You may have noticed that the four-dimensional gradient operator $\mathbf{\partial }\mathbf{/}\mathbf{\partial }{\mathbf{x}}^{\mathbf{\mu }}$ functions like a covariant 4-vector—in fact, it is often written${\mathbf{\partial }}_{\mathbf{\mu }}$ , for short. For instance, the continuity equation, ${\mathbf{\partial }}_{\mathbf{\mu }}{\mathbf{J}}^{\mathbf{\mu }}\mathbf{=}\mathbf{0}$, has the form of an invariant product of two vectors. The corresponding contravariant gradient would be${\mathbf{\partial }}^{\mathbf{\mu }}\mathbf{\equiv }\mathbf{\partial }\mathbf{/}\mathbf{\partial }{\mathbf{x}}_{\mathbf{\mu }}$ . Prove that${\mathbf{\partial }}^{\mathbf{\mu }}\mathbf{f}$ is a (contravariant) 4-vector, if$\mathbf{\varphi }$ is a scalar function, by working out its transformation law, using the chain rule.

Write the expression for the value of $\overline{{\partial }^0\varphi }$.

role="math" localid="1655877718000" $\begin{array}{c}\overline{{\partial }^0\mathrm{\varphi }}=\frac{\partial }{\partial {\overline{\mathrm{x}}}_0}\mathrm{\varphi }\\ \overline{{\partial }^0\mathrm{\varphi }}=-\frac{1}{\mathrm{c}}\frac{\partial }{\partial \mathrm{t}}\mathrm{\varphi }\\ \overline{{\partial }^0\mathrm{\varphi }}=-\frac{1}{\mathrm{c}}\left(\frac{\partial \mathrm{\varphi }}{\partial \mathrm{t}}\frac{\partial \mathrm{t}}{\partial \mathrm{t}}+\frac{\partial \mathrm{\varphi }}{\partial \mathrm{x}}\frac{\partial \mathrm{x}}{\partial \mathrm{t}}+\frac{\partial \mathrm{\varphi }}{\partial \mathrm{y}}\frac{\partial \mathrm{y}}{\partial \mathrm{t}}+\frac{\partial \mathrm{\varphi }}{\partial \mathrm{z}}\frac{\partial \mathrm{z}}{\partial \mathrm{t}}\right)\end{array}$ ……. (1)

It is known that:

$t=\gamma \left(\overline{t}+\frac{v}{c}\overline{x}\right)$

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

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

$\begin{array}{c}\frac{\partial t}{\partial t}=\gamma \\ x=\gamma (\overline{x}+v\overline{t})\\ \frac{\partial x}{\partial t}=\gamma v\\ y=\overline{y}\end{array}$

It is also known that:

$\begin{array}{c}z=\overline{z}\\ \frac{\partial y}{\partial t}=0\\ \frac{\partial z}{\partial t}=0\end{array}$

Substitute $\frac{\partial \mathrm{t}}{\partial \mathrm{t}}=\mathrm{\gamma }$, $x=\gamma \left(\overline{x}+v\overline{t}\right)$, $\frac{\partial x}{\partial t}=\gamma v$, $\frac{\partial y}{\partial t}=0$and $\frac{\partial \mathrm{z}}{\partial \mathrm{t}}=0$in equation (1)l.

$\begin{array}{c}\overline{{\partial }^0\varphi }=-\frac{1}{c}\left(\frac{\partial \varphi }{\partial t}\left(\gamma \right)+\frac{\partial \varphi }{\partial x}\left(\gamma v\right)+\frac{\partial \varphi }{\partial y}\left(0\right)+\frac{\partial \varphi }{\partial z}\left(0\right)\right)\\ \overline{{\partial }^0\varphi }=-\frac{1}{c}\left(\frac{\partial \varphi }{\partial t}\left(\gamma \right)+\frac{\partial \varphi }{\partial x}\left(\gamma v\right)\right)\\ \overline{{\partial }^0\varphi }=-\frac{1}{c}\gamma \left(-\frac{\partial \varphi }{\partial t}+v\frac{\partial \varphi }{\partial x}\right)\\ \overline{{\partial }^0\varphi }=\gamma \left[\frac{\partial \varphi }{\partial x_0}-\frac{v}{c}\frac{\partial \varphi }{\partial x^{\text{'}}}\right]\end{array}$

Here,

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

On further solving,

$\overline{{\partial }^0\varphi }=\gamma \left[{\partial }^0\varphi -\beta \left({\partial }^1\varphi \right)\right]$

Calculate the value of $\overline{{\partial }^1\varphi }$.

$\begin{array}{c}\overline{{\partial }^1\varphi }=\frac{\partial \varphi }{\partial x}\\ \overline{{\partial }^1\varphi }=\frac{\partial \varphi }{\partial t}\frac{\partial t}{\partial \overline{x}}+\frac{\partial \varphi }{\partial x}\frac{\partial x}{\partial \overline{x}}+\frac{\partial \varphi }{\partial y}\frac{\partial y}{\partial \overline{x}}+\frac{\partial \varphi }{\partial z}\frac{\partial z}{\partial \overline{x}}\end{array}$ …… (2)

It is known that:

$\begin{array}{c}\overline{x}=\gamma (\overline{x}+v\overline{t})\\ t=\gamma \left(\overline{t}+\frac{v}{c^2}\overline{x}\right)\\ \frac{\partial x}{\partial \overline{x}}=\gamma ,\frac{\partial y}{\partial \overline{x}}=0\\ \frac{\partial t}{\partial \overline{x}}=\frac{v}{c^2}\gamma ,\frac{\partial z}{\partial \overline{x}}=0\end{array}$

Substitute $\frac{\partial x}{\partial \overline{x}}=\gamma$, $\frac{\partial y}{\partial \overline{x}}=0$, $\frac{\partial t}{\partial \overline{x}}=\frac{v}{c^2}\gamma$ and $\frac{\partial z}{\partial \overline{x}}=0$in equation (2).

$\begin{array}{c}\overline{{\partial }^1\varphi }=\frac{\partial \varphi }{\partial t}\left(\frac{v}{c^2}\gamma \right)+\frac{\partial \varphi }{\partial x}\left(\gamma \right)+\frac{\partial \varphi }{\partial y}\left(0\right)+\frac{\partial \varphi }{\partial z}\left(0\right)\\ \overline{{\partial }^1\varphi }=\frac{\partial \varphi }{\partial t}\left(\frac{v}{c^2}\gamma \right)+\frac{\partial \varphi }{\partial x}\left(\gamma \right)\\ \overline{{\partial }^1\varphi }=\gamma \left[\frac{v}{c^2}\frac{\partial \varphi }{\partial t}+\frac{\partial \varphi }{\partial x}\right]\\ \overline{{\partial }^1\varphi }=\gamma [\left({\partial }^{\text{'}}\varphi \right)-\beta \left({\partial }^0\varphi \right)]\end{array}$

Calculate the value of $\overline{{\partial }^2\varphi }$.

$\begin{array}{c}\overline{{\partial }^2\varphi }=\frac{\partial \varphi }{\partial \overline{y}}\\ \overline{{\partial }^2\varphi }=\frac{\partial \varphi }{\partial t}\frac{\partial t}{\partial \overline{y}}+\frac{\partial \varphi }{\partial x}\frac{\partial x}{\partial \overline{y}}+\frac{\partial \varphi }{\partial y}\frac{\partial y}{\partial \overline{y}}+\frac{\partial \varphi }{\partial z}\frac{\partial z}{\partial \overline{y}}\\ \overline{{\partial }^2\varphi }={\partial }^2\varphi \end{array}$

Calculate the value of $\overline{{\partial }^3\varphi }$.

$\begin{array}{c}\overline{{\partial }^3\varphi }=\frac{\partial \varphi }{\partial \overline{z}}\\ \overline{{\partial }^3\varphi }=\frac{\partial \varphi }{\partial t}\frac{\partial t}{\partial \overline{z}}+\frac{\partial \varphi }{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial \varphi }{\partial y}\frac{\partial y}{\partial \overline{z}}+\frac{\partial \varphi }{\partial z}\frac{\partial z}{\partial \overline{z}}\\ \overline{{\partial }^3\varphi }=\frac{\partial \varphi }{\partial z}\end{array}$

Therefore,${\partial }^{\mu }\varphi$ is a contravariant 4-vector.