91Ó°ÊÓ

Write the expression for the value of $\stackrel{¯}{{∂}^0\mathrm{ϕ}}$.

role="math" localid="1655877718000" $\begin{array}{c}\stackrel{¯}{{∂}^0\mathrm{ϕ}}=\frac{∂}{∂{\stackrel{¯}{\mathrm{x}}}_0}\mathrm{ϕ}\\ \stackrel{¯}{{∂}^0\mathrm{ϕ}}=−\frac{1}{\mathrm{c}}\frac{∂}{∂\mathrm{t}}\mathrm{ϕ}\\ \stackrel{¯}{{∂}^0\mathrm{ϕ}}=−\frac{1}{\mathrm{c}}\left(\frac{∂\mathrm{ϕ}}{∂\mathrm{t}}\frac{∂\mathrm{t}}{∂\mathrm{t}}+\frac{∂\mathrm{ϕ}}{∂\mathrm{x}}\frac{∂\mathrm{x}}{∂\mathrm{t}}+\frac{∂\mathrm{ϕ}}{∂\mathrm{y}}\frac{∂\mathrm{y}}{∂\mathrm{t}}+\frac{∂\mathrm{ϕ}}{∂\mathrm{z}}\frac{∂\mathrm{z}}{∂\mathrm{t}}\right)\end{array}$ ……. (1)

It is known that:

$t=\mathrm{γ}\left(\stackrel{¯}{t}+\frac{v}{c}\stackrel{¯}{x}\right)$

Here, v is the velocity and c is the speed of light.

Using equation 12.19, write the expression for the transformation equations.

$\begin{array}{c}\frac{∂t}{∂t}=\mathrm{γ}\\ x=\mathrm{γ}(\stackrel{¯}{x}+v\stackrel{¯}{t})\\ \frac{∂x}{∂t}=\mathrm{γ}v\\ y=\stackrel{¯}{y}\end{array}$

It is also known that:

$\begin{array}{c}z=\stackrel{¯}{z}\\ \frac{∂y}{∂t}=0\\ \frac{∂z}{∂t}=0\end{array}$

Substitute $\frac{∂\mathrm{t}}{∂\mathrm{t}}=\mathrm{γ}$, $x=\mathrm{γ}\left(\stackrel{¯}{x}+v\stackrel{¯}{t}\right)$, $\frac{∂x}{∂t}=\mathrm{γ}v$, $\frac{∂y}{∂t}=0$and $\frac{∂\mathrm{z}}{∂\mathrm{t}}=0$in equation (1)l.

$\begin{array}{c}\stackrel{¯}{{∂}^0\mathrm{ϕ}}=−\frac{1}{c}\left(\frac{∂\mathrm{ϕ}}{∂t}\left(\mathrm{γ}\right)+\frac{∂\mathrm{ϕ}}{∂x}\left(\mathrm{γ}v\right)+\frac{∂\mathrm{ϕ}}{∂y}\left(0\right)+\frac{∂\mathrm{ϕ}}{∂z}\left(0\right)\right)\\ \stackrel{¯}{{∂}^0\mathrm{ϕ}}=−\frac{1}{c}\left(\frac{∂\mathrm{ϕ}}{∂t}\left(\mathrm{γ}\right)+\frac{∂\mathrm{ϕ}}{∂x}\left(\mathrm{γ}v\right)\right)\\ \stackrel{¯}{{∂}^0\mathrm{ϕ}}=−\frac{1}{c}\mathrm{γ}\left(−\frac{∂\mathrm{ϕ}}{∂t}+v\frac{∂\mathrm{ϕ}}{∂x}\right)\\ \stackrel{¯}{{∂}^0\mathrm{ϕ}}=\mathrm{γ}\left[\frac{∂\mathrm{ϕ}}{∂x_0}−\frac{v}{c}\frac{∂\mathrm{ϕ}}{∂x^{\text{'}}}\right]\end{array}$

Here,

$\mathrm{β}=\frac{\mathrm{v}}{\mathrm{c}}$

On further solving,

$\stackrel{¯}{{∂}^0\mathrm{ϕ}}=\mathrm{γ}\left[{∂}^0\mathrm{ϕ}−\mathrm{β}\left({∂}^1\mathrm{ϕ}\right)\right]$

Calculate the value of $\stackrel{¯}{{∂}^1\mathrm{ϕ}}$.

$\begin{array}{c}\stackrel{¯}{{∂}^1\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂x}\\ \stackrel{¯}{{∂}^1\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂t}\frac{∂t}{∂\stackrel{¯}{x}}+\frac{∂\mathrm{ϕ}}{∂x}\frac{∂x}{∂\stackrel{¯}{x}}+\frac{∂\mathrm{ϕ}}{∂y}\frac{∂y}{∂\stackrel{¯}{x}}+\frac{∂\mathrm{ϕ}}{∂z}\frac{∂z}{∂\stackrel{¯}{x}}\end{array}$ …… (2)

It is known that:

$\begin{array}{c}\stackrel{¯}{x}=\mathrm{γ}(\stackrel{¯}{x}+v\stackrel{¯}{t})\\ t=\mathrm{γ}\left(\stackrel{¯}{t}+\frac{v}{c^2}\stackrel{¯}{x}\right)\\ \frac{∂x}{∂\stackrel{¯}{x}}=\mathrm{γ},\frac{∂y}{∂\stackrel{¯}{x}}=0\\ \frac{∂t}{∂\stackrel{¯}{x}}=\frac{v}{c^2}\mathrm{γ},\frac{∂z}{∂\stackrel{¯}{x}}=0\end{array}$

Substitute $\frac{∂x}{∂\stackrel{¯}{x}}=\mathrm{γ}$, $\frac{∂y}{∂\stackrel{¯}{x}}=0$, $\frac{∂t}{∂\stackrel{¯}{x}}=\frac{v}{c^2}\mathrm{γ}$ and $\frac{∂z}{∂\stackrel{¯}{x}}=0$in equation (2).

$\begin{array}{c}\stackrel{¯}{{∂}^1\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂t}\left(\frac{v}{c^2}\mathrm{γ}\right)+\frac{∂\mathrm{ϕ}}{∂x}\left(\mathrm{γ}\right)+\frac{∂\mathrm{ϕ}}{∂y}\left(0\right)+\frac{∂\mathrm{ϕ}}{∂z}\left(0\right)\\ \stackrel{¯}{{∂}^1\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂t}\left(\frac{v}{c^2}\mathrm{γ}\right)+\frac{∂\mathrm{ϕ}}{∂x}\left(\mathrm{γ}\right)\\ \stackrel{¯}{{∂}^1\mathrm{ϕ}}=\mathrm{γ}\left[\frac{v}{c^2}\frac{∂\mathrm{ϕ}}{∂t}+\frac{∂\mathrm{ϕ}}{∂x}\right]\\ \stackrel{¯}{{∂}^1\mathrm{ϕ}}=\mathrm{γ}[\left({∂}^{\text{'}}\mathrm{ϕ}\right)−\mathrm{β}\left({∂}^0\mathrm{ϕ}\right)]\end{array}$

Calculate the value of $\stackrel{¯}{{∂}^2\mathrm{ϕ}}$.

$\begin{array}{c}\stackrel{¯}{{∂}^2\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂\stackrel{¯}{y}}\\ \stackrel{¯}{{∂}^2\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂t}\frac{∂t}{∂\stackrel{¯}{y}}+\frac{∂\mathrm{ϕ}}{∂x}\frac{∂x}{∂\stackrel{¯}{y}}+\frac{∂\mathrm{ϕ}}{∂y}\frac{∂y}{∂\stackrel{¯}{y}}+\frac{∂\mathrm{ϕ}}{∂z}\frac{∂z}{∂\stackrel{¯}{y}}\\ \stackrel{¯}{{∂}^2\mathrm{ϕ}}={∂}^2\mathrm{ϕ}\end{array}$

Calculate the value of $\stackrel{¯}{{∂}^3\mathrm{ϕ}}$.

$\begin{array}{c}\stackrel{¯}{{∂}^3\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂\stackrel{¯}{z}}\\ \stackrel{¯}{{∂}^3\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂t}\frac{∂t}{∂\stackrel{¯}{z}}+\frac{∂\mathrm{ϕ}}{∂x}\frac{∂x}{∂\stackrel{¯}{z}}+\frac{∂\mathrm{ϕ}}{∂y}\frac{∂y}{∂\stackrel{¯}{z}}+\frac{∂\mathrm{ϕ}}{∂z}\frac{∂z}{∂\stackrel{¯}{z}}\\ \stackrel{¯}{{∂}^3\mathrm{ϕ}}=\frac{∂\mathrm{ϕ}}{∂z}\end{array}$

Therefore,${∂}^{\mathrm{μ}}\mathrm{ϕ}$ is a contravariant 4-vector.