Difference between revisions of "4-gradient"

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<center><math>\textbf{\underline{Navigation}}</math>
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<center><math>\underline{\textbf{Navigation}}</math>
  
 
[[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]]
 
[[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]]
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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.
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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>
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<center><math>\mathbf \partial_\mu \equiv \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 x^0}\quad \frac{\partial}{\partial x^1}\quad \frac{\partial}{\partial x^2}\quad \frac{\partial}{\partial x^3}\Biggr ]</math></center>
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<center><math>\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></center>
  
  
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<center><math>\mathbf \partial^\mu \equiv \mathbf{x^{\mu}}=
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<center><math>\mathbf \partial^\mu \equiv  
 
\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}=
 +
\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}
 
\end{bmatrix}
 
</math></center>
 
</math></center>
  
  
 +
 +
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>\textbf{\underline{Navigation}}</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

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

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]

4-gradient

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


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


Where we have already defined the covariant term,

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

and the contravariant term

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


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


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

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]