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Since fermions like free electrons have an intrinsic spin magnetic dipole moment shouldn't they when subjected to an external static magnetic field experience a magnetic force even assuming they are stationary (stationary in a vacuum for example)?

I understand that because their angular momentum they may not be moved by an external magnet and resist any linear motion but saying that they will experience zero magnetic force $F_{(M)}=q(\mathbf{v} \times \mathbf{B})$ when they are stationary $v=0$, I find peculiar and had like to know the physical explanation for this if any?

Are there any cases, experiments or theory that prove otherwise?

Question Update 17 Feb 2022:

According to an answer here by @Hoody and other comments, a free unbound hypothetical stationary electron under a static homogeneous magnetic field would not experience a force that would force it to translational motion but nevertheless would experience a torque the would change its initial magnetic moment $\vec{m}$ orientation in space:

$\begin{array}{l} \vec{F}=q \vec{V} \times \vec{B}=0 \\ \vec{\tau}=\vec{m} \times \vec{B} \neq 0 \end{array}$

However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics, under this torque not forced to any translational motion (i.e. orbital or linear)?

Is this above paragraph actually true or just never experimentally or analytically verified up to now?

1) What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?

2) What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field? Will the same behavior occur with a homogeneous magnetic field and if not why?

If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?

Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?

Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?

Therefore the origin of the electromagnetism phenomenon in general must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?

As an implication of the above mentioned and conclusions reached is that the classical electromagnetism, magnetic force equation of the Lorentz force:

$$F_{(M)}=q(\mathbf{v} \times \mathbf{B})$$ for initial $v=0$ should be amended with a torque factor.

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  • $\begingroup$ When a force acts on an object, what happens to the body? $\endgroup$ Commented Feb 16, 2022 at 19:32
  • $\begingroup$ Hello. It accelerates unless it is hold by a larger force. $\endgroup$
    – Markoul11
    Commented Feb 16, 2022 at 19:34
  • $\begingroup$ So if the stationary fermion (like electron) is only in the presence of a magnetic field and nothing else, it does not accelerate, because that acceleration could only come from the Lorentz force, and that force is zero. Since it does not accelerate, we say that the force is zero. Is that peculiar? $\endgroup$ Commented Feb 16, 2022 at 19:37
  • $\begingroup$ Saying that force is zero because it does not accelerate is different from saying it does not experience any force at all. $\endgroup$
    – Markoul11
    Commented Feb 16, 2022 at 19:39
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    $\begingroup$ "IMO the Lorentz force theory is unproven unless it is experimentally applied at a truly stationary charge (...)" first of all, a truly stationary charge does not exist in quantum mechanics, as you are surely familiar with the Heisenberg uncertainty principle. So you have to clarify what you mean here. And, on the contrary, scientific theories can never be proven. They are not mathematical theorems. What you can do is try to disprove them. I am not aware of any experiment that disproves the Lorentz force, or the modern refinements like QED. $\endgroup$ Commented Feb 16, 2022 at 19:59

4 Answers 4

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Most generally, the force $\vec F_m$ and torque $\vec \tau$ due to the magnetic field $\vec B$ acting on a particle with a charge $q$ and magnetic moment $\vec\mu$, moving with velocity $\vec v$ can be written as \begin{align} \vec F_m &= q \vec v \times \vec B + \vec\nabla (\vec B\cdot\vec\mu), &&&&(1)\\ \vec\tau &= \vec r \times \vec F_m + \vec\mu \times \vec B. &&&&(2) \end{align} The first term in (1) is the standard Lorenz term, while the latter is usually not mentioned as a fundamental law since classical electrodynamics did not assume point-like particles with spin and magnetic moment but rather modeled magnetic dipoles by an electric current loop, behaviour of which in $\vec B$ can be derived just from the first term (focres act on parts of a the electric circuit because there are moving charges inside).

In Eq. (2), the first term $\vec r \times \vec F$ is the usual textbook definition of a torque in the case of a force acting on a point particle - it is just a tool to understand how will the tracejtory of the particle be curved due to $\vec F_m$ (i.e., it tells us something about orbital motion). Apparently, it depends on the choice of your coordinates. In contrary (but analogously to the previous paragraph), the second term in Eq. (2), $\vec \mu \times \vec B$, is an entirely new piece of physical law for a point-like particle. It is not present usual chapters on classical mechanics since they do not assume that that particles have spin (it is hard to imagine a rotating point-like particle). Again, if you modeled magnetic dipoles by an electric-current loop, you can derive the second term in (2) from $\vec F_m = q \vec v \times \vec B$.

However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics, under this torque not forced to any translational motion (i.e. orbital or linear)?

Now, let us have a look at what forces and torques actually do: Changes in motion of the center-of-mass of a body are always determined solely by the force, according to the famous Newton's law $\vec F = m \vec a$. Denoting the body's momentum by $\vec p=m\vec v$, this law can be also reformulated as $$\vec F = \frac{\mathrm d\vec p }{\mathrm dt}. \qquad (3)$$ In contrary, torque causes changes in angular momentum $\vec J$ (i.e., in "rotation"): $$ \vec \tau = \frac{\mathrm d \vec J}{\mathrm d t}. \qquad (4)$$ Generally, angular momentum of a moving particle with spin is a sum of two terms, $$\vec J = \vec L + \vec S$$ where $\vec L = \vec r \times \vec p = m\vec r \times \vec v$ is the usual orbital angular momentum while $\vec S$ is the particle's internal angular momentum, i.e., spin. Now, if $\vec F_m = 0$, Eq. (3) tells us that there will be no acceleration. This is perfectly consistent with Eq. (4) which yield $$ \vec \tau = \frac{\mathrm d\vec J}{\mathrm dt} = \frac{\mathrm d\vec L}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt} = \frac{\mathrm d(\vec r \times \vec p)}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt} =\vec v \times (m\vec v) + \vec r \times \vec F + \frac{\mathrm d\vec S}{\mathrm dt} =0+0+\frac{\mathrm d\vec S}{\mathrm dt}. $$ I.e., in the absence of a magnetic force, the magnetic torque acts solely on the spin of the particle. Using Eq. (2): $$\vec \mu \times \vec B =\vec \tau = \frac{\mathrm d\vec S}{\mathrm dt}. $$ Btw., notice that since the particle's magnetic moment $\vec \mu$ is proportional to its spin $\vec S$, the torque is always perpendicular to $\vec S$ and hence may cause a change in the orientation of the spin (i.e. precession) but not of its size (i.e. electrons always have spin of 1/2 $\hbar$ - but let's not dig into the quantum world in this discussion).

Is this above paragraph actually true or just never experimentally or analytically verified up to now?

It is true as far as classical (non-quantum) electrodynamics is true. As noted by others in the comments, a theory cannot be verified but only disproved. Electrodynamics is the most precisely measured part of physics whatsoever.

  1. What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?

Yes, diamagnetism is a property of all materials. But notice that the origin of the torque causing the orbital motion is not the magnetic moment of the electron but the first term, $\vec r \times \vec F_m.$

  1. What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field?

Yes.

Will the same behavior occur with a homogeneous magnetic field and if not why?

No. For a non-moving electron in a homogeneous magnetic field, both terms in Eq. (1) are zero. The electron might only undergo spin precession.

If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?

As explained above, this does not happen in vaccum. In the material, we must be careful about what we mean by stationary. In the quantum language, an electron in its atomic orbital is stationary; however, its classical analogue is that it is somehow circling around the nucleus and hence it undergoes an orbital motion eve initially before you apply the external magnetic field. Hence, the answer is NO.

Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?

Are you asking wheather we can compensate the curving effect of the magnetic force $q\vec v \times \vec B$ by the interaction of the electron's magnetic dipole $\mu$? I think the answer is NO, since both effects are directly proportional to the same field $\vec B$.

Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?

Indeed. If you accept the semi-classical picture of a point-like electron with non-zero spin and magnetic moment, then there are two sources of magnetism (moving charges and particle's spin), which also both respond to an external magnetic field (first and second terms in Eqs. (1) and (2).) Nevertheless, note that in relativistic quantum theories, both "sources of magnetism" naturally emerge from a single physical law (i.e. Lagrangian of quantum electrodynamics).

Therefore the origin of the electromagnetism phenomenon in general must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?

No, not all magnetic phenomena are caused by the magnetic dipoles of elementary particles - see the first terms in Eqs. (1) and (2). Note that pions are charged particles without spin or magnetic moment but their trajectories can still be bent by magnetic fields (and it is an experimental routine).

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  • $\begingroup$ Thank you for your canonical excellent answer to my question but in your equation (1), shouldn't it be there better a parenthesis in the first term? $q(\mathbf{v} \times \mathbf{B})$. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 7:47
  • $\begingroup$ Also are there any references in literature about equations (1) and (2) or are these your own combinatoric derivations? $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 7:52
  • $\begingroup$ quote: "Note that pions are charged particles without spin or magnetic moment but their trajectories can still be bent by magnetic fields (and it is an experimental routine" Doesn't count for combinatoric particles :). In case of combinatoric particles their elementary fermion constitutes like quarks can still be affected since they individually posses 1/2 spin. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 8:18
  • $\begingroup$ For $v=0$ we can delete the first term of eq,1. The second term $\vec\nabla (\vec B\cdot\vec\mu)$ tells us simply for an ideal homogeneous magnetic field that the spin magnetic moment of the electron $μ$ will align with the $B$ field vector and Larmor precess around it continuously. Since an ideal homogeneous B field has no gradient $∇$ you would normally expect that the electron would remain in place just precessing around $B$ vectror. However, the magnetic field $μ$ of the electron is non-homogeneous therefore second term of eq.1 will have a non-zero gradient and the $e$ will move forward. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 9:26
  • $\begingroup$ For the case of the non-homogeneous field $B$ from a permanent magnet the gradient changes continuously since the field has a non-zero curl and an initial hypothetical stationary electron will again Larmor precess and translate in forward motion towards the opposite permanent magnet pole. Free magnets always attract. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 9:34
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I think the confusion here is that you intuitively know that a small rare-earth magnet, released between the poles of a large horseshoe magnet, will be pulled linearly (and will snap into contact with the horseshoe magnet).

But which pole of the magnet will it be pulled to? The answer is the closest one. Regardless of which way the magnet was initially oriented.

What if you released the rare-earth magnet exactly half way between the poles of the large horseshoe magnet? This is difficult to perform because this position is unstable but the answer is that it would not be pulled linearly (not until some tiny perturbation knocks it slightly closer to one pole than the other).

The confusion is that in everyday experiences of a magnetic dipole in a magnetic field, that magnetic field also has a strong gradient. It is stronger near the poles. The only place where the field (of the horseshoe magnet) is approximately homogeneous is exactly half way between the two poles.

Classically, if you place a magnetic dipole in a uniform/homogeneous external magnetic field, you can think of the force upon the north end of the dipole being exactly balanced by the force upon the south end of the dipole, so that the net force is zero. (There may be some torque, to cause the dipole to align with the external field, but no linear net force results.) Whereas if the external magnetic field has a gradient, then the spatial extent of the dipole means that each of its poles will experience a different external field strength, and hence the forces will be slightly out of balance, resulting in a nonzero net force and a linear acceleration. The logic is similar in the quantum case, although without attributing a spatial extent to the dipole. The reason that the spin ½ charged fermion is not accelerated in an external static homogeneous magnetic field is because you constructed the field to be homogeneous, not any reason peculiar to fermions or quantum mechanics.

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  • $\begingroup$ Thank you for this clarification. Yes I believe you are right the Lorentz magnetic force equation I believe describes not the special case for $υ=0$ and $\mathbf{F}=\nabla(\mathbf{m} \cdot \mathbf{B})$ is the correct description for this special case. A complete 100% homogeneous field would generate non-translational motion to an electron charge in this case. $\endgroup$
    – Markoul11
    Commented Jan 20, 2023 at 8:13
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  • If something makes you accelerate, it is a force.
  • If it makes you rotate, it is a moment of force, i.e., a torgue.

A spin-1/2 particle in a homogeneous (!!!) magnetic field will experience a torgue but not a force.

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  • $\begingroup$ That is great news! So what is the F here: $\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}$ in our case? Any bibliography to backup your claim? $\endgroup$
    – Markoul11
    Commented Feb 16, 2022 at 20:07
  • $\begingroup$ If this is true, then the aged physics stereotype that magnetism is created only by moving charges is fallen and magnetism is an exclusive intrinsic property of the electron. $\endgroup$
    – Markoul11
    Commented Feb 16, 2022 at 20:23
  • $\begingroup$ If you imagine the electron as a point with an arrow sticking from in (representing the spin): Homogeneous magnetic field will indeed not move the whole electron but it will act on the arrow, changing its direction. That's what how we can imagine the torgue without a force. The formula $\vec \tau = \vec r \times \vec F$ holds for rigid body, it is inaplicable for a point-like object. $\endgroup$
    – Hoody
    Commented Feb 16, 2022 at 20:33
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    $\begingroup$ If you imagine the electron as a point-like object, you really have to admit that the magnetic dipole is an intrinsic property of the electron, though it does not mean that the whole magnetism is exclusively hidden there. $\endgroup$
    – Hoody
    Commented Feb 16, 2022 at 20:35
  • $\begingroup$ If you imagine the electron as a rotating ball, than it is the circling "pieces of the electron" which represent moving charges which acts like a small closed electric circuit loop, i.e., a dipole moment. These "pieces of the ball" create the electron's own magnetic field (it has a dipole moment indeed). Within this picture, you can imagine the magnetic force acting on the individual pieces of electron, forcing the whole ball to change rotation but not to move as a whole. Since it is ackward to describe electron as a non-zero-radius ball, we claim there is no magnetic force at all. $\endgroup$
    – Hoody
    Commented Feb 16, 2022 at 20:43
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I don't understand the question or the comments and answers. The magnetic force on a stationary electron is $F=\nabla(\mu\cdot{\bf B})$.

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  • $\begingroup$ Thanks for this equation. So if this is a force and a free electron in a conductor material has a mass, why then a stationary electron under an external applied magnetic field will only start to gyromagnetically rotate its spin magnetic moment vector around the B field vector and not start also to translate in space assuming it was initially stationary with $v=0$? The magnetic part of the Lorentz force says a stationary electron will not be forced to linear motion if it is stationary. This is a contradiction. If it is a force it should move even a stationary free electron in a vacuum. $\endgroup$
    – Markoul11
    Commented Feb 17, 2022 at 18:27
  • $\begingroup$ @Markoul11: there will be no magnetic force provided $\vec B$ is homogeneous, i.e. $\nabla \vec B =0$. If the field is non-homogeneous, the force will indeed be there. $\endgroup$
    – Hoody
    Commented Feb 18, 2022 at 6:18
  • $\begingroup$ @Hoody "there will be no magnetic force provided B⃗ is homogeneous, i.e. ∇B⃗ =0. If the field is non-homogeneous, the force will indeed be there." Very interesting note. Thank you. $\endgroup$
    – Markoul11
    Commented Feb 18, 2022 at 10:42
  • $\begingroup$ @Hoody the homogeneity or non-homogeneity of $\vec{B}$ has no bearing on whether it will induce a force on an electron, according to classical electrodynamics. If either the velocity of the electron, or the magnetic field at the location of the electron is 0, there will be no force (per Lorentz equation). If both the velocity and the magnetic field at the location of the electron are non-zero, then there will be a Lorentz force induced by the magnetic field. $\endgroup$ Commented Feb 18, 2022 at 20:04
  • $\begingroup$ @Jerold your equation is incorrect. The correct equation is the Lorentz force equation $\vec{F}=q(\vec{E}+(\vec{v} \times \vec{B})$. If $\vec{v}$ is $0$, the magnetic field causes no force upon the electron. $\endgroup$ Commented Feb 18, 2022 at 20:09

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