Conservative force in spherical coordinates

Click For Summary

Homework Help Overview

The discussion revolves around determining whether a given force vector in spherical coordinates is conservative. The force is expressed in terms of its spherical components, and there is a focus on the relationship between spherical and Cartesian coordinates.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of curl to assess the conservativeness of the force. There is a discussion about converting between spherical and Cartesian coordinates and the implications of using the chain rule in differentiation.

Discussion Status

The conversation includes attempts to clarify the correct representation of the force components and the appropriate method for calculating the curl. Some participants express confusion over the notation and conventions used, while others provide insights into the necessary mathematical processes.

Contextual Notes

There is mention of unconventional notation in the original poster's reference material, which may affect understanding and communication among participants. The discussion also highlights the importance of correctly identifying the components of the force vector in different coordinate systems.

Felipe Lincoln
Gold Member
Messages
99
Reaction score
11

Homework Statement


Is ##F=(F_r, F_\theta, F_\varphi)## a conservative force?
##F_r=ar\sin\theta\sin\varphi##
##F_\theta=ar\cos\theta\sin\varphi##
##F_\varphi=ar\cos\varphi##

Homework Equations


##\nabla\times F=0##

The Attempt at a Solution


In this case we have to use the curl for spherical coordinates, but since it's clear that every component of this force can be described by cartesian coordinates I'm wondering if there's a way of doing the curl of cartesian coordinates, can we do this conversion ?
 
Physics news on Phys.org
What you have listed is not the spherical components of the vector, but the representation of the cartesian components using the spherical angles. For example ##F_x=ar \sin\theta \cos \phi##, etc. So to find the curl, just take the appropriate derivatives and be sure to apply the chain rule of differentiation.

On edit: It looks like you have swapped the symbols and used ##\varphi## for the polar angle and ##\theta## for the azimuthal angle.
 
  • Like
Likes   Reactions: Charles Link and Felipe Lincoln
Right!
But why should I use the chain rule? I can just do some substitution like ##x=r\sin\varphi\cos\theta##
And yes, I swapped the symbols, I'm used to do this way.
Thank you for the answer
 
Wait, I din't understand why you can say that ##F_x=ar\sin\varphi\cos\theta## (please use the symbols I introduced, it's my book convension). What the problem gave me ins't the radial componente of the force, in the case of ##F_r##?
 
What I'm saying is that the same vector ##\vec F## is written in spherical coordinates as
##\vec F = F_r \hat r~+~F_{\varphi}\hat {\varphi}~+~F_{\theta}\hat {\theta}##
and in Cartesian coordinates as
##\vec F = F_x ~\hat x~+~F_{y}~\hat {y}~+~F_{z}~\hat {z}##
the last expression can be written in terms of the spherical angles as
##\vec F = F \sin\varphi \cos \theta~\hat x~+~F \sin\varphi \sin\theta~\hat {y}~+~F \cos\varphi~\hat {z}##
where ##F=\sqrt{F_x^2+F_y^2+F_z^2}=ar##.
I assumed you wanted to calculate the curl in the Cartesian representation, in which case you would have to find terms like ##\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}## which is where the chain rule comes in if you use the expressions of ##F_x## and ##F_y## that involve the spherical angles.

Note: Your book's convention is unconventional. Bear that in might if you use other books in the future.
 
  • Like
Likes   Reactions: Charles Link and Felipe Lincoln
I got it.
Thank you very much !
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
Replies
1
Views
1K