# *58 for general multiples (2023)

## introduction

We carried out electrical dipole, magnetic dipole and electrical radiation: quadrupol. What higher contributions? The techniques used so far in the really appropriate sand$- -$We have to be more systematic.Spherical vector harmonic, that appear in other areas, the dynamics of exposure fluids. We see that this Abelaw.jackson continues to find energy and angle impulse (Wedon does not do this here).Brown presented, theCommentThe"Jackson's algebra is more than one bit." These superior multipoles are not taken into account in Landaus Buch or Likharev Stonybrooks notes and they optionally opened. They were used more often in core physics than now.

However, for the highest multipoles we are interested in inspection fronts (but it is not symmetrically spherical!). In the distant radiation zone we will have waves that seem very similar to the waves of the family level.

It does not conclude that the$\stackrel{\to }{\mathrm{mi}}, So,\stackrel{\to }{H}$The fields have no components to spread$- -$It is fair that these components are getting smaller and smaller (factor$\sim 1/\mathrm{Riñonal}$) With regard to the transverse components during the wave, and these components play an important role in radiation of the angle impulse.

For more information on possible field configurations, we have to solve Maxwell's equations in spherical polar coordinates.

## Scalar wave

As a heating exercise (which turns out to be very relevant), we start with the spherical wave equation for AclimbCampos$\mathrm{Fi}\left(X, So,Y, So,z, So,T\right)$Before you switch to the most difficult vestocuations for electrical and magnetic fields

${\nabla }^{2}\mathrm{Fi}\left(X, So,Y, So,z, So,T\right)- -\frac{1}{{C}^{2}}\frac{{\partial }^{2}\mathrm{Fi}\left(X, So,Y, So,z, So,T\right)}{\partial {T}^{2}}=0$

in spherical polar coordinates.

As a dependence on the usual time${\mathrm{mi}}^{- -\mathrm{Ue}\mathrm{Vaya}T}, So,$

$\mathrm{Fi}\left(X, So,Y, So,z, So,T\right)=\mathrm{Fi}\left(X, So,Y, So,z\right){\mathrm{mi}}^{- -\mathrm{Ue}\mathrm{Vaya}T}, So,$

The wave equation is

$\left({\nabla }^{2}+ +{k}^{2}\right)\mathrm{Fi}\left(X, So,Y, So,z\right)=0, So,\text{ }k=\mathrm{Vaya}/C.$

Of course, this is identical to the equation that we have solved for the guides, but now we will use$\mathrm{Riñonal}, So,º, So,\varphi$Coordinates.

However, it should still be familiar:${\nabla }^{2}$In the spherical polar it is exactly the equation for the problem of hydrogen atom in the elementary quantum mechanics, which is dissolved by variable separation and the same trick works here: the standard ball extension

$\mathrm{Fi}\left(X, So,Y, So,z\right)=\sum _{\ell , So,\mathrm{Metro}}{F}_{\ell \mathrm{Metro}}\left(\mathrm{Riñonal}\right){Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right)$

(Video) Tudor Black Bay 58: 5 Things I wish I knew before buying

Where (remember)${Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right)=\sqrt{\frac{2\ell + +1}{4\mathrm{Pi}}\cdot \frac{\left(\ell - -\mathrm{Metro}\right)!}{\left(\ell + +\mathrm{Metro}\right)!}}\text{ }{\mathrm{Pag}}_{\ell }^{\mathrm{Metro}}\left(\mathrm{rib}º\right){\mathrm{mi}}^{\mathrm{Ue}\mathrm{Metro}\varphi }.$

## Radial wave equation: Bessel spherical functions

The radial function fulfills (regardless of$\mathrm{Metro}$)

$\left[\frac{{D}^{2}}{D{\mathrm{Riñonal}}^{2}}+ +\frac{2}{\mathrm{Riñonal}}\frac{D}{D\mathrm{Riñonal}}+ +{k}^{2}- -\frac{\ell \left(\ell + +1\right)}{{\mathrm{Riñonal}}^{2}}\right]{F}_{\ell }\left(\mathrm{Riñonal}\right)=0.$

At that time Jackson replaces${F}_{\ell }\left(\mathrm{Riñonal}\right)={\mathrm{Of}}_{\ell }\left(\mathrm{Riñonal}\right)/\sqrt{\mathrm{Riñonal}}, So,$To find that${\mathrm{Of}}_{\ell }\left(\mathrm{Riñonal}\right)$The beetle equation fulfills for a total average value$\mathrm{Norte}=\ell + +\frac{1}{2}:$

$\left[\frac{{D}^{2}}{D{\mathrm{Riñonal}}^{2}}+ +\frac{1}{\mathrm{Riñonal}}\frac{D}{D\mathrm{Riñonal}}+ +{k}^{2}- -\frac{{\left(\ell + +\frac{1}{2}\right)}^{2}}{{\mathrm{Riñonal}}^{2}}\right]{\mathrm{Of}}_{\ell }\left(\mathrm{Riñonal}\right)=0.$

So${\mathrm{Of}}_{\ell }\left(\mathrm{Riñonal}\right)$Fulfills the equation of Bessel$- -$except that$\ell$It is replaced by$\ell + +\frac{1}{2}.$Remember that the usual Bessel equation arises from the${\nabla }^{2}$Operator expressed incylindricalWe only derive coordinates the same equation, but with$\ell \to \ell + +\frac{1}{2}$von${\nabla }^{2}$EmsphericalCoordinates. The solutions for this new version are mentioned, not surprising thatsphericalFunctions of Bessel (Neumann, Hankel), written with small letters${J}_{\ell }\left(\mathrm{Riñonal}\right)$etc. and can be derived from the functions of Bessel (cylindrical) (Express series form) as follows:

$\begin{array}{l}{J}_{\ell }\left(\mathrm{Riñonal}\right)=\sqrt{\mathrm{Pi}/2\mathrm{Riñonal}}{J}_{\ell + +\frac{1}{2}}\left(\mathrm{Riñonal}\right), So,\\ {\mathrm{Norte}}_{\ell }\left(\mathrm{Riñonal}\right)=\sqrt{\mathrm{Pi}/2\mathrm{Riñonal}}{\mathrm{Norte}}_{\ell + +\frac{1}{2}}\left(\mathrm{Riñonal}\right), So,\\ {H}_{\ell }{}^{\left(1, So,2\right)}\left(\mathrm{Riñonal}\right)=\sqrt{\mathrm{Pi}/2\mathrm{Riñonal}}\left[{J}_{\ell + +\frac{1}{2}}\left(\mathrm{Riñonal}\right)±\mathrm{Ue}{\mathrm{Norte}}_{\ell + +\frac{1}{2}}\left(\mathrm{Riñonal}\right)\right].\end{array}$

That means we can find${J}_{\ell }\left(\mathrm{Riñonal}\right)$The expansions of the infinite series of functions of the Bessel theory, for example

${J}_{\mathrm{Norte}}\left(\mathrm{Riñonal}\right)={\left(\frac{\mathrm{Riñonal}}{2}\right)}^{\mathrm{Norte}}\sum _{J=0}^{\infty }\frac{{\left(- -1\right)}^{J}}{J!C\left(J+ +\mathrm{Norte}+ +1\right)}{\left(\frac{\mathrm{Riñonal}}{2}\right)}^{2J}$

and use$C\left(\frac{1}{2}\right)=\sqrt{\mathrm{Pi}}, So,\text{ }\text{ }C\left(\frac{3}{2}\right)=\frac{1}{2}\sqrt{\mathrm{Pi}}, So,\text{ }\text{ }C\left(\mathrm{Norte}\right)=\left(\mathrm{Norte}- -1\right)C\left(\mathrm{Norte}- -1\right)$We found:

${J}_{0}\left(\mathrm{Riñonal}\right)=\frac{\mathrm{sin}\mathrm{Riñonal}}{\mathrm{Riñonal}}, So,\text{ }{J}_{1}\left(\mathrm{Riñonal}\right)=\frac{\mathrm{sin}\mathrm{Riñonal}}{{\mathrm{Riñonal}}^{2}}- -\frac{\mathrm{rib}\mathrm{Riñonal}}{\mathrm{Riñonal}}, So,...$

Surprise: You are much easier than the functions of Bessel's theory!

ÖThe spherical functions of Bessel are finite polynomesEm$1/\mathrm{Riñonal}$With coefficient$\mathrm{sin}\mathrm{Riñonal}, So,\mathrm{rib}\mathrm{Riñonal}.$The only justification to derive it, as Wejet did, is to show why they are called Bessel functionsremovedeasier to derive them by writing${F}_{\ell }\left(\mathrm{Riñonal}\right)={\mathrm{Of}}_{\ell }\left(\mathrm{Riñonal}\right)/\mathrm{Riñonal}$In the original differential equation as an unexpected text of the introductory quantum mechanics!In particular, I indicate complete treatment (based on the representation of Landau) in my quantum mechanicsHere.

Asymptotic for great$\mathrm{Riñonal}, So,$

$\begin{array}{l}{J}_{\ell }\left(\mathrm{Riñonal}\right)\to \frac{1}{\mathrm{Riñonal}}\mathrm{sin}\left(\mathrm{Riñonal}- -\frac{\ell \mathrm{Pi}}{2}\right), So,\\ {\mathrm{Norte}}_{\ell }\left(\mathrm{Riñonal}\right)\to - -\frac{1}{\mathrm{Riñonal}}\mathrm{rib}\left(\mathrm{Riñonal}- -\frac{\ell \mathrm{Pi}}{2}\right), So,\\ {H}_{\ell }{}^{\left(1\right)}\left(\mathrm{Riñonal}\right)\to \frac{{\left(- -\mathrm{Ue}\right)}^{\ell + +1}{\mathrm{mi}}^{\mathrm{Ue}\mathrm{Riñonal}}}{\mathrm{Riñonal}}.\end{array}$

(Video) Subtraction with Regrouping - Math Video for 2nd Grade

That is, Hankel's spherical function${H}_{\ell }{}^{\left(1\right)}\left(\mathrm{Riñonal}\right)$It is the function that corresponds to the starting waves. We are interested in the inertia, so we are looking for solutions in the form$\sum _{\ell , So,\mathrm{Metro}}{A}_{\ell \mathrm{Metro}}\left(k\mathrm{Riñonal}\right){Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right).$Ö${Y}_{\ell \mathrm{Metro}}$(Cube is expressly close to the beginning of this section) are the angle family of the angular pulse operator$\stackrel{\to }{\mathrm{Ue}}=\left(1/\mathrm{Ue}\right)\left(\stackrel{\to }{\mathrm{Riñonal}}×\stackrel{\to }{\nabla }\right)$familiar with the quantum mechanics (of course without$\hslash$) Essentially the gradient operator on the spherical surface and of course, of course, of course,$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{Ue}}=0.$

That is,

$\begin{array}{c}{\mathrm{Ue}}^{2}{Y}_{\ell \mathrm{Metro}}=\ell \left(\ell + +1\right){Y}_{\ell \mathrm{Metro}}\\ {\mathrm{Ue}}_{+ +}{Y}_{\ell \mathrm{Metro}}=\sqrt{\left(\ell - -\mathrm{Metro}\right)\left(\ell + +\mathrm{Metro}+ +1\right)}{Y}_{\ell \text{ }\mathrm{Metro}+ +1}, So,\\ {\mathrm{Ue}}_{- -}{Y}_{\ell \mathrm{Metro}}=\sqrt{\left(\ell + +\mathrm{Metro}\right)\left(\ell - -\mathrm{Metro}+ +1\right)}{Y}_{\ell \text{ }\mathrm{Metro}- -1}, So,\\ {\mathrm{Ue}}_{z}{Y}_{\ell \mathrm{Metro}}=\mathrm{Metro}{Y}_{\ell \text{ }\mathrm{Metro}}.\end{array}$

Also,$\stackrel{\to }{\mathrm{Ue}}×\stackrel{\to }{\mathrm{Ue}}=\mathrm{Ue}\stackrel{\to }{\mathrm{Ue}}.$

## Multipolarer Formalismus

Maxwell's equations in Espaço Empio (over time${\mathrm{mi}}^{- -\mathrm{Ue}\mathrm{Vaya}T}, So,\text{ }\mathrm{Vaya}=Ck$),

$\begin{array}{c}\stackrel{\to }{\nabla }×\stackrel{\to }{\mathrm{mi}}=\mathrm{Ue}k{Z}_{0}\stackrel{\to }{H}, So,\text{ }\text{ }\stackrel{\to }{\nabla }×\stackrel{\to }{H}=- -\mathrm{Ue}k\stackrel{\to }{\mathrm{mi}}/{Z}_{0}\\ \stackrel{\to }{\nabla }\cdot \stackrel{\to }{\mathrm{mi}}=0, So,\text{ }\text{ }\text{ }\text{ }\stackrel{\to }{\nabla }\cdot \stackrel{\to }{H}=0\end{array}$

$\left({\nabla }^{2}+ +{k}^{2}\right)\stackrel{\to }{\mathrm{mi}}=0, So,\text{ }\stackrel{\to }{\nabla }\cdot \stackrel{\to }{\mathrm{mi}}=0, So,$

And the same equations keep it for$\stackrel{\to }{H}.$These equations can be dissolved for the three in standard fashionCartesian $\left(X, So,Y, So,z\right)$Coordinates,ButWe are interested in spherical starting waves and there isTrivial separationof these vectors in spherical coordinates$\left(\mathrm{Riñonal}, So,º, So,\varphi \right).$How do we go on?

Jackson uses an intelligent solution from Bouwkamp and Casimir (1954): The equations for theclimbAmounts$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{mi}}, So,\text{ }\text{ }\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}.$This copies the standard approach for inwaveguides waves, whereby the approach should first solve the field component for spreading.${\mathrm{mi}}_{z}$Ö${H}_{z}.$

In this spherical case,

${\nabla }^{2}\left(\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{mi}}\right)=\stackrel{\to }{\mathrm{Riñonal}}\cdot {\nabla }^{2}\stackrel{\to }{\mathrm{mi}}+ +2\stackrel{\to }{\nabla }\cdot \stackrel{\to }{\mathrm{mi}}, So,$

and in the same way too$\stackrel{\to }{H}, So,$Then from$\stackrel{\to }{\nabla }\cdot \stackrel{\to }{\mathrm{mi}}=0, So,\text{ }\text{ }\stackrel{\to }{\nabla }\cdot \stackrel{\to }{H}=0$, mi${\nabla }^{2}\stackrel{\to }{\mathrm{mi}}=- -{k}^{2}\stackrel{\to }{\mathrm{mi}}, So,$

$\left({\nabla }^{2}+ +{k}^{2}\right)\left(\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{mi}}\right)=0, So,\text{ }\text{ }\left({\nabla }^{2}+ +{k}^{2}\right)\left(\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}\right)=0.$

So$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{mi}}$EsA solution for the wave equation! (How is it$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}.$)

(Video) Least Common Multiple (LCM) || Tagalog Explanation

The general starting solution is a series in spherical harmonious with Hankel functions that are accompanied.

$\mathrm{Fi}\left(\stackrel{\to }{\mathrm{Riñonal}}\right)=\sum _{\ell , So,\mathrm{Metro}}\left[{A}_{\ell \mathrm{Metro}}^{\left(1\right)}{H}_{\ell }^{\left(1\right)}\left(k\mathrm{Riñonal}\right)+ +{A}_{\ell \mathrm{Metro}}^{\left(2\right)}{H}_{\ell }^{\left(2\right)}\left(k\mathrm{Riñonal}\right)\right]{Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right).$

To see how this works, we will select a certain multipolus.

### Magnetic (and electrical) field fields

After Jackson we followdefineA multipolar magnetic field of$\left(\ell , So,\mathrm{Metro}\right)$For conditions

$\begin{array}{c}\stackrel{\to }{\mathrm{Riñonal}}\cdot {\stackrel{\to }{H}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}=\frac{\ell \left(\ell + +1\right)}{k}{\mathrm{Gramm}}_{\ell }\left(k\mathrm{Riñonal}\right){Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right), So,\\ \stackrel{\to }{\mathrm{Riñonal}}\cdot {\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}=0, So,\end{array}$

(Then this is analogous to a wave guide mode) where${\mathrm{Gramm}}_{\ell }\left(k\mathrm{Riñonal}\right)={H}_{\ell }^{\left(1\right)}\left(k\mathrm{Riñonal}\right)$For the starting waves. We know that this is a solution to the wave equation, and we can visualize the component of the magnetic field, which points vertically on a spherical surface with an angle pattern${Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right)$As a wave that stands in a spherical balloon (and for larger spherical surfaces that drops to a size like the inverter).

The electrical field is only tangential, but has the same standard: from Maxwell'Sefations,

${Z}_{0}k\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}=- -\mathrm{Ue}\stackrel{\to }{\mathrm{Riñonal}}\cdot \left(\stackrel{\to }{\nabla }×\stackrel{\to }{\mathrm{mi}}\right)=- -\mathrm{Ue}\left(\stackrel{\to }{\mathrm{Riñonal}}×\stackrel{\to }{\nabla }\right)\cdot \stackrel{\to }{\mathrm{mi}}=\stackrel{\to }{\mathrm{Ue}}\cdot \stackrel{\to }{\mathrm{mi}}.$

Then with$\left(\mathrm{METRO}\right)$Describes that this is a magnetic multipolus)

$\stackrel{\to }{\mathrm{Ue}}\cdot {\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}\propto \stackrel{\to }{\mathrm{Riñonal}}\cdot {\stackrel{\to }{H}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}\propto {Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right).$

But when we expand it$\stackrel{\to }{\mathrm{mi}}$In spherical harmonious of the operators$\stackrel{\to }{\mathrm{Ue}}$will create other neighboring

So that cannot be correctunless $\stackrel{\to }{\mathrm{mi}}\left(º, So,\varphi \right)$It is a properties of$\stackrel{\to }{\mathrm{Ue}}$:

$\begin{array}{l}{\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}={Z}_{0}{\mathrm{Gramm}}_{\ell }\left(k\mathrm{Riñonal}\right)\stackrel{\to }{\mathrm{Ue}}{Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right), So,\text{ }\\ {H}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}=- -\left(\mathrm{Ue}/k{Z}_{0}\right)\stackrel{\to }{\nabla }×{\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}, So,\end{array}$

Then$\stackrel{\to }{\mathrm{Ue}}\cdot {\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{METRO}\right)}\propto {\mathrm{Ue}}^{2}{Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right).$

Remember that the magnetic field standard can determine

(Video) Trigonometry 43: General solution to sinƟ using arcsin including multiple solutions

$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}\propto \stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\nabla }×\stackrel{\to }{\mathrm{mi}}\propto \stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\nabla }×\stackrel{\to }{\mathrm{Ue}}{Y}_{\ell \mathrm{Metro}}=\left(\stackrel{\to }{\mathrm{Riñonal}}×\stackrel{\to }{\nabla }\right)\cdot \stackrel{\to }{\mathrm{Ue}}{Y}_{\ell \mathrm{Metro}}={\mathrm{Ue}}^{2}{Y}_{\ell \mathrm{Metro}}=\ell \left(\ell + +1\right){Y}_{\ell \mathrm{Metro}}.$

ÖelectricThe multipolar fields are defined in the same way, whereby the fields are replaced:

$\begin{array}{l}\stackrel{\to }{\mathrm{Riñonal}}\cdot {\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{mi}\right)}=- -{Z}_{0}\left(\frac{\ell \left(\ell + +1\right)}{k}\right){F}_{\ell }\left(k\mathrm{Riñonal}\right){Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right), So,\\ \stackrel{\to }{\mathrm{Riñonal}}\cdot {\stackrel{\to }{H}}_{\ell \mathrm{Metro}}^{\left(\mathrm{mi}\right)}=0.\end{array}$

The electrical multipolar fields are

$\begin{array}{l}{\stackrel{\to }{H}}_{\ell \mathrm{Metro}}^{\left(\mathrm{mi}\right)}={F}_{\ell }\left(k\mathrm{Riñonal}\right)\stackrel{\to }{\mathrm{Ue}}{Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right), So,\\ {\stackrel{\to }{\mathrm{mi}}}_{\ell \mathrm{Metro}}^{\left(\mathrm{mi}\right)}=\frac{\mathrm{Ue}{Z}_{0}}{k}\stackrel{\to }{\nabla }×{\stackrel{\to }{H}}_{\ell \mathrm{Metro}}^{\left(\mathrm{mi}\right)}.\end{array}$

### Spherical vector harmonic

It is useful to introduce a little more annotation, the spherical harmonious ones that are called:

${\stackrel{\to }{X}}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right)=\frac{1}{\sqrt{\ell \left(\ell + +1\right)}}\stackrel{\to }{\mathrm{Ue}}{Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right).$

These have simple orthogonality properties:

$\begin{array}{l}\int {\stackrel{\to }{X}}^{\ast }{}_{{\ell }^{\text{'}}{\mathrm{Metro}}^{\text{'}}}\cdot {\stackrel{\to }{X}}_{\ell \mathrm{Metro}}={D}_{\ell {\ell }^{\text{'}}}{D}_{\mathrm{Metro}{\mathrm{Metro}}^{\text{'}}}, So,\\ \int {\stackrel{\to }{X}}^{\ast }{}_{{\ell }^{\text{'}}{\mathrm{Metro}}^{\text{'}}}\cdot \left(\stackrel{\to }{\mathrm{Riñonal}}×{\stackrel{\to }{X}}_{\ell \mathrm{Metro}}\right)D\mathrm{Oh}=0.\end{array}$

At this point, Jackson states (but not proves) that these two types of waves form a complete set of vector solutions for Maxwell's equations at the origin of the origin. This is the general solution

$\begin{array}{l}\stackrel{\to }{H}=\sum _{\ell \mathrm{Metro}}\left[{A}_{\mathrm{mi}}\left(\ell , So,\mathrm{Metro}\right){F}_{\ell }\left(k\mathrm{Riñonal}\right){\stackrel{\to }{X}}_{\ell \mathrm{Metro}}- -\frac{\mathrm{Ue}}{k}{A}_{\mathrm{METRO}}\left(\ell , So,\mathrm{Metro}\right)\stackrel{\to }{\nabla }×{\mathrm{Gramm}}_{\ell }\left(k\mathrm{Riñonal}\right){\stackrel{\to }{X}}_{\ell \mathrm{Metro}}\right]\\ \stackrel{\to }{\mathrm{mi}}={Z}_{0}\sum _{\ell \mathrm{Metro}}\left[\frac{\mathrm{Ue}}{k}{A}_{\mathrm{mi}}\left(\ell , So,\mathrm{Metro}\right)\stackrel{\to }{\nabla }×{F}_{\ell }\left(k\mathrm{Riñonal}\right){\stackrel{\to }{X}}_{\ell \mathrm{Metro}}+ +{A}_{\mathrm{METRO}}\left(\ell , So,\mathrm{Metro}\right){\mathrm{Gramm}}_{\ell }\left(k\mathrm{Riñonal}\right){\stackrel{\to }{X}}_{\ell \mathrm{Metro}}\right]\end{array}$

Wo${A}_{\mathrm{mi}}\left(\ell , So,\mathrm{Metro}\right), So,\text{ }{A}_{\mathrm{METRO}}\left(\ell , So,\mathrm{Metro}\right)$These are the amplitudes of the electrical and magnetic multipolar fields.$\mathrm{Fi}\left(\stackrel{\to }{\mathrm{Riñonal}}\right)=\sum _{\ell , So,\mathrm{Metro}}\left[{A}_{\ell \mathrm{Metro}}^{\left(1\right)}{H}_{\ell }^{\left(1\right)}\left(k\mathrm{Riñonal}\right)+ +{A}_{\ell \mathrm{Metro}}^{\left(2\right)}{H}_{\ell }^{\left(2\right)}\left(k\mathrm{Riñonal}\right)\right]{Y}_{\ell \mathrm{Metro}}\left(º, So,\varphi \right).$

Coefficients can be determined if$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}$mi$\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{mi}}$You are known

${A}_{\mathrm{METRO}}\left(\ell , So,\mathrm{Metro}\right)=\frac{k}{\sqrt{\ell \left(\ell + +1\right)}}\int \begin{array}{l}{Y}_{\ell \mathrm{Metro}}^{*}\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}D\mathrm{Oh}\\ \end{array}$

$\begin{array}{l}{A}_{\mathrm{METRO}}\left(\ell , So,\mathrm{Metro}\right){\mathrm{Gramm}}_{\ell }\left(k\mathrm{Riñonal}\right)=\frac{k}{\sqrt{\ell \left(\ell + +1\right)}}\int {Y}_{\ell \mathrm{Metro}}^{*}\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{H}D\mathrm{Oh}, So,\\ {Z}_{0}{A}_{\mathrm{mi}}\left(\ell , So,\mathrm{Metro}\right){F}_{\ell }\left(k\mathrm{Riñonal}\right)=- -\frac{k}{\sqrt{\ell \left(\ell + +1\right)}}\int {Y}_{\ell \mathrm{Metro}}^{*}\stackrel{\to }{\mathrm{Riñonal}}\cdot \stackrel{\to }{\mathrm{mi}}D\mathrm{Oh}.\end{array}$

(Video) Multiply a Decimal by a Decimal | Math with Mr. J

### With the result

In the rest of Chapter 9, Jackson uses the impressions of vonxes to find the energy and angle motif of the multipolar radiation, the strangers, the seeds and a linear antenna that is fed by the center. We will not appeal to this material.

FrontIndex next

## Videos

1. Write first five common multiples of 3 and 4
(Chauhan Basic concepts of Maths)
2. Dave Portnoy Sells Barstool For \$610M, Calls KSI Irrelevant, Admits He Lied To Rogan: IMPAULSIVE 369
(IMPAULSIVE)
3. Controlling Your Dopamine For Motivation, Focus & Satisfaction | Huberman Lab Podcast #39
(Andrew Huberman)
4. Using Play to Rewire & Improve Your Brain | Huberman Lab Podcast #58
(Andrew Huberman)
5. The Murdaugh Mysteries | Full Episode
(48 Hours)
6. If the least common multiple of 56, 57 and 58 is k, then what will be the LCM of 56, 57, 58 and 59 ?
(owl classes)
Top Articles
Latest Posts
Article information

Author: Arielle Torp

Last Updated: 02/02/2023

Views: 6553

Rating: 4 / 5 (61 voted)

Author information

Name: Arielle Torp

Birthday: 1997-09-20

Address: 87313 Erdman Vista, North Dustinborough, WA 37563

Phone: +97216742823598

Job: Central Technology Officer

Hobby: Taekwondo, Macrame, Foreign language learning, Kite flying, Cooking, Skiing, Computer programming

Introduction: My name is Arielle Torp, I am a comfortable, kind, zealous, lovely, jolly, colorful, adventurous person who loves writing and wants to share my knowledge and understanding with you.