91Ó°ÊÓ

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

${\mathit{K}}^{\mathbf{μ}}\mathbf{=}\mathit{q}{\mathit{η}}_{\mathbf{v}}{\mathit{F}}^{\mathbf{μ}\mathbf{ν}}$ …… (1)

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

Substitute$\mathrm{μ}=0$in equation (1).

$K^0=q{\mathrm{η}}_v{F}^{0v}$

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

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

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

$F^{\mathrm{μ}v}=\frac{∂A^v}{∂x_{\mathrm{μ}}}-\frac{∂A^{\mathrm{μ}}}{∂x_v}$ …… (3)

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

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

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

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).

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

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$d\mathcal{b}$ .

$d\mathcal{b}=\frac{1}{\mathrm{γ}}dt$

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

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

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.