# Deduction of $\mathbf H =\dfrac{\mathbf B}{\mu_0}-\mathbf M$

merry Christmas to all the users!

I want to get to $$\mathbf H =\dfrac{\mathbf B}{\mu_0}-\mathbf M$$ from the superposition principle, like some texts have done in electrostatics with $$\mathbf D=\epsilon_0\mathbf E+\mathbf P$$. In magnetostatic I'm stuck.

For example, in electrostatics, by the superposition principle, the potential outside of a polarized body must be the sum of the potentials due to free and bound charges. When applying the gradient it results $$\begin{array}{cl}\boldsymbol\nabla V=\boldsymbol\nabla V_\text{l}+\boldsymbol\nabla V_\text{p}&(1)\end{array}$$

The total field will be given by the total electric potential (again, by the ppio of superposition), so that $$\mathbf E=-\boldsymbol\nabla V$$. If we apply the gradient (with respect to the coordinates of $$\mathbf r$$) to $$$$V_\text{p}\left(\mathbf r\right)=k_e \displaystyle\int_V\dfrac{\mathbf P\left(\mathbf r^{\prime}\right)\cdot\hat{\mathbf R}}{R^2}\ \mathrm{d}\tau^{\prime}$$$$

we will obtain the electric field due to the polarization: $$\begin{array}{rcl}\boldsymbol\nabla V_p&=&\boldsymbol\nabla\left(k_e \displaystyle\int_V\dfrac{\mathbf P\left(\mathbf r^{\prime}\right)\cdot\hat{\mathbf R}}{R^2}\mathrm{d}\tau^{\prime}\right)=k_e \displaystyle\int_V\mathbf P\left(\mathbf r^{\prime}\right)\cdot\underbrace{\boldsymbol\nabla\left(\dfrac{\hat{\mathbf R}}{R^2}\right)}_{4\pi\delta^3\left(\mathbf R\right)}\mathrm{d}\tau^{\prime}\\&=&\underbrace{4\pi k_e }_{1/\epsilon_0}\displaystyle\int_V\mathbf P\left(\mathbf r^{\prime}\right)\delta^3\left(\mathbf r-\mathbf r^{\prime}\right)\mathrm{d}\tau^{\prime}=\dfrac{\mathbf P\left(\mathbf r\right)}{\epsilon_0}\end{array}$$

Substituting in $$(1)$$ and multiplying the equation by $$\epsilon_0$$ results $$$$\epsilon_0\mathbf E+\mathbf P=-\epsilon_0\boldsymbol\nabla V_l$$$$

The member on the left is usually abbreviated by $$$$\mathbf D=\epsilon_0\mathbf E+\mathbf P$$$$ which we called displacement vector.

In magnetostatics I have not seen in books that they do that in this way, if not by adding the currents of magnetization and free. Although this serves I would like to do it by the superposition principle and I would want to do the same:

Starting from $$\begin{array}\mathbf A_\text{m}\left(\mathbf r\right)=k_m\displaystyle\int_V\dfrac{\mathbf M\left(\mathbf r^{\prime}\right)\wedge\hat{\mathbf R}}{R^2}\ \mathrm{d}\tau^{\prime},&(2)\end{array}$$ by the superposition principle, the vector potential outside a magnetized body must be the sum of the vector potentials due to free currents and magnetization. When applying the curl we obtain $$\begin{array}{cl}\boldsymbol\nabla\wedge\mathbf A=\boldsymbol\nabla\wedge\mathbf A_\text{l}+\boldsymbol\nabla\wedge\mathbf A_\text{m}&(3)\end{array}$$

The total magnetic field will result from applying the curl to the total vector potential (sup. pple. again), so that $$\mathbf B=\boldsymbol\nabla\wedge\mathbf A$$. If we apply the curl (with respect to the coordinates of $$\mathbf r$$) to $$(2)$$ we will obtain the magnetic field due to the magnetization of the material: $$$$\boldsymbol\nabla\wedge\mathbf A_\text{m}=\boldsymbol\nabla\wedge\left(k_m\displaystyle\int_V\dfrac{\mathbf M\left(\mathbf r^{\prime}\right)\wedge\hat{\mathbf R}}{R^2}\ \mathrm{d}\tau^{\prime}\right)=k_m\displaystyle\int_V\boldsymbol\nabla\wedge\left(\mathbf M\left(\mathbf r^{\prime}\right)\wedge\dfrac{\hat{\mathbf R}}{R^2}\right)\ \mathrm{d}\tau^{\prime}.$$$$ Expanding the integrand: $$$$\boldsymbol\nabla\wedge\left(\mathbf M\left(\mathbf r^{\prime}\right)\wedge\dfrac{\hat{\mathbf R}}{R^2}\right)=\underbrace{\left(\mathbf M\cdot\boldsymbol\nabla\right)\dfrac{\hat{\mathbf R}}{R^2}}_{(\text{a})} -\underbrace{\left(\dfrac{\hat{\mathbf R}}{R^2}\cdot\boldsymbol\nabla\right)\mathbf M}_{(\text{b})}+\underbrace{\dfrac{\hat{\mathbf R}}{R^2}\left(\boldsymbol\nabla\cdot\mathbf M\right)}_{(\text{c})}-\underbrace{\mathbf M\left(\boldsymbol\nabla\cdot\dfrac{\hat{\mathbf R}}{R^2}\right)}_{(\text{d})}$$$$

Everything that derives from the vector magnetization is null because it only depends on $$\mathbf r^{\prime}$$, so $$(\text{b})$$ and $$(\text{c})$$ are canceled. The term $$(\text{d})$$ is the one that interests: $$$$k_m\mathbf M\left(\boldsymbol\nabla\cdot\dfrac{\hat{\mathbf R}}{R^2}\right)=4\pi k_m \mathbf M\left(\mathbf r^{\prime}\right)\delta^3\left(\mathbf R\right)=\mu_0 \mathbf M\left(\mathbf r^{\prime}\right)\delta^3\left(\mathbf R\right)$$$$

I thought that the term $$(\text{a})$$ would be canceled out but it does not give me null: $$\begin{array}{rcl}\left(\mathbf M\cdot\boldsymbol\nabla\right)\dfrac{\hat{\mathbf R}}{R^2}&=&\displaystyle\sum_{i=1}^3M_i\frac{\partial }{\partial x_i}\displaystyle\sum_{j=1}^3\frac{R_j}{R^3}\mathbf e_j=\displaystyle\sum_{i,j=1}^3M_i\mathbf e_j\frac{\partial }{\partial x_i}\left(\frac{R_j}{R^3}\right)\\&=&\displaystyle\sum_{i,j=1}^3\dfrac{M_i\mathbf e_j}{R^6}\left[R^3\frac{\partial R_j}{\partial x_i}-R_j\frac{\partial R^3}{\partial x_i}\right]\end{array}$$

Separately (With $$\mathbf R=\mathbf r-\mathbf r^\prime=R_1\hat{\mathbf e}_1+R_2\hat{\mathbf e}_2+R_3\hat{\mathbf e}_3$$ and $$R_i=x_i-x_i^\prime$$): $$\begin{array}{rcl}\dfrac{\partial R_j}{\partial x_i}&=&\dfrac{\partial x_j}{\partial x_i}=\delta_{ij}\\\dfrac{\partial R^3}{\partial x_i}&=&\dfrac{\partial R^3}{\partial R}\dfrac{\partial R}{\partial x_i}=3R^2\cdot\dfrac{R_i}{R}=3RR_i\end{array}$$ Substituting: $$\begin{array}{rcl}\left(\mathbf M\cdot\boldsymbol\nabla\right)\dfrac{\hat{\mathbf R}}{R^2}&=&\displaystyle\sum_{i,j=1}^3\dfrac{M_i\mathbf e_j}{R^6}\left[R^3\delta_{ij}-R_j3RR_i\right]=\displaystyle\sum_{i,j=1}^3\dfrac{M_i\mathbf e_j}{R^6}R^3\delta_{ij}-\displaystyle\sum_{i,j=1}^3\dfrac{M_i\mathbf e_j}{R^6}R_j3RR_i\\&=&\dfrac{1}{R^3}\displaystyle\sum_{i=1}^3M_i\mathbf e_i-\dfrac{3}{R^5}\displaystyle\sum_{i=1}^3M_iR_i\displaystyle\sum_{j=1}^3R_j\mathbf e_j=\dfrac{\mathbf M}{R^3}-\dfrac{3}{R^5}\left(\mathbf M\cdot\mathbf R\right)\mathbf R\\&=&\dfrac{1}{R^3}\left[\mathbf M-3\left(\mathbf M\cdot\hat{\mathbf R}\right)\hat{\mathbf R}\right]\end{array}$$

Substituting above and integrating (taking into account that $$\mathbf m=\int_V\mathbf M\ \mathrm{d}\tau^\prime$$) I obtain:

$$$$\mathbf B_\text{m}=-\dfrac{k_m}{R^3}\left[3\left(\mathbf m\cdot\hat{\mathbf R}\right)\hat{\mathbf R}-\mathbf m\right]-\mu_0\mathbf M$$$$ The first term coincides with the magnetic field of a magnetic dipole. I don't understand the reason it is there, in electrostatics we don't have the electric field of a electric dipole.

If $$(\text{a})$$ was null, the latter would only integrate the term $$(\text{d})$$ and obtain $$\boldsymbol\nabla\wedge\mathbf A_\text{m}=-\mu_0\mathbf M\left(\mathbf r\right)$$, so substituting on $$(3)$$ would be $$\mathbf B+\mu_0\mathbf M=\boldsymbol\nabla\wedge\mathbf A_\text{l}$$. But I'd need a minus sign there so that it would be $$\mu_0\mathbf H$$. Yes, I'm definetly sutck.

Well, life is hard and I do not get that, does anyone see the failure?

PS: Who did it, thanks for reading this tedious speech ;).

• If you apply a Lorentz transformation to the electrical relation, you get the relation immediately. – my2cts Dec 28 '18 at 12:43
• @David G. I think, in title the wronh sign. It should be H = B/\mu_0 - M – Sergio Dec 28 '18 at 14:19
• My understanding is that's the definition of $\mathbf{H}$, similarly for $\mathbf{D}$. You can bring in higher order multi-poles into this, but they're usually really small. Jackson's textbook has a decent discussion of this. – Sean E. Lake Dec 28 '18 at 14:40
• using the Helmholtz theorem (en.wikipedia.org/wiki/Helmholtz_decomposition) your question is worked out in detail in Brown: Magnetostatic Principles in Ferromagnetism, pp 18-25 – hyportnex Dec 28 '18 at 14:41
• Under Lorentz boosts D, E and P transform into H, B, M respectively. – my2cts Dec 29 '18 at 23:45

I can go back to sleep calmly, after days of arduous search ... I found it! The calculations were not really bad, the additional term is due to the magnetic potential $$V_\text{m}$$ (the scalar, not the potential vector $$\mathbf A$$).