordinates, specifically questioning the validity of certain expressions for these derivatives.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the expression for the partial derivative in spherical coordinates can be represented as a simple sum of the partial derivatives with respect to each coordinate.

Discussion Status

There is a mix of confusion and clarification regarding the definitions and expressions for partial derivatives in different coordinate systems. Some participants are seeking verification of their understanding, while others are questioning the terminology used in the original post.

Contextual Notes

Participants note the distinction between simple partial derivatives and those related to divergence in vector fields, indicating a need for clarity in terminology and understanding of the concepts involved.

yungman
Messages
5,741
Reaction score
291
Is partial derivative of ##u(x,y,z)## equals to
[tex]\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}[/tex]
Is partial derivative of ##u(r,\theta,\phi)## in Spherical Coordinates equals to
[tex]\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}[/tex]

Thanks
 
Physics news on Phys.org
I'm slightly confused about the question, but if you're just changing co-ordinates and taking partials it looks fine.
 
Zondrina said:
I'm slightly confused about the question, but if you're just changing co-ordinates and taking partials it looks fine.

Thanks for the reply. I just want to verify partial derivative in TRUE Spherical Coordinates system is [tex]\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}[/tex]

Not the spherical amplitude representation of (x,y,z) where
[tex]\vec r=\hat x x +\hat y y+\hat z z\;\hbox { to which}\; x=r\cos\phi\sin\theta,\;y=r\sin\phi\sin\theta, \;z=r\cos\theta[/tex]
In the ordinary Calculus III class.

Thanks
 
You are just differentiating wrt different functions, any calculations done by either method should give you same end results.
 
Thanks
 
yungman said:
Is partial derivative of ##u(x,y,z)## equals to
[tex]\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}[/tex]
You need to get your terminology straight if you want people to understand you. The expression you wrote above is for the [strike]divergence[/strike] directional derivative of u in the (1,1,1) direction times the square root of 3. Each term is a partial derivative. So the answer to your question is no.

Is partial derivative of ##u(r,\theta,\phi)## in Spherical Coordinates equals to
[tex]\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}[/tex]
No.
 
Last edited:
This expression is not a divergence. The divergence is defined for a vector field. In Cartesian coordinates it reads
[tex]\vec{\nabla} \cdot \vec{V}=\frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z}.[/tex]
 
D'oh!
 
Yes, I know Divergence, Gradient and Laplace/Poisson equation in Spherical. Just the simple partial derivative.
 

Similar threads

Replies
4
Views
1K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
10
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K