Reference: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.55.

The gradient of a scalar function ${\phi}$ is a covariant vector since it transforms as

$\displaystyle \frac{\partial\phi}{\partial\bar{x}{}^{a}}=\frac{\partial\phi}{\partial x^{i}}\frac{\partial x^{i}}{\partial\bar{x}{}^{a}} \ \ \ \ \ (1)$

We can therefore regard the gradient operator ${\partial_{a}}$ on its own as a covariant vector, so it should have a contravariant counterpart. In flat space, the only change in switching from covariant to contravariant is that the time component changes sign. Given that the Lorentz transformation for a contravariant four-vector is

 $\displaystyle \bar{x}^{0}$ $\displaystyle =$ $\displaystyle \gamma\left(x^{0}-\beta x^{1}\right)\ \ \ \ \ (2)$ $\displaystyle \bar{x}^{1}$ $\displaystyle =$ $\displaystyle \gamma\left(x^{1}-\beta x^{0}\right)\ \ \ \ \ (3)$ $\displaystyle \bar{x}^{2}$ $\displaystyle =$ $\displaystyle x^{2}\ \ \ \ \ (4)$ $\displaystyle \bar{x}^{3}$ $\displaystyle =$ $\displaystyle x^{3} \ \ \ \ \ (5)$

the transformations for the covariant four-vector are obtained by lowering all indices and replacing the time components by their negatives:

 $\displaystyle \bar{x}_{0}$ $\displaystyle =$ $\displaystyle \gamma\left(x_{0}+\beta x_{1}\right)\ \ \ \ \ (6)$ $\displaystyle \bar{x}_{1}$ $\displaystyle =$ $\displaystyle \gamma\left(x_{1}+\beta x_{0}\right)\ \ \ \ \ (7)$ $\displaystyle \bar{x}_{2}$ $\displaystyle =$ $\displaystyle x_{2}\ \ \ \ \ (8)$ $\displaystyle \bar{x}_{3}$ $\displaystyle =$ $\displaystyle x_{3} \ \ \ \ \ (9)$

where we multiplied the ${\bar{x}_{0}}$ equation through by ${-1}$. The corresponding inverse transformations are obtained by replacing ${\beta}$ by ${-\beta}$:

 $\displaystyle x_{0}$ $\displaystyle =$ $\displaystyle \gamma\left(\bar{x}_{0}-\beta\bar{x}_{1}\right)\ \ \ \ \ (10)$ $\displaystyle x_{1}$ $\displaystyle =$ $\displaystyle \gamma\left(\bar{x}_{1}-\beta\bar{x}_{0}\right)\ \ \ \ \ (11)$ $\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle \bar{x}_{2}\ \ \ \ \ (12)$ $\displaystyle x_{3}$ $\displaystyle =$ $\displaystyle \bar{x}_{3} \ \ \ \ \ (13)$

Thus the inverse covariant transformations are the same as the forward contravariant transformations.

The contravariant gradient is ${\partial^{i}\phi=\frac{\partial\phi}{\partial x_{i}}}$ so

 $\displaystyle \overline{\partial^{i}\phi}$ $\displaystyle =$ $\displaystyle \frac{\partial\phi}{\partial\bar{x}_{i}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\phi}{\partial x_{k}}\frac{\partial x_{k}}{\partial\bar{x}_{i}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial^{k}\phi\frac{\partial x_{k}}{\partial\bar{x}_{i}} \ \ \ \ \ (16)$

The transformations for each value of ${i}$ are then

 $\displaystyle \overline{\partial^{0}\phi}$ $\displaystyle =$ $\displaystyle \gamma\partial^{0}\phi-\beta\gamma\partial^{1}\phi\ \ \ \ \ (17)$ $\displaystyle \overline{\partial^{1}\phi}$ $\displaystyle =$ $\displaystyle \gamma\partial^{1}\phi-\beta\gamma\partial^{0}\phi\ \ \ \ \ (18)$ $\displaystyle \overline{\partial^{2}\phi}$ $\displaystyle =$ $\displaystyle \partial^{2}\phi\ \ \ \ \ (19)$ $\displaystyle \overline{\partial^{3}\phi}$ $\displaystyle =$ $\displaystyle \partial^{3}\phi \ \ \ \ \ (20)$

Thus ${\partial^{i}\phi}$ transforms like a contravariant vector.