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I am given a vector field $\vec{A} = A_\rho \space \hat{e_\rho} + A_\phi \space \hat{e_\phi} + A_z \space \hat{e_z}$, and I am supposed to use the unit vectors (provided below) in cylindrical coordinates to calculate the gradient in cylindrical coordinates.

\begin{align*} \hat{\rho} = <\cos(\phi), \sin(\phi),0> \end{align*} \begin{align*} \hat{\phi} = <-\sin(\phi), \cos(\phi),0> \end{align*} \begin{align*} \hat{z} = <0,0,1> \end{align*}

Then, I am asked to use the form of the gradient operator in cylindrical coordinates to compute $ \nabla \times \vec{A} $, where the gradient operator in cylindrical coordinates, found as:

\begin{align*} \nabla\times\vec{F} = (\frac{1}{\rho} \frac{\delta A_z}{\delta \phi} - \frac{\delta A_\phi}{\delta z}) \hat{\rho} + (\frac{\delta A_\rho}{\delta z} - \frac{\delta A_z}{\delta \rho})\hat{\phi} + \frac{1}{\rho}(\frac{\delta(\rho A_\phi)}{\delta \rho} - \frac{\delta A_\rho}{\delta \phi} )\hat{z}\end{align*}

Now, the professor also provides a hint:

"Given the forms of the unit vectors, be careful when $\delta/\delta\phi$ hits either $\hat{e_\rho}, \hat{e_\phi}$ -- these terms will be non trivial.

The issue I'm having is how is the above expression not the final solution? Doesn't that represent the gradient ($\nabla \times \vec{A}$) in cylindrical coordinates for the given vector? Or is there a way to simplify the expression?

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    $\begingroup$ That is the expression; your professor is talking about deriving it. $\endgroup$
    – Muphrid
    Mar 30, 2014 at 23:12
  • $\begingroup$ Ah, totally missed that. Thank you. $\endgroup$
    – A4Treok
    Mar 30, 2014 at 23:44
  • $\begingroup$ The question is closely related to those. The gradient operator in cylindrical coordinates takes the form $\nabla = \hat{\rho} \frac{\delta}{\delta\rho} + \hat{\phi} \frac{1}{\rho} \frac{\delta}{\delta\phi} + \hat{z} \frac{\delta}{\delta z}$ with your notations. $\endgroup$
    – EditPiAf
    Feb 24, 2017 at 17:22

2 Answers 2

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My attempt at the solution (using $<r,\phi,z>$)

Curve of constant r

First consider the curve of constant r, provided above. For $C_1$ and $C_3$, we have:

$\int_{C_1} F\cdot \hat{t} = -F_\phi (r,\phi, z + \Delta z/2)(r + \Delta r/2)\Delta \phi$

$\int_{C_3} F\cdot \hat{t} = F_\phi (r,\phi, z - \Delta z/2)(r - \Delta r/2)\Delta \phi$

Thus, knowing the change in surface for constant r is $\delta S_r = r \delta \phi \delta z$

\begin{align*} \frac{1}{\Delta S}\int_{C_1 + C_3} F\cdot \hat{t} = -\frac{\Delta \phi}{r \Delta \phi \ \Delta z}[ F_\phi (r,\phi, z + \Delta z/2)(r + \Delta r/2) - F_\phi (r,\phi, z - \Delta z/2)(r - \Delta r/2)] \end{align*}

Which taking the limits for $\Delta \phi, \Delta z \rightarrow 0$, we get

\begin{align*} -\frac{\delta F_\phi}{\delta z}\end{align*}

For $C_2$ and $C_4$:

$\int_{C_2} F\cdot \hat{t} = F_z (r,\phi + \Delta \phi/2, z)\Delta z$

$\int_{C_4} F\cdot \hat{t} = F_z (r,\phi - \Delta \phi/2, z)\Delta z$

Thus,

\begin{align*} \frac{1}{\Delta S}\int_{C_2 + C_4} F\cdot \hat{t} = \frac{\Delta \phi}{r \Delta \phi \ \Delta z}[F_z(r,\phi + \Delta \phi/2, z) - F_z(r,\phi - \Delta \phi/2, z)] \end{align*}

Which taking the limits for $\Delta \phi, \Delta z \rightarrow 0$, we get

\begin{align*} \frac{1}{r} \frac{\delta F_z}{\delta \phi}\end{align*}


Curve of constant $\phi$

First consider the curve of constant $\phi$, provided above. For $C_1$ and $C_3$, we have:

$\int_{C_1} F\cdot \hat{t} = F_r (r,\phi, z + \Delta z/2)\Delta r$

$\int_{C_3} F\cdot \hat{t} = -F_r (r,\phi, z - \Delta z/2)\Delta r$

Change in surface for constant $\phi$ is $\delta S_r = r \phi \delta z$

\begin{align*} \frac{1}{\Delta S}\int_{C_1 + C_3} F\cdot \hat{t} = \frac{\Delta r}{\Delta r \ \Delta z}[F_r (r,\phi, z + \Delta z/2) - F_r (r,\phi, z - \Delta z/2)] \end{align*}

Which taking the limits for $\Delta r, \Delta z \rightarrow 0$, we get

\begin{align*} \frac{\delta F_r}{\delta z}\end{align*}

For $C_2$ and $C_4$:

$\int_{C_2} F\cdot \hat{t} = -F_z (r + \Delta r/2, \phi, z)\Delta z$

$\int_{C_4} F\cdot \hat{t} = F_z (r - \Delta r/2, \phi, z)\Delta z$

Thus,

\begin{align*} \frac{1}{\Delta S}\int_{C_2 + C_4} F\cdot \hat{t} = -\frac{\Delta z}{\Delta r \ \Delta z}[F_z(r + \Delta r/2, \phi, z)-F_z (r - \Delta r/2, \phi, z)] \end{align*}

Which taking the limits for $\Delta r, \Delta z \rightarrow 0$, we get

\begin{align*} \frac{\delta F_z}{\delta r}\end{align*}


enter image description here

The derivation for the final curve, of constant $z$, shown above, can be found in Schey's "Div Grad Curl and All That", but it follows the same procedure as above.

The result for the curve of constant $z$ is

\begin{align*} (\nabla \times F)_z =\frac{1}{r}\frac{\delta}{\delta r} (rF_\phi) - \frac{1}{r} \frac{\delta F_r}{\delta \phi} \end{align*}

Combining the results of the above, we reach

\begin{align*} \nabla\times\vec{F} = (\frac{1}{r} \frac{\delta F_z}{\delta \phi} - \frac{\delta F_\phi}{\delta z}) \hat{r} + (\frac{\delta F_r}{\delta z} - \frac{\delta F_z}{\delta r})\hat{\phi} + \frac{1}{r}(\frac{\delta(r F_\phi)}{\delta r} - \frac{\delta F_r}{\delta \phi} )\hat{z}\end{align*}

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I think I know what he's getting at.

By the chain rule $df= \frac{\partial f}{\partial r}dr+\frac{\partial f}{\partial \theta}d\theta+\frac{\partial f}{\partial z}dz$..

In cylindrical coordinates the line element is $\vec{ds}=dr\hat{r} +rd\theta\hat{\theta} + dz \hat{k}$

We have the property of the gradient that $df=\nabla f \cdot \vec{ds}$.

To get the dot product to come match the chain rule, we need $\nabla f = \frac{\partial f}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial f}{\partial \theta}\hat{\theta} + \frac{\partial f}{\partial z}\hat{k}$

There's an integral method to find the curl, but I find the integrals confusing. You can use the gradient to follow it in the following way:

$\vec{E} = \sum_i E_i \hat{e_i}$

$\nabla \times \vec{E} = \sum_i \nabla(E_i) \times \hat{e_i} + \sum_i E_i (\nabla \times \hat{e_i})$ (note the gradient in the left sum)

So $\nabla \times \hat{E}= \nabla(E_r) \times \hat{e_r} + \nabla(E_\theta)\times \hat{e_\theta}+ \nabla(E_z) \times \hat{k}+ E_r (\nabla \times \hat{r}) + E_\theta(\nabla \times \hat{\theta}) + E_z(\nabla \times \hat{k})$

$\nabla \times \nabla f =0, $ and $\hat{r}=\nabla r, \hat{k}=\nabla z$, so $\nabla \times \hat{r}=0$ and $ \nabla \times \hat{k}$=0.

$\nabla \theta = \frac{1}{r} \hat{\theta}$

$\nabla \times (\nabla \theta) =0=\frac{-1}{r^2} \hat{k} +\frac{1}{r} (\nabla \times \hat{\theta})$

$\nabla \times \hat{\theta} = \hat{k}/r$

So $\nabla \times \hat{E} = \frac{-1}{r}\frac{\partial E_r}{\partial \theta}\hat {k}+ \frac{\partial E_r}{\partial z}\hat{\theta} +\frac{\partial E_\theta}{\partial r}\hat{k} - \frac{\partial E_\theta}{\partial z}\hat{r} - \frac{\partial E_z}{\partial r} \hat{\theta}+\frac{1}{r}\frac{\partial E_z}{\partial \theta}\hat {r} +E_\theta/r \hat{k}$

$\nabla \times \vec{E}=(\frac{1}{r}\frac{\partial E_z}{\partial \theta}-\frac{\partial E_\theta}{\partial z})\hat{r}+(\frac{\partial E_r}{\partial z}-\frac{\partial E_z}{\partial r})\hat{\theta}+\frac{1}{r}(\frac{\partial (rE_\theta)}{\partial r}-\frac{\partial E_r}{\partial \theta})\hat{k}$

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