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Let's consider a single charged elementary particle (i.e. NO internal structure, such as the electron) in its rest frame. Does it produce a magnetic field because of its spin? Would a neutral elementary particle with non-zero spin at rest produce any magnetic field? If yes, what expression should the magnetic field have? Does it make any difference if the particle is a fermion or boson?

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  • $\begingroup$ whether is not it produces the magnetic field depend on your frame of reference. Can you be specific about your frame of reference. $\endgroup$ – Young Kindaichi Oct 25 at 19:03
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    $\begingroup$ "Particle at rest" means "in the frame of reference where the particle is not moving" $\endgroup$ – Antonio19932806 Oct 25 at 19:24
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    $\begingroup$ @YoungKindaichi The question was very clear: in its rest frame. $\endgroup$ – Pieter Oct 25 at 19:25
  • $\begingroup$ #YoungKinDaichi There is no inertial frame in which the magnetic field of a spin, if it has any, vanishes. $\endgroup$ – my2cts Oct 25 at 20:56
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Both electrons and neutrons at rest produce magnetic fields because they have non-zero magnetic moments.

The electron moment is $$ {\boldsymbol \mu}= \frac{eg}{2m} {\bf S} $$ where the spin ${\bf S}$ has magnitude $|{\bf S}|=\hbar/2$ and $g\approx 2$. A point dipole with magnetic moment ${\boldsymbol \mu}$ produces a field with spherical polar components $$ B_r= \frac{\mu_0}{2\pi} |{\boldsymbol \mu}|\frac{\cos\theta}{r^3}\\ B_\theta = \frac{\mu_0}{2\pi} |{\boldsymbol \mu}|\frac{\sin\theta}{r^3} $$ where the moment is aligned along the $z$ axis.

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    $\begingroup$ Because it has a Dirac moment of ${\boldsymbol \mu}= eg\hbar/ 4 m$ with $g\approx 2$. Anything with a magnetic moment produces a field. You can measure the moment through the torque it feels, but the field it produces is there whether you apply an external field or not. What is perhaps surprising is that the neutral neutron also produces a a magnetic field. But then the neutron has charged constituents. $\endgroup$ – mike stone Oct 25 at 19:36
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    $\begingroup$ As OP asked, Can you give an expression for the field produce by the electron? $\endgroup$ – Young Kindaichi Oct 25 at 19:40
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    $\begingroup$ @YoungKindaichi The OP asked five questions, which we discourage. $\endgroup$ – G. Smith Oct 25 at 19:46
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    $\begingroup$ Is the formula you wrote valid also for neutral elementary particles (i.e. no substructure)? Is it valid also for bosons? $\endgroup$ – Antonio19932806 Oct 25 at 20:43
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    $\begingroup$ If you mention electrons and neutrons you should probably mention protons as well, even if they are not "elementary". With respect to that, one might wonder whether quarks have a magnetic moment (if the question makes sense at all). $\endgroup$ – Peter - Reinstate Monica Oct 26 at 12:58
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Elementary particles are never at rest in Quantum Mechanics or QED.

If you take a plane wave of a "free" electron $\text{e}^{\text{i}\bf{p}\bf{x}}$, the rest frame corresponds to $\bf{p}=0$ with arbitrary $\bf{x}$. So the dipole field given above corresponds to a highly localized calssical magnetic dipole, and not to a free elementary particle at rest.

In QM the bound particles with small velocities have interaction terms like $\propto \bf{S}\bf{L}$ and $\propto \bf{S_1}\bf{S_2}$. And although the relative positions of constituent particles are small and "finite", the center of mass of a bound system is still a plane wave with its position incertainty when $\bf{P}=0$.

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Both electrons and protons generate an electric field, and electrons, protons and neutrons generate a magnetic field.

The equivalent of this statement is the following. Electrons and protons have an electric field and electrons, protons and neutrons have a magnetic field. The second statement better points to the fact that both fields are intrinsic (existing under any circumstances).

What is decisive is that the electron was first identified as the carrier of the electric field. And the daily usefulness of the electron is given by the separation of the (electric) charges by a potential difference.

What is often ignored is the fact that all potential differences we create today to obtain a current are based on the interaction between the magnetic dipole of the electron (its magnetic field) and the magnetic field of generators. The electrons magnetic field is involved in our daily live as well as the electric field.

The simple fact that the magnetic dipoles could be aligned by subatomic particles and that in some materials this alignment is self-holding, shows us that subatomic particles have a magnetic field at rest. Read about permanent magnets.

The calculation of a magnetic moment by the rotation of an electron was an unsuccessful attempt. The idea for such an approach came from the knowledge of magnetic field induction. Accelerated electrons of a current in a coil generate a magnetic field.

What we accept for a permanent magnet - the alignment of magnetic dipoles - is overshadowed for accelerated electrons. The acceleration forces the magnetic dipoles of the electrons to align.

Both the spintronic and any chemical process is accompanied by the interaction of the magnetic dipoles of the subatomic particles involved. Call it spin or call it magnetic dipoles, the comparison between these two contents for each phenomenon of electromagnetism will show you the usefulness of this or that concept.

In short, subatomic particles have an intrinsic magnetic dipole. Protons and electrons are both, charges and magnets.

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  • $\begingroup$ Thank you for the answer about electrons, but I'd like to know also about neutral elementary (NO substructure) particles with spin (imagine a long lived $Z^o$ boson). Protons and neutrons are not elementary. $\endgroup$ – Antonio19932806 Oct 26 at 11:37
  • $\begingroup$ Antonio, I’ve read the question to fast. Sorry. $\endgroup$ – HolgerFiedler Oct 26 at 12:53

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