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It is known that an electron in a magnetic field will undergo gyromagnetic rotation of its spin magnetic dipole moment around the $B$ vector of the external magnetic field.

Also in addition to the gyromagnetic rotation of its magnetic moment, if the electron has an initial translational velocity vector $v$ perpendicular to the magnetic field vectors it will be deflected by the magnetic part of the Lorentz force $F_{M}$ to a spiral trajectory path motion inside the magnetic field.

However I have like to know?

Assuming the electron is inserted in a static external homogeneous magnetic field inside a vacuum environment after the field is switched on and made somehow to stand still (hold in place) $v=0$ and then released, what kind of trajectory will it make?:

1) Circular loop motion under a homogeneous external static magnetic field.

2) Spiral motion attracted to one of the magnetic poles N or S of the external homogeneous magnetic field source.

3) Nothing, no translational motion at all. It will remain in its initial place immovable and only gyromagnetically rotate its spin magnetic moment axis at the Larmor frequency $ω=-γΒ$.

4) Same cases as above but this time the external static magnetic field is non-homogeneous for example the field on one of the poles of a permanent magnet?

Analysis of your explanations would be welcomed.

free electron vs magnet

Update 19 Feb 2022:

I don't know the history of the Lorentz force equation, but I believe it is a pure theoretical and never really truly verified by experiment for the condition $v=0$ since it is very hard to maybe impossible to make an electron to stand still. The physics IMO were made to match this product mathematical equation (i.e. concerning magnetic part of the Lorentz equation) for $v=0$. That's all. However, although a classical equation it does not match classical physics. A permanent magnet pole (i.e. non-homogeneous field) given enough strength should overcome the intrinsic spin angular momentum of a single stationary electron in a vacuum resisting linear motion and attract it. This should be a very important QED experiment because it would prove magnetism as being not an emergent phenomenon of electron's translational motion but that both electric and magnetic phenomenon thus electromagnetism, originate from the intrinsic unknown mechanics of the electron and both being intrinsic phenomena and properties of the electron.

Electromagnetism IMO is an intrinsic phenomenon of the electron.

Update 20 Feb 2022:

Still no a definitive physical explanation of why the stationary electron (i.e. Quantum Magnet) will be or will be not attracted towards the magnet shown. Both electron and magnet are stationary.

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    $\begingroup$ If you're talking about electron spin, I assume you're going to need to invoke quantum mechanics, and the idea of a "path" that the electron takes becomes somewhat fraught. $\endgroup$ Commented Feb 18, 2022 at 19:39
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    $\begingroup$ Regarding your timestamped update: we have a tool for that information. Not a big enough deal for an edit by itself, but if you happen to change the post again, you might clean that up. $\endgroup$
    – rob
    Commented Feb 19, 2022 at 17:32

3 Answers 3

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A magnetic field exerts no force upon a stationary charge such as an electron. If I understand your initial conditions correctly, they involve exactly that, a magnetic field and a stationary charge. Thus, there will be no motion.

The force exerted upon a charge by a magnetic field is given by

$$\vec{F}_{magnetic} = q(\vec{v} \times \vec{B})$$

where $q$ is the charge, $\vec{v}$ is the velocity vector of the charge, and $\vec{B}$ is the magnetic field.

The above formula is a direct result of the equation for the Lorentz force, i.e. the force upon a charge by a combination of Electric and Magnetic fields.

$$\vec{F} = q(\vec{E}+(\vec{v} \times \vec{B}))$$

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  • $\begingroup$ You are forgeting torquehttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html $\endgroup$
    – anna v
    Commented Feb 19, 2022 at 12:08
  • $\begingroup$ @annav Unless I learned my mechanics incorrectly, torque will not cause a initially stationary particle to have a non-stationary trajectory. $\endgroup$ Commented Feb 19, 2022 at 12:29
  • $\begingroup$ @ it is not only mechanics, but electrodynamics (classical). That is how domains are formed in permanent magnets.nationalmaglab.org/education/magnet-academy/watch-play/… . hyperphysics.phy-astr.gsu.edu/hbase/Solids/ferro.html $\endgroup$
    – anna v
    Commented Feb 19, 2022 at 12:49
  • $\begingroup$ Yes I second on that Lorentz force holds for macroscopic electromagnetism but my question is referring mainly to QED. In the case I have described as mentioned also previously by @Hoody, $\begin{array}{l} \vec{F}=q \vec{V} \times \vec{B}=0 \\ \vec{\tau}=\vec{m} \times \vec{B} \neq 0 \end{array}$. My question is if this non-zero torque besides the gyromagnetic rotation of the spin magnetic dipole moment of the electron around the z-axis $B$ field vectors will make the electron at least to move with an additional orbital, circular loop, motion inside the $B$ field? Αbout a non-homogeneous Β? $\endgroup$
    – Markoul11
    Commented Feb 19, 2022 at 13:11
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    $\begingroup$ @Markoul11 The Lorentz force equation has never been verified with an electron perfectly at rest because we have never seen an electron perfectly at rest. In that sense, applying the equation to an electron at perfect rest is an extrapolation. However, any theory of what happens to an electron at perfect rest will also necessarily be an extrapolation unless we actually see electrons at perfect rest. Is there another theory, with a different equation that shows correct results for all observables, but when extrapolated to perfect rest gives a different result? I can't see how. $\endgroup$ Commented Feb 19, 2022 at 16:32
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To deal with charged particles in a magnetic field in quantum mechanics you need to consider Landau's analysis. Basically the system breaks up into a bunch of harmonic oscillator levels with the cyclotron frequency $\omega_c$. If you want to consider a 'stationary electron' this would presumably be the lowest harmonic oscillator level, but as usual this has some zero point energy $\hbar \omega_c/2$ which is associated to the cyclotron frequency.

As far as the spin is concerned, it works the same way as the case where you don't consider the particle's position and momentum. The spin up and down states (relative to the direction of the magnetic field) are eigenstates with different energies, so a superposition of them will exhibit Larmor precession. The only quirk is that the energy gap between the spin states is precisely equal to the difference between the Landau levels $\hbar \omega_c$ (for an electron with g-factor $g=2$).

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  • $\begingroup$ Fine, but this is all known. I'm asking specifically here if there will be any kind of translational motion of the electron? I cannot find any experiment with the experimental conditions described here, carried out under a very strong magnetic field that confirms the $v=0$ -> $F_{M}=0$ -> zero translational motion, conclusion of the Lorentz equation? $\endgroup$
    – Markoul11
    Commented Feb 19, 2022 at 13:29
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    $\begingroup$ @Markoul11, I guess you need to read behind the lines a little bit. Classically anything moving in a uniform magnetic field spirals around with the cyclotron frequency. Quantum mechanically, it is ill defined what you mean by stationary electron due to the Heisenberg principle, and you find that the very lowest energy state has an energy related to the cyclotron frequency. Zero-point motion = translational motion, that bit of the energy wouldn't be there if the magnetic field were only coupled to the spin $\endgroup$
    – octonion
    Commented Feb 19, 2022 at 21:02
  • $\begingroup$ @Markoul11, Also it is rather unclear to me now whether you are asking about classical or quantum physics. I assumed you meant quantum, and that is the answer I gave. $\endgroup$
    – octonion
    Commented Feb 19, 2022 at 21:06
  • $\begingroup$ Why should it be two different answers? I want the actual physics. $\endgroup$
    – Markoul11
    Commented Feb 20, 2022 at 1:29
  • $\begingroup$ @Markoul11 There are problems with making a real electron perfectly stationary. As you reduce its momentum you also need to reduce the uncertainty in its momentum. But then the HUP says the uncertainty in its position increases. So as your electron's momentum approaches zero its position becomes more and more delocalised. $\endgroup$
    – PM 2Ring
    Commented Feb 20, 2022 at 2:49
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Electromagnetism IMO is an intrinsic phenomenon of the electron.
We are the victim of our own historical conceptions of the electron as primarily an electric charge. In fact, however, the electron has an inherent constant of a magnetic moment (NIST / Fundamental Physical Constants / electron magnetic moment).

It is known that an electron in a magnetic field will undergo gyromagnetic rotation of its spin magnetic dipole moment around the 𝐵 vector of the external magnetic field.
I wouldn't be so sure about that. For a permanent gyroscopic rotation of the spin magnetic dipole moment around the 𝐵 vector of the external magnetic field, an energy source is needed.

You yourself associate spin and magnetic dipoles and this is also plausible considering the above mentioned fundamental and intrinsic magnetic field of the electron. The simplest conclusion is that when an electron remains in an invariant magnetic field, the magnetic dipole of the electron simply aligns with the external magnetic field.

… if the electron has an initial translational velocity vector perpendicular to the magnetic field vectors it will be deflected by the magnetic part of the Lorentz force to a spiral trajectory path motion inside the magnetic field. However I have like to know?

The following happens:

  1. the magnetic dipole of the moving electron is aligned in the direction of the external magnetic field.
  2. this leads to the emission of electromagnetic radiation.
  3. the emitting photons, which are known to have a moment, deflect the photon sideways and the alignment of the magnetic dipole is disturbed.
    This process repeats periodically until the entire kinetic energy of the electron is consumed.

… to a spiral trajectory path motion inside the magnetic field.
Even more detailed spiral path is a path that consists of nothing but slices of tangerine.

Assuming the electron is inserted in a static external homogeneous magnetic field inside a vacuum environment after the field is switched on and made somehow to stand still (hold in place) and then released, what kind of trajectory will it make?
3) Nothing, no translational motion at all. It will remain in its initial place immovable and nothing rotates gyromagnetically.

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  • $\begingroup$ The electron gyro magnetically rotates its spin magnetic moment vector continuously around the vector of the z-axis component of the external B field on the perimeter of rotational cone (spin up or spin down) with the cone surface set fixed at the magic angle $θ_{m}=54.7$ degrees to the z-axis. This gyromagnetic rotation goes forever as long the electron is inside the B field. This is common knowledge and experimentally verified. It is what is referred a "alignment" of the electron's magnetic moment with the external magnetic field B but the two vectors never actually coincide in space. $\endgroup$
    – Markoul11
    Commented Feb 19, 2022 at 19:10
  • $\begingroup$ Magic Angle: en.wikipedia.org/wiki/Magic_angle $\endgroup$
    – Markoul11
    Commented Feb 19, 2022 at 19:12
  • $\begingroup$ Hi Markoul11. Are you sure that these are free electrons? Aren't they electrons that are bound to atomic structures? Besides, MRI techniques are always about switching the magnetic field on and off and the relaxation reaction of the subatomic particles. Or did I misunderstand your question and you are referring to bound electrons? $\endgroup$ Commented Feb 19, 2022 at 19:32
  • $\begingroup$ The magic angle 54.7 referenced previously is an intrinsic property of the electron and its quantum 1/2 spin and is independent of if the electron is in a bound atomic state or not thus free: qsstudy.com/… . Only the Larmor frequency changes according to B field strength $ω=-γΒ$ but this angle remains fixed for fermions (i.e. 1/2 spin). Nobody has a physical explanation for this fixed angle intrinsic property of the electron? $\endgroup$
    – Markoul11
    Commented Feb 19, 2022 at 19:43
  • $\begingroup$ I have now read carefully. It is about bound electrons, or is it not? $\endgroup$ Commented Feb 19, 2022 at 20:45

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