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There are two separate issues here. (1) Why does it make sense to consider a dipole moment as a vector? (2) Given that it's a vector, why does it make sense to say that it points in this particular direction, rather than the opposite direction. Intuitively, it makes sense to define a dipole as a vector because when we put it in a field, it aligns itself ...

19

The electron is a fundamental point particle. It does not have sub-particles “inside”. However, its quantum interactions with other particles should give it a small electric dipole moment, according to the Standard Model of particle physics. (It is a very difficult calculation but there are estimates for it.) Some people like to picture this in their minds ...

15

The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = -\...

13

A 'moment' is quite a general term, and its use ranges from electrostatics (e.g. dipole and other multipole moments) to mechanics (moment of force but also moment of inertia) to huge stretches of statistics. The general intuition is that you have some amount of 'stuff' (charge, force, mass, probability) with some distribution function $s(x)$, and the various ...

13

I went through unanswered questions, and stumbled over this... Did you find the original books? The mistake should be in your formula for the $\mu$ of a hollow sphere; the value with $1/5$ you gave is that of a solid sphere... The problem gets more simple I think, if you compare the two things directly: You get both, the angular momentum and $\mu$, from ...

11

It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results. Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ \... 11 The reason we believe the electron to be an elementary particle is precisely because people do high precision tests like the one you reference, looking for some kind of sub-structure which would indicate that the electron is made up of more fundamental bits, and come up empty-handed. Of course, such tests can never definitively rule out an EDM or anything ... 8 As the neutron is not point-like, consider it has a continuous distribution of charge \rho(\mathbf{r}) confined in a volume \Omega. The dipole electric moment is then given by \mathbf{D}(\mathbf{r})=\int_\Omega \rho(\mathbf{r}')\delta(\mathbf{r}-\mathbf{r}')d^3r' where the coordinates are measured from the centre of mass of the ... 8 The mystery is really why molecules can have electric dipole moments. The strong nuclear force and the electromagnetic interaction both conserve energy, angular momentum, and parity, and that's why we usually label states of nuclei with those quantum numbers, e.g., the 2^- ground state of a particular odd-odd nucleus. If these quantum numbers are ... 7 Currently our best model of the electron is that it is a point particle. Measurements give an upper bound of 10^{-22} meters on its radius. A point particle has no radius, and thus cannot rotate. Therefore we do not consider an electron with spin as a "spinning" electron in any classical sense. The spin and the magnetic moment of the electrons are instead ... 7 You are asking if the neutron can have an "induced neutron electric dipole moment" (quotes are for search terms): yes, it can. Whether is does is unknown. Before addressing that, note that a permanent electric dipole moment of the neutron (nEDM) is an active area of research, both theoretically and experimentally. Since and electric dipole is parity odd and ... 6 The electric dipole moment is defined as$$p = \int r \;\mathrm dq$$In the case of a pair of charges for which both charges are of the same magnitude, the choice of the origin turns out to be irrelevant:$$ p = \mathbf{r_1} q - \mathbf{r_2} q = q(\mathbf{r_1} - \mathbf{r_2}) = q\mathbf{d}where \mathbf{d} is the distance between the charges. ... 6 The integral of a vector, as you have written it, is just shorthand notation for a vector of integrals. Concretely, if we write \vec r= (x,y,z) in cartesian coordinates, then \begin{align} \int d^3r\, \vec r\,\rho(r) &= \left(\int d^3 r\, x\,\rho(r),\int d^3 r\, y\,\rho(r),\int d^3 r\, z\,\rho(r) \right) \end{align} Now, we simply note the ... 6 It's a matter of choice. You can set the potential energy to be any value at any angle. You don't even have to have a zero-value at all; you could make U purely positive or purely negative if you're feeling adventurous. But the advantage for U(\pi/2)=0 is, as you said, the simple expression U(\theta)=-pE\cos\theta = -\vec p \cdot \vec E instead of U(\... 6 Are point dipoles a thing, i.e. a physical object that you can find in the world? Well, not really, but does it matter? In particular, let me flip the question around for a bit: are point charges physically realizable? Well, so far as we can tell from the experimental data, yes, electrons and quarks are pointlike particles, but then again that's not ... 6 As explained in the comments, this is due to the Wigner-Eckart Theorem. This is a bit of a hard one to really grok, but what it really says is that if you have a quantum-mechanical system with well-defined directional characteristics (in the sense that it is in a state with a well-defined angular momentum), and you're studying the properties of an observable ... 6 Hint: Formally one should introduce testfunctions to deal with distributions. Another more physical approach is to regularize the dipole potential \Phi_{\varepsilon}~=~ \frac{\vec{p}\cdot\vec{r}}{(r^2+\varepsilon)^{3/2}}, \tag{1} $$similar to my Phys.SE answer here. The regularized dipole potential \Phi_{\varepsilon}\in C^{\infty}(\mathbb{R}^3) is ... 5 Your question is based on a misapprehension. The heating in microwave ovens is not a resonant process. See the answers to Does a domestic microwave work by emitting an electromagnetic wave at the same frequency as a OH bond in water? for a discussion of this. But let's leave this aside, because there is some interesting physics in your question. Suppose ... 5 Well you want to go from QFT to Classical mechanics. Let's do this in three steps 1. QED to Dirac Equation QED lagrangian with electric dipole is \mathcal{L} = \bar{\psi}\left(\gamma\cdot\Pi - \frac{\mathrm{i}d}{2}\sigma^{\mu\nu}\gamma^5 F_{\mu\nu}\right)\psi\\ Where \Pi\equiv \partial - \mathrm{i}eA. This implies the hamiltonian$$\mathcal{H} = \...

3

The Rydberg electron - the electron in the high n level - is highly polarizable and very weakly held. The binding energy of the electron is very small. A small electric field will distort the wavefunction of the Rydberg electron so that it spends more time on one side of the atom than the other. The result is the formation of dipole. Rydberg electrons in ...

3

The dipole has its least potential energy when it is in equilibrium orientation, which is when its momentum is lined up with Electric field (then $\tau$ = 0) It has greater potential energy in all other orientations. We are free to define the zero potential energy configuration in a perfectly arbitrary way , because only difference in potential energy ...

3

Neoh's answer is very thorough and mathematical. Here is a less mathematical, but hopefully more intuitive way to look at it. Take the two exterme cases of $\theta = 0, 180^\circ$. Then $V = \pm PE$, and you just need to fix the sign. To do that, recall the convention that systems move from higher states of potential energy to lower states of potential ...

3

The only real reason is because of how they derive the expression from physical equations. Verbal arguments simply won't make the cut here. Consider an electric dipole consists of two opposite charges of equal magnitude q, separated by a distance of d as figure below: The vector d is directed from negative charge to positive charge. When placed in a static ...

3

Here is a possible elementary and almost completely classical answer to my own question, but I don't know if it's right. Hnizdo 2011 discusses the field of a dipole moving at $v\ll c$. He gives references to papers that discuss the ultrarelativistic case, but those are all paywalled. However, he points out that the electric and magnetic polarizations $(-\... 3 Well, I recommend always use the definition to prove things. So, take the definition of superposition:$\psi_{net} = \psi_1 + \psi_2$. So, let two potentials be$V_1$and$V_2$. Since potentials obey superposition principle, the net potential is:$V_{net} = V_1 + V_2$. Now.. let two dipole moments be$p_1$and$p_2$. The net dipole:$$p = \int_{V_1+V_2}\... 3 How the electron's magnetic dipole moment gets influenced when electrons are moving through a magnetic field? In a classical context, the dipole moment of a fundamental particle is fixed: it doesn't change when in an electromagnetic field. If Nature were classical, the dipole moment of electrons would be$\boldsymbol\mu=2\mu_B\boldsymbol S$, where$\mu_B$... 3 The expectation value of the dipole moment operator for states of definite parity is zero. This is because the dipole moment operator is odd operator. 3 Electric field lines of an electric dipole do not form closed loops .See here ,they begin at positive charges and end at negative charges. But magnetic field lines form closed loops ,see the difference Magnetic fields form closed loops because there are no magnetic monopoles. See this post for more information Is there a magnetic line that is a Eucliden ... 3 Yes. Classically, (as mentioned by garyp) it would stay in that position forever if the angle between the electric field vector and the dipole moment is the exactly$\pi\$ radians. The dipole is in unstable equilibrium, so a slight change in position will cause it to align with the electric field, thus decrease its potential energy.

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