# Thread: Divergence in cylindrical coordinates

1. ## Divergence in cylindrical coordinates

Hello, I would like to know how do we get from (2) to (3):

$\displaystyle (1)\ \ \nabla \cdot B = 0$

$\displaystyle (2)\ \ \frac{1}{r}\frac{\partial r B_r}{\partial r}\frac{\partial B_z}{\partial z} = 0$

$\displaystyle (3)\ \ rB_r = -\int_{0}^{r} r \frac{\partial B_r}{\partial z} dr$

After (3) I know how to integrate, I just want to know how do we get from (2) to (3), if someone can do it slow with comments and as much steps.

Thanks.

2. ## Re: Divergence in cylindrical coordinates

I suspect that line (2) is equivalent to

$\dfrac 1 r \dfrac{\partial}{\partial r}(rB_r)+\dfrac{\partial B_z}{\partial z} = 0$

$\dfrac 1 r \left(B_r + r \dfrac{\partial B_r}{\partial r}\right)+ \dfrac{\partial B_z}{\partial z} = 0$

work that through and see where it goes.

3. ## Re: Divergence in cylindrical coordinates

Oh sorry, there is a mistake. The second equation is missing a plus sign: Originally Posted by Nforce Hello, I would like to know how do we get from (2) to (3):

$\displaystyle (1)\ \ \nabla \cdot B = 0$

$\displaystyle (2)\ \ \frac{1}{r}\frac{\partial r B_r}{\partial r} + \frac{\partial B_z}{\partial z} = 0$

$\displaystyle (3)\ \ rB_r = -\int_{0}^{r} r \frac{\partial B_r}{\partial z} dr$

After (3) I know how to integrate, I just want to know how do we get from (2) to (3), if someone can do it slow with comments and as much steps.

Thanks.

4. ## Re: Divergence in cylindrical coordinates

This seems pretty straightforward actually

$\dfrac 1 r \dfrac{\partial}{\partial r}(r B_r) + \dfrac{\partial B_z}{\partial z} = 0$

$\dfrac{\partial}{\partial r}(r B_r) = -r\dfrac{\partial B_z}{\partial z}$

$rB_r = - \displaystyle \int_0^R r\dfrac{\partial B_z}{\partial z}~dr$

$R$ is some constant that comes from the specific problem. I don't think it's intended to be the variable $r$

5. ## Re: Divergence in cylindrical coordinates

If I understand correctly then here:

$\dfrac{\partial}{\partial r}(r B_r)$

If we integrate:

$$\int \dfrac{\partial}{\partial r}(r B_r) dr$$

Then by the fundamental theorem of calculus, if we integrate a derivative we get the integrand?

Does this apply also to partial derivatives? I always thought that then we have to expose the partial derivative operator. So, in this case, the partial derivative is a normal derivative.

6. ## Re: Divergence in cylindrical coordinates

Sorry, did I ask something stupid?

7. ## Re: Divergence in cylindrical coordinates Originally Posted by Nforce Sorry, did I ask something stupid?
sorry missed the question.

You can always integrate w/respect to any variable of a multivariable function (provided that integral exists).
In this case they are just integrating with respect to the variable $r$

8. ## Re: Divergence in cylindrical coordinates

Why don't we write then:

$$\int \dfrac{\partial}{\partial r}(r B_r) \partial r$$

?

9. ## Re: Divergence in cylindrical coordinates

I just don't know Jim.... I'm only a country doctor....

10. ## Re: Divergence in cylindrical coordinates Originally Posted by Nforce Why don't we write then:

$$\int \dfrac{\partial}{\partial r}(r B_r) \partial r$$

?
My two cents:

If r can be written as a function of, say, two other variables p and q then
$\displaystyle dr = \dfrac{\partial r}{\partial p} dp + \dfrac{\partial r}{\partial q} dq$

We can't do that with $\displaystyle \partial r$.

I realize that that isn't the case here but it possibly explains the difference in notation.

-Dan