Find all elements of the Maxwell stress tensor for a monochromatic plane wave traveling in the z direction and linearly polarized in the x direction (Eq. 9.48). Does your answer make sense? (Remember that $\mathbf{-}\overleftrightarrow{\mathbf{T}}$represents the momentum flux density.) How is the momentum flux density related to the energy density, in this case?

Write the expression for the electric field.

$\mathbf{E}\left(\mathbf{z}\mathbf{,}\mathbf{t}\right)\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\mathbf{cos}\left(\mathbf{kz}\mathbf{-}\mathbf{\omega t}\right)\hat{\mathbf{x}}$ ……. (1)

Write the expression for the magnetic field.

$\mathbf{B}\left(\mathbf{z}\mathbf{,}\mathbf{t}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}}{\mathbf{E}}_{\mathbf{0}}\mathbf{cos}\left(\mathbf{kz}\mathbf{-}\mathbf{\omega t}\right)\hat{\mathbf{y}}$ ……. (2)

The momentum flux density ${\mathit{T}}_{\mathbf{i}\mathbf{j}}$is given by,

${\mathit{T}}_{\mathbf{ij}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}{\mathbf{E}}_{\mathbf{i}}{\mathbf{E}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}{\mathbf{B}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$

With the fields in Eq. 9.48, E has only an x component, and B only has a y component. So, all the “off-diagonal” $\mathbf{(}\mathbf{i}\mathbf{\ne }\mathbf{j}\mathbf{)}$terms will be zero.

As for the “diagonal” elements:

$$\begin{gathered} {\mathit{T}}_{\mathbf{x}\mathbf{x}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\left(E_x{E}_x-\frac{1}{2}{E}^2\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\left(-\frac{1}{2}{B}^2\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left({\epsilon }_0{E}^2-\frac{1}{{\mu }_0}{B}^2\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

Solve for second diagonal element.

$$\begin{gathered} {\mathit{T}}_{\mathbf{yy}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}{\mathbf{B}}_{\mathbf{y}}{\mathbf{B}}_{\mathbf{y}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{-}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{E}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

Solve for third diagonal element.

$$\begin{gathered} {\mathit{T}}_{\mathbf{u}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathit{u} \end{gathered}$$

So, ${\mathit{T}}_{\mathbf{z}\mathbf{z}}\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{E}}_{\mathbf{0}}^{\mathbf{2}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\left(kz-\omega t+\delta \right)$(all other elements zero).

The momentum of these fields is in the z direction, and it is being transported in the z direction, so yes, it does make sense that ${\mathit{T}}_{\mathbf{zz}}$should be the only nonzero element in ${\mathit{T}}_{\mathbf{ij}}$.

It is known that localid="1658405424229" $\mathbf{-}\overleftrightarrow{\mathbf{T}}\mathbf{·}\mathit{d}\mathit{a}$is the rate at which momentum crosses an area $\mathit{d}\mathit{a}$. Here we have no momentum crossing areas oriented in the x or y-direction.

The momentum per unit time per unit area flowing across a surface oriented in thez-direction is,

$$\begin{gathered} \mathbf{-}{\mathit{T}}_{\mathbf{zz}}\mathbf{=}\mathit{u} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{g}\mathit{c} \end{gathered}$$

Therefore,

$$\begin{gathered} \mathbf{∆}\mathit{p}\mathbf{=}\mathit{g}\mathit{c}\mathit{A}\mathbf{∆}\mathit{T} \\ \frac{\mathbf{∆}\mathbf{p}}{\mathbf{∆}\mathbf{T}}\mathbf{=}\mathit{g}\mathit{c}\mathit{A} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{m}\mathit{o}\mathit{m}\mathit{e}\mathit{n}\mathit{t}\mathit{u}\mathit{m}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{u}\mathit{n}\mathit{i}\mathit{t}\mathbf{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathbf{}\mathit{c}\mathit{r}\mathit{o}\mathit{s}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{a}\mathit{r}\mathit{e}\mathit{a}\mathbf{}\mathit{A} \end{gathered}$$

It is known that momentum flux density is equal to energy density. Therefore,

$$\begin{gathered} \mathit{g}\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{E}}_{\mathbf{0}}^{\mathbf{2}}{\mathbf{cos}}^{\mathbf{2}}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega t}\mathbf{+}\mathbf{\delta }\mathbf{)}\hat{\mathbf{z}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}}\mathit{u}\hat{\mathbf{z}} \end{gathered}$$

Therefore, the all the elements ${\mathit{T}}_{\mathbf{zz}}$are 0, where ${\mathit{T}}_{\mathbf{zz}}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{E}}_{\mathbf{0}}^{\mathbf{2}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega t}\mathbf{+}\mathbf{\delta }\mathbf{)}$, the answer make sense as the direction of the field is in the z direction. The relation between momentum flux density and the energy is $\frac{\mathbf{1}}{\mathbf{c}}\mathit{u}\hat{\mathbf{z}}$.