Question

(a) Construct a tensor ${\mathbf{D}}^{\mathbf{\mu \upsilon }}$(analogous to ${\mathbf{F}}^{\mathbf{\mu \upsilon }}$) out of $\mathbf{D}$and $\mathbf{H}$. Use it to express Maxwell's equations inside matter in terms of the free current density ${\mathbf{J}}_{\mathbf{f}}^{\mathbf{\mu }}$.

(b) Construct the dual tensor ${\mathbf{H}}^{\mathbf{\mu \upsilon }}$(analogous to ${\mathbf{G}}^{\mathbf{\mu \upsilon }}$)

(c) Minkowski proposed the relativistic constitutive relations for linear media:

${\mathbf{D}}^{\mathbf{\mu \upsilon }}{\mathbf{\eta }}_{\mathbf{\upsilon }}\mathbf{=}{\mathbf{c}}^{\mathbf{2}}{\mathbf{\epsilon F}}^{\mathbf{\mu \upsilon }}{\mathbf{\eta }}_{\mathbf{\upsilon }}$ and${\mathbf{H}}^{\mathbf{\mu \upsilon }}{\mathbf{\eta }}_{\mathbf{\upsilon }}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\mu }}{\mathbf{G}}^{\mathbf{\mu \upsilon }}{\mathbf{\eta }}_{\mathbf{\upsilon }}$

Where $\mathbf{\epsilon }$is the proper permittivity, $\mathbf{\mu }$is the proper permeability, and${\mathbf{\eta }}^{\mathbf{\upsilon }}$ is the 4-velocity of the material. Show that Minkowski's formulas reproduce Eqs. 4.32 and 6.31, when the material is at rest.

(d) Work out the formulas relating D and H to E and B for a medium moving with (ordinary) velocity u.

Short answer

a. Maxwell's equation inside the matter in terms of free current density is: ${\mathrm{D}}^{{}_{01}}={\mathrm{cD}}_{\mathrm{x}}$, ${\mathrm{D}}^{12}={\mathrm{H}}_{\mathrm{z}}$, etc., $\frac{\partial {\text{D}}^{\mathrm{\mu \upsilon }}}{\partial {\mathrm{x}}^{\mathrm{\upsilon }}}={\mathrm{J}}_{\mathrm{f}}^{\mathrm{\mu }}$

b. The dual sensor${\mathrm{H}}^{\mathrm{\mu \upsilon }}$ is:${\mathrm{H}}^{\mathrm{\mu \upsilon }}=\left[\begin{array}{cccc}0& {\mathrm{H}}_{\mathrm{x}}& {\mathrm{H}}_{\mathrm{y}}& {\mathrm{H}}_{\mathrm{z}}\\ -{\mathrm{H}}_{\mathrm{x}}& 0& -{\mathrm{cD}}_{\mathrm{z}}& {\mathrm{cD}}_{\mathrm{y}}\\ -{\mathrm{H}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{z}}& 0& -{\mathrm{cD}}_{\mathrm{x}}\\ -{\mathrm{H}}_{\mathrm{z}}& -{\mathrm{cD}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{x}}& 0\end{array}\right]$

c. It has been shown that Minkowski's formulas reproduce $D=\epsilon E$ and $\mathrm{H}=\frac{\mathrm{B}}{\mathrm{\mu }}$

d. The formulae relating to D and H are:

$D={\gamma }^2\epsilon \left\{\left(1-\frac{u^2{v}^2}{c^4}\right)E+\left(1-\frac{v^2}{c^2}\right)\left[(u\times B)-\frac{1}{c^2}(E.u)u\right]\right\}$ and

$\mathrm{H}=\frac{{\mathrm{\gamma }}^2}{\mathrm{\mu }}\left\{\left(1-\frac{{\mathrm{u}}^2}{{\mathrm{v}}^2}\right)\mathrm{B}+\left(\frac{1}{{\mathrm{v}}^2}-\frac{1}{{\mathrm{c}}^2}\right)(\mathrm{u}\times \mathrm{E})-(\mathrm{B}.\mathrm{u})\mathrm{u}\right\}$

Step-by-step solution

Define the Tensor.

In mathematical terminology, a tensor is an algebraic object that defines the relationship between the sets of algebraic objects and vector space. Tensors are those quantities that are neither vector nor scaler, meaning they have magnitude and direction but do not follow the vector law of addition.

Expression of Maxwell's Equation in terms of the free current density Jfμ.

(a)

It's known that:

$\mathrm{D}={\mathrm{\epsilon }}_0\mathrm{E}+\mathrm{P}$ which suggests $\mathrm{E}\to \frac{\mathrm{D}}{{\mathrm{\epsilon }}_0}$and $\mathrm{H}=\frac{1}{{\mathrm{\mu }}_0}\mathrm{B}-\mathrm{M}$which suggests $\mathrm{B}\to {\mathrm{\mu }}_0\mathrm{H}$

Now, by dividing the equations by,${\mathrm{\mu }}_0$we get:

$\mathrm{E}\to \frac{1}{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}\mathrm{D}={\mathrm{c}}^2\mathrm{D}$and $\mathrm{B}\to \mathrm{H}$

Therefore, it's evident from here that,

${\mathrm{D}}^{\mathrm{\mu \upsilon }}=\left[\begin{array}{cccc}0& {\mathrm{cD}}_{\mathrm{x}}& {\mathrm{cD}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{z}}\\ -{\mathrm{cD}}_{\mathrm{x}}& 0& {\mathrm{H}}_{\mathrm{z}}& -{\mathrm{H}}_{\mathrm{y}}\\ -{\mathrm{cD}}_{\mathrm{y}}& -{\mathrm{H}}_{\mathrm{z}}& 0& {\mathrm{H}}_{\mathrm{z}}\\ -{\mathrm{cD}}_{\mathrm{z}}& {\mathrm{H}}_{\mathrm{y}}& -{\mathrm{H}}_{\mathrm{x}}& 0\end{array}\right]\text{}$

Then from section 12.3.4, if the derivative is performed, we get,

$\begin{array}{c}\frac{\partial }{\partial {\mathrm{x}}^{\mathrm{\upsilon }}}{\mathrm{D}}^{0\mathrm{\upsilon }}=\mathrm{c}\nabla .\mathrm{D}\\ ={\mathrm{c\rho }}_{\mathrm{f}}\\ ={\mathrm{J}}_{\mathrm{f}}^0\end{array}$

Also, it can be obtained after performing derivative that,

$\begin{array}{c}\frac{\partial }{\partial x^{\upsilon }}{D}^{1\upsilon }=\frac{1}{c}\frac{\partial }{\partial t}(-c{D}_x)+{(\nabla \times H)}_x\\ ={\left(J_f\right)}_x\end{array}$

Therefore, we get from the above equations that,

$\frac{\partial {\mathrm{D}}_{\mathrm{\mu \upsilon }}}{\partial {\mathrm{x}}^{\mathrm{\upsilon }}}={\mathrm{J}}_{\mathrm{f}}^{\mathrm{\mu }}$

It has been known that:

${\mathrm{J}}_{\mathrm{f}}^{\mathrm{\mu }}=({\mathrm{c\rho }}_{\mathrm{f}},{\mathrm{J}}_{\mathrm{f}})$Meanwhile, the homogeneous Maxwell equations

$(\nabla \cdot \mathrm{B}=0,\mathrm{E}=-\partial \mathrm{t}/\partial \mathrm{B})\text{}$are unchanged, hence,$\frac{\partial {\mathrm{G}}^{\mathrm{\mu \upsilon }}}{\partial {\mathrm{x}}^{\mathrm{\upsilon }}}=0$

Step 3:  Construction of the dual tensor Hμυ

(b)

It is known that the tensor matrix can be determined from the equation 12.120 as:

${\mathrm{H}}^{\mathrm{\mu \upsilon }}=\left[\begin{array}{cccc}0& {\mathrm{H}}_{\mathrm{x}}& {\mathrm{H}}_{\mathrm{y}}& {\mathrm{H}}_{\mathrm{z}}\\ -{\mathrm{H}}_{\mathrm{x}}& 0& -{\mathrm{cD}}_{\mathrm{z}}& {\mathrm{cD}}_{\mathrm{y}}\\ -{\mathrm{H}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{z}}& 0& -{\mathrm{cD}}_{\mathrm{x}}\\ -{\mathrm{H}}_{\mathrm{z}}& -{\mathrm{cD}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{x}}& 0\end{array}\right]$

To show the equations 4.32 and 6.31 using Minkowski's formulas.

(c)

Its been known that if the material is at rest, then it is true that ${\eta }_{\upsilon }=(-c,0,0,0)$the sum over $\upsilon$ collapses to a single term, which can be written as:

$\begin{array}{c}{\mathrm{D}}^{{\mathrm{\mu }}_0}{\mathrm{\eta }}_0\text{}={\mathrm{c}}^2{\mathrm{\epsilon F}}^{{\mathrm{\mu }}_0}{\mathrm{\eta }}_0\text{}\\ {\mathrm{D}}^{{\mathrm{\mu }}_0}={\mathrm{c}}^2{\mathrm{\epsilon F}}^{{\mathrm{\mu }}_0}\\ -\mathrm{cD}=-{\mathrm{c}}^2\mathrm{\epsilon }\frac{\mathrm{E}}{\mathrm{c}}\text{}\\ \mathrm{D}=\mathrm{\epsilon E}\end{array}$

Under the above condition, it can also be determined that:

$\begin{array}{c}{\mathrm{H}}^{{\mathrm{\mu }}_0}{\mathrm{\eta }}_{0\text{}}=\frac{1\text{}}{\mathrm{\mu }}{\mathrm{G}}^{{\mathrm{\mu }}_0}{\mathrm{\eta }}_0\text{}\\ {\mathrm{H}}^{{\mathrm{\mu }}_0}{\mathrm{\eta }}_{0\text{}}=\frac{1\text{}}{\mathrm{\mu }}{\mathrm{G}}^{{\mathrm{\mu }}_0}\\ -\mathrm{H}=-\frac{1\text{}}{\mathrm{\mu }}\text{}\mathrm{B}\\ \mathrm{H}=\frac{1\text{}}{\mathrm{\mu }}\text{}\mathrm{B}\end{array}$

Determination of the formulae that relate D and H to E and B for a medium moving with (ordinary) velocity u.

(d)

In general, ${\eta }_{\nu }=\gamma (-c,u)$ hence, for $\mathrm{\mu }=0$

$\begin{array}{c}{\mathrm{D}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }}={\mathrm{D}}^{01}{{\mathrm{\eta }}_1}_{\text{}}+{\mathrm{D}}^{02}{\mathrm{\eta }}_2\text{}+{\mathrm{D}}^{03}{\mathrm{\eta }}_3\text{}\\ ={\mathrm{cD}}_{\mathrm{x}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{x}\text{}}\right)+{\mathrm{cD}}_{\mathrm{y}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{y}}\text{}\right)+{\mathrm{cD}}_{\mathrm{z}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{z}}\text{}\right)\\ =\mathrm{\gamma c}(\mathrm{D}\cdot \mathrm{u})\end{array}$

Similarly,

$\begin{array}{c}{\mathrm{F}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }}={\mathrm{F}}^{01}{{\mathrm{\eta }}_1}_{\text{}}+{\mathrm{F}}^{02}{\mathrm{\eta }}_2\text{}+{\mathrm{F}}^{03}{\mathrm{\eta }}_3\text{}\\ =\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{x}\text{}}\right)+\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{y}}\right)+\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{z}}\right)\\ =\frac{\mathrm{\gamma }}{\mathrm{c}}(\mathrm{E}\cdot \mathrm{u})\end{array}$

So,

$\begin{array}{c}{D}^{0\upsilon }{\eta }_{\upsilon \text{}}=c^2\epsilon F^{0\upsilon }{\eta }_{\upsilon }\text{}\\ \gamma c(D\cdot u)=c^2\epsilon \left(\frac{\gamma }{c}\right)(E\cdot u)\\ D\cdot u=\epsilon (E\cdot u)\end{array}$…… (i)

Similarly,

$\begin{array}{c}{\mathrm{D}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }\text{}}={\mathrm{c}}^2{\mathrm{\epsilon F}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }}\text{}\\ \mathrm{\gamma c}(\mathrm{D}\cdot \mathrm{u})={\mathrm{c}}^2\mathrm{\epsilon }\left(\frac{\mathrm{\gamma }}{\mathrm{c}}\right)(\mathrm{E}\cdot \mathrm{u})\\ \mathrm{D}\cdot \mathrm{u}=\mathrm{\epsilon }(\mathrm{E}\cdot \mathrm{u})\end{array}$

Again, under the same condition, we get,

$\begin{array}{c}{\mathrm{H}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }}={\mathrm{H}}^{01}{{\mathrm{\eta }}_1}_{\text{}}+{\mathrm{H}}^{02}{\mathrm{\eta }}_2\text{}+{\mathrm{H}}^{03}{\mathrm{\eta }}_3\text{}\\ ={\mathrm{H}}_{\mathrm{x}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{x}\text{}}\right)+{\mathrm{H}}_{\mathrm{y}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{y}}\right)+{\mathrm{H}}_{\mathrm{z}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{z}}\right)\\ =\mathrm{\gamma }(\mathrm{H}\cdot \mathrm{u})\end{array}$

Similarly,

$\begin{array}{c}{\mathrm{G}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }}={\mathrm{G}}^{01}{{\mathrm{\eta }}_1}_{\text{}}+{\mathrm{G}}^{02}{\mathrm{\eta }}_2\text{}+{\mathrm{G}}^{03}{\mathrm{\eta }}_3\text{}\\ ={\mathrm{B}}_{\mathrm{x}}\text{}{\mathrm{\gamma u}}_{\mathrm{x}\text{}}+{\mathrm{B}}_{\mathrm{y}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{y}}\right)+{\mathrm{B}}_{\mathrm{z}}\text{}\left({\mathrm{\gamma u}}_{\mathrm{z}}\right)\\ =\mathrm{\gamma }(\mathrm{B}\cdot \mathrm{u})\end{array}$

So, the required equation is:

$\begin{array}{c}{\mathrm{H}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }\text{}}=\frac{1}{\mathrm{\mu }}{\mathrm{G}}^{0\mathrm{\upsilon }}{\mathrm{\eta }}_{\mathrm{\upsilon }}\text{}\\ \mathrm{\gamma }(\mathrm{H}\cdot \mathrm{u})=\frac{1}{\mathrm{\mu }}\mathrm{\gamma }(\mathrm{B}\cdot \mathrm{u})\\ \mathrm{H}\cdot \mathrm{u}=\frac{1}{\mathrm{\mu }}(\mathrm{B}\cdot \mathrm{u})\end{array}$ …… (ii)

Using Eq. [i] to rewrite ($(u\cdot D)$):

$\begin{array}{c}D\left(1-\frac{u^2}{c^2}\right)=-\frac{\epsilon }{c^2}(E.u)u+\epsilon [E+(u\times B)]-\frac{1}{\mu c^4}[(E.u)u-u^{2}E]\\ D={\gamma }^2\epsilon \left\{\left(1-\frac{u^2{v}^2}{c^4}\right)E+\left(1-\frac{v^2}{c^2}\right)\left[(u\times B)-\frac{1}{c^2}(E.u)u\right]\right\}\end{array}$

Using Eq. [ii] to rewrite $\left(u.D\right)$:

$\begin{array}{c}\mathrm{H}\left(1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)=-\frac{1}{{\mathrm{\mu c}}^2}(\mathrm{B}.\mathrm{u})\mathrm{u}+\frac{1}{\mathrm{\mu }}\mathrm{B}-\frac{1}{{\mathrm{c}}^2}(\mathrm{u}\times \mathrm{E})+\mathrm{\epsilon }(\mathrm{u}\times \mathrm{E})+\mathrm{\epsilon }[(\mathrm{B}.\mathrm{u})\mathrm{u}-{\mathrm{u}}^2\mathrm{B}]\\ \mathrm{H}=\frac{{\mathrm{\gamma }}^2}{\mathrm{\mu }}\left\{\left(1-\frac{{\mathrm{u}}^2}{{\mathrm{v}}^2}\right)\mathrm{B}+\left(\frac{1}{{\mathrm{v}}^2}-\frac{1}{{\mathrm{c}}^2}\right)[(\mathrm{u}\times \mathrm{E})-(\mathrm{B}.\mathrm{u})\mathrm{u}]\right\}\end{array}$