40
The answer is that it doesn't matter.
The distance at which fields resemble that from a point charge is also the distance at which it does not matter where that point is located within the structure. The change of field due to switching the origin within the conductor will be comparable to the corrections to the point charge approximation, both arising from ...
35
Based on some of the back-and-forth I see, I think you're asking the wrong question. I think the question you want to ask is "Given a charge distribution $\rho(\mathbf{r})$, where should I place a point source so that the exact potential $\phi(\mathbf{r}) = \int \rho(\mathbf{r}')/|\mathbf{r}-\mathbf{r}'| dv'$ is most closely approximated by the potential ...
32
A monopole (gravitational) of a system is basically the amount of mass-energy the system has.
A dipole is a measure of how the mass is distributed away from some center.
The quadrupole moment describes how stretched out the mass distribution is along an axis. Quadrupole would be zero for a sphere, but non-zero for a rod, for instance. It is also non-zero ...
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},$$
where $r$ is a displacement of the charge around some fiducial point.
You don't get a radiation field unless the charged particle is accelerating and ...
18
It seems that within the standard model of particle physics
A permanent electric dipole moment of a fundamental particle violates both parity (P) and time reversal symmetry (T). These violations can be understood by examining the neutron's magnetic dipole moment and hypothetical electric dipole moment. Under time reversal, the magnetic dipole moment ...
17
Sometimes you can
The obvious example is a purely dipolar charge which has been displaced from the origin, such as a dipolar gaussian
$$
\rho(\mathbf r)
=p_z(z-z_0)\frac{e^{-(\mathbf{r}-\mathbf{r}_0)^2/2\sigma^2}}{\sigma^5(2\pi)^{3/2}} .
$$
This system is neutral, and it has a nonzero dipole moment $p_z=\int z\rho(\mathbf r)\mathrm d\mathbf r$ along the $...
14
I usually find it easier to use model multipoles that are surface charges on a sphere, rather than point charges on some polyhedron's vertices. These charge densities are given in general by
$$\sigma_{lm}(\theta,\phi)=N\cos(m\phi)P_l^m(\cos(\theta)),$$
where $P_l^m$ is an associated Legendre function (so $\sigma_{lm}$ is just a real-valued spherical ...
12
Simple reason: an oscillating monopole field in a region isolated from currents would violate charge conservation. Note a monopole field is not the same as an oscillating monopole charge, which, as Rob Jeffries's answer discusses, actually produces a dipolar field.
Let $(r,\,\theta,\phi)$ be the standard spherical co-ordinates, with corresponding ...
10
You are correct that in electrodynamics the only real sources of radiation are non-uniformly moving charges. However, when you solve for the potentials, you get some intricate expressions, the so-called Liénard-Wiechert potentials, for which the fields become very complicated expressions when calculated from them. Moreover, decomposing an arbitrary system ...
9
In the nonrelativistic limit the energy lost by the system due to gravitational radiation is defined by the third time derivative of quadrupole moment:
$$- \frac{d E}{dt} = \frac{G}{45 c^5}\dddot{D}^2_{ij}.$$
Where indices $i$, $j$ correspond to (flat) 3D space, and dot denotes time derivative. This equation is taken from Landau & Lifshitz' 'Classical ...
9
The function $\sin 3\theta$ on the unit sphere is not an eigenfunction of the Laplacian on the sphere, i.e. the angular part of the Laplacian, i.e. of $L^2$, so it is not convenient a basis vector in problems whose Hamiltonian involves the Laplacian.
The function $\sin 3\theta$ may be written as a combination of spherical harmonics $Y_{lm}$ with many ...
8
There are no sextupoles in the expansion! All $n$-poles have numbers $n$ which are powers of two, i.e. $n=1,2,4,8,16,32,$ and so on.
The case $n=1$ is the monopole, the actual charge. Its electric field goes like $1/r^2$ – let's adopt the electric notation.
The $n=2$ dipole is a pair of $n=1$ charges of opposite signs, shifted relatively to each other. The ...
8
In a multipole expansion of the electric potential, outside of some charge charge distribution $\rho(\mathbf r,t)$, the monopole term is simply
$$V_{mp}(\mathbf r) = \frac{Q_{total}}{4\pi \epsilon_0 r}$$
The associated electric field is then
$$\vec E_{mp} = \frac{Q_{total}}{4\pi \epsilon_0 r^2}\hat r$$
For this term to be time varying at some fixed $r$, ...
8
The multipole expansion is a useful approximation to low order (mono-, di, - quadrupole) if the diameter of the charge distribution $d$ is much smaller than the distance at which you observe the field or potential $r$.
That observation distance is with respect to an origin, which you conveniently place somewhere inside the charge distribution. Emphasis on ...
7
Imagine having a mass distribution $\rho(x,y,z)$ around the origin O and we want to calculate the potential energy and force at a certain point P on the z-axis. The potential energy can easily expressed by the integral: $$U=-GM\int_{V}\frac{\rho(x,y,z)}{R}dv$$
However this intgral might be difficult to calculate and it's often easier to express the integrand ...
7
The expression you found in Sakurai's textbook is the correct one and can be further expanded in spherical harmonics, if needed.
The difference with electrostatics is due to the different position the "dipole" emerges in the case of quantum treatment of light-matter interaction.
In electrostatic is the term $l=1$ of the Legendre polynomial expansion of the ...
6
First of all, don't think of multipole moments as separate things that have their own individual meaning. Instead, think of them as parts of one thing. Once we have all the parts written down, we can start naming and organizing each one to determine its contribution to the whole.
Now, for your question
Is there a physical interpretation for multipole ...
6
You are implicitly assuming that sufficiently far away from a bounded charge distribution, the electric field becomes perfectly radial, i.e. every electric field line would meet at a single point if you just projected it in a straight line back to the neighborhood of the charge distribution. But this isn't the case; if you could measure the direction of the ...
6
The low $l$ power in the CMB spectrum is poorly defined because of cosmic variance - essentially there are a limited number of "samples" that can be used to characterise the temperature variance on large angular scales. The state-of-the-art measurements from Planck (see below) show that the slope of this region is actually undefined. That most experiments ...
5
(this is a partial answer)
One example is the preceding work
Barrett, J. W. "The asymmetric monopole and non-newtonian forces." Nature 341.6238 (1989): 131-132. doi:10.1038/341131a0
which is Ref. 13 in the Connes et al. paper.
This paper contains one example of asymmetric monopole produced by rotating figure shown
about the horizontal axis passing ...
5
The solution I found is to rotate the gradient. If the gradient is defined by
$$\vec{\nabla}B=\frac{\partial B}{\partial x} \hat{x}+\frac{\partial B}{\partial y} \hat{y}+\frac{\partial B}{\partial z} \hat{z}$$
And we use the rotation matrix on that
$$R\cdot \vec{\nabla}B=\left(
\begin{array}{ccc}
\cos (\phi ) & -\sin (\phi ) & 0 \\
\sin (\phi ) &...
answered Jul 15 '14 at 9:02
The Quantum Physicist
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5
I think the easiest way to visualise the various multipoles is through their connection with the spherical harmonics. These are probably best known as the functions that give the shape of the atomic orbitals in the hydrogen atom.
The $p$ orbitals correspond to a dipole, the $d$ orbitals to quadrupole and the $f$ orbitals to an octapole, that is a charge ...
4
For the $l$-th term you take totally symmetric tensors of rank $l$ which are totally traceless under contractions of each index. This comes because that is how the multipole moments, or rather the spherical harmonics, are built from the cartesian coordinates. For example $l=2$ has $5$ components that look like $x_i x_j -\frac{1}{3}\delta_{ij} x^2$ which is a ...
4
Monopole = point charge; the potential falls off like 1/r.
X
Dipole: duplicate the point charge and make it the opposite sign. Separate the two charges by a small distance. The potential falls off like 1/r^2 to leading order.
X O
Quadrupole: duplicate the dipole and reverse the signs on the duplicate. Separate the two dipoles by a small distance. (...
4
I think it's easier to see in Cartesian coordinates. Define the "primitive" moments
\begin{align}
q & = \int \rho(\mathbf r)\, \text{d}^3\mathbf{r}\\
p_i & = \int r_i \rho(\mathbf r)\, \text{d}^3 \mathbf{r}\\
Q_{ij} & = \int r_i r_j \rho(\mathbf r)\, \text{d}^3 \mathbf{r}
\end{align}
Assume $q = 0$, $p_i$ not all 0.
If you displace the origin ...
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
Spherical Multipole Expansion
The first nonvanishing moment is the quadrupole moment. Assuming that you are looking for an exterior multipole expansion, the potential from the charge distribution $\rho$ can be expanded as
$$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\sum_{L=0}^\infty\sum_{m=-L}^LI_{Lm}(\mathbf{r})\langle R_{Lm},\rho\rangle$$
where $I_{Lm},R_{Lm}...
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