4-gradient

[math]\displaystyle{ \underline{\textbf{Navigation}} }[/math]

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]

4-gradient

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

[math]\displaystyle{ \mathbf x_{\mu} \equiv \eta_{\mu}^{\mu} \mathbf x^{\mu} }[/math]

Where we have already defined the covariant term,

[math]\displaystyle{ \mathbf{x_{\mu}}= \begin{bmatrix} x_0 & -x_1 & -x_2 & -x_3 \end{bmatrix} }[/math]

and the contravariant term

[math]\displaystyle{ \mathbf{x^{\mu}}= \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix} }[/math]

From quantum mechanics we know that partial differential is a linear operator. 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.

[math]\displaystyle{ \partial_{\mu}=\frac{\partial}{\partial x^{\mu}} }[/math]

[math]\displaystyle{ \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 ] }[/math]

[math]\displaystyle{ \partial^{\mu}=\frac{\partial}{\partial x_{\mu}} }[/math]

[math]\displaystyle{ \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} }[/math]

Since it is an operator, the dot product of two partial differentials yields an operator known as the D'Alembert operator.

[math]\displaystyle{ \partial^{\mu} \partial_{\mu}= \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 ]\cdot \begin{bmatrix} \frac{\partial}{\partial x_0} \\ \\ \frac{\partial}{\partial x_1} \\ \\ \frac{\partial}{\partial x_2} \\ \\ \frac{\partial}{\partial x_3} \end{bmatrix}= \frac{\partial^2}{\partial t^2}-\nabla^2\equiv \Box }[/math]

---

[math]\displaystyle{ \underline{\textbf{Navigation}} }[/math]

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]