Total angular momentum of EM fields

[MMS]

Homework Statement

The problem deals with a charged (Q) rotating sphere around its axis (Ω_0) z^^ (z hat) of radius a.
I'm asked to find the total angular momentum of the EM fields.

2. The attempt at a solution

There is a solution posted to this question and I was just wondering why my calculation aren't right. Please give it a look: http://docdro.id/WcBxAKH

I understand the calculation itself and how they've gotten to their final answer and all that but what I'm unable to get yet is why write theta in Cartesian coordinates in the first place? Why can't I keep it as the unit vector theta? Am I not allowed to take it (unit vector theta) out the integral and then calculate?

Or in general, when am I allowed to take unit vectors out the integrals when calculating something and when am I not?

Thanks in advance.

 

[TSny]

Spherical coordinate unit vectors are not "constant" vectors, they change their directions as you move around in space. For example, $\hat{r}$ points in the $\hat{i}$ direction for points on the positive x axis. But $\hat{r}$ points in the $\hat{j}$ direction for points on the positive y axis.

 

[MMS]

Spherical coordinate unit vectors are not "constant" vectors, they change their directions as you move around in space. For example, $\hat{r}$ points in the $\hat{i}$ direction for points on the positive x axis. But $\hat{r}$ points in the $\hat{j}$ direction for points on the positive y axis.

First, Thanks for the reply.
I do understand that but I can also say the opposite, right? Not just that, why can't I say that every unit vector isn't constant? I mean, there are formulas for unit vectors that transform us from Cartesian to spherical and vice-versa.
Say for example we had z^^ and not theta^^ in some integral (not particularly this one). Do I plug z^^=cos(theta) r^^ - sin(theta) theta^^ or keep it as z^^?

 

[TSny]

If you want to be able to pull the unit vector out of the integral, then you need to keep it as $\hat{z}$. The Cartesian unit vectors do not change direction as you move around in space.

The integral represents a summation of vectors at different regions of space.

Suppose you have a vector $\vec{v}_1 = 2 \hat{r}$ corresponding to a point on the positive x-axis and a vector $\vec{v}_2 = 3 \hat{r}$ corresponding to a point on the positive y axis. The sum would be $\vec{v}_1+ \vec{v}_2= 2\hat{r} + 3 \hat{r}$. You cannot factor out the $\hat{r}$ and express this as $\vec{v}_1 +\vec{v}_2= (2 + 3) \hat{r} = 5 \hat{r}$ [wrong]. This is because the unit vector $\hat{r}$ for a point on the x-axis is not equal to the unit vector $\hat{r}$ for a point on the y axis. They are perpendicular to each other.

But suppose you have a vector $\vec{v}_1 = 2 \hat{i}$ corresponding to a point on the positive x-axis and a vector $\vec{v}_2 = 3 \hat{i}$ corresponding to a point on the positive y axis. Now when you sum the vectors you can factor out the unit vector and get a correct answer of $5 \hat{i}$.

 

[MMS]

If you want to be able to pull the unit vector out of the integral, then you need to keep it as $\hat{z}$. The Cartesian unit vectors do not change direction as you move around in space.

The integral represents a summation of vectors at different regions of space.

Suppose you have a vector $\vec{v}_1 = 2 \hat{r}$ corresponding to a point on the positive x-axis and a vector $\vec{v}_2 = 3 \hat{r}$ corresponding to a point on the positive y axis. The sum would be $\vec{v}_1+ \vec{v}_2= 2\hat{r} + 3 \hat{r}$. You cannot factor out the $\hat{r}$ and express this as $\vec{v}_1 +\vec{v}_2= (2 + 3) \hat{r} = 5 \hat{r}$ [wrong]. This is because the unit vector $\hat{r}$ for a point on the x-axis is not equal to the unit vector $\hat{r}$ for a point on the y axis. They are perpendicular to each other.

But suppose you have a vector $\vec{v}_1 = 2 \hat{i}$ corresponding to a point on the positive x-axis and a vector $\vec{v}_2 = 3 \hat{i}$ corresponding to a point on the positive y axis. Now when you sum the vectors you can factor out the unit vector and get a correct answer of $5 \hat{i}$.

Great explanation. I believe I get it better now.

Thank you once again!