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##
 

Charles Link

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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

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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

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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

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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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Charles Link

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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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