Many physics textbooks and websites state that electric current is the source of all magnetism. They go on to support this statement by saying that electrons move around nuclei AND they have spin. Later, and almost without exception, the book or site will warn us that the term spin is not to be taken too literally because the electron is not actually spinning like a top, but is instead an intrinsic property of the electron. My question is: aren't these sources contradicting themselves? If spin is not associated with moving charge, then how can it be claimed that all magnetism arises from electric current? It seems that if spin is an intrinsic property of electrons (and other particles), then in this case magnetism is also just an intrinsic property of them.


2 Answers 2


There are much easier, classical counterexamples to this claim. For example, a magnetic field can be created by a time-varying electric field, through induction. Context matters. These books are saying things like this in the context of a presentation of classical electromagnetism (in which there is no spin), and they're starting from magnetostatics (so there is no induction).

Similarly, a freshman physics textbook will present Newton's law of gravity without apologizing and saying that it's superseded by general relativity. They can't tell you everything before they even get started, and they're also mirroring the way that science advances, with new theories staying backward-compatible with older ones through the correspondence principle.

By the way, to me the word "source" has the connotation that it's a place where field lines begin or end. In this sense, current is not actually a source of the magnetic field, but you could say it's the cause of it.

  • $\begingroup$ Isn't a time-varying electric field inevitably caused by a current, though? Like, moving around charges? $\endgroup$ Commented Mar 17, 2018 at 1:41
  • $\begingroup$ @JahanClaes: No. For example, we have solutions to Maxwell's equations that consist of electromagnetic waves in a vacuum. $\endgroup$
    – user4552
    Commented Mar 17, 2018 at 15:11
  • 1
    $\begingroup$ Right, but in a finite system those waves had to come from somewhere. Moving charges. $\endgroup$ Commented Mar 17, 2018 at 17:25

It depends what you mean by current. If by this you mean the movement of charges, then certainly yes, it is wrong.

Current for electromagnetism, in a very general way, means the derivative of the interaction term with respect to the EM potential. For spinor fields, this corresponds to the vector current of the field

$$J^\mu = -\frac em \bar \psi \gamma^\mu \psi$$

which features in the Maxwell equation as

$${F^{\mu\nu}}_{,\mu} = \frac{e\mu_0}{m} \bar \psi \gamma^\nu \psi$$

The induction of a magnetic field from this is given by the relation $B_{i}=-{\frac {1}{2}}\epsilon _{ijk}F^{jk}$, which from

\begin{eqnarray} {F^{\mu j}}_{,\mu} &=& \frac{e\mu_0}{m} \bar \psi \gamma^j \psi\\ &=& - {F^{tj}}_{,t} + {F^{ij}}_{,i}\\ &=& \frac{\partial E}{\partial t} + \vec \nabla\times \vec B \end{eqnarray}

gives us the proper expression for the Ampère law, with the current $\dfrac{e\mu_0}{m} \bar \psi \gamma^j \psi$. We can decompose this expression using the Gordon decomposition :

$$J^\mu = -\frac{e\hbar}{m^2} \left[ \frac i2 (\frac{\partial \bar \psi }{\partial x^\nu}\psi - \bar \psi \frac{\partial \psi }{\partial x^\nu}) + \frac{\partial}{\partial x^\nu} (\bar \psi \sigma^{\mu\nu} \psi) \right] $$

Or, for the spacelike part,

$$\vec J = -\frac{e\hbar}{m^2} \left[ \frac i2 ((\vec \nabla \bar \psi)\psi - \bar \psi \vec\nabla \psi ) - \frac{\partial}{\partial t} (\bar \psi \sigma^{i0} \psi) + \partial_j (\bar \psi {\sigma}^{ij} \psi) \right] $$

The first part of the current is simply the current, as is generally accepted (it is the same expression as the current in the Schrodinger equation, for instance). The second part is the time variation of the polarization tensor, while the third part is the derivative of magnetization tensor. The magnetization tensor is the part containing the dependance on spin, which is the source of the magnetic field from the spin. It can be rewritten in a more readable way as

$$\vec J = -\frac{e\hbar}{m^2} \left[ \vec{j}_{\text{free}} - \frac{\partial}{\partial t} \vec{P} + \vec{\nabla} \times \vec{M} \right] $$

Of course, it is still wrong to say that the magnetic field stems only from currents. There are at least two cases where this is not the case : sourceless electromagnetic waves and, if the underlying spacetime isn't simply connected, it is possible to have EM fields stemming from the geometry of the spacetime directly (this was the basis of geometrodynamics).

  • $\begingroup$ How can a sourceless EM wave exist in the real world? Aren't these waves always "born" from a source, this being electric charge or current? This is what the Maxwell equations tell us. $\endgroup$
    – David S
    Commented Mar 18, 2018 at 3:00
  • $\begingroup$ No, the Maxwell equations have perfectly reasonable solutions where some EM waves have no source at all. Then it only depends on the initial conditions of the universe. $\endgroup$
    – Slereah
    Commented Mar 18, 2018 at 8:34

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