4-gradient

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4-gradient

From the use of the Minkowski metric, converting between contravariant and covariant

$\mathbf x_{\mu} \equiv \eta_{\mu}^{\mu} \mathbf x^{\mu}$

Where we have already defined the covariant term,

$\mathbf{x_{\mu}}= \begin{bmatrix} x_0 & -x_1 & -x_2 & -x_3 \end{bmatrix}$

and the contravariant term

$\mathbf{x^{\mu}}= \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}$

Following the rules of matrix multiplication this implies that the derivative with respect to a contravariant coordinate transforms as a covariant 4-vector, and the derivative with respect to a covariant coordinate transforms as a contravariant vector.

$\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}$

$\mathbf \partial_\mu \equiv \Biggl [\frac{\partial}{\partial x^0}\quad -\frac{\partial}{\partial x^1}\quad -\frac{\partial}{\partial x^2}\quad -\frac{\partial}{\partial x^3}\Biggr ]=\Biggl [ \frac{\partial}{\partial t}\quad -\frac{\partial}{\partial x}\quad -\frac{\partial}{\partial y}\quad -\frac{\partial}{\partial z}\Biggr ]=\Biggl [\frac{\partial}{\partial t}\quad -\nabla \Biggr ]$

$\partial^{\mu}=\frac{\partial}{\partial x_{\mu}}$

$\mathbf \partial^\mu \equiv \begin{bmatrix} \frac{\partial}{\partial x_0} \\ \frac{\partial}{\partial x_1} \\ \frac{\partial}{\partial x_2} \\ \frac{\partial}{\partial x_3} \end{bmatrix}= \begin{bmatrix} \frac{\partial}{\partial t} \\ \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}= \begin{bmatrix} \frac{\partial}{\partial t} \\ \nabla \end{bmatrix}$

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