Question

Show that the second equation in Eq. 12.127 can be expressed in terms of the field tensor ${\mathbf{F}}^{\mathbf{\mu \nu }}$as follows:

localid="1654746948628" $\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{\mu \nu }}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{\lambda }}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{\nu \lambda }}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{\mu }}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{\lambda \mu }}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{\nu }}}\mathbf{=}\mathbf{0}$

Short answer

The second equation $\frac{\partial G^{\mu \nu }}{\partial x^{\nu }}=0$can be expressed in terms of field tensor as

$\frac{\partial {\mathrm{F}}_{\mathrm{\mu \nu }}}{\partial x^{\lambda }}+\frac{\partial {\mathrm{F}}_{\mathrm{\nu \lambda }}}{\partial x^{\mu }}+\frac{\partial {\mathrm{F}}_{\mathrm{\lambda \mu }}}{\partial x^{\nu }}=0$

Step-by-step solution

Expression for Maxwell’s equation:

Write the expression for Maxwell’s equation.

$\frac{\partial G^{\mu \nu }}{\partial x^{\nu }}=0$

....................(1)

$$\begin{gathered} \mathrm{If}\mathrm{the}\mathrm{\mu }-0\mathrm{then},\mathrm{equation}\left(1\right)\mathrm{becomes}, \\ \frac{\partial {\mathrm{G}}^{\mathrm{o\nu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=\frac{\partial {\mathrm{G}}^{00}}{\partial {\mathrm{x}}^0}+\frac{\partial {\mathrm{G}}^{01}}{\partial {\mathrm{x}}^1}+\frac{\partial {\mathrm{G}}^{02}}{\partial {\mathrm{x}}^2}+\frac{\partial {\mathrm{G}}^{03}}{\partial {\mathrm{x}}^3} \\ \frac{\partial {\mathrm{G}}^{\mathrm{o\nu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{B}}_{\mathrm{Z}}}{\partial \mathrm{z}} \\ \frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{B}}_{\mathrm{Z}}}{\partial \mathrm{z}}=\nabla .\mathrm{B} \\ \nabla .\mathrm{B}=0 \\ \mathrm{If}\mathrm{\mu }=1\mathrm{then}\mathrm{equation}\left(1\right)\mathrm{becomes}, \\ \frac{\partial {\mathrm{G}}^{1\mathrm{\nu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=\frac{\partial {\mathrm{G}}^{10}}{\partial {\mathrm{x}}^0}+\frac{\partial {\mathrm{G}}^{11}}{\partial {\mathrm{x}}^1}+\frac{\partial {\mathrm{G}}^{12}}{\partial {\mathrm{x}}^2}+\frac{\partial {\mathrm{G}}^{13}}{\partial {\mathrm{x}}^3} \\ \frac{\partial {\mathrm{G}}^{1\mathrm{\nu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=-\frac{1}{\mathrm{c}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{t}}-\frac{1}{\mathrm{c}}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{t}}+\frac{1}{\mathrm{c}}\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{z}} \\ -\frac{1}{\mathrm{c}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{t}}-\frac{1}{\mathrm{c}}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{t}}+\frac{1}{\mathrm{c}}\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{z}}=-\frac{1}{\mathrm{c}}\left(\frac{\partial \mathrm{B}}{\partial \mathrm{t}}+\nabla \times \mathrm{E}\right) \\ \nabla \times \mathrm{E}=-\frac{\partial \mathrm{B}}{\partial \mathrm{t}} \end{gathered}$$

Show that∂Fμν∂xλ+∂Fνλ∂xμ+∂Fμν∂xν=0

$\mathrm{Take}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{spatial}\mathrm{components}.$

$$\begin{gathered} \mathrm{If}\mathrm{\mu }=1,\mathrm{v}=2\mathrm{and}\mathrm{\lambda }=3,\mathrm{then},\mathrm{the}\mathrm{equation}\left(2\right)\mathrm{becomes}, \\ \frac{\partial {\mathrm{F}}_{12}}{\partial {\mathrm{x}}^3}+\frac{\partial {\mathrm{F}}_{23}}{\partial {\mathrm{x}}^1}+\frac{\partial {\mathrm{F}}_{31}}{\partial {\mathrm{x}}^2}=\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}+\frac{\partial {\mathrm{F}}_{\mathrm{x}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{F}}_{\mathrm{y}}}{\partial \mathrm{y}} \\ \nabla .\mathrm{B}=0 \\ \mathrm{If}\mathrm{\mu }=0,\mathrm{v}=1\mathrm{and}\mathrm{\lambda }=2,\mathrm{the}\mathrm{z}‐\mathrm{component}\mathrm{from}\mathrm{equation}\left(2\right)\mathrm{becomes}, \\ \frac{\partial {\mathrm{F}}_{01}}{\partial {\mathrm{x}}^2}+\frac{\partial {\mathrm{F}}_{12}}{\partial {\mathrm{x}}^1}+\frac{\partial {\mathrm{F}}_{30}}{\partial {\mathrm{x}}^2}=\frac{\partial \left({\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\right)}{\partial \mathrm{y}}+\frac{{\mathrm{B}}_{\mathrm{z}}}{\partial \left(\mathrm{ct}\right)}+\frac{\partial \left({\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\right)}{\partial \mathrm{x}} \\ \nabla \times \mathrm{E}=-\frac{\partial \mathrm{B}}{\partial \mathrm{t}} \\ \mathrm{If}\mathrm{\mu }=0,\mathrm{v}=2\mathrm{and}\mathrm{\lambda }=3,\mathrm{the}\mathrm{x}‐\mathrm{and}\mathrm{y}\mathrm{component}\mathrm{from}\mathrm{equation}\left(2\right)\mathrm{becomes}, \\ \frac{\partial {\mathrm{F}}_{02}}{\partial {\mathrm{x}}^3}+\frac{\partial {\mathrm{F}}_{23}}{\partial {\mathrm{x}}^0}+\frac{\partial {\mathrm{F}}_{30}}{\partial {\mathrm{x}}^2}=\frac{\partial \left({\mathrm{E}}_{\mathrm{y}}/\mathrm{c}\right)}{\partial {\mathrm{x}}^3}+\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \left(\mathrm{ct}\right)}+\frac{\partial \left({\mathrm{E}}_{\mathrm{z}}/\mathrm{c}\right)}{\partial {\mathrm{x}}^2} \\ \nabla \times \mathrm{E}=-\frac{\partial \mathrm{B}}{\partial \mathrm{t}} \\ \mathrm{So},\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{thet}\mathrm{the}\mathrm{function}\frac{\partial {\mathrm{G}}^{\mathrm{\mu \nu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=0\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{field} \\ \mathrm{tensor}\mathrm{as}\frac{\partial {\mathrm{F}}_{\mathrm{\mu \nu }}}{\partial {\mathrm{x}}^{\mathrm{\lambda }}}+\frac{\partial {\mathrm{F}}_{\mathrm{\nu \lambda }}}{\partial {\mathrm{x}}^{\mathrm{\mu }}}+\frac{\partial {\mathrm{F}}_{\mathrm{\lambda \mu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=0 \\ \mathrm{Therefore},\mathrm{the}\mathrm{second}\mathrm{equation}\frac{\partial {\mathrm{G}}^{\mathrm{\mu \nu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=0\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{field} \\ \mathrm{tensor}\mathrm{as}\frac{\partial {\mathrm{F}}_{\mathrm{\mu \nu }}}{\partial {\mathrm{x}}^{\mathrm{\lambda }}}+\frac{\partial {\mathrm{F}}_{\mathrm{\nu \lambda }}}{\partial {\mathrm{x}}^{\mathrm{\mu }}}+\frac{\partial {\mathrm{F}}_{\mathrm{\lambda \mu }}}{\partial {\mathrm{x}}^{\mathrm{\nu }}}=0 \end{gathered}$$