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This seems rather incredible that these two seemingly conflicting
announcements come on the same day.
The pre-print for the Nature paper by the BMW group was placed on arXiv in 2020 around the same time as the muon g-2 Theory Initiative paper (submitted on 8 Jun 2020 and last revised 13 Nov 2020, published here) that the Fermilab collaboration referenced in ...
15
The $g$ factor describes the magnetic moment of a spinning particle. The $g$ factor for a classically spinning particle is equal to 1, but in the "basic" (ie, non-interacting) quantum field theory of spin-1/2 particles you would expect this number to come out to be 2, or in other words you would expect $g-2$ to be zero. It turns out that in "...
11
Steve's intuition for the measurement operator is in the right direction, but it's not quite the right interaction potential. Getting that right is simple, but it requires the complex task of taking the naive answer to its ultimate consequences and then understanding what it's really saying. The hamiltonian is the same as always: a direct coupling between ...
10
The neutron is not an elementary particle. Its overall charge is zero, but depending on the energy available the fact that it is composed out of charged quarks becomes important. Thus both the strong force and the electromagnetic will play a role depending on the energies of the scatters and the targets.
For low energies , like the ones coming from fission ...
10
Has anyone tried to incorporate the electrons magnetic dipole moment into the atomic orbital theory?
To be crystal clear:
Has anyone tried to incorporate the electrons magnetic dipole moment into the atomic orbital theory?
YES. They've tried and they've succeeded. The electron's spin magnetic dipole is a standard part of atomic physics and quantum chemistry. Anybody attempting to claim that this isn't the case is simply describing their own ignorance ...
9
An electron is not a spinning ball of charge and the intrinsic spin of particles cannot be understood in such terms. Not only is it difficult to make sense of what it means for a pointlike particle to spin, but also when treating the electron as a spinning ball of charge one finds a value of the ratio between the magnetic moment and the angular momentum that ...
9
The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop. However, the more general case is that of a a volume current $\vec{J}$ in some finite region of space. In this case, the the ...
9
It is an experimental fact that neutrons have no net charge. They have a magnetic dipole moment, which points to having charged constituents (quarks), the charges of which algebraically sum to $0$. (It is not the only evidence for constituents in the neutron: scattering experiments also indicate the presence of constituents, and exclude the possibility of ...
9
Ampere's law says that
$$\nabla \times \boldsymbol{B} = \mu_0 \boldsymbol{J} + \epsilon_0\mu_0 \frac{\partial}{\partial t} \boldsymbol{E}$$
so "far away from sources" means that the current density $\boldsymbol{J}$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be ...
9
In the Stern-Gerlach experiment you want the atoms
to be deflected depending on the direction of their magnetic dipole moment.
But you get a net force on the magnetic dipole moment
only if the magnetic field is non-homogenous.
(image from "Force and Torque on a Magnetic Dipole" (page 26))
If the magnetic field would be homogenous,
then the forces (...
9
@Thomas Fritsch uploaded a nice picture that provides intuition about the dynamics of the situation. I would just like to add that the force $\textbf{F}$ exerted on this infinitesimal loop pictured (which is how we model a single silver atom in the Stern-Gerlach experiment), is classically given by:
$$\textbf{F} := \nabla \left( \boldsymbol{\mu} \cdot \...
8
Elementary particles have quantum mechanical spin. This induces a spin magnetic moment, independent of the presence (or, indeed, absence) of a (net) electric charge. This is how the neutron attains its magnetic moment (as you already mentioned).
The case of the neutrino magnetic moment is slightly confusing, as they are not completely understood yet. ...
8
The current loop creates a magnetic field. This field looks just like that of a magnet, which has two poles (a dipole).
7
There are two important interactions in magnetic materials: the dipolar interaction and the exchange interaction.
The dipolar interaction corresponds to the magnetic fields from the Maxwell equations. This field is dominant on the macroscopic scale. Therefore, if we look to the macroscopic bar magnets, they nicely follow the behavior of the magnetic field ...
7
This part-per-million measurement of the muon's anomalous magnetic moment is a part-per-billion measurement of the muons total magnetic moment. Every part-per-billion measurement is hard. This one is, fundamentally, a measurement of a frequency: the frequency at which the muon's spin precesses in a magnetic field.
In order to measure a frequency with ...
7
Suppose your "particle" is actually a positively-charged sphere. Spin the sphere about some axis and the moving charges generate a magnetic field. You can use the Biot-Savart law to figure out that the direction of the magnetic field along the rotation axis will be parallel to the angular momentum.
Switch to a negatively-charged sphere, and you ...
6
It is zero. If the magnetic dipole moment $\boldsymbol \mu$ were different from zero, the particle would have a preferred direction in space, which is not possible for a scalar.
Recall that, in general,
$$
\boldsymbol \mu\propto\boldsymbol S
$$
which, for a scalar, implies $\boldsymbol\mu\equiv\boldsymbol 0$.
answered Jan 2 '17 at 18:18
AccidentalFourierTransform
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6
As far as we know, electrons are point particles, so you don't want to try to calculate anything when they are at zero separation. You'd just get infinite forces.
Classically, the magnetostatic dipole-dipole interaction dominates over the electrostatic interaction when two electrons are separated by less than about one Compton wavelength. This is because ...
6
It is not necessary for them to move. Even an electron at rest has a magnetic field. This is related to its intrinsic angular momentum or “spin”. It is not actually spinning like a little ball, and this magnetic field cannot be understood as arising from a current.
6
First of all, let us consider only a solenoid (without iron core) of length ${l}$ and area of cross-section ${A}$. The magnetic moment of the solenoid is given by:- $${M}_{solenoid}=({n}{l}){I}{A}$$
Where,n=number of turns per unit length of solenoid and
${I}$ = current flowing in the solenoid
Now, the net magnetic field is:-
$${B}={\mu_0}{n}{I}$$
Now if an ...
6
The spin magnetic moment of a fundamental particle with mass $m$, charge $q$, and spin 1/2 is
$$\vec\mu=g\frac{q}{2m}\vec S$$
where $\vec S$ is its spin vector. The "$g$-factor" is a dimensionless number. The Dirac equation predicts that it should be exactly 2.
However, the complications of quantum field theory cause $g$ to differ slightly from 2, ...
5
Let's first look at the classical situation. A charged particle moving round a circular loop had an angular momentum and because it is also a current, it produces a magnetic moment. Therefore, it can be considered to be a magnetic dipole with a moment $\vec{m}$. The magnetic moment and the angular momentum are proportional to each other with the constant of ...
5
The magnetic moment, in general, is an indication of the torque that will be experienced by an object when subjected to a magnetic field. $\mu = I\cdot A$ implies that it scales with the area (of a current loop) and the magnitude of the current. You can easily convince yourself that this is so by drawing a simple rectangular current loop, with an axis of ...
5
The Earth's magnetic field is caused by eddy currents in the liquid parts of the planet's interior. We believe the field is not due to a permanent magnet because: (1) Its direction and strength change over time, and (2) the planet's interior is hotter than the Curie temperature of its elements, and so a permanent magnet would not retain its magnetism.
...
5
$\mathbf{S}$ is the spin operator. It is a vector operator that acts on spinors. It will have three components $(S_x, S_y, S_z)$ and for example if you take the $z$ axis as your spin measurement axis, you define spin up and down as the two eigenstates of $S_z$.
It can be shown that in matrix form $S_i$ is proportional to the Pauli matrix $\sigma_i$.
...
5
An electron acts like a very small magnet. We say "magnetic moment" rather than "magnetic field" because a field is a local property (move further from the electron and the field gets weaker), while the magnetic moment is a property of the electron from which you can deduce the field at any point, and from which you can calculate the energy required to align ...
5
Since gravity uses energy to push us down to earth
This is incorrect. Gravity does not use energy to pull in us.
If we started falling, then yes, gravity used energy do make us move. But that is only in the special case where gravity makes us move. In general, gravity spends no energy pulling in us.
In general, a force spends no energy. An apple lying on ...
5
The electron can be seen as a small magnet with the field of a magnetic dipole of strength $2\mu_B \vec S$. So my answer is yes in both cases. I interpret as at rest a state in which the electron has zero momentum expectation value. Note that there is no accepted model of spin as a rotation, although spin is an angular moment to every effect.
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