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- 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##

##\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##