basically, we approximate all the charges as one point charge at the origin. all you need to describe that is the charge $q,$ which is a scalar quantity.

now, suppose that the net charge is zero. then there's no such contribution to the electric field. but that doesn't mean that you can just say $E=0$ far away; instead, one has $\vec{E}$ in terms of the dipole moment $\vec{p}$ of the configuration

i forget the exact expression, but it's one which involves $\vec{p}\cdot \hat{r}$. point is, the controlling quantity is now the vector $\vec{p}$.

which makes clear that $\vec{p}$, in particular its orientation, determines what the field looks like far away.

now, suppose that $\vec{p}$ vanishes as well.

so now you can't even determine a direction from far enough away. but once again, the electric field doesn't just disappear far away.

to describe how it behaves far away now, though, will require something more sophisticated than just a scalar charge $q$ or a vector dipole moment $\vec{p}$.

you'll need something that in some sense tracks multiple directions of influence at once

(handwavey, yes)

and when you formalize that, you end up describing the far-field behavior in terms of the quadrupole moment, which is a rank-2 tensor i.e. a matrix.

if that vanished as well, you'd need an even more complicated object: the octupole moment, a rank-3 tensor.

My head hurts

What is a quadrupole moment?

(comment: I should not have described this as a simple physics explanation. that was too blithe a description)

i can describe it better in terms of what it isn't, at least physically.

suppose i had a field configuration that looks like this: goo.gl/images/olScqY

that configuration can be described by a single quantity, the charge $q$ (i.e. the density of field lines and their direction).

all i need is a scalar quantity to describe that. yes?

Depends

(ignore the N/S labels and pretend they're just +/n charges, btw)

If the charge is uniformly distributed, yes

eh, if the field is exactly as its drawn in there---all field lines radiating outwards or inwards---then even that's not an issue, since you'd need a point charge to make that work.

which is to say, take that picture fairly literally for the moment: if I have that field configuration everywhere in space, i know i've got a single point charge.

Okay I take it literally then

okay. then from that picture i'd conclude that my electric field is something like $\vec{E}=k\frac{q}{r^2}\hat{r}$ and the only thing I need to know to describe it fully is the charge $q$.

For large enough r?

well, if it's a true point charge, it's for all $r$. but for a real charge distribution, yes.

point being, in that simple case, all I need is one single number in order to describe the field.

on the other hand, if i have field configurations like the ones shown here: upload.wikimedia.org/wikipedia/commons/a/aa/…

a single number won't do, since those configurations all have some notion of orientation

Quick question

sure.

actually, in the interest of not bowling over the rest of the chat, let's move this quick

did you figure out your question?

Is $\nabla \times \vec{E} \neq \mathbf{0}$?

depends. do you know faraday's law?

That was a while ago

Something about integrating over the flux

yeah. the differential form of it is written as $\nabla \times \vec{E}=\frac{\partial }{\partial t}\vec{B}$

Or

Oh

Neat

so the curl of the electric field is controlled by the time variation of the magnetic field

That is a cool result

in particular, if there's never a magnetic field, then the electric field is curl-free

Or if it's constant in time?

yeah.

which is useful, since the earth does have a magnetic field.

it means that, to the extent that the earth's magnetic field is constant, we can still treat the electric field as being curl-free when doing experiments

So if it's irrotational, then the force is conservative?

right. if there's no time-varying magnetic flux, you can write the electric field as $\vec{E}=-\nabla \phi$

Neat, but why the minus sign?

so that when you go to the force $\vec{F}=q\vec{E}=-q\nabla\phi =-\nabla U$, the force will point from higher potential energy to lower potential energy.

if the minus sign wasn't there, the force would cause the particle to move in the direction of increasing potential energy. not good.

(in elementary terms: you increase the potential energy of something by doing work against the corresponding conservative force)

Yeah okay makes sense

mmkay

anyways

shall we go back to the multipole stuff? (sidenote: one name for the charge $q$ is the monopole moment)

yes

okay. so if our field configuration can be well-described by just a set of lines radiating inward/outwards from a point, then all we need is the scalar charge $q$ in order to describe the electric field.

yes

but if we have a configuration like the ones in the animation i linked before, that won't do

that picture doesn't just have a density of field lines, it also has an overall orientation

yeah

if i rotated it 90 degrees, it'd still be the same type of configuration but the field at each point in space would be different.

so i can't describe that picture using just a scalar quantity. i need something with an inherent orientation, and that's precisely what a vector has.

once i've got the dipole moment $\vec{p}$, though, i do have enough info to get the field.

quick note on definitions: to compute the net charge $q$, i'd write $q=\int \rho(\vec{r})\,dV$ i.e. i integrate the charge density over all of space. (if $\rho$ were the usual density i.e. mass per unit volume, i'd get the total mass. same idea)

← 34 messages moved from Mathematics

thx @Danu

np

the dipole moment $\vec{p}$, on the other hand, is defined as $\vec{p}=\int \vec{r} \rho(\vec{r})\,dV$

technicality but shouldnt that be a triple integral?

yeah. i'm being lazy :p

oh okay, go ahead

(it's actually a bit more complicated to define than that, but that definition works if there's no net charge so w/e)

anyways!

suppose now I have a field configuration like this:

goo.gl/images/wmefQJ

if i was to look at the very top of the image alone, i might think that it looks like a dipole oriented vertically upwards. but that doesn't match the bottom, which looks like a dipole pointed vertically downwards!

similarly with the left- and right-hand sides of the image.

point being, there's no simple notion of orientation here. there is some preference to its directions, but it's not something that can be summed up as "it points in this direction."

let me know when you've absorbed that and can continue

but isn't the quadrupule and the dipole both anti-symmetric along the x- or y-axis?

take a closer look at the directions of the lines

with a dipole, you'd expect it to be going out in one direction and coming in along another

but with that picture, field lines are coming out from both the top and bottom

okay i was only looking at one quadrant at a time

yeah. key is to look at all of them

yes okay

another thing to note, i guess, is that the dipole field had two 'lobes' whereas the quadrupole shown has four, one in each quadrant

also, note that that means that the quadrupole shown is symmetric along both axes (not antisymmetric)

so the quadrupole is like 2 dipoles orthogonal to each other?

not orthogonal. if that were the case, you'd add them together and get another dipole

instead, it's like 2 dipoles in opposite directions

yeah i see that now :P

yeah

now, you could also get a similar picture if you rotate it by 45 degrees.

and i think you could also arrange for a quadrupole configuration such that the four lobes don't lie in the same plane

not finding a picture of that right now, though.

point is, there's more information required to describe this configuration than just orientation and density of field lines.

to properly describe it, you need not a vector quantity (3-by-1 in this case) but a matrix quantity (3-by-3)

yeah okay

so there's a further jump in required data

moreover, you can continue this pattern. in order to make the dipole evident, i had to make sure there was no net charge ($q=0$). in order to make the quadrupole evident, i had to make sure there was no net dipole ($\vec{p}=0$)

side note: the quadrupole moment is defined by $Q_{ij}=\int x_i x_j \rho(\vec{r})\,dV$. it's got two indices, so it's definitely a rank-2 tensor.

ugh

not a nice integral

that's the correct reaction, yes.

it's not entirely terrible if you've got some symmetries you can take advantage of

but in general it is pretty miserable.

granted, you notice those symmetries :P

quite

maybe you could use the divergence theorem if you could somehow manipulate the integrand to the divergence of some vector field

not sure if that would simplify things

looks like i got the definition a bit wrong. should be $Q_{ij}=\int (3x_i x_j -|\vec{r}|^2\delta_{ij})\rho(\vec{r})\,dV$

which looks worse, but has the advantage of making $Q_{ij}$ a traceless matrix.

traceless? is that when the the trace is equal to 0?

right.

okay

which is actually pretty simple here: $Q_{ii}=\int (3x_i x_i -r^2 \delta_{ii})\rho(r)\,dV=0$ since $\delta_{ii}=3$ and $x_i x_i=r^2$ per summation convention

aaaanyways

point is, you need an integral with two indices to define $Q$, making it a rank-2 tensor. (strictly speaking i should also say something about how it transforms under rotations, but nooope)

to finish up, there are charge configurations where all three of $q,p_i, Q_{ij}$ vanish identically

in that case, you have to go to the so-called octupole moment

which (ignoring any requirements re: the trace) would look like $C_{ijk}=\int x_i x_j x_k \,\rho(\vec{r})\,dV$

which now has 3 indices and so is a rank-3 tensor.

it's pretty horrible.

The volume in this case

Is 3-dimensional?

yeah.

all integrals here are over the charge density as distributed in 3D space.

Oh yeah okay

so each index can be 1,2,3 (i.e. x,y,z)

so it's got a total of 3*3*3=27 components.

But you need an additional $x_k$ component

i had to introduce that third factor in the integrand, yeah.

So the overall message would be that, in this context, the various tensors with their corresponding ranks (0 for monopole moment i.e. charge, 1 for vector dipole moment, 2 for quadrupole moment, 3 for octupole moment) are all ways of representing more and more involved information about direction and orientation (that's a bit handwavey, but i hope it makes intuitive sense)

and of encapsulating the info required to describe the corresponding electric field.

yeah I think I kinda understand

at its grossest level, a tensor is just a list of quantities with some collection of indices. but typically we also have some geometric sensibility for what they represent

for instance, if i were to take the very first picture (with field lines coming in/out) and rotate it, I get the same picture back

Is $\epsilon$ a tensor?

you mean, the Levi-Civita symbol?

yeah

it is, yes.

and the Kronecker delta $\delta_{ij}$ is a tensor as well---namely, it's the identity matrix.

difference is, of course, that $\delta_{ij}$ is a symmetric rank-2 tensor and $\epsilon_{ijk}$ is rank-3 tensor which is antisymmetric in its indices

why isn't kronecker delta rank-3? it has 3 linearly independent columns?

that's a terminology thing

there's the matrix rank, i.e. the number of linearly independent columns, and that is 3

yeah

but the tensor rank is the number of indices

ah I see, that makes sense

and since $\delta_{ij}$ has two indices, it's a rank-2 tensor

kind've annouying, but usually you wouldn't see matrix rank discussed in an engineering context whereas you could see tensor rank

we try to stay away from the mathy stuff

yeah. if you were taking a differential geometry course you'd probably be much more careful about the terminology

the place where things get subtle/interesting is how those various definitions change if you make a change of coordinates

The basic lesson being: "things are complicated when you try to talk about multiple nonzero moments at once"

i.e. it's hard enough talking about the quadrupole moment when there's no net charge or net dipole moment. it's even thornier when either of these are still around in the system.

oh I hope I won't have to do that...

are you in physics or mathematics?

this stuff also comes up if you want to describe radiation patterns from antenna. (not really surprising: you're trying to see what the resulting electric fields look like if you're standing far away from the source)

physics, though i'm a mathy sort

I don't do a lot of this stuff myself, I should confess, but I did have a graduate level electromagnetism course which described this stuff

Jackson's book on Classical Electrodynamics is the usual physics reference for it

and it's pretty painful to slog through.

i think we only have one course in theoretical electrical engineering

Ah.

That may or may not touch on some of this stuff

we do have quantum and flud mechanics though

which is pretty painful

if it's more about circuits, you'll probably see things like Laplace transforms and impedance

yeah we have that in our differential equations course

if it has anything about antennas, though, you'll see some of this stuff (though possibly greatly distilled)

i think you could also see tensors in the context of fluid mechanics and deformations of solids. it tends to show up there, since applying a force in some direction can result in a response in some other direction (i.e. two directions = two indices = rank-2 tensor)

seems like I'm gonna have a lot of fun this spring :P

it sort've shows up in quantum mechanics when you do addition of angular momentum, but that usually doesn't get stated in terms of indices.

lots of fun :P

The general rule is that the more realistic you make your system, the more degrees of freedom you need to allow

so the more likely you'll need tensors to properly describe how the system responds to various applied stresses.

anyways, i've rambled on enough and I shouldn't linger.

yeah I gotta study some integral theorems now

thanks for the help though

np

I think I gained some more insight

glad to hear it