Jackson 5.10 topic problem

[dingo_d]

Homework Statement

There is one thing I'm not getting in this problem. It's about uniformly magnetized sphere. In solving the problem with formula for magnetic potential:

$$\phi_M=-\nabla\cdot\int\frac{\vec{M}}{|\vec{r}-\vec{r}'|}d^3r'$$

At one point a change of variable for differentiation is made

$$\frac{\partial}{\partial z}=\frac{\partial r}{\partial z}\frac{\partial}{\partial r}$$

And he says that $$\frac{\partial r}{\partial z}=\cos\theta$$. But I can't see that. If I'm using the formula for spherical coordinate system transformation: $$z=r\cos\theta$$ I don't get just cosine, I get 1/cosine :\

So what formula is he using?

Thanks

 

[rafaelpol]

He is using $r = \sqrt {x^2 + y^2+z^2}$, and then
$\frac{\partial r}{\partial z} = \frac{z}{r} = cos{(\theta)}$

 

[fzero]

Since $$r = \sqrt{ x^2 + y^2 + z^2}$$,

$$\frac{\partial r }{\partial z} = \frac{z}{r},$$

which leads to the formula from the text. I think you're making a mistake because the partial derivatives in spherical and Cartesian coordinates are related by a 3x3 matrix, so it's not enough to merely compare reciprocals.

 

[dingo_d]

oh! I see, thanks :D

 

[paranormal]

for bringing this problem to my attention. It seems like there may be a misunderstanding about the use of spherical coordinates in this problem. The formula for the transformation from Cartesian to spherical coordinates is indeed z=r\cos\theta, but this is only for the z-component of a vector. In this case, we are interested in the change of variable for the differentiation operator, which is a scalar quantity. The correct formula for this is \frac{\partial}{\partial z}=\frac{1}{r}\frac{\partial}{\partial r}, which can be derived from the chain rule. This means that the change of variable for differentiation in this problem should be \frac{\partial}{\partial z}=\frac{1}{r}\frac{\partial}{\partial r}, not \frac{\partial}{\partial z}=\frac{\partial r}{\partial z}\frac{\partial}{\partial r}. I hope this clarifies the confusion and helps you solve the problem correctly.