|
|
| Line 75: |
Line 75: |
| \frac{\partial}{\partial x_2} \\ | | \frac{\partial}{\partial x_2} \\ |
| \frac{\partial}{\partial x_3} | | \frac{\partial}{\partial x_3} |
| \end{\bmatrix}= | | \end{bmatrix}= |
| \frac{\partial^2}{\partial t^2}-\nabla^2 | | \frac{\partial^2}{\partial t^2}-\nabla^2 |
| </math></center> | | </math></center> |
Revision as of 14:32, 10 July 2017
[math]\displaystyle{ \textbf{\underline{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
}[/math]
[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]