Div an curl in different coordinate systems

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  • #1
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To calculate the divergence of a vectorfield in cartesian coordinates, you can think of it as a dot product, and to calculate the curl, you can think of it as a cross product. But how can you calculate the div and curl when you have spherical or cylindrical coordinates, without explicitely converting them to the cartesian system?

For example the radial unity vector e_r = (sin a * cos b, sin a * sin b, cos a)
(standard: a = Theta, b = phi) should affect the curl and div by more than the scaling factor (h)

P.S Is it possible to use tex when writing a post?
 
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  • #3
quasar987
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I think a good starting point is to define the nabla operator as the operator [itex]\nabla[/itex] such that

[tex]df=\nabla f\cdot d\vec{r}[/tex]

This allows us to recover easily the form of [itex]\nabla[/itex] in any coordinate system.
 
  • #4
quasar987
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For exemple, in cylindrical coordinate: Consider a function [itex]f = f(r,\theta,z)[/itex].Then

[tex]df = \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \theta}d\theta + \frac{\partial f}{\partial z }dz[/tex]

and

[tex]d\vec{r} = \hat{r}dr + \hat{\theta}rd\theta + \hat{z}dz[/tex]

So after messing around a bit you find that

[tex]df=\nabla f\cdot d\vec{r} \Leftrightarrow \nabla = \hat{r}\frac{\partial}{\partial r} + \frac{\hat{\theta}}{r}\frac{\partial}{\partial \theta}+ \hat{z}\frac{\partial}{\partial z }[/tex]
 
  • #5
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quasar987 said:
[tex]d\vec{r} = \hat{r}dr + \hat{\theta}rd\theta + \hat{z}dz[/tex]

So after messing around a bit you find that

[tex]df=\nabla f\cdot d\vec{r} \Leftrightarrow \nabla = \hat{r}\frac{\partial}{\partial r} + \frac{\hat{\theta}}{r}\frac{\partial}{\partial \theta}+ \hat{z}\frac{\partial}{\partial z }[/tex]
Hm, shouldn't it be
[tex]d\vec{r} = \hat{r}dr + \mathbf{\dfrac{\hat{\theta}}{r}}d\theta + \hat{z}dz[/tex]
if you divide with the scalefactor?

Otherwise the second equation would become
[tex]df=\nabla f\cdot d\vec{r} \Leftrightarrow \nabla = \hat{r}\frac{\partial}{\partial r} + \hat{\theta}r\frac{\partial}{\partial \theta}+ \hat{z}\frac{\partial}{\partial z }[/tex]

I'm still not sure if I've got it right, but I'll think over this, and return with more questions if they arise. Thanks for the help :)
 
  • #6
quasar987
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Sunshine said:
Hm, shouldn't it be
[tex]d\vec{r} = \hat{r}dr + \mathbf{\dfrac{\hat{\theta}}{r}}d\theta + \hat{z}dz[/tex]
if you divide with the scalefactor?
Mmh.. the position vector in cylindrical coordinate is just [itex]\vec{r}=r\hat{r}+z\hat{z}[/itex]. If you differentiate that with respect to an hypothetical variable t, you get

[tex]\frac{d\vec{r}}{dt} = \frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}+\frac{dz}{dt}\hat{z}+\z\frac{d\hat{z}}{dt} = \frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{d\theta}\frac{d\theta}{dt}+\frac{dz}{dt}\hat{z}+0 = \frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}+\frac{dz}{dt}\hat{z}[/tex]

And multiplying by dt gives [itex]d\vec{r}[/itex].

Sunshine said:
Otherwise the second equation would become
[tex]df=\nabla f\cdot d\vec{r} \Leftrightarrow \nabla = \hat{r}\frac{\partial}{\partial r} + \hat{\theta}r\frac{\partial}{\partial \theta}+ \hat{z}\frac{\partial}{\partial z }[/tex]
Better double check your calculations.
 
  • #7
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You're right, my mistake

However my biggest problem remains.

Let's look at the position vector
[tex]\vec{r} = (r cos \theta, r sin \theta, z)[/tex]

[tex]\frac{\partial \vec{r}}{\partial r} = (cos \theta, sin \theta, 0) [/tex]

[tex]h_r = \sqrt{cos^2 \theta + sin^2 \theta} = 1 [/tex]

and according to the definition

[tex]\hat{r} = \frac{1}{h_r} \frac{\partial \vec{r}}{\partial r} = (cos \theta, sin \theta, 0)[/tex]

the queston is, which of the equations below are correct

[tex] \nabla \cdot \hat{r} = \nabla \cdot (1,0,0)[/tex]
OR
[tex] \nabla \cdot \hat{r} = \nabla \cdot (cos \theta, sin \theta, 0)[/tex]

...if I use the [tex]\nabla[/tex] that you gave for cylindrical coordinates.
If it's #1, would #2 apply if the [tex]\nabla[/tex] was defined on another way, i.e not as

[tex] \nabla = \hat{r}\frac{\partial}{\partial r} + \hat{\theta}r\frac{\partial}{\partial \theta}+ \hat{z}\frac{\partial}{\partial z }[/tex]
 
  • #8
quasar987
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Both

[tex] \nabla \cdot \hat{r} = \nabla \cdot (1,0,0)[/tex]
AND
[tex] \nabla \cdot \hat{r} = \nabla \cdot (cos \theta, sin \theta, 0)[/tex]

are correct, if you notice the distinction that in #1, (1,0,0) means [itex](r=1,\theta=0,z=0)[/itex] (i.e. (1,0,0) represents the vector [itex]\hat{r} = 1\hat{r}+0\hat{\theta}+0\hat{z}[/itex]), while in #2, [itex](cos\theta,sin\theta,0)[/itex] means [itex](x=cos\theta,y=sin\theta,z=0) [/itex] (i.e. represents the vector [itex]\hat{r}=cos\theta \hat{x} + sin\theta \hat{y} + 0\hat{z}[/itex]

The base of [itex]\mathbb{R}^3[/itex] used in #1 is different than that used in #2. Therfor if you're gonna take the dot product of either of them with nabla, you gotta express nabla in the same base.
 
  • #9
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so you mean that
[tex] (\frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z }) \cdot (cos \theta, sin \theta, 0) = (\frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta},\frac{\partial}{\partial z }) \cdot (1,0,0)[/tex]

Well, I see that it's true in this case where both sides are zero but is it so for an arbitrary vector field?

And is curl similar?
 
  • #10
dextercioby
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Of course.Remember that the divergence of a vector field is a scalar wrt general coordinates transformations,so eveluating it in cartesian is no difference than eveluating it in cylindrical or spherical...:wink:

As for the curl,well,it gives a pseudotensor of a higher rank,so it's not invariant under general coordinate transformations...

Daniel.
 
  • #12
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I've turned those formulas upside down by now, and still very confused... :confused:

The thing is that I'm supposed to calculate [tex]\nabla \cdot \vec{e_x}[/tex] and [tex]\nabla \times \vec{e_x}[/tex] where [tex]\vec{e_x}[/tex] are the unit vectors with [tex]x = r, \rho, \theta, \phi[/tex] (cylindrical and spherical coordinates).
It means a total of 8 "questions" and all of them (!) becomes 0 if I use any of the equations in my last post (you always get 0 when differentiating a scalar)... that's why I've been suspecting that I'm supposed to use
[tex]\vec{e_r} = (cos \theta, sin \theta, 0)[/tex]

[tex]\vec{e_{\theta}} = (-sin \theta, cos \theta, 0)[/tex]

etc. and differentiate with respect to [tex]r, \theta, \phi ...[/tex]

But that isn't the case...?
 

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