Electromagnetic Fields in a Spherical Coordinate System

Table 6.7: The $TE$ system: All components of any $TE$ e.m. field $(\vec E,\vec B)$ , as well as those of any four-vector $TE$ potential $(\vec A, \phi)$ , are derived from a single master scalar function $\Phi ^{TE}$ . Its source scalar $S^{TE}$ determines the vectorial charge flux vector field. It is purely transverse: it is tangent to the set of nested two-spheres.

$TE$ Potential
$\hat A_\theta$	$\hat A_\phi$	$\hat A_r$	$\phi$
$\frac{1}{r\sin\theta}\frac{\partial \Phi^{TE}}{\partial \phi}$	$-\frac{1}{r}\frac{\partial \Phi^{TE}}{\partial \theta}$	0	0
$TE$ Electric Field
$\hat E_\theta$	$\hat E_\phi$	$\hat E_r$
$-\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi}\frac{\partial \Phi^{TE}}{\partial t }$	$\frac{1}{r}\frac{\partial }{\partial \theta}\frac{\partial \Phi^{TE}}{\partial t}$	0
$TE$ Magnetic Field
$\hat B_\theta$	$\hat B_\phi$	$\hat B_r$
$\frac{1}{r}\frac{\partial }{\partial \theta} \frac{\partial \Phi^{TE} }{\partial r}$	$\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi} \frac{\partial \Phi^{TE}}{\partial r}$	$-\frac{1}{r^2}\left( \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \sin... ...a} + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right) \Phi^{TE}$
$TE$ Source
$\hat J_\theta$	$\hat J_\phi$	$\hat J_r$	$\rho$
$\frac{1}{r\sin\theta}\frac{\partial S^{TE}}{\partial \phi}$	$-\frac{1}{r}\frac{\partial S^{TE}}{\partial \theta}$	0	0

Table 6.8: The $TM$ system: All components of any $TM$ e.m. field $(\vec E,\vec B)$ , as well as those of any four-vector $TM$ potential $(\vec A, \phi)$ , are derived from a single master scalar function $\Phi ^{TM}$ . Its source scalar $S^{TM}$ determines the vectorial charge flux vector field, which is purely longitudinal.

$TM$ Potential
$\hat A_\theta$	$\hat A_\phi$	$\hat A_r$	$\phi$
$0$	$0$	$\frac{\partial \Phi^{TM}}{\partial t}$	$-\frac{\partial \Phi^{TM}}{\partial r}$
$TM$ Electric Field
$\hat E_\theta$	$\hat E_\phi$	$\hat E_r$
$\frac{1}{r}\frac{\partial }{\partial \theta} \frac{\partial \Phi^{TM} }{\partial r}$	$\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi} \frac{\partial \Phi^{TM}}{\partial r}$	$\left( \frac{\partial^2}{\partial r^2 } -\frac{\partial^2}{\partial t^2} \right)\Phi^{TM}$
$TM$ Magnetic Field
$\hat B_\theta$	$\hat B_\phi$	$\hat B_r$
$\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi}\frac{\partial \Phi^{TM}}{\partial t }$	$-\frac{1}{r}\frac{\partial }{\partial \theta}\frac{\partial \Phi^{TM}}{\partial t}$	0
$TM$ Source
$\hat J_\theta$	$\hat J_\phi$	$\hat J_r$	$\rho$
0	0	$\frac{1}{r^2}\frac{\partial r^2 S^{TM}}{\partial t}$	$-\frac{1}{r^2}\frac{\partial r^2 S^{TM}}{\partial r}$

Table 6.9: The $TEM$ system: All components of any $TEM$ e.m. field $(\vec E,\vec B)$ are derived from a single master scalar function, the difference $\Psi -\Phi$ between the two scalar functions. Even though both, separately, are necessary for the definition of the $TEM$ vector potential $(\vec A, \phi)$ , it is only their difference which is determined by an inhomogeneous Poisson equation and an inhomogeneous wave equation, Eqs.(6.106) and (6.107).

$TEM$ Potential
$\hat A_\theta$	$\hat A_\phi$	$\hat A_r$	$\phi$
$\frac{1}{r}\frac{\partial\Psi}{\partial\theta}$	$\frac{1}{r\sin\theta}\frac{\partial\Psi}{\partial\phi}$	$\frac{\partial \Phi}{\partial r}$	$- \frac{\partial \Phi}{\partial t}$
$TEM$ Electric Field
$\hat E_\theta$	$\hat E_\phi$	$\hat E_r$
$\frac{1}{r}\frac{\partial }{\partial \theta} \frac{\partial (\Phi-\Psi)}{\partial t}$	$\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi} \frac{\partial (\Phi-\Psi)}{\partial t}$	0
$TEM$ Magnetic Field
$\hat B_\theta$	$\hat B_\phi$	$\hat B_r$
$\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi}\frac{\partial (\Phi-\Psi)}{\partial r }$	$-\frac{1}{r}\frac{\partial }{\partial \theta} \frac{\partial (\Phi-\Psi)}{\partial r}$	0
$TEM$ Source
$\hat J_\theta$	$\hat J_\phi$	$\hat J_r$	$\rho$
$\frac{1}{r}\frac{\partial I^{~}}{\partial \theta}\frac{1}{r^2}$	$\frac{1}{r\sin\theta}\frac{\partial I^{~}}{\partial \phi}\frac{1}{r^2}$	$\frac{\partial J^{~}}{\partial r}\frac{1}{r^2}$	$-\frac{\partial J^{~}}{\partial t}\frac{1}{r^2}$

A spherical coordinate system induces a decomposition into a set of nested transverse manifolds (concentric spheres) spanned by the angular coordinates and a longitudinal manifold spanned by the radial and the time cordinates.

Such a coordinate decomposition induces a corresponding one in the Maxwell field equation. Following our experience with cylindrical coordinates, we make a corresponding transition to spherical coordinates according to the following heuristic replacement recipe:

$$dx \rightarrow r d\theta; \quad dy \rightarrow r\sin\theta\, d\phi; \quad dz \rightarrow dr; \quad dt \rightarrow dt$$

$$\frac{\partial}{\partial x} \rightarrow \frac{1}{r}\frac{\partial}{\partial\theta}; \quad \frac{\partial}{\partial y} \rightarrow \frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}; \quad \frac{\partial}{\partial z} \rightarrow \frac{\partial}{\partial r}; \quad \frac{\partial}{\partial t} \rightarrow \frac{\partial}{\partial t}$$

and

$$\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} \rightarrow\, \frac{1}{r^2}\left( \frac{1}{\sin\theta}\frac{\part... ...artial\theta} + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right)$$

As already mentioned, once the cartesian components of Maxwell's $TE$ , $TM$ , and $TEM$ system have been exhibited explicitly, one can apply this recipe also to spherical coordinates. The results are given in Tables 6.7, , and respectively.

The recipe guarantees that all electric and magnetic field components in these tables satisfy the first half, Eq.()-(6.40), of Maxwell's field equations. Furthermore, the application of this recipe to the $TE$ , $TM$ , and $TEM$ cartesian master scalar Eqs.(6.71), (6.72), (6.73), and (6.74), yields Eqs.()-(6.107), the corresponding master equations relative to spherical coordinates. The above replacement recipe applies to the e.m. field and its vector potential.

However, the relation between the concentric spheres introduces the squared radius as a conformal factor between their squared elements of arclength and hence their areas. This conformal factor enters only into the the TM source and the longitudinal part of the TEM source, and hence does not seem to be under the purview of the above recipe. It is, however, taken into account by the explicit calculations that lead to Eqs.(6.100)-(6.107).

Exercise 62.2 (Existence and Uniqueness of the 2+2 Decomposition)

a)

Exhibit the partial differential equation which each of the scalars $\Phi,\cdots,\Phi^{TM}$ satisfies, point out why each solution is unique and hence why

$$[\phi,A_z,A_x,A_y]^T\leftrightarrow (\Phi,\Phi^{TE},\Psi,\Phi^{TM})$$ (6109)

is a one-to-one mapping.

Solution: Given $[\phi,A_z,A_x,A_y]^T$ , one has the following system of equations for the scalars $\Phi$ and $\Phi ^{TE}$ ,

$$\frac{\partial \Phi}{\partial x}+\frac{\partial \Phi^{TE}}{\partial y} =A_x$$ (6110)

$$\frac{\partial \Phi}{\partial y}-\frac{\partial \Phi^{TE}}{\partial x} =A_y~.$$ (6111)

Taking the two-dimensional curl and divergence of this system, one finds

$$\frac{\partial^2\Phi^{TE}}{\partial x^2}+ \frac{\partial^2\Phi^{TE}}{\partial y^2} = \frac{\partial A_x}{\partial y}-\frac{\partial A_y}{\partial x}$$ (6112)

$$\frac{\partial^2\Phi}{\partial x^2}+ \frac{\partial^2\Phi}{\partial y^2} = \frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}~.$$ (6113)

With appropriate boundary conditions in the $x$ -$y$ plane, these 2-d Poisson equations have unique scalar solutions $\Phi ^{TE}$ and $\Phi$ . Similarly one obtains

$$\frac{\partial^2\Phi^{TM}}{\partial t^2}- \frac{\partial^2\Phi^{TM}}{\partial z^2} = \frac{\partial A_z}{\partial t}+\frac{\partial\phi}{\partial z}$$ (6114)

$$\frac{\partial^2\Psi}{\partial z^2}- \frac{\partial^2\Psi}{\partial t^2} = \frac{\partial A_z}{\partial z}+\frac{\partial\phi}{\partial t}~.$$ (6115)

With appropriate initial conditions in the $z$ -$t$ plane, these inhomogeneous 2-d wave equations have unique scalar solutions $\Phi ^{TM}$ and $\Psi$ .

On the other hand, given the four scalar fields, Eq.(6.63) implies the unique four-vector field $(\phi,\vec A)$ . Thus Eq.(6.109) is a one-to-one mapping indeed.

b)

Point out why the four vectors

$$\left\{ \left[ \begin{array}{r} 0\\ 0\\ \partial_y\\ -\partial_x ... ...[ \begin{array}{r} 0\\ 0\\ \partial_x\\ \partial_y \end{array} \right] \right\}$$ (6116)

form a linearly independent set, i.e. why the only solution to

$$\left[ \begin{array}{r} 0\\ 0\\ \partial_y\\ -\partial_x \end{arr... ...d{array} \right]\Psi =\left[ \begin{array}{r} 0\\ 0\\ 0\\ 0 \end{array} \right]$$ (6117)

is the trivial one, $\Phi^{TE}=\Phi^{TM}=\Phi=\Psi\equiv 0~.$

c)

Show that the set of vectors

$$\left\{ \vec {\mathcal V}^{(1)} =\left[ \begin{array}{r} 0\\ 0\\ ... ...\partial_t\\ \partial_z\\ \partial_x\\ \partial_y \end{array} \right] \right\},$$ (6118)

where

$$c =\partial_x^2+\partial_y^2$$ (6119)
and

$$d =\partial_z^2-\partial_t^2~,$$ (6120)

also forms a linearly independent set.

Exercise 62.3 (TE SCALAR WAVE EQUATION: ITS MAXWELL ORIGIN)
Consider a TE e.m. potential and its source,

$$[\phi,A_z,A_x,A_y] =[0,0,\partial_y\Phi^{TE},-\partial_x\Phi^{TE}]$$ (6121)

$$[\rho,J_z,J_x,J_y] =[0,0,\partial_y S^{TE},-\partial_x S^{TE}]~.$$ (6122)

a)

Which two of the Maxwell field equations

$$\nabla \cdot \vec E =4\pi\rho$$

$$\nabla\times\vec B -\partial_t \vec E =4\pi \vec J$$

are satisfied trivially ($0=0$ ), and which imply the nontrivial result

$$\frac{\partial}{\partial y}\{ \cdots \}^{TE} =0$$

$$\frac{\partial}{\partial x}\{ \cdots \}^{TE} =0~?$$

b)

What is $\{ \cdots \}^{TE}$ ?

 

Solution.
a) Introducing the $\vec E$ -field, the $\vec B$ -field, and the charge flux-density $(\vec J,\rho)$ into Eqs.(6.41) and (6.42) yields the following results:

$$\nabla \cdot \vec E =4\pi\rho :~ 0=0$$

$$(\nabla \times \vec B)_z -\frac{\partial E_z}{\partial t} =4\pi J_z :~ 0=0$$

$${ (\nabla \times \vec B)_x -\frac{\partial E_x}{\partial t}=4\pi J_x~:}$$

$$\frac{\partial}{\partial y}(-)\left( \frac{\partial^2\Phi^{TE}}{\... ...}\left( \frac{\partial}{\partial y} \frac{\partial\Phi^{TE}}{\partial z}\right)$$

$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad~~~ -\frac... ...c{\partial\Phi^{TE}}{\partial t}\right)=4\pi \frac{\partial S^{TE}}{\partial y}$$

or equivalently

$$\frac{\partial }{\partial y}\left\{ (\partial_x^2+\partial_y^2+\partial_z^2-\partial_t^2)\Phi^{TE} +4\pi S^{TE}\right\}=0~.$$ (6123)

Similarly, and finally,

$$(\nabla \times \vec B)_y -\frac{\partial E_y}{\partial t}=4\pi J_y$$

yields

$$\frac{\partial }{\partial x}\left\{ (\partial_x^2+\partial_y^2+\partial_z^2-\partial_t^2)\Phi^{TE} +4\pi S^{TE}\right\}=0.$$ (6124)

b) $\{ \cdots \}^{TE}\equiv \left(\partial_x^2+\partial_y^2+\partial_z^2-\partial_t^2 \right) \Phi^{TE}+4\pi S^{TE}$ .

Exercise 62.4 (TM SCALAR WAVE EQUATION: ITS MAXWELL ORIGIN)
Consider a TM e.m. potential and its source,

$$[\phi,A_z,A_x,A_y] =[0,0,\partial_y\Phi^{TM},-\partial_x\Phi^{TM}]$$ (6125)

$$[\rho,J_z,J_x,J_y] =[0,0,\partial_y S^{TM},-\partial_x S^{TM}]~.$$ (6126)

a)

Which two of the Maxwell field equations

$$\nabla \cdot \vec E =4\pi\rho$$

$$\nabla\times\vec B -\partial_t \vec E =4\pi \vec J$$

are satisfied trivially ($0=0$ ), and which imply the nontrivial result

$$\frac{\partial}{\partial z}\{ \cdots \}^{TM} =0$$

$$\frac{\partial}{\partial t}\{ \cdots \}^{TM} =0~?$$

b)

What is $\{ \cdots \}^{TM}$ ?

 

Solution.
a) Introducing the $\vec E$ -field, the $\vec B$ -field, and the charge flux-density $(\vec J,\rho)$ into Eqs.(6.41) and (6.42) yields the following result:

$$\nabla \cdot \vec E =4\pi\rho :~ \frac{\partial}{\partial z} \left\{ \frac{\partial^2}{\partial x^... ...ial^2}{\partial t^2} \right\}\Phi^{TM}=-4\pi \frac{\partial S^{TM}}{\partial z}$$
or equivalently

$$~ ~ ~ \frac{\partial }{\partial z}\left\{ (\partial_x^2+\partial_y^2+\partial_z^2-\partial_t^2)\Phi^{TM} +4\pi S^{TM}\right\}=0~.$$ (6127)

$$(\nabla \times \vec B)_x -\frac{\partial E_x}{\partial t} =4\pi J_x :~ ~~0=0$$

$$(\nabla \times \vec B)_y -\frac{\partial E_y}{\partial t} =4\pi J_y :~ ~~0=0$$

$${ (\nabla \times \vec B)_z -\frac{\partial E_z}{\partial t}=4\pi J_z~:}$$

$$\frac{\partial}{\partial x} (-)\frac{\partial^2\Phi^{TM}}{\partia... ...\frac{\partial}{\partial y} \frac{\partial^2\Phi^{TM}}{\partial y \,\partial t}$$

$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad~~~ -\frac... ...artial^2}{\partial t^2}\right)\Phi^{TM}=4\pi \frac{\partial S^{TM}}{\partial t}$$

or equivalently

$$\frac{\partial }{\partial t}\left\{ (\partial_x^2+\partial_y^2+\partial_z^2-\partial_t^2)\Phi^{TM} +4\pi S^{TM}\right\}=0~.$$ (6128)

b) $\{ \cdots \}^{TM}\equiv \left(\partial_x^2+\partial_y^2+\partial_z^2-\partial_t^2 \right) \Phi^{TM}+4\pi S^{TM}$ .

Exercise 62.5 (TEM SCALAR WAVE EQUATIONS: THEIR MAXWELL ORIGIN)
Consider a TEM e.m. potential and its source,

$$[\phi,A_z,A_x,A_y] =[-\partial_t \Phi,\partial_z \Phi, \partial_x\Psi, \partial_y\Psi]$$ (6129)

$$[\rho,J_z,J_x,J_y] =[-\partial_t J,\partial_z J, \partial_x I, \partial_y I]~.$$ (6130)

a)

Which two of the Maxwell field equations

$$\nabla \cdot \vec E =4\pi\rho$$

$$\nabla\times\vec B -\partial_t \vec E =4\pi \vec J$$

imply

$$\frac{\partial}{\partial t}\{ \cdots \}^{TEM} =0$$

$$\frac{\partial}{\partial z}\{ \cdots \}^{TEM} =0~,$$
and which two imply

$$\frac{\partial}{\partial x} [ \cdots ]^{TEM} =0$$

$$\frac{\partial}{\partial y} [ \cdots ]^{TEM} =0~?$$

b)

What is $\{ \cdots \}^{TEM}$ ? What is $[ \cdots ]^{TEM}$ ?

 

Solution.
a) Introducing the $\vec E$ -field, the $\vec B$ -field, and the charge flux-density $(\vec J,\rho)$ into Eqs.(6.41) and (6.42) yields the following result:

$$\nabla \cdot \vec E =4\pi\rho :~ \frac{\partial}{\partial t} \left( \frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} \right) (\Phi-\Psi) =4\pi(-)\frac{\partial J^{~}}{\partial t}$$ (6131)

$$(\nabla \times \vec B)_z -\frac{\partial E_z}{\partial t} =4\pi J_z:~ -\frac{\partial}{\partial z} \left( \frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} \right) (\Phi-\Psi) =4\pi\frac{\partial J^{~}}{\partial z}$$ (6132)

$$(\nabla \times \vec B)_x -\frac{\partial E_x}{\partial t} =4\pi J_x:~ \frac{\partial}{\partial x} \left( \frac{\partial^2}{\partial z^2} -\frac{\partial^2}{\partial t^2} \right) (\Phi-\Psi) =4\pi\frac{\partial I^{~}}{\partial x}$$ (6133)

$$(\nabla \times \vec B)_y -\frac{\partial E_y}{\partial t} =4\pi J_y:~ \frac{\partial}{\partial y} \left( \frac{\partial^2}{\partial z^2} -\frac{\partial^2}{\partial t^2} \right) (\Phi-\Psi) =4\pi\frac{\partial I^{~}}{\partial y}$$ (6134)

b) $\{ \cdots \}^{TEM}\equiv \left(\partial_x^2+\partial_y^2\right) (\Phi -\Psi )+4\pi J$ ;
$~~~~~~[ \cdots ]^{TEM}\equiv \left( \partial_z^2-\partial_t^2 \right)(\Phi -\Psi )-4\pi I$ .

Exercise 62.6 (TE, TM, AND TEM SCALAR WAVE EQUATIONS)
Point out why the previous three exercises imply

$$\left( \partial^2_x+\partial^2_y+\partial^2_z-\partial^2_t \right) \Phi^{TE} =-4\pi S^{TE}$$ (6135)

$$\left( \partial^2_x+\partial^2_y+\partial^2_z-\partial^2_t \right) \Phi^{TM} =-4\pi S^{TM}$$ (6136)
and

$$(\partial_x^2+\partial_y^2)(\Phi-\Psi) =-4\pi J$$ (6137)

$$(\partial_z^2-\partial_t^2)(\Phi-\Psi) =+4\pi I ~.$$ (6138)

Exercise 62.7 (TEM MASTER SCALAR SYSTEM IS INTEGRABLE)
Show that if $I$ and $J$ satisfy Eq.() then there exists a scalar, call it $\Phi -\Psi$ , such that Eqs.(6.138) and (6.139) are satisfied.
Hint: Use Green's function.

Exercise 62.8 (MAGNETIC DIPOLE MOMENT AS A TE FIELD SOURCE)
The total energy of a charge flux and charge density distribution $(\vec J,\rho)$ interacting with the electromagnetic potential $(\vec A, \phi)$ is

$$W=\frac{1}{2}\int\int\int (\vec J \cdot \vec A +\rho \phi )d^3x~.$$

(Nota bene: This energy is the work which an external agent expends to assemble such a distribution against the quasistatic electric and magnetic force fields generated by the distribution at any moment of time.)

a) Show that for the Transverse Electric vector potential in Eq.(6.75), this energy is

$$W=\frac{1}{2}\int\int\int S^{TE} B_z d^3x~.$$

Comment. If one assumes that the $TE$ source density $S^{TE}$ is localized to such a small region that the magnetic field $B_z$ is constant across it, then

$$W=\frac{1}{2}B_z M~.$$

Here

$$M=\int\int\int S^{TE} d^3x$$

is called the magnetic dipole moment along the z-axis, and $S^{TE}$ is the magnetic moment density, also known as the magnetization along the z-direction.

b) Let

$$\theta(x) =\left\{ \begin{array}{c} 1\quad 0\le x\\ 0\quad x<0 \end{array} \right.$$ (6139)
be the Heaviside unit step function so that

$$\frac{d \theta}{dx} =\delta (x)$$

is the Dirac delta function.

Consider a charge flux distribution confined to the boundary of a rectangular cylinder,

$$J_x= +J(z,t)\theta(x)\theta(L_1-x) \left[ \delta(y)\theta(L_2-y)-\theta(y)\delta(L_2-y)\right]$$

$$J_y= -J(z,t)\theta(y)\theta(L_2-y) \left[ \delta(x)\theta(L_1-x)-\theta(x)\delta(L_1-x)\right]$$

$$J_z=$$ 0

$$\rho= 0~.$$

Figure 6.5: Current distributed in the form of a rectangular loop of aerea $L_1 L_2$ . This loop has linear current density $J(z,t)=I(t)\,\delta(z)$ .

(i)

Show that it satisfies the conservation law, Eq.().

(ii)

Find the magnetic dipole density $S^{TE}$ .

Answer: $S^{TE}=J(z,t)~\theta(x)\theta(L_1-x)\theta(y)\theta(L_2-y)$ .

(iii)

Point out why the magnetic moment is

$$M=I\times Area$$

where $Area=L_1L_2$ and

$$I(t)=\int J(z,t)\,dz$$

is the current circulating around the rectangular boundary. The linear current density exemplified in Figure 6.5 has the form $J(z,t)=I(t)\,\delta(z)$ .

Exercise 62.9 (ELECTRIC DIPOLE AS A SOURCE OF TM RADIATION)

Consider the rate at which a given external agent does work on two charges $q_1$ and $q_2$ in order to keep them on their symmetrically placed trajectories

$$\vec X(t)=(0,0,Z(t))$$ (6140)
and

$$-\vec X(t)=(0,0,-Z(t))~.$$ (6141)

Figure 6.6: Spacetime trajectories of two charges, $q_1 =-q$ and $q_2=q$ , symmetrically placed and moving into opposite directions.Their dipole moment is $q\times (separation)~=~2qZ(t)$ .

Given that they move in an environment having an electric field $\vec E(x,y,z,t)$ , the power expended by this agent is

$$\frac{d\textrm{(Energy)}}{dt} = q_2\,\frac{d\vec X(t)}{dt}\cdot \vec E(0,0,Z(t),t) -q_1\,\frac{d\vec X(t)}{dt}\cdot \vec E(0,0,-Z(t),t)$$ (6142)

$$=\int\int\int \vec J(x,y,z,t)\cdot \vec E(x,y,z,t)\, d^3 x$$ (6143)
where

$$\vec J(x,y,z,t) =\delta(x)\delta(y) \left[ q_2\,\delta\left(z-Z(t)\right)\frac{d\vec X}{dt} -q_1\,\delta\left(z+Z(t)\right)\frac{d\vec X}{dt} \right]~.$$ (6144)

is the total charge flux vector due to these two charges.

a)

Taking advantage of linear superposition, find the charge flux-density four-vector $(J_x,J_y,J_z,\rho)$ such that it expresses the conservation of charge,

$$\vec\nabla \cdot \vec J +\frac{\partial \rho}{\partial t}=0$$

$$\textit{Answer:}~J_x =0$$

$$J_y =0$$

$$J_z =q_2\,\frac{d Z(t)}{dt}\,\delta(x)\delta(y)\delta(z-Z(t))- q_1\,\frac{d Z(t)}{dt}\,\delta(x)\delta(y)\delta(z+Z(t))$$

$$\rho =q_2\,~~\delta(x)\delta(y)\delta(z-Z(t))\quad+ \quad q_1 \quad\delta(x)\delta(y)\delta(z+Z(t))$$

Note: Even though we shall (in compliance with physical observations) ultimately set

$$q_2=-q_1\equiv q~,$$ (6145)

it is somewhat easier to keep track of distinguishing contributions to the charge flux vector $\vec J$ by assigning correspondingly distinguishing labels to them.

b)

Show that there exists a function $S^{TM}$ such that

$$J_z=\frac{\partial S^{TM}}{\partial t}~, \quad \rho=-\frac{\partial S^{TM}}{\partial z}~.$$

Do this by expressing the answer in terms of the Heaviside unit step function $\theta$ , Eq.(6.140), on page .

Answer: $S^{TM}=-\delta(x)\delta(y) \left[ q_2\,\theta\left(z-Z(t)\right)+q_1\,\theta\left(z+Z(t)\right) \right]$

c)

In compliance with the observation of many cases of interest, assume that the fractional temporal rate of change of $E_z$ is neglegibly small compared to that of $Z(t)$ :

$$\frac{1}{E_z} \frac{\partial E_z}{\partial t} \ll \frac{1}{Z}\frac{\partial Z}{\partial t} =\frac{1}{S^{TM}}\frac{\partial S^{TM}}{\partial t}$$

In light of this observation, point out why the power expended by the external agent, Eq.(6.144), can be written as

$$\frac{d\textrm{(Energy)}}{dt} =\frac{d}{dt}\int\int\int S^{TM} E_z\, d^3 x~.$$ (6146)

d)

Designating $\int\int\int S^{TM} E_z\, d^3 x$ as the energy of the $(q_1,q_2)$ -system, exhibit

(i)

its form whenever $q_2=-q_1\equiv q$ , Eq.(6.146), as well as

(ii)

its explicit value when in addition $E_z$ is constant on $[-Z,Z]$ , the support of $S^{TM}$ , i.e.

$$\frac{\partial E_z}{\partial z}\vert Z(t)\vert\ll E_z ~.$$ (6147)

Answer:
(i) $\int\int\int S^{TM} E_z\, d^3 x= \int\int\int (-)\delta(x)\delta(y)q \left[ \theta\left(z-Z(t)\right)-\theta\left(z+Z(t)\right) \right]E_z\, d^3 x$

(ii) $\left[\int\int\int S^{TM}\, d^3 x\right] E_z(0,0,0,t_0)$

where $\int\int\int S^{TM}\, d^3 x=2Z(t)q$ is the ``dipole moment" of the system.

Comment: The quantity $\int\int\int S^{TM}\, d^3 x$ is called the dipole moment of the ``microscopic'' [as identified by the inequality (6.148)] $(q_1,q_2)$ -system. The function

$$S^{TM}(x,y,z,t)=\frac{\textrm{(dipole moment)}}{\textrm{(volume)}}$$

is called its dipole moment density, and $\vec E_z(0,0,0,t_0)$ is the electric field at the location of the system.

Comment: If one has an aggregate of such systems, then their total energy is their sum. Under suitable circumstances it can be approximated by the integral

$$\int\int\int S^{TM}(x,y,z,t)E_z(x,y,z,t_0)\, d^3 x \quad \left(=\int\int\int \vec J\cdot \vec E \, d^3 x \right)$$

where $E_z(x,y,z,t_0)$ refers to the average electric field associated with the microscopic dipole moment centered around $(x,y,z)$ .

[references_for_chapter6]