91Ó°ÊÓ

Using equation 12.127, write the expression for Maxwell’s equation.

$\frac{∂{\mathrm{F}}^{\mathrm{μ\pm ¹}}}{∂{\mathrm{x}}^{\mathrm{v}}}={\mathrm{μ}}_0{\mathrm{J}}^{\mathrm{μ}}$ …… (1)

It is known that:

$\frac{∂}{∂x^v}={∂}_v$

Substitute ${∂}_v$for $\frac{∂}{∂x^v}$in equation (1).

${∂}_v{F}^{\mathrm{μ}v}={\mathrm{μ}}_0{J}^{\mathrm{μ}}$ …… (2)

Differentiate the equation (2).

${∂}_v{∂}_{\mathrm{μ}}{F}^{\mathrm{μ}v}={\mathrm{μ}}_0{∂}_{\mathrm{μ}}{J}^{\mathrm{μ}}$

From the above equation, observe the symmetric and anti-symmetric combination.

$\begin{array}{c}{∂}_v{∂}_{\mathrm{μ}}={∂}_{\mathrm{μ}}{∂}_v\text{ â¶Ä‰â¶Ä‰}\left(\text{Symmetric}\right)\\ F^{\mathrm{μ}v}=−F^{\mathrm{μ}v}\text{ â¶Ä‰â¶Ä‰}\left(\text{Anti-symmetric}\right)\end{array}$

Since it is known that:

${∂}_{\mathrm{μ}}{∂}_v{F}^{\mathrm{μ}v}=0$

As the above indices are summed from 0 to 3, the term $\mathrm{μ}$and v can be pronounced as the same. Hence,

$\begin{array}{c}{∂}_{\mathrm{μ}}{∂}_v{F}^{\mathrm{μ}v}={∂}_v{∂}_{\mathrm{μ}}{F}^{\mathrm{μ}v}\\ ={∂}_{\mathrm{μ}}{∂}_v(−F^{\mathrm{μ}v})\\ =−{∂}_{\mathrm{μ}}{∂}_v{F}^{\mathrm{μ}v}\end{array}$

Now, as the above quantity is equal to minus itself, it must be zero. Hence,$\frac{∂J^{\mathrm{μ}}}{∂x^{\mathrm{μ}}}=0$

Therefore, the continuity equation is obtained as$\frac{∂J^{\mathrm{μ}}}{∂x^{\mathrm{μ}}}=0$ .