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The simple Newton-like explanation of dipole gravitational radiation unexistence is following. The gravitational analog of electric dipole moment is $$\mathbf d = \sum_{\text{particles}}m_{p}\mathbf r_{p}$$ The first time derivative $$\dot{\mathbf d} =\sum_{\text{particles}}\mathbf p_{p},$$ while the second one is $$\ddot{\mathbf d} = \sum_{\text{... 21 The smallest radiating unit is an accelerating dipole moment. That can of course be produced by an accelerated single charge, which can be made equivalent to an oscillating dipole.$$ \ddot{p} = q\ddot{r},$$where r is a displacement of the charge around some fiducial point. You don't get a radiation field unless the charged particle is accelerating and ... 18 It seems that within the standard model of particle physics A permanent electric dipole moment of a fundamental particle violates both parity (P) and time reversal symmetry (T). These violations can be understood by examining the neutron's magnetic dipole moment and hypothetical electric dipole moment. Under time reversal, the magnetic dipole moment ... 17 Especially the hydrogen atom, with a proton in the nucleus and an electron revolving acting as a dipole This is a problematic way of understanding the hydrogen atom ─ it basically tries to insist on treating it within classical mechanics, and this is doomed to fail. Instead, the hydrogen atom must be treated within quantum mechanics. This introduces a bunch ... 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 = -\... 12 Simple reason: an oscillating monopole field in a region isolated from currents would violate charge conservation. Note a monopole field is not the same as an oscillating monopole charge, which, as Rob Jeffries's answer discusses, actually produces a dipolar field. Let (r,\,\theta,\phi) be the standard spherical co-ordinates, with corresponding ... 11 You are correct that in electrodynamics the only real sources of radiation are non-uniformly moving charges. However, when you solve for the potentials, you get some intricate expressions, the so-called Liénard-Wiechert potentials, for which the fields become very complicated expressions when calculated from them. Moreover, decomposing an arbitrary system ... 9 I don't think you need quantum mechanics to understand what's going on in dipole-induced dipole interaction. The basic mechanism is quite simple and just the details of the calculations change by switching to a quantum description. Polarizable molecule in an external field So first things first. Let us consider a simple model of polarizable molecule as ... 8 What do we mean with magnetic monopole and dipole? I can not find a way to relate magnetic monopoles and dipoles with electric ones. I do not understand their outcomes. Luckily, there exists a truly amazing one-to-one correspondence between magnetism and electricity. Monopole in magnetism is analogous to charge in electrostatics/electricity. Just like ... 7 Very simply, the field of the positive and negative elements of the dipole "almost" cancel out - but not quite. It is because they are some small distance away that there is a residual (third order) term. You can see this by taking two charges +q and -q at a distance 2d, and look at the field a distance r from the center of the two (on the same axis).... 7 neutral atom, Most chemistry happens between neutral atoms. There are the so called "spill over" forces , like the van der Waals ones, which allow for electromagnetic bondings between atoms and molecules. This is because the orbitals of the electrons in the neutral atoms have "shapes" which allow the positive regions around the atom, from the positive ... 7 Summary: A dipole moving through space radiates. Specifically, the power radiated depends both on the dipole's acceleration and its jerk. Finding the potentials: Consider an idealized dipole \vec{p} moving along a trajectory \vec{w}(t). We assume that this dipole has a constant magnitude and direction in an inertial frame. This is somewhat ... 7 Well, let’s start with a static electric field: the electrons would move in the opposite direction of it and the nucleus in the same direction. But the nucleus’ and the electrons’ attractive force counteracts this process, thus forming a (steady) dipole. I know, the picture I painted is very classical, but that’s sufficient. To solve it quantum mechanically ... 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}$$where \mathbf{d} is the distance between the charges. ... 6 Earth's magnetic field isn't really a dipole, but a dynamic field due to the convection occurring in the planet's core (consists of molten iron). The model below shows a simulation of the magnetic field (blue is pointing towards the core while yellow points away), the cluster of curves in the middle is the planet's core. The geomagnetic pole is the location ... 6 The force on a dipole placed in an electrical field is given by \mathbf{F} = (\mathbf{p}\cdot \nabla)\mathbf{E} (see, e.g., Griffiths, 3rd edition, eq. 4.5). Recall that,$$ \nabla(\mathbf{p}\cdot\mathbf{E}) = \mathbf{p}\times (\nabla\times \mathbf{E}) + \mathbf{E}\times(\nabla\times \mathbf{p})+(\mathbf{p}\cdot\nabla)\mathbf{E} + (\mathbf{E}\cdot\nabla)\...

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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 The water molecule is neutral on overall basis, i.e: the water molecule as a whole has no net charge. The water molecule is not linear rather it has a bent shape with two hydrogens on the same side. This happens because of the lone-pair-bond-pair repulsions. The oxygen has is a more electronegative element than hydrogen, i.e: oxygen has high electron-... 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 It is true that there is no (electrostatic) force between an electrified body and a body not electrified. (Let's ignore gravitational force for now.) It is also true that all bodies (in earth or earth-like environment) are electrified or will be electrified if approached by another electrified body. But in general, not all bodies can be electrified. For ... 5 You are making the mistake of thinking of the gradient as a regular one dimensional function where you pop in a value and it throws out an output. You can't take the gradient of a number (in your case,$0$). You take the gradient of a function (which can have as many dimensions as you like -- a multivariable function, that is). Now, the potential function of ... 5 Yes, this is perfectly possible The way you do this is by choosing a charge distribution which is 'completely dipolar' in some suitable sense, and this will produce an electric field which is also a pure dipole. In more technical terms, all you need to do is use a charge distribution with a separable angular dependence (i.e.$\rho(\mathbf r) = f(r) g(\theta,...

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The origin of magnetism is an complex problem in solid physics, (may be the the most lasting discussion in condensed matter physics). My point is: if caring about interaction of spin in system, we need to be very careful and dipole-dipole interactions sometimes are not the leading term. I will list four kinds magnetic origin which may help you. dipole-...

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Classically a non-pointlike spinning charged object possesses a magnetic dipole moment due to the fact that charged particles in the object are spinning around some axis. In contrast, the electron has a dipole moment that arises from its intrinsic spin angular momentum. As you point out, the electron has no internal structure, so the spin does not refer to ...

4

Hint: Interaction energy of two dipoles : $$U=\frac{1}{4\pi \epsilon_0r^3}\left( \mathbf{p}_1.\mathbf{p}_2-3\left ( \mathbf{p}_1.\hat r )(\mathbf{p}_2.\hat r\right) \right)$$

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Here's one way to think about it (though it isn't mathematically rigorous). From very far away the dipole would appear to have zero charge and thus there wouldn't be an electric field at all. However, you also know that the electric field falls off as $1/r$, so from very far away you'd expect the electric field to be small. The additional charge ...

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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 ...

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Imagine you have a single charge (not a dipole). We say it has zero potential energy at infinity. If we bring it close to another charge, it will end up with some non-zero PE: $$V_+ = \frac{Q}{4\pi \epsilon_0 r}$$ Now if we have a second charge with opposite polarity, and we move it to the same distance $r$, it would have potential energy V_- = -\frac{Q}...

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You may be confusing torque and force. The force $\vec{\mathrm{F}}$ is given by $q\, \vec{\mathrm{E}}$, so you can clearly see that the force is in different directions for the positive and negative charges, and is either parallel or antiparallel to the electric field. The torque $\vec{\tau}$ about any point $O$ is given by $\vec{\mathrm{r}} \times \vec{\... 4 If you look at this diagram: then it should be obvious that while$V$is constant along the horizontal line it varies along the vertical line. That means the horizontal component of$\nabla V$is zero and the vertical component is non-zero. So on the horizontal line$\mathbf E\$ is a vector directed vertically upwards.

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