Divergence/Curl of \vec{E}=α\frac{\vec{r}}{r^2}

[oddjobmj]

Homework Statement

Calculate the divergence and curl of $\vec{E}$=α$\frac{\vec{r}}{r^2}$

Homework Equations

Div($\vec{E}$)=$\vec{∇}$°$\vec{E}$
Div($\vec{E}$)=$\vec{∇}$x$\vec{E}$

Table of coordinate conversions, div, and curl:
http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

The Attempt at a Solution

My confusion is stemming from the spherical coordinates. I believe I have the divergence down because the result is a scalar so the direction doesn't matter (phi & theta) so it's just a matter of the magnitude of the field at any given point which is dependent only on the distance from the origin. Or can I not make this assumption? I'm not sure how to bring phi and theta into the mix when dealing with the curl.

For the divergence I know that $\vec{r}$=r$\hat{r}$ so $\vec{E}$ simplifies to:

α$\frac{\hat{r}}{r}$

Using the table linked above to find the form of the divergence in spherical coordinates I believe I can ignore the theta and phi 'contributions' because divergence will depend only on r.

div(E)=$\frac{α}{r^2}$$\frac{∂(r)}{∂r}$=$\frac{α}{r^2}$

I think the curl should be zero by observation but I don't know how to show this. Can I do the same as with divergence and ignore the theta and phi contributions because this is simply a function of r?

Thank you!

 

[ZetaOfThree]

Just use the formula in the table for the curl in spherical coordinates. Then just make sure you do the derivatives correctly. Easy.

 

[oddjobmj]

I'm sorry for not being more clear. Referencing the table in specific I don't know what A_$\theta$ and A_$\phi$ are. That's what is confusing me.

 

[ZetaOfThree]

Those are the components of $\vec{A}$ in the $\hat{\theta}$ and $\hat{\phi}$ directions. In your case, $\vec{A}=\vec{E}=a\frac{\vec{r}}{r^2}$.

 

[oddjobmj]

edit: sorry, accidental double post

 

[oddjobmj]

Those are the components of $\vec{A}$ in the $\hat{\theta}$ and $\hat{\phi}$ directions. In your case, $\vec{A}=\vec{E}=a\frac{\vec{r}}{r^2}$.

Right, I understand that A is a placeholder for E in the table. It is the components of E in those directions that I don't know how to find.

 

[nrqed]

Right, I understand that A is a placeholder for E in the table. It is the components of E in those directions that I don't know how to find.

In general, $\vec{E} = E_r \hat{r} + E_\theta \hat{\theta} + E_\phi \hat{\phi}$

 

[oddjobmj]

In general, $\vec{E} = E_r \hat{r} + E_\theta \hat{\theta} + E_\phi \hat{\phi}$

Right, I'm trying to find out what E_r$\hat{r}$ and so on are functionally (starting from the value given for $\vec{E}$ in the problem). I can't take the partial derivative of the general form for E in the direction of theta without breaking E down into the relevant components first. I am given an actual value for E which is a function of r.

 

[ZetaOfThree]

Right, I'm trying to find out what E_r$\hat{r}$ and so on are functionally (starting from the value given for $\vec{E}$ in the problem). I can't take the partial derivative of the general form for E in the direction of theta without breaking E down into the relevant components first. I am given an actual value for E which is a function of r.

You should read up on spherical coordinates, it will be very helpful. The $\theta$ and $\phi$ components of $\vec{E}$ are zero because it only points in the radial direction and $E_r=\frac{a}{r}$.

 

[nrqed]

In general, $\vec{E} = E_r \hat{r} + E_\theta \hat{\theta} + E_\phi \hat{\phi}$

You are told that $\vec{E}= a \frac{\vec{r}}{r^2}$. By comparing with the general form, what do you conclude about $E_\theta$ and $E_\phi$ ?

 

[oddjobmj]

You should read up on spherical coordinates, it will be very helpful. The $\theta$ and $\phi$ components of $\vec{E}$ are zero because it only points in the radial direction.

Which is why I assumed by observation that the curl would be zero. I just wanted to verify that I didn't have to consider a general representation of phi and theta, to be complete, as they vary for any arbitrary r. For some reason I was thinking back to r in cartesian coordinates and projecting it onto some equivalent to x, y, and z for theta and phi. Sorry for the confusion!

Thank you

 

[ZetaOfThree]

I believe I have the divergence down because the result is a scalar so the direction doesn't matter (phi & theta) so it's just a matter of the magnitude of the field at any given point which is dependent only on the distance from the origin. Or can I not make this assumption? I'm not sure how to bring phi and theta into the mix when dealing with the curl.

This is wrong. Divergence absolutely depends of the direction of $\vec{E}$.

You really shouldn't intuit your way through these exercises, you need to do them nrqed's way to find the components of $\vec{E}$ and then plug them into your formulas from the table.

For some reason I was thinking back to r in cartesian coordinates and projecting it onto some equivalent to x, y, and z for theta and phi.

Spherical coordinates is an orthogonal coordinate system, meaning that $\hat{r}$, $\hat{\theta}$, and $\hat{\phi}$ are perpendicular, so their projections onto each other are zero.

 

[oddjobmj]

This is wrong. Divergence absolutely depends of the direction of $\vec{E}$.

You really shouldn't intuit your way through these exercises, you need to do them nrqed's way to find the components of $\vec{E}$ and then plug them into your formulas from the table.

This seems to contradict your earlier statement. In this case E only depends on r. I considered E_r which is effectively the function they gave me for E and used the partial derivative equation from the table and the only non-zero component was the one dependent on r. In other words, the partial with respect to theta is zero because only r shows up in the equation which is not a function of theta and thus E is a constant with respect to theta (and phi). The partial of a constant is zero. The calculated divergence in this case is listed above in my original post (non-zero). Do you disagree with the result?

I didn't -just- intuit my way through the problem but I did try to think about the problem and compare that to my results to make sure that I understand what is going on and that the results make sense.