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24

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: \begin{equation} 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 ...


10

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

Because the black area is half the box below. To explain: move the dipole from an area of no field to an area of field strength E. As you do, there's a force proportional to the dipole moment and to the gradient of E. For a fixed dipole, this force depends only on the gradient (horizontal dashed line). But for an induced dipole, the dipole moment depends on ...


7

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

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

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

Dipole moment is a vector and can be calculated using formula $$\vec{p} = \sum_i q_i \vec{r}_i.$$ It can be shown easily using the formula above that in case of two charges separated by distance $d$ $$\vec{p} = q \vec{d},$$ where vector $\vec{d}$ goes starts at negative ends at positive charge. http://en.wikipedia.org/wiki/Electric_dipole_moment#...


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)\...


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 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

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

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)....


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,...


4

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


4

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

The potential energy in this case should be $U=+\vec{m}.\vec{B}$, hence the potential energy is minimized, as it should be. Here is the explanation: Let’s look at the derivation of interaction energy between magnetic dipole and magnetic field carefully. The dipole energy $U=-\vec{m}.\vec{B}$ is derived using principle of virtual work with an assumption ...


4

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 ...


4

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.


4

It is just a consequence of the fact that the multipole expansion of an arbitrary magnetic field configuration doesn't have a monopole term due to absence of magnetic charges, and the first non-zero term in the expansion is of order of $\frac{1}{r^3}$. Generically, electric and gravitational fields also have analogous terms in their $\frac{1}{r}$ expansions, ...


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