Question: (a) Carry through the argument in Sect. 8.1.2, starting with Eq. 8.6, but using $J_f$in place of J. Show that the Poynting vector becomes $S=E\times H$and the rate of change of the energy density in the fields is$\frac{\partial u}{\partial t}=E·\frac{\partial D}{\partial t}+H·\frac{\partial B}{\partial t}·$

For linear media, show that

$u=\frac{1}{2}\left(E·D+B·H\right)$.

(b) In the same spirit, reproduce the argument in Sect. 8.2.2, starting with Eq. 8.15, with ${\rho }_f$and in$J_f$place of $\rho$and J. Don’t bother to construct the Maxwell stress tensor, but do show that the momentum density is$g=D\times B.$

.

(a)

Write the expression for the rate at which work is done on all free charges.

$\frac{\partial W}{\partial t}={\int }_v\left(E\times J_f\right)d\zeta$ …… (1)

Write the expression for .

$J_f=∆·H-\frac{\partial D}{\partial t}$ …… (2)

The product of $E·J_f$.

$E·J_f=E·\left(\nabla \times H\right)-E·\frac{\partial D}{\partial t}$

From the product rule,

$\nabla ·\left(E\times H\right)=H\left(\nabla \times E\right)-E·\left(\nabla \times H\right)$

Here,$\nabla \times E=-\frac{\partial B}{\partial t}$ . Therefore,

$E·\left(\nabla \times H\right)=-H·\frac{\partial B}{\partial t}-\nabla ·\left(E\times H\right)$

Determine $E·J_f$as follows.

$E·J_f=-H·\frac{\partial B}{\partial t}-\nabla ·\left(E\times H\right)-E·\frac{\partial D}{\partial t}$

Substitute all the values in equation (1).

$$\begin{gathered} \frac{dW}{dt}=-{\int }_v\left(E·\frac{\partial D}{\partial t}+H·\frac{\partial B}{\partial t}\right)d\zeta -{\int }_v\nabla ·\left(E\times H\right)\partial \zeta \\ \frac{dW}{dt}=-{\int }_v\left(E·\frac{\partial D}{\partial t}+H·\frac{\partial B}{\partial t}\right)d\zeta -\boxed{{\int }_v}\left(E\times H\right)·da \end{gathered}$$

The Poynting vector is the power per unit area, that is $S=E\times H$, and the rate of change of electromagnetic energy density is $\frac{\partial u}{\partial t}=E·\frac{\partial D}{\partial t}+H·\frac{\partial B}{\partial t}$.

For linear media, it is known that:

$$\begin{gathered} \mathit{D}\mathbf{=}\mathit{\epsilon }\mathit{E} \\ \mathit{H}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\mu }}\mathit{B} \end{gathered}$$

Here, $\epsilon$and $\mu$are constant in time.

Find the expression for $u$as follows.

$$\begin{gathered} \frac{\partial u}{\partial t}=\epsilon \mathit{E}·\frac{\partial \mathit{E}}{\partial t}+\frac{\mathit{B}}{\mu }·\frac{\partial \mathit{B}}{\partial t} \\ \frac{\partial u}{\partial t}=\frac{1}{2}\epsilon \frac{\partial }{\partial t}\left(\mathit{E}\mathbf{·}\mathit{E}\right)+\frac{1}{2\mu }\frac{\partial }{\partial t}\left(\mathit{B}\mathbf{·}\mathit{B}\right) \\ \frac{\partial u}{\partial t}=\frac{1}{2}\frac{\partial }{\partial t}\left(\mathit{E}\mathbf{·}\mathit{D}\mathbf{+}\mathit{B}\mathbf{·}\mathit{H}\right) \\ u=\frac{1}{2}\left(\mathit{E}\mathbf{·}\mathit{D}\mathbf{+}\mathit{B}\mathbf{·}\mathit{H}\right) \end{gathered}$$

Therefore, it is proved that $u=\frac{1}{2}\left(\mathit{E}\mathbf{·}\mathit{D}\mathbf{+}\mathit{B}\mathbf{·}\mathit{H}\right)$.

(b)

The force in free changes is,

$f={\rho }_f\mathit{E}+{\mathit{J}}_f\times \mathit{B}$ …… (3)

It is known that and ${\rho }_f=\nabla ·\mathit{D}$and ${\mathit{J}}_f=\nabla \times \mathit{H}-\frac{\partial \mathit{D}}{\partial t}$.

Substitute all the values in equation (3).

$f=\mathit{E}\left(\nabla ·\mathit{D}\right)+\left(\nabla \times \mathit{H}\right)\times \mathit{B}-\frac{\partial \mathit{D}}{\partial t}\times \mathit{B}$ …… (4)

The value of $\frac{\partial }{\partial t}\left(\mathit{D}\times \mathit{B}\right)$is known as:

role="math" localid="1653384426174" $$\begin{gathered} \frac{\partial }{\partial t}\left(\mathit{D}\mathbf{\times }\mathit{B}\right)=\frac{\partial \mathit{D}}{\partial t}\times \mathit{B}+\mathit{D}\times \frac{\partial \mathit{B}}{\partial t} \\ \frac{\partial \mathit{B}}{\partial t}=-\nabla \times \mathit{E} \end{gathered}$$

So, $\frac{\partial \mathit{D}}{\partial t}\times \mathit{B}$will be as follows:

$\frac{\partial \mathit{D}}{\partial t}\times \mathit{B}=\frac{\partial }{\partial t}\left(\mathit{D}\times \mathit{B}\right)+D\times \left(\nabla \times \mathit{E}\right)$

Substitute all known values in equation (4).

$$\begin{gathered} f=\mathit{E}\left(\nabla ·\mathit{D}\right)-\mathit{D}\times \left(\nabla \times \mathit{E}\right)-\mathit{B}\times \left(\nabla \times \mathit{H}\right)-\frac{\partial }{\partial t}\left(\mathit{D}\times \mathit{B}\right) \\ =\left\{\left(\mathit{E}\left(\nabla ·\mathit{D}\right)-\mathit{D}\times \left(\nabla \times \mathit{E}\right)\right)+\left[H\left(\nabla ·\mathit{B}\right)-\mathit{B}\times \left(\nabla \times \mathit{H}\right)\right]\right\}-\frac{\partial }{\partial t}\left(\mathit{D}\times \mathit{B}\right) \end{gathered}$$

Here, the term $\left\{\left(\mathit{E}\left(V·\mathit{D}\right)-\mathit{D}\times \left(\nabla \times \mathit{E}\right)\right)+\left[H\left(\nabla ·\mathit{B}\right)-\mathit{B}\times \left(\nabla \times \mathit{H}\right)\right]\right\}$indicates stress tensor and the term $\frac{\partial }{\partial t}\left(\mathit{D}\times \mathit{B}\right)$is the rate of change of momentum density. Therefore, the momentum density is $g=\mathit{D}\times \mathit{B}$.

Therefore, it is proved that $g=\mathit{D}\times \mathit{B}$.