Curl in cylindrical coordinates

  • #1

Main Question or Discussion Point

So, let me derive the curl in the cylindrical coordinate system so I can showcase what I get. Let ##x=p\cos\phi##, ##y=p\sin\phi## and ##z=z##. This gives us a line element of ##ds^2 = {dp}^2+p^2{d\phi}^2+{dz}^2## Given that this is an orthogonal coordinate system, our gradient is then ##\nabla = d = \partial_{p} dp + \frac{1}{p}\partial_{\phi} d\phi + \partial_{z} dz## and ##V= V_p dp + V_{\phi} d\phi + V_z dz## Thus, the curl is then ##\nabla \times V = \star dV## So, going through with this, we see that $$dV = \frac{\partial V_{\phi}}{\partial p} dp \wedge d\phi + \frac{\partial V_z}{\partial p}dp \wedge dz + \frac{1}{p}\frac{\partial V_p}{\partial \phi}d\phi \wedge dp + \frac{1}{p}\frac{\partial V_z}{\partial \phi}d\phi \wedge dz + \frac{\partial V_p}{\partial z}dz \wedge dp + \frac{\partial \phi}{\partial z}dz \wedge d\phi$$

Using the anti-symmetric property of wedge products, we see this become:
$$ [\frac{\partial V_{\phi}}{\partial p} - \frac{1}{p}\frac{\partial V_p}{\partial \phi}] dp \wedge d\phi + [\frac{\partial V_z}{\partial p} - \frac{\partial V_p}{\partial z}] dp \wedge dz + [\frac{1}{p}\frac{\partial V_z}{\partial \phi} - \frac{\partial \phi}{\partial z}] d\phi \wedge dz$$ Applying the hodge dual, we see this become $$\nabla \times V = [\frac{\partial V_{\phi}}{\partial p} - \frac{1}{p}\frac{\partial V_p}{\partial \phi}] dz - [\frac{\partial V_z}{\partial p} - \frac{\partial V_p}{\partial z}] d\phi + [\frac{1}{p}\frac{\partial V_z}{\partial \phi} - \frac{\partial \phi}{\partial z}] dp $$

However, looking at cylinderical coordinate curl, sometimes I will see sometimes people set it up like this:
$$\nabla \times V = [\frac{1}{p}\frac{\partial V_z}{\partial \phi}-\frac{\partial V_{\phi}}{\partial z}] \hat{p}+ [\frac{\partial V_p}{\partial z}-\frac{\partial V_z}{\partial p }] \hat{\phi} + \frac{1}{p}[\frac{\partial (p V_{\phi})}{\partial p} - \frac{\partial V_p}{\partial \phi}] \hat{z}$$


So, zooming onto the ##\frac{1}{p}[\frac{\partial (p V_{\phi})}{\partial p} - \frac{\partial V_p}{\partial \phi}] \hat{z}## we see a difference in notation! I keep my ##\hat{z}## term like this ##[\frac{\partial V_{\phi}}{\partial p} - \frac{1}{p}\frac{\partial V_p}{\partial \phi}]##. While they factored out a ##\frac{1}{p}##.

So two things; How do I show that ##\frac{1}{p}\frac{\partial (p V_{\phi})}{\partial p} = \frac{\partial V_{\phi}}{\partial p}## I know this is true, but i forget the property. I don't believe it's just a product rule? Maybe it's linearity? Not sure... would appreciate some insight.

Second thing, what's the benefit of this notation vs the one i derived?

Thanks for the help.
 
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Answers and Replies

  • #2
Ahh, what a brain fart! I think ##\frac{1}{p}\frac{(\partial (pV_{\phi})}{\partial p}=\frac{\partial V_{ \phi}}{\partial p} ## because p would be a radius, and thus constant? so that is why we are able to bring it inside the differential because d(const*f(x))=constant*d(f(x)).
 
  • #3
Charles Link
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Ahh, what a brain fart! I think ##\frac{1}{p}\frac{(\partial (pV_{\phi})}{\partial p}=\frac{\partial V_{ \phi}}{\partial p} ## because p would be a radius, and thus constant? so that is why we are able to bring it inside the differential because d(const*f(x))=constant*d(f(x)).
Incorrect. ## \frac{\partial{(r V_{\phi}})}{\partial{r}}=V_{\phi}+r \frac{\partial{V_{\phi}}}{\partial{r}} ##. I think you have something incorrect in your initial equations, but I don't know exactly what.
 
  • #4
Incorrect. ## \frac{\partial{(r V_{\phi}})}{\partial{r}}=V_{\phi}+r \frac{\partial{V_{\phi}}}{\partial{r}} ##. I think you have something incorrect in your initial equations, but I don't know exactly what.
That's what i thought as well, but was trying to force it. Let me try to re-derive it as I was doing this at at around 1:30 AM last night!
 
  • #5
So, I can see my issue was not taking the hodge dual right, and forgetting about my scale factors for my V. I still have an issue. So, this might be a long post...

Let's derive it for an arbitrary orthogonal coordinate system.
##\omega = h_uh_vh_w du \wedge dv \wedge dw## and our ##V=h_uV^udu+h_vV^vdv+h_wV^wdw## We also know that ##\nabla = d = \frac{1}{h_u} \partial_u h_u du+ \frac{1}{h_v} \partial_v h_v dv + \frac{1}{h_w}\partial_w h_w dw## which i write this way was it will be useful later. Now we ask what the curl is.
$$\nabla \times V = \star dV = \star[\frac{1}{h_u}\frac{ \partial h_vV^v }{ \partial u } h_u du \wedge dv +\frac{1}{h_u}\frac{ \partial h_wV^w }{ \partial u } h_u du \wedge dw + \frac{1}{h_v}\frac{ \partial h_uV^u }{ \partial v } h_v dv \wedge du + \frac{1}{h_v}\frac{ \partial h_wV^w }{ \partial v } h_v dv \wedge dw + \frac{1}{h_w}\frac{ \partial h_uV^u }{ \partial w } h_w dw \wedge du + \frac{1}{h_w}\frac{ \partial h_vV^v }{ \partial w } h_w dw \wedge dv] $$

So, now in order to factor the wedge products we must multiply them by some other scale factor and basis. Doing so, we see our equation now become..

$$\star[(\frac{ \partial h_vV^v }{ \partial u } - \frac{ \partial h_uV^u }{ \partial v })\frac{h_u h_v du \wedge dv}{h_uh_v} + (\frac{ \partial h_wV^w }{ \partial u } - \frac{ \partial h_uV^u }{ \partial w })\frac{h_u h_w du \wedge dw}{h_uh_w} + (\frac{ \partial h_wV^w }{ \partial v } - \frac{ \partial h_vV^v }{ \partial w }) \frac{h_v h_w dv \wedge dw}{h_vh_w}$$

So there is where I kind of cheat the math, I can't actually show that ##\star (h_uh_v du \wedge dv) = dw## because as far as I'm aware, it should be ## \star (h_uh_v du \wedge dv) = h_w dw## But I had a feeling that it was just ##dw##, and if I do it that way, I see that my curl is...

$$ \star dV = (\frac{ \partial h_vV^v }{ \partial u } - \frac{ \partial h_uV^u }{ \partial v }) \frac{ dw }{h_uh_v} - (\frac{ \partial h_wV^w }{ \partial u } - \frac{ \partial h_uV^u }{ \partial w })\frac{dv}{h_uh_w} + (\frac{ \partial h_wV^w }{ \partial v } - \frac{ \partial h_vV^v }{ \partial w })\frac{du}{h_vh_w} $$

Now, we can verify this on our curl in cylindrical formula. Let ##ds^2 = dr^2 + r^2d\theta^2+dz^2##, thus ##h_u=h_z=1## and ##h_{\phi} = r##. So using this in our newly found formula, we see that $$ \star dV = (\frac{ \partial (rV^{\phi}) }{ \partial r } - \frac{ \partial V^r }{ \partial \phi }) \frac{ dz }{r} - (\frac{ \partial V^z }{ \partial r } - \frac{ \partial V^r }{ \partial z })d\phi + (\frac{ \partial V^z }{ \partial \phi } - \frac{ \partial (rV^{\phi}) }{ \partial z })\frac{dr}{r}$$ Which is what we want! And the trick I wanted to use above, I can actually use here on the last time to make our equation more like the standard one above! That is...
$$(\frac{ \partial V^z }{ \partial \phi } - \frac{ \partial rV^{\phi} }{ \partial z })\frac{dr}{r} \rightarrow (\frac{1}{r}\frac{ \partial V^z }{ \partial \phi } - \frac{r \partial V^{\phi} }{r \partial z })dr \rightarrow (\frac{1}{r}\frac{ \partial V^z }{ \partial \phi } - \frac{ \partial V^{\phi} }{ \partial z })dr$$ So finally, we can write...
$$\nabla \times V = \star dF = (\frac{1}{r}\frac{ \partial V^z }{ \partial \phi } - \frac{ \partial V^{\phi} }{ \partial z })dr - (\frac{ \partial V^z }{ \partial r } - \frac{ \partial V^r }{ \partial z })d\phi + \frac{1}{r}(\frac{ \partial (rV^{\phi}) }{ \partial r } - \frac{ \partial V^r }{ \partial \phi }) dz$$
 

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