Volume Integral : $\int \mathrm d^3\mathbf{r}' \frac{\nabla \cdot \mathbf{M}(\mathbf{r'})}{|\mathbf{r'}-\mathbf{r'}|}$

I am trying to understand the following claims. I would appreciate if you could help me, as I am still unable to understand it.

The problem asks us to find the following integral: $$I = \int \mathrm d^3\mathbf{r}'\frac{\nabla \cdot \mathbf{M}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}$$

Assuming $$\mathbf{M}$$ does not blow up, we can use the following identity:

$$\frac{\nabla \cdot \mathbf{M}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} = \nabla \cdot \frac{ \mathbf{M}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} + \mathbf{M(r')} \cdot \nabla_{r'} \frac{ 1}{|\mathbf{r}-\mathbf{r'}|}$$

Question 01: It then says that if we use this expression in the integral, the first term of the sum vanishes because the localisation (in german "Lokalisation") of M vanishes after using the Gauss theorem. Why should it vanish? what does localisation mean?

Then, the problem says that we are left with the following: $$I = \nabla_{r} \int \mathrm d^3\mathbf{r}' \frac{\mathbf{M}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}$$

Question 02: How is it that we can take the gradient out? I mean, we have a vector M inside, so it doesn't make any sense to me. Is this a problem of the notation used?

At the end, for a $$\mathbf{M}=M_0 \mathbf{e_z}$$ we get:

$$I = M_0 \frac{\mathrm d}{\mathrm dz} \int \mathrm d^3\mathbf{r}' \frac{1}{|\mathbf{r}-\mathbf{r'}|}$$

By the way, this was taken from the book: Grundkurs Theoretische Physik 3 by Wolfgang Nolting page 214

Thanks!

• You can take the del out because it is with respect to the unprimed variable. Commented Jan 3, 2022 at 2:59
• Thanks! I did not notice it at first. Commented Jan 3, 2022 at 15:51

Q1: In my Classical Electrodynamics book (Jackson), the derivation is a little different. By localized in Jackson, he means that there exists an object with magnetization within some fixed boundary, and $$M(r) = 0$$ outside (and on) that boundary. So the first integral is $$\int_V \nabla \cdot \frac{M(r')}{|r-r'|}$$ dV, right? The divergence theorem then says this integral is equal to $$\int_s \frac{M(r')}{|r-r'|} \cdot n\, dS$$ . The idea then is that M(r') vanishes on this boundary. If we want to go a little deeper, I believe you can choose the volume of integration $$V$$ s.t. the surface of $$V$$ lies outside the magnet where $$M(r) = 0$$.