91Ó°ÊÓ

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.

(a)

It's known that:

$\mathrm{D}={\mathrm{Î\mu }}_0\mathrm{E}+\mathrm{P}$ which suggests $\mathrm{E}→\frac{\mathrm{D}}{{\mathrm{Î\mu }}_0}$and $\mathrm{H}=\frac{1}{{\mathrm{μ}}_0}\mathrm{B}−\mathrm{M}$which suggests $\mathrm{B}→{\mathrm{μ}}_0\mathrm{H}$

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

$\mathrm{E}→\frac{1}{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}\mathrm{D}={\mathrm{c}}^2\mathrm{D}$and $\mathrm{B}→\mathrm{H}$

Therefore, it's evident from here that,

${\mathrm{D}}^{\mathrm{μÏ\dots }}=\left[\begin{array}{cccc}0& {\mathrm{cD}}_{\mathrm{x}}& {\mathrm{cD}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{z}}\\ −{\mathrm{cD}}_{\mathrm{x}}& 0& {\mathrm{H}}_{\mathrm{z}}& −{\mathrm{H}}_{\mathrm{y}}\\ −{\mathrm{cD}}_{\mathrm{y}}& −{\mathrm{H}}_{\mathrm{z}}& 0& {\mathrm{H}}_{\mathrm{z}}\\ −{\mathrm{cD}}_{\mathrm{z}}& {\mathrm{H}}_{\mathrm{y}}& −{\mathrm{H}}_{\mathrm{x}}& 0\end{array}\right]\text{​}$

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

$\begin{array}{c}\frac{∂}{∂{\mathrm{x}}^{\mathrm{Ï\dots }}}{\mathrm{D}}^{0\mathrm{Ï\dots }}=\mathrm{c}∇.\mathrm{D}\\ ={\mathrm{³¦ÒÏ}}_{\mathrm{f}}\\ ={\mathrm{J}}_{\mathrm{f}}^0\end{array}$

Also, it can be obtained after performing derivative that,

$\begin{array}{c}\frac{∂}{∂x^{\mathrm{Ï\dots }}}{D}^{1\mathrm{Ï\dots }}=\frac{1}{c}\frac{∂}{∂t}(−c{D}_x)+{(∇×H)}_x\\ ={\left(J_f\right)}_x\end{array}$

Therefore, we get from the above equations that,

$\frac{∂{\mathrm{D}}_{\mathrm{μÏ\dots }}}{∂{\mathrm{x}}^{\mathrm{Ï\dots }}}={\mathrm{J}}_{\mathrm{f}}^{\mathrm{μ}}$

It has been known that:

${\mathrm{J}}_{\mathrm{f}}^{\mathrm{μ}}=({\mathrm{³¦ÒÏ}}_{\mathrm{f}},{\mathrm{J}}_{\mathrm{f}})$Meanwhile, the homogeneous Maxwell equations

$(∇â‹\dots \mathrm{B}=0,\mathrm{E}=−∂\mathrm{t}/∂\mathrm{B})\text{​}$are unchanged, hence,$\frac{∂{\mathrm{G}}^{\mathrm{μÏ\dots }}}{∂{\mathrm{x}}^{\mathrm{Ï\dots }}}=0$

(b)

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

${\mathrm{H}}^{\mathrm{μÏ\dots }}=\left[\begin{array}{cccc}0& {\mathrm{H}}_{\mathrm{x}}& {\mathrm{H}}_{\mathrm{y}}& {\mathrm{H}}_{\mathrm{z}}\\ −{\mathrm{H}}_{\mathrm{x}}& 0& −{\mathrm{cD}}_{\mathrm{z}}& {\mathrm{cD}}_{\mathrm{y}}\\ −{\mathrm{H}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{z}}& 0& −{\mathrm{cD}}_{\mathrm{x}}\\ −{\mathrm{H}}_{\mathrm{z}}& −{\mathrm{cD}}_{\mathrm{y}}& {\mathrm{cD}}_{\mathrm{x}}& 0\end{array}\right]$

(c)

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

$\begin{array}{c}{\mathrm{D}}^{{\mathrm{μ}}_0}{\mathrm{η}}_0\text{​}={\mathrm{c}}^2{\mathrm{Î\mu ¹ó}}^{{\mathrm{μ}}_0}{\mathrm{η}}_0\text{​}\\ {\mathrm{D}}^{{\mathrm{μ}}_0}={\mathrm{c}}^2{\mathrm{Î\mu ¹ó}}^{{\mathrm{μ}}_0}\\ −\mathrm{cD}=−{\mathrm{c}}^2\mathrm{Î\mu }\frac{\mathrm{E}}{\mathrm{c}}\text{​}\\ \mathrm{D}=\mathrm{Î\mu ·¡}\end{array}$

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

$\begin{array}{c}{\mathrm{H}}^{{\mathrm{μ}}_0}{\mathrm{η}}_{0\text{​}}=\frac{1\text{​}}{\mathrm{μ}}{\mathrm{G}}^{{\mathrm{μ}}_0}{\mathrm{η}}_0\text{​}\\ {\mathrm{H}}^{{\mathrm{μ}}_0}{\mathrm{η}}_{0\text{​}}=\frac{1\text{​}}{\mathrm{μ}}{\mathrm{G}}^{{\mathrm{μ}}_0}\\ −\mathrm{H}=−\frac{1\text{​}}{\mathrm{μ}}\text{​}\mathrm{B}\\ \mathrm{H}=\frac{1\text{​}}{\mathrm{μ}}\text{​}\mathrm{B}\end{array}$

(d)

In general, ${\mathrm{η}}_{\mathrm{ν}}=\mathrm{γ}(−c,u)$ hence, for $\mathrm{μ}=0$

$\begin{array}{c}{\mathrm{D}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}={\mathrm{D}}^{01}{{\mathrm{η}}_1}_{\text{​}}+{\mathrm{D}}^{02}{\mathrm{η}}_2\text{​}+{\mathrm{D}}^{03}{\mathrm{η}}_3\text{​}\\ ={\mathrm{cD}}_{\mathrm{x}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{x}\text{​}}\right)+{\mathrm{cD}}_{\mathrm{y}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{y}}\text{​}\right)+{\mathrm{cD}}_{\mathrm{z}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{z}}\text{​}\right)\\ =\mathrm{㳦}(\mathrm{D}â‹\dots \mathrm{u})\end{array}$

Similarly,

$\begin{array}{c}{\mathrm{F}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}={\mathrm{F}}^{01}{{\mathrm{η}}_1}_{\text{​}}+{\mathrm{F}}^{02}{\mathrm{η}}_2\text{​}+{\mathrm{F}}^{03}{\mathrm{η}}_3\text{​}\\ =\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{x}\text{​}}\right)+\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\text{​â¶Ä‹}\left({\mathrm{γ³Ü}}_{\mathrm{y}}\right)+\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\text{​â¶Ä‹}\left({\mathrm{γ³Ü}}_{\mathrm{z}}\right)\\ =\frac{\mathrm{γ}}{\mathrm{c}}(\mathrm{E}â‹\dots \mathrm{u})\end{array}$

So,

$\begin{array}{c}{D}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }\text{​}}=c^2\mathrm{Î\mu }{F}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}\text{​}\\ \mathrm{γ}c(Dâ‹\dots u)=c^2\mathrm{Î\mu }\left(\frac{\mathrm{γ}}{c}\right)(Eâ‹\dots u)\\ Dâ‹\dots u=\mathrm{Î\mu }(Eâ‹\dots u)\end{array}$…… (i)

Similarly,

$\begin{array}{c}{\mathrm{D}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }\text{​}}={\mathrm{c}}^2{\mathrm{Î\mu ¹ó}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}\text{​}\\ \mathrm{㳦}(\mathrm{D}â‹\dots \mathrm{u})={\mathrm{c}}^2\mathrm{Î\mu }\left(\frac{\mathrm{γ}}{\mathrm{c}}\right)(\mathrm{E}â‹\dots \mathrm{u})\\ \mathrm{D}â‹\dots \mathrm{u}=\mathrm{Î\mu }(\mathrm{E}â‹\dots \mathrm{u})\end{array}$

Again, under the same condition, we get,

$\begin{array}{c}{\mathrm{H}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}={\mathrm{H}}^{01}{{\mathrm{η}}_1}_{\text{​}}+{\mathrm{H}}^{02}{\mathrm{η}}_2\text{​}+{\mathrm{H}}^{03}{\mathrm{η}}_3\text{​}\\ ={\mathrm{H}}_{\mathrm{x}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{x}\text{​}}\right)+{\mathrm{H}}_{\mathrm{y}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{y}}\right)+{\mathrm{H}}_{\mathrm{z}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{z}}\right)\\ =\mathrm{γ}(\mathrm{H}â‹\dots \mathrm{u})\end{array}$

Similarly,

$\begin{array}{c}{\mathrm{G}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}={\mathrm{G}}^{01}{{\mathrm{η}}_1}_{\text{​}}+{\mathrm{G}}^{02}{\mathrm{η}}_2\text{​}+{\mathrm{G}}^{03}{\mathrm{η}}_3\text{​}\\ ={\mathrm{B}}_{\mathrm{x}}\text{​}{\mathrm{γ³Ü}}_{\mathrm{x}\text{​}}+{\mathrm{B}}_{\mathrm{y}}\text{​â¶Ä‹}\left({\mathrm{γ³Ü}}_{\mathrm{y}}\right)+{\mathrm{B}}_{\mathrm{z}}\text{​}\left({\mathrm{γ³Ü}}_{\mathrm{z}}\right)\\ =\mathrm{γ}(\mathrm{B}â‹\dots \mathrm{u})\end{array}$

So, the required equation is:

$\begin{array}{c}{\mathrm{H}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }\text{​}}=\frac{1}{\mathrm{μ}}{\mathrm{G}}^{0\mathrm{Ï\dots }}{\mathrm{η}}_{\mathrm{Ï\dots }}\text{​}\\ \mathrm{γ}(\mathrm{H}â‹\dots \mathrm{u})=\frac{1}{\mathrm{μ}}\mathrm{γ}(\mathrm{B}â‹\dots \mathrm{u})\\ \mathrm{H}â‹\dots \mathrm{u}=\frac{1}{\mathrm{μ}}(\mathrm{B}â‹\dots \mathrm{u})\end{array}$ …… (ii)

Using Eq. [i] to rewrite ($(uâ‹\dots D)$):

$\begin{array}{c}D\left(1−\frac{u^2}{c^2}\right)=−\frac{\mathrm{Î\mu }}{c^2}(E.u)u+\mathrm{Î\mu }[E+(u×B)]−\frac{1}{\mathrm{μ}{c}^4}[(E.u)u−u^{2}E]\\ D={\mathrm{γ}}^2\mathrm{Î\mu }\left\{\left(1−\frac{u^2{v}^2}{c^4}\right)E+\left(1−\frac{v^2}{c^2}\right)\left[(u×B)−\frac{1}{c^2}(E.u)u\right]\right\}\end{array}$

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

$\begin{array}{c}\mathrm{H}\left(1−\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)=−\frac{1}{{\mathrm{쳦}}^2}(\mathrm{B}.\mathrm{u})\mathrm{u}+\frac{1}{\mathrm{μ}}\mathrm{B}−\frac{1}{{\mathrm{c}}^2}(\mathrm{u}×\mathrm{E})+\mathrm{Î\mu }(\mathrm{u}×\mathrm{E})+\mathrm{Î\mu }[(\mathrm{B}.\mathrm{u})\mathrm{u}−{\mathrm{u}}^2\mathrm{B}]\\ \mathrm{H}=\frac{{\mathrm{γ}}^2}{\mathrm{μ}}\left\{\left(1−\frac{{\mathrm{u}}^2}{{\mathrm{v}}^2}\right)\mathrm{B}+\left(\frac{1}{{\mathrm{v}}^2}−\frac{1}{{\mathrm{c}}^2}\right)[(\mathrm{u}×\mathrm{E})−(\mathrm{B}.\mathrm{u})\mathrm{u}]\right\}\end{array}$