Force field in spherical polar coordinates

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SUMMARY

The discussion focuses on determining whether a given force field is conservative by analyzing it in spherical polar coordinates. Participants confirm that since the curl of the force field, ##\nabla \times \vec F##, equals zero, a potential exists. They explore the integration of the differential path element ##d\vec{\lambda} = \hat r dr + r \hat \theta d \theta + r \sin \theta \hat \phi d \phi## along a unit circle, concluding that if the circle is centered at the origin, both ##dr## and ##d\theta## are zero, leading to a final answer of zero for the integral.

PREREQUISITES
  • Spherical polar coordinates
  • Vector calculus, specifically curl and line integrals
  • Stokes' Theorem
  • Understanding of conservative force fields
NEXT STEPS
  • Study Stokes' Theorem and its applications in vector calculus
  • Learn about conservative vector fields and potential functions
  • Explore the parametrization of curves in spherical coordinates
  • Review dipole fields and their characteristics in physics
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism or advanced calculus, will benefit from this discussion. It is especially relevant for those tackling problems involving conservative force fields and spherical coordinate systems.

MatinSAR
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Homework Statement
A certain force field is given to me and I should do the following tasks to find out is it a conservative field or not.
Relevant Equations
pls see below.
Picture of question:
1702943544674.png

Part (a) : ##\nabla \times \vec F = 0## so a Potensial exists. I don't have problem with this part.
Part (b) : what I've done :
1702943893692.png

First experssion is 0 because ##\theta = \dfrac {\pi} {2}##. I don't know how to integrate over ##\theta ## when it is a constant.
 
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Go back to where you wrote correctly $$d\vec{\lambda} = \hat r dr + r \hat \theta d \theta + r \sin \theta \hat \phi d \phi$$
Simplify this for integrating along the given unit circle.
 
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The problem doesn't state whether or not the unit circle is centered at the origin of the coordinate system. So, I don't know if you are meant to assume that it is.
 
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TSny said:
Go back to where you wrote correctly $$d\vec{\lambda} = \hat r dr + r \hat \theta d \theta + r \sin \theta \hat \phi d \phi$$
Simplify this for integrating along the given unit circle.
Is it the only way? I haven't done this before and I cannot understand what you mean ...
TSny said:
The problem doesn't state whether or not the unit circle is centered at the origin of the coordinate system. So, I don't know if you are meant to assume that it is.
I guess I don't want to assume that. According to the book final answer should be 0.

1702945890697.png
 
MatinSAR said:
Is it the only way? I haven't done this before and I cannot understand what you mean ...

I guess I don't want to assume that. According to the book final answer should be 0.
Let's assume the unit circle is centered at the origin. For ##d\vec{\lambda}## along this circle, what can you say about the values of ##r##, ##dr##, ##d\theta## and ##\sin \theta##?

Thus, what does ##d\vec{\lambda}## simplify to?
 
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TSny said:
Let's assume the unit circle is centered at the origin. For ##d\vec{\lambda}## along this circle, what can you say about the values of ##r##, ##dr##, ##d\theta## and ##\sin \theta##?

Thus, what does ##d\vec{\lambda}## simplify to?
I have checked my book again yet I could not find sth similar to this. I guess:
##r=1##
##\sin \theta=1##
##dr=dr##
##d \theta= 0##

If I'm right the second expression should be 0 to and I will get 0 as final answer.
 
MatinSAR said:
I have checked my book again yet I could not find sth similar to this. I guess:
##r=1##
##\sin \theta=1##
##dr=dr##
##d \theta= 0##
OK, these look right. But you should be able to say more about ##dr##. Then you should be able to simplify the expression for ##d \vec \lambda## to a very simple result (which should make sense intuitively).

Then you can go on to think about the expression ##\vec F \cdot d\vec \lambda##
 
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TSny said:
OK, these look right. But you should be able to say more about ##dr##. Then you should be able to simplify the expression for ##d \vec \lambda## to a very simple result (which should make sense intuitively).

Then you can go on to think about the expression ##\vec F \cdot d\vec \lambda##
Sorry for taking your time ... Should ##dr## be 0 for the circle?

Can't I answer without ##dr##? The first expression was zero and ##dr## doesn't change anything.
 
MatinSAR said:
Sorry for taking your time ... Should ##dr## be 0 for the circle?
MatinSAR said:
Homework Statement: A certain force field is given to me and I should do the following tasks to find out is it a conservative field or not.
Relevant Equations: pls see below.

Picture of question:
View attachment 337396
Part (a) : ##\nabla \times \vec F = 0## so a Potensial exists. I don't have problem with this part.
Part (b) : what I've done :
View attachment 337397
First experssion is 0 because ##\theta = \dfrac {\pi} {2}##. I don't know how to integrate over ##\theta ## when it is a constant.
Is this from a published source?? If so, please identify.
 
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  • #10
hutchphd said:
Is this from a published source?? If so, please identify.
Yes.
This book is Arfken mathematical methods for physicists.
 
  • #11
MatinSAR said:
Should dr be 0 for the circle?
You tell us. What is the definition of dr?
 
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  • #12
Orodruin said:
You tell us. What is the definition of dr?
Radial spacing element If I have translated correctly.
Actually translating it to english is harder than its explanation for me.
 
  • #13
MatinSAR said:
Radial spacing element If I have translated correctly.
Actually translating it to english is harder than its explanation for me.
So how does radius change along the circle if r=1?
 
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  • #14
Orodruin said:
So how does radius change along the circle if r=1?
It doesn't change since radius is constant for a circle.
 
  • #15
Do you have access to Griffiths? Chapter 3.4.4. This field looks a lot like a dipole pointing in the z-direction. Just sayin’

That should answer part C.

Edit: A lot of dipole questions this week.
 
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  • #16
PhDeezNutz said:
Do you have access to Griffiths? Chapter 3.4.4. This field looks a lot like a dipole pointing in the z-direction. Just sayin’

That should answer part C.
Yes. I will check. Thanks.
 
  • #17
For part B can’t you use part A……via Stokes Theorem. Although I do think it is instructive to do the line integral directly.
 
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  • #18
MatinSAR said:
It doesn't change since radius is constant for a circle.
And therefore dr is …
 
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  • #19
PhDeezNutz said:
For part B can’t you use part A……via Stokes Theorem. Although I do think it is instructive to do the line integral directly.
Good idea!
Orodruin said:
And therefore dr is …
0 i think.
 
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  • #20
MatinSAR said:
0 i think.
Indeed.

You can also just use the parametrization ##\varphi = t## along with ##r=1## and ##\theta =\pi/2##. By definition
$$
dr = \frac{dr}{dt} dt
$$
and it should be pretty clear that ##dr/dt = 0##.
 
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