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Using the magnetic current model, the force on a magnetic dipole, commonly derived in textbooks, is found to be: $$ \mathbf{F} = \nabla(\mathbf{m} \cdot \mathbf{B}) \tag{1} $$ If the magnetic pole ("Gilbert") model is used, the force is instead found to be: $$ \mathbf{F} = (\mathbf{m} \cdot\nabla) \mathbf{B}\tag{2} $$ (similar to that for the electric dipole).

Using the vector identity $$ \nabla(\mathbf{m} \cdot \mathbf{B}) = \mathbf{m} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{m}) + (\mathbf{m} \cdot\nabla) \mathbf{B} + (\mathbf{B} \cdot\nabla) \mathbf{m} $$ if $\mathbf{m}$ has no spatial dependence (i.e. $\mathbf{m} \neq \mathbf{m}(\mathbf{r})$) and $\nabla \times \mathbf{B} = \mathbf{0}$ then the two expressions are equivalent. However, if $\mathbf{m}$ did have spatial dependence which is the correct equation to use? In fact, in general which is the "correct" formula (I suspect $(1)$ since the magnetic poles do not exist)?

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The use of these two force equations depends how you're modelling your system. There are two ways you can do this$^1$.

  1. If the magnetic dipole is modeled as a current loop (which is how it's usually done) then $$\bf F=\nabla(m\cdot B)$$

  2. If the magnetic dipole is modelled as two separated "magnetic charges" then $$\bf F=(m\cdot\nabla)B$$ where $\bf m$ is the magnetic moment.

The second expression is analogous to the electric force between two separated electrically charged particles (two electric monopoles) in an electric field where $$\bf F=(p\cdot \nabla\mathbf) E$$ and $\bf p$ is the electric dipole moment and there is a spatial dependence as you say.

Your concern that there exists no magnetic monopoles is justified, but as state earlier, which equation you use depends on what type of system you're modelling.

$^1$A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles as the size of the source is reduced to zero while keeping the magnetic moment constant. It is a magnetic analogue of the electric dipole, but the analogy is not perfect. In particular, a true magnetic monopole, the magnetic analogue of an electric charge, has never been observed in nature. However, magnetic monopole quasiparticles have been observed as emergent properties of certain condensed matter systems.

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You seem to have a misconception about what $\nabla (m \cdot B)$ means, which is understandable because the notation is confusing. The intent is that $m$ is a constant with respect to $\nabla$, and the differentiation only applies to $B$. To be more precise, suppose we have some material with a magnetization $M$, which may vary over its volume. To get the total force, you have to integrate $\nabla(M \cdot B)$ over space. This actually means

$$F = \int \nabla(M(x) \cdot B) \, \mathrm{d}x$$

That is, at each point $x$, we form the function $g(y) = M(x) \cdot B(y)$ and take the gradient of $g$ with respect to $y$. With respect to $y$, $M(x)$ is a constant and only $B$ varies.

If you differentiate $M$ as well, you unambiguously end up with the wrong answer (regardless of whether you have Ampère or Gilbert dipoles).

So in fact it doesn't matter whether $M$ varies with space or not: such variation does not give rise to any difference between the two formulas $\nabla(M \cdot B)$ and $(M \cdot \nabla)B$, as long as you interpret the first one correctly.

The only difference comes from the $m \times (\nabla \times B)$ term, so if you can do an experiment that lets you measure the force on a magnetic dipole in a region of space where $J \ne 0$, you can determine whether $\nabla(m \cdot B)$ or $(m \cdot \nabla)B$ is actually the correct formula. For the neutron, it is in fact $\nabla(m \cdot B)$ that gives the correct result, lending strong support to the hypothesis that the neutron is an Ampère dipole (not Gilbert). See Griffiths, 1992.

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  • $\begingroup$ So there is no way of knowing a priori which model is correct for a given physical situation and it has to be verified by experiment? $\endgroup$
    – user246795
    Mar 31, 2022 at 13:25
  • $\begingroup$ @user246795 It's like you said in the question. For Ampère dipoles, the first formula is correct (as you said, it is commonly derived in textbooks). For Gilbert dipoles, the second formula is correct by analogy with the electric case, which is also commonly derived in textbooks. However, we have no a priori way of knowing whether particles found in nature are Ampère or Gilbert dipoles. $\endgroup$
    – Brian Bi
    Mar 31, 2022 at 13:27
  • $\begingroup$ That makes sense, thanks. So just to check, you shouldn't differentiate $\textbf{m}$ in the form $\nabla(\textbf{m} \cdot \textbf{B})$ even if it is a function of position - is it more general to just say that $F_i = m_j \frac{\partial B_j}{\partial x_i}$? $\endgroup$
    – user246795
    Mar 31, 2022 at 13:30
  • $\begingroup$ @user246795 Yes, exactly, that would be a way of writing it that makes it clear. $\endgroup$
    – Brian Bi
    Mar 31, 2022 at 13:39
  • $\begingroup$ @user246795 I'm afraid I don't understand. $\endgroup$
    – Brian Bi
    Mar 31, 2022 at 13:51

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