I Why am I getting Maxwell's second equation wrong?

Summary
I am going somewhere wrong in my calculations (maybe somewhere in the highlighted text). Please point out my mistakes.
While going through an article titled "Reflections in Maxwell's treatise" a misunderstanding popped out at page 227 and 228. Consider the following equations $(23\ a)$ and $(23\ c)$ in the article (avoiding the surface integral):

$\displaystyle \psi_m (\mathbf{r})=-\dfrac{1}{4 \pi} \int_V \dfrac{\nabla' \cdot \mathbf{M} (\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV'=\dfrac{1}{4 \pi} \int_V \dfrac{\rho_m}{|\mathbf{r}-\mathbf{r'}|} dV' \tag{23a}$

$\displaystyle \mathbf{H}(\mathbf{r})=-\dfrac{1}{4 \pi} \int_V \nabla' \cdot \mathbf{M} (\mathbf{r'}) \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV'=\dfrac{1}{4 \pi} \int_V \rho_m \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag{23b}$

$\displaystyle \mu_0\mathbf{H}(\mathbf{r})=\dfrac{\mu_0}{4 \pi} \int_V \rho_m \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV'$

where $\rho_m=-\nabla' \cdot \mathbf{M} (\mathbf{r'})$

Using Gauss law and divergence theorem and noting that the divergence due to the surface integral (in the article) is zero:

$\nabla \cdot \mu_0\mathbf{H}(\mathbf{r})=\mu_0\ \rho_m=-\mu_0 \nabla' \cdot \mathbf{M} (\mathbf{r'})$

Using the above result:

$\nabla \cdot \mathbf{B}(\mathbf{r}) =\nabla \cdot (\mu_0\mathbf{H}(\mathbf{r})+\mu_0\mathbf{M} (\mathbf{r'}))=-\mu_0 \nabla' \cdot \mathbf{M} (\mathbf{r'})+\bbox[yellow]{\mu_0 \nabla \cdot \mathbf{M} (\mathbf{r'})}$

$\bbox[yellow]{\text{In the second term, since divergence is with respect to field coordinates, the second term is zero.}}$ Therefore:

$\nabla \cdot \mathbf{B}(\mathbf{r})=-\mu_0 \nabla' \cdot \mathbf{M} (\mathbf{r'})$

But it should be zero (equation $32$ in the article). There must be something wrong in my calculation. Please explain why am I getting $\nabla \cdot \mathbf{B}(\mathbf{r}) \neq 0$

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Homework Helper
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I think this one is simple: You have $\nabla \cdot H(r)=\rho_m$, but that is actually $\rho_m(r)$. Your previous equation with $\rho_m$ was $\rho_m(r')= -\nabla' \cdot M(r')$. You can't throw away the understood coordinate parameter that the $\rho_m$ carries with it. With the $\nabla \cdot H(r)$, that is $\rho_m(r)$ and not $\rho_m(r')$.

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Another difficulty which arises is the following:

$\displaystyle \nabla \cdot \mathbf{H}(\mathbf{r})=\dfrac{d\Phi}{dV} =\dfrac{d\ \displaystyle\oint_{A} \mathbf{H}(\mathbf{r}) \cdot \mathbf{\hat{n}}\ dA }{dV}=\dfrac{dq'_m}{dV} \tag1$

But we should be getting $\nabla \cdot \mathbf{H}(\mathbf{r})=\dfrac{dq_m}{dV} \tag2$

Why are we getting $(1)$ instead of $(2)$?

vanhees71

Gold Member
I don't understand your notation. You cannot take a derivative with respect to a volume.

I guess, what you mean is (using Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{H}(\vec{r})=\frac{1}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 r' \rho(\vec{r}') \vec{\nabla} \cdot \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}.$$
Now the nabla operator is with respect to $\vec{r}$, and
$$\vec{E}_{\text{coulomb}}(\vec{r})=\frac{1}{4 \pi} \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3},$$
is the field of a unit point charge sitting at $\vec{r}'$, i.e.,
$$\vec{\nabla} \cdot \vec{E}(\vec{r})=\delta(\vec{r}-\vec{r}').$$
Thus you find
$$\vec{\nabla} \cdot \vec{H}(\vec{r})=\int_{\mathbb{R}^3} \mathrm{d}^3 r' \rho_m(\vec{r}') \delta^{(3)}(\vec{r}-\vec{r}')=\rho_m(\vec{r}),$$
as it should be.

You have to be more "pedantic" concerning clearly dinstinguishing betwee $\vec{r}$ and $\vec{r}'$ and more carefully observe, to which position vector your nabla oparator refers!

I don't understand your notation. You cannot take a derivative with respect to a volume.
In equation (1), I just used the formal definition of divergence.

By denoting the source charge with prime, I get flux through a closed surface $(\Phi)= \displaystyle\oint_{A} \mathbf{H}(\mathbf{r}) \cdot \mathbf{\hat{n}}\ dA =q'_m$

And now using the formal definition of divergence:

$\displaystyle \nabla \cdot \mathbf{H}(\mathbf{r})=\dfrac{d\Phi}{dV} = \dfrac{dq'_m}{dV} \tag3$

In order to get $\nabla \cdot \mathbf{H}(\mathbf{r})=\rho_m (\mathbf{r})$, we should be having:

$\nabla \cdot \mathbf{H}(\mathbf{r})=\dfrac{dq_m}{dV} \tag4$

instead of equation (3). Where am I going wrong?

vanhees71

Gold Member
As I said, you have to be more pedantic concerning your notation. The coordinate-independent definition of the divergence is as follows: The divergence of a vector field at point $\vec{r}$, $\vec{\nabla} \cdot \vec{V}(\vec{r})$ is given by the following limit
$$\vec{\nabla} \cdot \vec{V}(\vec{r}) = \lim_{\Delta V \rightarrow \{\vec{r} \}} \frac{1}{\Delta V} \int_{\partial \Delta V} \mathrm{d}^2 \vec{f}' \cdot \vec{V}(\vec{r}').$$
Here $\Delta V$ is some volume containing the point $\vec{x}$, $\partial \Delta V$ is its boundary surface, by convention oriented such that the surface-normal vectors along the surface all point out of the volume. The limit means to "contract" $\Delta V$ to the point $\vec{r}$. You can of course easily derive the differential form for any given set of coordinates. E.g., for Cartesian coordinates you get
$$\vec{\nabla} \cdot \vec{V}(\vec{r})=\sum_{j=1}^3 \frac{\partial}{\partial r^j} V^j(\vec{r}).$$

It is clear that the only free position variable is $\vec{r}$. You integrate over the $\vec{r}'$.

The notation $\mathrm{d q_m}{\mathrm{d} V}$ doesn't make any sense. I've shown, how to derive the correct formula
$$\vec{\nabla} \cdot \vec{H}(\vec{r})=\rho_m(\vec{r})$$
in my previous posting.

Of course, it's easier to derive directly from Maxwell's equations for the magnetostatic case, assuming that only magnetization, no current densities are present.
$$\vec{\nabla} \times \vec{H}=0, \quad \vec{\nabla} \cdot \vec{B}=\vec{\nabla} (\vec{H}+\vec{M})=0.$$
From the first equation you get that there exists a magnetostatic potential,
$$\vec{H}=-\vec{\nabla} \phi,$$
and from the 2nd equation
$$\vec{\nabla} \cdot \vec{H}=-\Delta \phi=-\vec{\nabla} \cdot \vec{M}=\rho_m.$$
The solution is known from electrostatics (or more formally by using the Green's function of the Laplace operator),
$$\phi(\vec{r})=\int_{\mathbb{R}} \mathrm{d}^3 r' \frac{\rho_m(\vec{r}')}{4 \pi |\vec{r}-\vec{r}'|}.$$

Thanks for showing another way to derive $\vec{\nabla} \cdot \vec{H}(\vec{r})=\rho_m(\vec{r})$. However I am eager to derive it using Gauss divergence theorem. Can you please show how to properly do it?

vanhees71

Gold Member
You always come from the local form of the Maxwell equations to their integral form by using the integral theorems. In this case from $\vec{\nabla} \cdot \vec{H}=\rho_m$ you get by using Gauss's integral theorem [corrected in view of #9]
$$q_{mV}=\int_V \mathrm{d}^3 r' \vec{\nabla}' \cdot \vec{H}(\vec{r}') = \int_{\partial V} \mathrm{d}^2 \vec{f}' \cdot \vec{H}(\vec{r}').$$
Now, assuming that $\rho_m$ is a sufficiently smooth function (particularly no surface-charge distribution or even a point-charge distribution which are generalized functions ("distributions" in the functional-analygiscal sense) rather then proper functions!). Then in making $V$ be a small volume $\Delta V$ around $\vec{r}$, you can approximate
$$q_{m \Delta V}=\Delta V \rho_m(\vec{r})$$
Then dividing the equation by $\Delta V$ and making $\Delta V \rightarrow \{\vec{r} \}$ you immediately get
$$\vec{\nabla} \cdot \vec{H}(\vec{r})=\rho_m(\vec{r}).$$
Of course in some sense, defining the divergence in this very clever coordinate-indpendent way via the surface integral and the limit process described, Gauss's theorem is merely a tautology!

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Homework Helper
Gold Member
2018 Award
You always come from the local form of the Maxwell equations to their integral form by using the integral theorems. In this case from $\vec{\nabla} \cdot \vec{H}=\rho_m$ you get by using Gauss's integral theorem
$$q_{mV}=\int_V \mathrm{d}^3 r' \vec{\nabla}' \cdot \vec{V}(\vec{r}') = \int_{\partial V} \mathrm{d}^2 \vec{f}' \cdot \vec{H}(\vec{r}').$$
Now, assuming that $\rho_m$ is a sufficiently smooth function (particularly no surface-charge distribution or even a point-charge distribution which are generalized functions ("distributions" in the functional-analygiscal sense) rather then proper functions!). Then in making $V$ be a small volume $\Delta V$ around $\vec{r}$, you can approximate
$$q_{m \Delta V}=\Delta V \rho_m(\vec{r})$$
Then dividing the equation by $\Delta V$ and making $\Delta V \rightarrow \{\vec{r} \}$ you immediately get
$$\vec{\nabla} \cdot \vec{H}(\vec{r})=\rho_m(\vec{r}).$$
Of course in some sense, defining the divergence in this very clever coordinate-indpendent way via the surface integral and the limit process described, Gauss's theorem is merely a tautology!
One simple correction: I believe that needs to be a $\nabla' \cdot H(r')$ instead of a $\nabla' \cdot V(r')$ in the top middle equation above. Otherwise, an excellent proof !

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vanhees71

Gold Member
Sure, I'll correct it immediately! Thanks!

"Why am I getting Maxwell's second equation wrong?"

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