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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 \begin{equation} 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} \end{equation}

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 \begin{equation}\epsilon_0\mathbf E+\mathbf P=-\epsilon_0\boldsymbol\nabla V_l\end{equation}

The member on the left is usually abbreviated by \begin{equation}\mathbf D=\epsilon_0\mathbf E+\mathbf P\end{equation} 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: \begin{equation}\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}.\end{equation} Expanding the integrand: \begin{equation}\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})}\end{equation}

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: \begin{equation}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)\end{equation}

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:

\begin{equation}\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 \end{equation} 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 ;).

  • 1
    $\begingroup$ If you apply a Lorentz transformation to the electrical relation, you get the relation immediately. $\endgroup$
    – my2cts
    Commented Dec 28, 2018 at 12:43
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    $\begingroup$ @David G. I think, in title the wronh sign. It should be H = B/\mu_0 - M $\endgroup$
    – Sergio
    Commented Dec 28, 2018 at 14:19
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    $\begingroup$ 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. $\endgroup$ Commented Dec 28, 2018 at 14:40
  • 2
    $\begingroup$ 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 $\endgroup$
    – hyportnex
    Commented Dec 28, 2018 at 14:41
  • 1
    $\begingroup$ Under Lorentz boosts D, E and P transform into H, B, M respectively. $\endgroup$
    – my2cts
    Commented Dec 29, 2018 at 23:45

1 Answer 1


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 $\mathbf A$).

Whoever has been thinking about this like me, can consult López Rodríguez, V., Montoya Lirola, M. &Pancorbo Castro, M. (2016). Electromagnetismo II. Madrid: Universidad Nacional de Educación a Distancia. Tema 1. CAMPO MAGNÉTICO EN MATERIALES - 6.1 Potencial escalar magnético.

PS:Sorry, the only one I found was in spanish. I did not see it in any english textbooks. But, it has a derivation practically equal to mine.

Thank you very much to anyone who had the trouble of sending an answer. Happy New Year to everyone!


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