Difference between revisions of "4-gradient"
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− | <center><math>\ | + | <center><math>\underline{\textbf{Navigation}}</math> |
[[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]] | [[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]] | ||
Line 32: | Line 32: | ||
− | 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. | + | 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. |
<center><math>\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}</math></center> | <center><math>\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}</math></center> | ||
Line 48: | Line 48: | ||
\begin{bmatrix} | \begin{bmatrix} | ||
\frac{\partial}{\partial x_0} \\ | \frac{\partial}{\partial x_0} \\ | ||
+ | \\ | ||
\frac{\partial}{\partial x_1} \\ | \frac{\partial}{\partial x_1} \\ | ||
+ | \\ | ||
\frac{\partial}{\partial x_2} \\ | \frac{\partial}{\partial x_2} \\ | ||
+ | \\ | ||
\frac{\partial}{\partial x_3} | \frac{\partial}{\partial x_3} | ||
\end{bmatrix}= | \end{bmatrix}= | ||
\begin{bmatrix} | \begin{bmatrix} | ||
\frac{\partial}{\partial t} \\ | \frac{\partial}{\partial t} \\ | ||
+ | \\ | ||
\frac{\partial}{\partial x} \\ | \frac{\partial}{\partial x} \\ | ||
+ | \\ | ||
\frac{\partial}{\partial y} \\ | \frac{\partial}{\partial y} \\ | ||
+ | \\ | ||
\frac{\partial}{\partial z} | \frac{\partial}{\partial z} | ||
\end{bmatrix}= | \end{bmatrix}= | ||
Line 65: | Line 71: | ||
+ | |||
+ | Since it is an operator, the dot product of two partial differentials yields an operator known as the D'Alembert operator. | ||
+ | |||
+ | <center><math>\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></center> | ||
---- | ---- | ||
− | <center><math>\ | + | <center><math>\underline{\textbf{Navigation}}</math> |
[[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]] | [[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]] |
Latest revision as of 18:47, 15 May 2018
4-gradient
From the use of the Minkowski metric, converting between contravariant and covariant
Where we have already defined the covariant term,
and the contravariant term
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.
Since it is an operator, the dot product of two partial differentials yields an operator known as the D'Alembert operator.