Plotting the Poynting vector of a radiating electric dipole [matlab]

[PhDeezNutz]

Did I inadvertently solve the octupole while trying to solve the quadrupole? I mean what is going on here.

 

[PhDeezNutz]

@TSny I may be terribly misguided but this is what I'm thinking. If you look on page 416 (Figure 9.2) of Jackson he talks about a system that is a little bit different than the one we're dealing with but the impression I get is that each vertical plane has a quadrupole (4 lobe pattern)...my picture has that feature so that is reassuring.

Granted Jackson's radiation pattern comes from the time averaged poynting vector radial flux...but I think the magnitude of $\vec{S}$ is closely connected to that...in a way I can't quite articulate.

 

[TSny]

@TSny I may be terribly misguided but this is what I'm thinking. If you look on page 416 (Figure 9.2) of Jackson he talks about a system that is a little bit different than the one we're dealing with but the impression I get is that each vertical plane has a quadrupole (4 lobe pattern)...

Jackson is assuming a charge distribution with cylindrical symmetry, such as a spheroid (ellipsoid of revolution). At the beginning of the paragraph containing equation (9.50) he assumes "an oscillating spheroidal distribution of charge". Figure 9.2 refers to this case.

 

[PhDeezNutz]

Jackson is assuming a charge distribution with cylindrical symmetry, such as a spheroid (ellipsoid of revolution). At the beginning of the paragraph containing equation (9.50) he assumes "an oscillating spheroidal distribution of charge". Figure 9.2 refers to this case.

I’m having trouble imagining what that means. Would that mean

$\rho (\vec{r}’) = \frac{3Q}{4 \pi R^3} e^{-i \omega t}$

It seems not much different than a point charge and I was under the impression that point charges don’t radiate unless accelerated.

 

[PhDeezNutz]

I feel like I'm getting the exact opposite of what I'm trying to get. In my opinion the level surfaces should connect close to the origin for a specified level. Instead they are connecting near the outside.

 

[PhDeezNutz]

Finally got it. Was missing a factor of $\frac{1}{r^3}$ in my vector potential for the oscillating quadrupole.

 

[TSny]

Yes, your last plot looks good to me.

I’m having trouble imagining what that means. Would that mean

$\rho (\vec{r}’) = \frac{3Q}{4 \pi R^3} e^{-i \omega t}$

It seems not much different than a point charge and I was under the impression that point charges don’t radiate unless accelerated.

A "spheroid" is an ellipsoid of revolution. See here. As Jackson mentions, the nonzero components of the quadrupole moment tensor are the diagonal elements. These satisfy $Q_{11} = Q_{22}= -\frac{1}{2}Q_{33}$. We imagine that somehow these components vary harmonically in time.

 

[PhDeezNutz]

I am very happy with this, got to change the viewing angle to catch the near field behavior but all in all I'm very happy.

I think this is the first gif of its kind. I've seen many quadrupole animations online but they are all confined to a plane. That could be because creating a 3D version is a trivial extension...but I'm proud of it.

 

[TSny]

Yes, that looks very nice. It would be interesting to see some of the near field behavior. Good work!

 

[PhDeezNutz]

Yes, that looks very nice. It would be interesting to see some of the near field behavior. Good work!

I tried changing the viewing angle but it didn’t show much. It showed something but not as much as I would like. I think the best approach will be to take slices.

I’ll have it by tonight hopefully. Count on it!

 

[PhDeezNutz]

@TSny do we dare tackle the octupole next? (I should probably get back to all my class work that I've been neglecting but the idea of tackling the octupole is very enticing)

I am very happy right now.

 

[PhDeezNutz]

@TSny I’ve thanked you many times in this thread but I really can’t thank you enough. You truly went out of your way to help me, maybe one day I’ll be able to pay it forward. I look forward to that day.

 

[PhDeezNutz]

@TSny may I ask what the definition of the (primitive, non-traceless) Magnetic Quadrupole moment is?

Is it

$M_{ij} = \sum I_{\ell} (a_i)(a_j)$

Where $a_i$ is the #ith#-component of the area vector.

 

[TSny]

I've never worked with magnetic quadrupoles. Jackson doesn't appear to give any explicit expressions for magnetic quadrupole moments. I did find the following expression in Morse and Feshbach's Methods of Mathematical Physics for a general distribution of current density $\vec J$:

$\mathbf{M} = \frac{1}{2}\int\left[ \vec r (\vec r \times \vec J) + (\vec r \times \vec J) \vec r \right] d^3r$

This is dyadic notation. Thus to get ${M}_{xy}$, you dot the integrand from the left with a unit vector in the x-direction and dot the integrand from the right with a unit vector in the y-direction. So,

$M_{xy} = \frac{1}{2} \int \left[x(zJ_x-xJ_z) + (yJ_z-zJ_y)y\right]d^3r$
$\,\,\,\,\,\,\,\,\,\,\,\,= \frac{1}{2} \int \left[ (y^2-x^2)J_z + xzJ_x - yzJ_y \right] d^3x$

$M_{zz} = \int \left( zxJ_y - zyJ_x \right) d^3r$

etc.

The dimensions of $M$ are current times distance cubed; whereas, the dimensions of your expression for $M_{ij}$ are current times distance to the fourth.

I applied this dyadic definition to the case of a magnetic quadrupole formed from two current loops with opposite currents.

I got the same result for the magnetic quadrupole moment components as given in this link. See pages 9 through 11. The final result is at the bottom of page 11.

The vector potential $\mathbf A$ produced by this quadrupole is given by the last term shown in the equation at the top of page 11. But this is only for the far field. I would imagine that the near field would be very tedious to work out.

 

[PhDeezNutz]

Dyadic notation is something that is new to me. But it seems like I must learn it if I am to create a gif of the radiating magnetic quadrupole/electric octupole.

I truly would be lost without you.

Thank you very very much.

 

[TSny]

A dyad is sometimes denoted with a double arrow, such as $\overset{\leftrightarrow}{\mathbf Q}$

Two vectors $\mathbf a$ and $\mathbf b$ can form a dyad: $\overset{\leftrightarrow}{\mathbf Q} = \mathbf a \mathbf b$. The two vectors are simply written side by side without any product symbol. (See the integrand of $\mathbf M$ in post #104.) The components of the dyad are

$Q_{xx} = a_xb_x$
$Q_{xy} = a_xb_y$
$Q_{yx} = a_yb_x$
etc.

 

[PhDeezNutz]

I've never worked with magnetic quadrupoles. Jackson doesn't appear to give any explicit expressions for magnetic quadrupole moments. I did find the following expression in Morse and Feshbach's Methods of Mathematical Physics for a general distribution of current density $\vec J$:

$\mathbf{M} = \frac{1}{2}\int\left[ \vec r (\vec r \times \vec J) + (\vec r \times \vec J) \vec r \right] d^3r$

This is dyadic notation. Thus to get ${M}_{xy}$, you dot the integrand from the left with a unit vector in the x-direction and dot the integrand from the right with a unit vector in the y-direction. So,

$M_{xy} = \frac{1}{2} \int \left[x(zJ_x-xJ_z) + (yJ_z-zJ_y)y\right]d^3r$
$\,\,\,\,\,\,\,\,\,\,\,\,= \frac{1}{2} \int \left[ (y^2-x^2)J_z + xzJ_x - yzJ_y \right] d^3x$

$M_{zz} = \int \left( zxJ_y - zyJ_x \right) d^3r$

etc.

The dimensions of $M$ are current times distance cubed; whereas, the dimensions of your expression for $M_{ij}$ are current times distance to the fourth.

I applied this dyadic definition to the case of a magnetic quadrupole formed from two current loops with opposite currents.

View attachment 257972

I got the same result for the magnetic quadrupole moment components as given in this link. See pages 9 through 11. The final result is at the bottom of page 11.

The vector potential $\mathbf A$ produced by this quadrupole is given by the last term shown in the equation at the top of page 11. But this is only for the far field. I would imagine that the near field would be very tedious to work out.

I'm having a brain fart, I cannot see how the $M$ you posted has units of current times distance cubed. I'm seeing current times distance squared.

But I would logically conclude that $M$ does indeed have units current times distance cubed because it comes right after the magnetic dipole which is known to have units of current times distance squared.

 

[TSny]

Current density $J$ has dimensions of current divided by area.

 

[PhDeezNutz]

@TSny is this the radiation pattern you would expect from a time harmonic oscillating magnetic quadrupole?

arranged as

Again, these are level surfaces of the Poynting vector magnitude.

Also, I hope you are having a good day and are staying safe during this pandemic.

 

[TSny]

Nice. That could be right. I don't have much experience with this. But look at the figure on page 8 here. You can see a sketch of the radiation pattern for the two circular current loops. There is no radiation in the horizontal direction (in the xy plane) and no radiation in the z-direction. That looks like what you are getting.

 

[PhDeezNutz]

Nice. That could be right. I don't have much experience with this. But look at the figure on page 8 here. You can see a sketch of the radiation pattern for the two circular current loops. There is no radiation in the horizontal direction (in the xy plane) and no radiation in the z-direction. That looks like what you are getting.

Yeah I stumbled across that link earlier and was happy to see that mine corroborated theirs but wasn't sure. For the next case (magnetic octupole and electric hexadecapole) I'm probably going to have to learn some tensor calculus.

Also setting up configurations and having MATLAB do moment calculations for me because there's no way I'm doing 27, 81, or 243 calculations by hand lol. I'm going to post one more gif in this thread (the radiating electric octupole) and then after that I will start a new thread when I need to. I know the mods don't like it when threads get too big. It will be good practice in doing (trivial) tensor multiplications like $Q_{ijk}r_k r_j$ because goodness knows I will need it later.

 

[PhDeezNutz]

Ok. I am satisfied with the magnetic quadrupole animation but the electric quadrupole is giving me considerable difficulty. Which is weird because

1) They are subsumed in the same term ($\ell = 2$) in the expansion of Jackson 9.11.

2) The electric multipoles are much easier to define and recognize

If anything I thought it would be the magnetic quadrupole that would give me trouble but I digress. The configuration I'm trying to model is the following.

With the coordinates of the charges being (the origin is at the center of the cube)

I'm using the definition of

$Q_{ijk} = \sum_{\ell}^{8} \sum_{i,j,k}^{3} q_{\ell} r_{i ,\ell} r_{j ,\ell} r_{k ,\ell}$

The result I got for the configuration above (using my program) is the following

With the following code, PCQ (is the 4x8 array pictured above)

for i = 1:3 for j = 1:3 for k = 1:3 Q(i,j,k) = 0; for i1 = 1:length(PCQ) Q(i,j,k) = Q(i,j,k) + PCQ(4,i1)*PCQ(i,i1)*PCQ(j,i1)*PCQ(k,i1); end end end end I think I may have populated the electric octupole tensor wrong because I'm getting all negatives.

 

[PhDeezNutz]

Vaguely an electric octupole. I'm going to go back over my analytical expressions for the vector potential and see if I mad any mistakes.

One thing I know for certain is that if we expand Jackson 9.11 with the $\ell = 2$ term we should get the quantities $M_{quad}\vec{r} \times \vec{r}$ and $Q_{oct} \vec{r}\vec{r}$ in the mix.

 

[PhDeezNutz]

Getting better but intuitively I would think a radiating octupole would have 8 lobes. I appear to have 12. There are some residual terms in my analytical expressions that I have not incorporated yet. I hope including them will fix my figure. *Crosses fingers*

 

[PhDeezNutz]

Here's the latest one. Pretty much the same thing as my last picture but a little sharper. My advisor said my pattern looks reasonable; despite there being 12 lobes total there are 8 lobes when you look from the top. That doesn't quite make sense to me but here's the way I think of it;

Each edge of the cube is a dipole. As we know radiation patterns are zero on the axis of each dipole.

Counting the four vertical dipoles and the four horizontal dipoles we get the disappearance of the radiation pattern at the vertices/corners. (8 triangular holes shown below).

The radiation pattern vanishes on the faces of the cubes because that area has field contributions from two opposing dipoles (6 square holes).

For a total of 14 holes

I know that is extremely "hand wavy" but that's the best I can do with my current understanding of multipoles. I think the meaning of the octupole is a little bit more nuanced than "8 lobes".

@TSny may I have your opinion on this matter? Even if you are not certain I feel like your input would still be valuable.

Edit: My advisor originally told me that the radiating magnetic quadrupole is unsolved and that Bethe and Bouwkamp contrived wrong solutions to it that somehow worked. So you can imagine that I was really happy when I solved the magnetic quadrupole with minimal assumptions. So I inquired with my advisor further and he elaborated by saying something to the effect of "the radiation pattern of the magnetic quadrupole isn't unsolved in general but rather in regards to diffraction". I'm just getting involved in research and from what I understand apertures can be modeled with "magnetic currents" and these magnetic currents have multipoles. The dipole term of said magnetic current is understood, however the quadrupole term is not. That's what I took away from it anyway, but enough of my rambling.

Edit2: I am finding that the vector potential dependence on $Q_{ijk}$ is a little bit more complicated than $Q_{ijk}r_{i}r_{j}$. I'm getting the same units but not nice expressions. I also have residual terms that I have not accounted for and are difficult to manipulate.

 

[TSny]

Your "hand-wavy" arguments sound good to me.

I agree that there are 12 lobes. I decided to try using the approach to multipole radiation fields given in chapter 16 of Jackson's 2nd edition. I believe this material is covered in chapter 9 of Jackson's 3rd edition in a somewhat condensed form compared to the 2nd edition.

Using this approach I attempted to produce a plot showing the time-averaged power radiated in different directions. For the "cube" octupole I get the following surface. (If you pick a point on the surface, the power radiated in the direction from the origin to that point is proportional to the distance of the point from the origin.) So, the power is mostly radiated into 4 "upper" lobes, 4 "lower" lobes, and 4 "horizontal" lobes. The lobes point toward the midpoints of the edges of the cube.

There is no radiation in the 14 directions that you found. So, this seems to corroborate your results.

 

[PhDeezNutz]

Your "hand-wavy" arguments sound good to me.

I agree that there are 12 lobes. I decided to try using the approach to multipole radiation fields given in chapter 16 of Jackson's 2nd edition. I believe this material is covered in chapter 9 of Jackson's 3rd edition in a somewhat condensed form compared to the 2nd edition.

Using this approach I attempted to produce a plot showing the time-averaged power radiated in different directions. For the "cube" octupole I get the following surface. (If you pick a point on the surface, the power radiated in the direction from the origin to that point is proportional to the distance of the point from the origin.) So, the power is mostly radiated into 4 "upper" lobes, 4 "lower" lobes, and 4 "horizontal" lobes. The lobes point toward the midpoints of the edges of the cube.

View attachment 259830

There is no radiation in the 14 directions that you found. So, this seems to corroborate your results.

Very nice. Thank you.

I wish I had the 2nd edition to see it in it's full glory.

Was your approach to expand out

→A(→r,t)=e−iωtμ4π∫→J(→r′)eik∣∣→r−→r′∣∣∣∣→r−→r′∣∣d3r′A→(r→,t)=e−iωtμ4π∫J→(r′→)eik|r→−r′→||r→−r′→|d3r′​

Using the well known identity

eik∣∣→r−→r′∣∣∣∣→r−→r′∣∣=4πik∞∑ℓ,mh(1)ℓ(kr)jℓ(kr′)ℓ∑m=−ℓYm∗ℓ(θ′,ϕ′)Ymℓ(θ,ϕ)eik|r→−r′→||r→−r′→|=4πik∑ℓ,m∞hℓ(1)(kr)jℓ(kr′)∑m=−ℓℓYℓm∗(θ′,ϕ′)Yℓm(θ,ϕ)​

I don't see how you would do this entire thing numerically. I had to do integration by parts several times in order to bring out expressions involving $M_{ij}$ and $Q_{ijk}$ analytically. I guess you could solve the continuity equation $\nabla \cdot \vec{J} = i \omega \rho$ and get 8 delta functions. At least in my experience Matlab has difficulty integrating delta functions and that may not be the case with Mathematica.

I also see the formula

dPdΩ=Z02k2|a(ℓ,m)|2∣∣→Xℓ,m∣∣2dPdΩ=Z02k2|a(ℓ,m)|2|X→ℓ,m|2​

So maybe that is what you're referring to. Unfortunately I know absolutely nothing about vector spherical harmonics.

I think I'm going to switch gears to diffraction (equivalent aperture dipoles and what not). I know next to nothing about it but I find it interesting and it appears to be an active area of research (with regards to the aperture equivalent quadrupoles). If I have any questions (which I most definitely will) I will start a new thread. I'm happy with the progress made in this thread. Thank you again.

I don't know why the Latex rendering isn't working. It renders as I type it but doesn't show up when I press post.

 

[TSny]

I used the result for $\frac{dP}{d \Omega}$ that is given in equation (9.150) in the 3rd edition. Other key equations are (9.119), (9.169), and (9.170). So, I didn't calculate the E and B fields.

I didn't study this material in any detail. I just accepted these equations and let Mathematica do the rest.

 

[PhDeezNutz]

Does anyone know how the magnetic octupole would be defined?