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I would like to know if the $H$ and $M$ fields are always in the same (or opposite) direction. If not, are there special conditions where it is always true?

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The relationship between $\mathbf H,\mathbf M,$ and $\mathbf B$ is $$\mathbf B = \mu_0(\mathbf H+\mathbf M)$$

Under certain conditions, some systems exhibit linear response, in which $\mathbf M = \chi \mathbf H$ for some matrix $\chi$ called the magnetic susceptibility tensor. In other words, $$\pmatrix{M_x\\M_y\\M_z} = \pmatrix{\chi_{xx}& \chi_{xy} & \chi_{xz} \\ \chi_{yx}& \chi_{yy} & \chi_{yz} \\ \chi_{zx}& \chi_{zy} & \chi_{zz}}\pmatrix{H_x\\H_y\\H_z}$$

In turn, some of those systems have a magnetic susceptibility tensor of the form $$\chi = \pmatrix{\chi_0 & 0 & 0\\ 0 & \chi_0 & 0 \\ 0 & 0 & \chi_0} =\chi_0 \pmatrix{1&0&0\\0&1&0\\0&0&1}$$ where $\chi_0$ is called the scalar susceptibility. It is only for these systems that $\mathbf M$ and $\mathbf H$ (and therefore $\mathbf B$) are (anti)parallel.


So in summary, $\mathbf H, \mathbf M,$ and $\mathbf B$ are all (anti)parallel if we are considering a system which exhibits linear response response and the magnetic susceptibility tensor is equal to a scalar times the identity matrix, a phenomenon called magnetic isotropy. In all other cases, $\mathbf H, \mathbf M,$ and $\mathbf B$ are in different directions.

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  • $\begingroup$ can you have an isotropic second order response? I have 3 rank-4 isotropic tensors in my notes: $\delta_{ij}\delta_{kl}, \delta_{ik}\delta_{jl}, \delta_{il}\delta_{jk}$, 15 rank-6, and either 91 or 105 rank-8...depending on the FICT or Motzkin number. $\endgroup$
    – JEB
    Nov 5, 2023 at 16:45
  • $\begingroup$ Hello @JEB, no that is not possible. Consider the expression $A_i = \chi^{(2)}_{ijk}H_j H_k$. If $\chi^{(2)}$ is isotropic, then it is invariant under rotations. Without loss of generality, take $\mathbf H=H \hat z$, so $A_i = \chi^{(2)}_{izz}H^2$. If we perform any rotation about the $\hat z$ axis, $\mathbf H$ is left unchanged, which implies that $A_x=A_y=0$. If we perform a $\pi$ rotation about $\hat x$, then $\mathbf H\mapsto -\mathbf H$ and $\mathbf A \mapsto -\mathbf A$. But the RHS of the prior expression (and so $\mathbf A$) is left unchanged. The only possibility is $\mathbf A=0$. $\endgroup$
    – J. Murray
    Nov 9, 2023 at 2:44
  • $\begingroup$ @JEB More generally (but in the same way), one can convince themselves that if $\mathbf A = \mathbf T(\mathbf B,\mathbf C)$ for some tensor $\mathbf T$ which is invariant under rotations, then $T_{ijk} \propto \epsilon_{ijk}$. Extending the argument further, it is not possible to have an isotropic $n^{th}$ order response with $n$ even. $\endgroup$
    – J. Murray
    Nov 9, 2023 at 2:47
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Contrast the vacuum equation $$ \nabla\times\mathbf B = \frac 1{\mu_0}\left( \mathbf J + \epsilon_0\frac{\partial\mathbf E}{\partial t} \right) $$ against the matter equation $$ \nabla\times\mathbf H = \mathbf J_f + \frac{\partial\mathbf D}{\partial t}. $$

The definitions of the matter fields \begin{align} \mathbf D &= \epsilon_0 \mathbf E + \mathbf P \\ \mathbf H &= \frac1{\mu_0}\mathbf B -\mathbf M \end{align} put all of the matter effects into the matter response fields $\mathbf P$ and $\mathbf M$. The fields $\mathbf D$ and $\mathbf H$ depend only on the "free" charges and currents that an experimentalist can control, and are independent of the "bound" charges and currents which manifest as electric or magnetic polarization.

Imagine putting a compass needle in a magnetic coil. If the coil is off, we have $\mathbf H=0$ but $\mathbf M\neq0$. When you turn the coil on, your nonzero $\mathbf H$ exerts a torque on the compass needle — but it may be a torque you can counteract with your finger, to make the compass point the "wrong" way. If you release the needle and it sproings parallel to $\mathbf H=0$, that demonstrates that you had previously maintained $\mathbf M$ nonparallel to $\mathbf H$.

As other answers have pointed out, there exist "linear" materials where the matter response is to create parallel (or antiparallel) polarizations on timescales which are fast enough to ignore in macroscopic experiments. The timescales aren't zero, however, which is part of the reason that indices of refraction depend on frequency.

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They are for e.g. diamagnets and paramagnets. However, ferromagnets (and antiferromagnets) maintain their magnetization, even when the magnetic field goes to zero (hysteresis). Therefore, in those materials, no such restriction can be made.

TL;DR: for some materials yes, for others (such as iron) no.

I hope this helps!

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  • $\begingroup$ thank you for your answer! So about the ferromagnets, I think we have $H=0$ when there is no external field, so they can not be ‘parallel’. However, what about the case when there is an external field? Are they parallel then? $\endgroup$
    – Riemann
    Nov 5, 2023 at 12:06
  • $\begingroup$ I was also thinking about the volume magnetic susceptibility. Does $M= \chi_v H$ imply that $M$ and $H$ are parallel? Does this equation not hold for ferromagnets? $\endgroup$
    – Riemann
    Nov 5, 2023 at 12:08

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