Question

Recall that a covariant 4-vector is obtained from a contravariant one by changing the sign of the zeroth component. The same goes for tensors: When you “lower an index” to make it covariant, you change the sign if that index is zero. Compute the tensor invariants

${\mathbf{F}}^{\mathbf{\mu v}}\mathbf{}{\mathbf{F}}_{\mathbf{\mu v}\mathbf{}\mathbf{,}}{\mathbf{G}}^{\mathbf{\mu v}}\mathbf{}{\mathbf{G}}_{{}^{\mathbf{\mu v}}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{F}}^{\mathbf{\mu v}}\mathbf{}{\mathbf{G}}_{{}^{\mathbf{\mu v}}}$

in terms of E and B. Compare Prob. 12.47.

Short answer

The tensor invariants are ${\mathrm{F}}^{\mathrm{\mu v}}{\mathrm{F}}_{\mathrm{\mu v}}=2\left({\mathrm{B}}^2-\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}\right){\mathrm{G}}^{\mathrm{\mu v}}{\mathrm{G}}_{\mathrm{\mu v}}=2\left(\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}-{\mathrm{B}}^2\right)\mathrm{and}$${\mathrm{F}}^{\mathrm{\mu v}}{\mathrm{G}}_{\mathrm{\mu v}}=-\frac{4}{\mathrm{c}}\left(\mathbf{E}\mathbf{.}\mathbf{B}\right)$

Step-by-step solution

Expression for the Product of  and :

Write the expression for the product of ${\mathrm{F}}^{\mathrm{\mu v}}{\mathrm{F}}_{\mathrm{\mu v}}$

$$\begin{gathered} {\mathrm{F}}^{\mathrm{\mu v}}{\mathrm{F}}_{\mathrm{\mu v}}=\left[\begin{array}{ccccc}{\mathrm{F}}^{00}{\mathrm{F}}^{00}& -{\mathrm{F}}^{01}{\mathrm{F}}^{01}& -{\mathrm{F}}^{02}{\mathrm{F}}^{02}& -{\mathrm{F}}^{03}{\mathrm{F}}^{03}& -\\ {\mathrm{F}}^{10}{\mathrm{F}}^{10}& -{\mathrm{F}}^{20}{\mathrm{F}}^{20}& -{\mathrm{F}}^{30}{\mathrm{F}}^{30}& +& \\ {\mathrm{F}}^{11}{\mathrm{F}}^{11}& +{\mathrm{F}}^{12}{\mathrm{F}}^{12}& +{\mathrm{F}}^{13}{\mathrm{F}}^{13}& +{\mathrm{F}}^{21}{\mathrm{F}}^{21}& +{\mathrm{F}}^{22}{\mathrm{F}}^{22}+\\ {\mathrm{F}}^{23}{\mathrm{F}}^{23}& +{\mathrm{F}}^{31}{\mathrm{F}}^{31}& +{\mathrm{F}}^{32}{\mathrm{F}}^{32}& +{\mathrm{F}}^{33}{\mathrm{F}}^{33}& \end{array}\right] \\ \mathrm{Here},{\mathrm{F}}^{00}=0,{\mathrm{F}}^{01}=\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}},{\mathrm{F}}^{02}=\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}},{\mathrm{F}}^{03}=\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}},{\mathrm{F}}^{12}={\mathrm{B}}_{\mathrm{z}},{\mathrm{F}}^{31}={\mathrm{B}}_{\mathrm{y}}\mathrm{and} \\ {\mathrm{F}}^{23}={\mathrm{B}}_{\mathrm{x}} \end{gathered}$$ .......(1)

Write the expression for the product of $G^{\mathrm{\mu v}}{\mathrm{G}}_{\mathrm{\mu v}}$

$$\begin{gathered} G^{\mathrm{\mu v}}{\mathrm{G}}_{\mathrm{\mu v}}=\left[\begin{array}{ccccc}{G}^{00}{G}^{00}& -G^{01}{G}^{01}& -G^{02}{G}^{02}& -G^{03}{G}^{03}& -\\ G^{10}{\mathrm{G}}^{10}& -G^{20}{G}^{20}& -G^{30}{G}^{30}& +& \\ G^{11}{G}^{11}& +G^{12}{G}^{12}& +G^{13}{G}^{13}& +G^{21}{G}^{21}& +G^{22}{G}^{22}+\\ G^{23}{G}^{23}& +G^{31}{G}^{31}& +G^{32}{G}^{32}& +G^{33}{G}^{33}& \end{array}\right] \\ \mathrm{Here},G^{00}=0,G^{01},G^{12}={\mathrm{B}}_{\mathrm{z}},G^{31}={\mathrm{B}}_{\mathrm{y}},{\mathrm{G}}^{23}={\mathrm{B}}_{\mathrm{x}},\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}},{\mathrm{G}}^{02}=\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\mathrm{and} \\ {\mathrm{G}}^{03}=\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}} \end{gathered}$$ ......(2)

Determine the Product of :

Substitute ${\mathrm{F}}^{00}=0,{\mathrm{F}}^{01}=\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}},{\mathrm{F}}^{02}=\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}},{\mathrm{F}}^{03}=\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}},{\mathrm{F}}^{12}={\mathrm{B}}_{\mathrm{z}},{\mathrm{F}}^{31}={\mathrm{B}}_{\mathrm{y}}\mathrm{and}{\mathrm{F}}^{23}={\mathrm{B}}_{\mathrm{x}}$in equation (1).

$$\begin{gathered} {\mathrm{F}}^{\mathrm{\mu v}}{\mathrm{F}}_{\mathrm{\mu v}}=\left[\begin{array}{c}0-{\left(\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\right)}^2--{\left(\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\right)}^2-{\left(\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\right)}^2-{\left(\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\right)}^2-{\left(\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\right)}^2-{\left(\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\right)}^2+0+\\ {\mathrm{B}}_{\mathrm{z}}^2+{\mathrm{B}}_{\mathrm{y}}^2+{\mathrm{B}}_{\mathrm{x}}^2+0+{\mathrm{B}}_{\mathrm{z}}^2+{\mathrm{B}}_{\mathrm{y}}^2+{\mathrm{B}}_{\mathrm{x}}^2+0\end{array}\right] \\ {\mathrm{F}}^{\mathrm{\mu v}}{\mathrm{F}}_{\mathrm{\mu v}}=\left[\begin{array}{c}\\ \end{array}-\frac{2{{\mathrm{E}}_{\mathrm{x}}}^2}{{\mathrm{c}}^2}-\frac{2{{\mathrm{E}}_{\mathrm{y}}}^2}{{\mathrm{c}}^2}-\frac{2{{\mathrm{E}}_{\mathrm{z}}}^2}{{\mathrm{c}}^2}+2{\mathrm{B}}_{\mathrm{x}}^2+2{\mathrm{B}}_{\mathrm{y}}^2+2{\mathrm{B}}_{\mathrm{z}}^2\right] \\ \mathrm{Here},{\mathbf{B}}_{\mathbf{x}}\mathbf{+}{\mathbf{B}}_{\mathbf{y}}\mathbf{+}{\mathbf{B}}_{\mathbf{z}}\mathbf{=}\mathbf{B}\mathrm{and}{\mathbf{E}}_{\mathbf{x}}\mathbf{+}\mathbf{}{\mathbf{E}}_{\mathbf{x}}\mathbf{+}\mathbf{}{\mathbf{E}}_{\mathbf{x}}\mathbf{=}\mathbf{E} \end{gathered}$$

So, the above equation becomes,

$$\begin{gathered} {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=\left[-\frac{2{\mathrm{E}}_{\mathrm{x}}^2}{{\mathrm{c}}^2}-\frac{2{\mathrm{E}}_{\mathrm{y}}^2}{{\mathrm{c}}^2}-\frac{2{\mathrm{E}}_{\mathrm{z}}^2}{{\mathrm{c}}^2}+2{\mathrm{B}}_{\mathrm{x}}^2+2{\mathrm{B}}_{\mathrm{y}}^2+2{\mathrm{B}}_{\mathrm{z}}^2\right] \\ {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=\left[-\frac{2}{{\mathrm{c}}^2}\left({\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2+{\mathrm{E}}_{\mathrm{z}}^2\right)+2\left({\mathrm{B}}_{\mathrm{x}}^2+{\mathrm{B}}_{\mathrm{y}}^2+{\mathrm{B}}_{\mathrm{z}}^2\right)\right] \\ {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=\left(-\frac{2{\mathrm{E}}^2}{{\mathrm{c}}^2}+2{\mathrm{B}}^2\right) \\ {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=2\left({\mathrm{B}}^2\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}\right) \end{gathered}$$

Determine the Product of :

Substitute ${\mathrm{G}}^{00}=0,{\mathrm{G}}^{01}={\mathrm{B}}_{\mathrm{x}},{\mathrm{G}}^{01}={\mathrm{B}}_{\mathrm{y}},{\mathrm{G}}^{01}={\mathrm{B}}_{\mathrm{z}},{\mathrm{G}}^{12}=\frac{{\mathrm{E}}_{\mathrm{Z}}}{\mathrm{c}},{\mathrm{G}}^{31}=\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\mathrm{and}{\mathrm{G}}^{31}=\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}$in equation (2).

$$\begin{gathered} {\mathrm{G}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=\left[\begin{array}{c}0-{\mathrm{B}}_{\mathrm{x}}^2-{\mathrm{B}}_{\mathrm{x}}^2-{\mathrm{B}}_{\mathrm{x}}^2-{\mathrm{B}}_{\mathrm{x}}^2-{\mathrm{B}}_{\mathrm{x}}^2-{\mathrm{B}}_{\mathrm{x}}^2+0+{\left(-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\right)}^2+\\ \left(-\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\right)+\left(-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\right)+0+\left(-\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\right)+\left(-\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\right)+\left(-\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\right)+0\end{array}\right] \\ {\mathrm{G}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=\left[-2{\mathrm{B}}_{\mathrm{x}}^2-2{\mathrm{B}}_{\mathrm{x}}^2-2{\mathrm{B}}_{\mathrm{x}}^2+\frac{{\mathrm{E}}_{\mathrm{z}}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{E}}_{\mathrm{y}}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{E}}_{\mathrm{z}}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{E}}_{\mathrm{x}}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{E}}_{\mathrm{z}}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{E}}_{\mathrm{x}}^2}{{\mathrm{c}}^2}\right] \\ {\mathrm{G}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=\left[-2\left({\mathrm{B}}_{\mathrm{x}}^2+{\mathrm{B}}_{\mathrm{y}}^2+{\mathrm{B}}_{\mathrm{z}}^2\right)+\frac{2}{{\mathrm{c}}^2}\left({\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2+{\mathrm{E}}_{\mathrm{z}}^2\right)\right] \end{gathered}$$

On further solving,

$$\begin{gathered} {\mathrm{G}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=\left(-2{\mathrm{B}}^2+\frac{2}{{\mathrm{c}}^2}{\mathrm{E}}^2\right) \\ {\mathrm{G}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=2\left(\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}-{\mathrm{B}}^2\right) \end{gathered}$$

Determine the Product of :

Write the expression for the product of ${\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}$

role="math" localid="1654678337382" ${\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=-2\left({\mathrm{F}}^{01}{\mathrm{G}}^{01}+{\mathrm{F}}^{02}{\mathrm{G}}^{02}+{\mathrm{F}}^{03}{\mathrm{G}}^{03}\right)+2\left({\mathrm{F}}^{12}{\mathrm{G}}^{12}+{\mathrm{F}}^{13}{\mathrm{G}}^{13}+{\mathrm{F}}^{23}{\mathrm{G}}^{23}\right)$

Substitute, role="math" localid="1654679206876" ${\mathrm{F}}^{01}=\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}},{\mathrm{F}}^{02}=\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}},{\mathrm{F}}^{03}=\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}},{\mathrm{F}}^{12}={\mathrm{B}}_{\mathrm{z}},{\mathrm{F}}^{31}={\mathrm{B}}_{\mathrm{z}}\mathrm{and}{\mathrm{F}}^{23}={\mathrm{B}}_{\mathrm{x}}$

${\mathrm{G}}^{01}={\mathrm{B}}_{\mathrm{x}},{\mathrm{G}}^{02}={\mathrm{B}}_{\mathrm{y}},{\mathrm{G}}^{03}={\mathrm{B}}_{\mathrm{z}}{\mathrm{G}}^{12}=-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}},{\mathrm{G}}^{31}=-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\mathrm{and}{\mathrm{G}}^{23}=-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}$in the above expression.

$$\begin{gathered} {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=\left[\begin{array}{c}2\left(\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\right)\left({\mathrm{B}}_{\mathrm{x}}\right)+\left(\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\right)\left({\mathrm{B}}_{\mathrm{x}}\right)+\left(\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\right)\left({\mathrm{B}}_{\mathrm{x}}\right)\\ +2\left({\mathrm{B}}_{\mathrm{z}}\left(\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{c}}\right)+{\mathrm{B}}_{\mathrm{y}}\left(\frac{{\mathrm{E}}_{\mathrm{y}}}{\mathrm{c}}\right)+{\mathrm{B}}_{\mathrm{x}}\left(\frac{{\mathrm{E}}_{\mathrm{x}}}{\mathrm{c}}\right)\right)\end{array}\right] \\ {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=-\frac{2}{\mathrm{c}}\left({\mathrm{E}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{x}}+{\mathrm{E}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{y}}+{\mathrm{E}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{z}}\right)-\frac{2}{\mathrm{c}}\left({\mathrm{E}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{x}}+{\mathrm{E}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{y}}+{\mathrm{E}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{z}}\right) \\ {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=-\frac{2}{\mathrm{c}}\left(\mathbf{E}\mathbf{.}\mathbf{B}\right)-\frac{2}{\mathrm{c}}\left(\mathbf{E}\mathbf{.}\mathbf{B}\right) \\ {\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=-\frac{4}{\mathrm{c}}\left(\mathbf{E}\mathbf{.}\mathbf{B}\right) \end{gathered}$$

Therefore, the tensor invariants are${\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{F}}_{\mathrm{\mu \nu }}=2\left({\mathrm{B}}^2-\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}\right){\mathrm{G}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=2\left(\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}-{\mathrm{B}}^2\right)\mathrm{and}{\mathrm{F}}^{\mathrm{\mu \nu }}{\mathrm{G}}_{\mathrm{\mu \nu }}=\frac{4}{\mathrm{c}}\left(\mathbf{E}\mathbf{.}\mathbf{B}\right)$.