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## 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.

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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