Work out, and interpret physically, the$\mathit{\mu }\mathbf{=}\mathbf{0}$ component of the electromagnetic force law, Eq. 12.128.

Write the expression for the Minkowski force on a charge q.

${\mathit{K}}^{\mathbf{\mu }}\mathbf{=}\mathit{q}{\mathit{\eta }}_{\mathbf{v}}{\mathit{F}}^{\mathbf{\mu }\mathbf{\nu }}$ …… (1)

Here, q is the charge and${\eta }_v$ is the proper velocity.

Substitute$\mu =0$in equation (1).

$K^0=q{\eta }_v{F}^{0v}$

Write the above equation up to 0 to 3 variable terms.

$K^0=q\left[{\eta }_1{F}^{01}+{\eta }_2{F}^{02}+{\eta }_3{F}^{03}\right]$ …… (2)

Write the equation for the field-tensor in terms of four-vector potential.

$F^{\mu v}=\frac{\partial A^v}{\partial x_{\mu }}-\frac{\partial A^{\mu }}{\partial x_v}$ …… (3)

For $F^{01}$, equation (3) becomes,

$F^{01}=\frac{\partial A^1}{\partial x_0}-\frac{\partial A^0}{\partial x_1}$ …… (4)

Here localid="1653996612820" $x_0=ct,x_1=x,A^1=-A_x$and $A^0=\frac{v}{c}$.

Substitute the above values in equation (4).

$$\begin{gathered} F^{01}=\frac{\partial A_x}{\partial \left(ct\right)}-\frac{\partial \left({\displaystyle \frac{v}{c}}\right)}{\partial x} \\ {F}^{01}=\frac{-\partial A_x}{\partial \left(ct\right)}-\frac{1}{c}\frac{\partial v}{\partial x} \\ {F}^{01}=-\frac{1}{c}\left(\frac{\partial A_x}{\partial t}+\nabla v\right) \\ {F}^{01}=-\frac{E_x}{c} \end{gathered}$$

Similarly, for $F^{02}$and$F^{03}$:

$$\begin{gathered} F^{02}=-\frac{E_y}{c} \\ {F}^{03}=-\frac{E_z}{c} \end{gathered}$$

Substitute $F^{01}=\frac{E_x}{c},F^{02}=\frac{E_y}{c}$and$F^{03}=\frac{E_z}{c}$in equation (2).

$$\begin{gathered} K^0=-q\left[{\eta }_1\frac{E_x}{c}+{\eta }^2\frac{E_y}{c}+{\eta }_3\frac{E_z}{c}\right] \\ {K}^0=q\frac{\left(\eta ·\mathit{E}\right)}{c} \\ {K}^0=q\frac{\gamma \left(\mathit{u}\mathbf{·}\mathit{E}\right)}{c} \end{gathered}$$

It is also known that:

$K^0=\frac{1}{c}\frac{dW}{d\mathcal{b}}$ ……. (5)

Here, W is the energy of a particle.

Write the equation for$d\mathcal{b}$ .

$d\mathcal{b}=\frac{1}{\gamma }dt$

Substitute$d\mathcal{b}=\frac{1}{\gamma }dt$ and$K^0=q\frac{\gamma \left(\mathit{u}\mathbf{·}\mathit{E}\right)}{c}$ in equation (5).

$$\begin{gathered} q\frac{\gamma \left(\mathit{u}\mathbf{·}\mathit{E}\right)}{c}=\frac{1}{c}\frac{dW}{\left({\displaystyle \frac{1}{\gamma }}dt\right)} \\ \frac{dW}{dt}=q\left(\mathit{u}\mathbf{·}\mathit{E}\right) \end{gathered}$$

The above equation says that power given to the particle is equal to the product of charge and electric field, i.e., force and the velocity u.

Therefore, the power delivered to the particle is force qE times velocity u.