Difference between revisions of "4-gradient"

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Using matrix multiplication, we have already defined the covariant term,  
 
Using matrix multiplication, we have already defined the covariant term,  
 
<center><math>\mathbf{x_{\mu}}= \begin{bmatrix}
 
<center><math>\mathbf{x_{\mu}}= \begin{bmatrix}
dx_0 & -dx_1 & -dx_2 & -dx_3
+
x_0 & -x_1 & -x_2 & -x_3
 
\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
  
Line 14: Line 14:
 
<center><math>\mathbf{x^{\mu}}=
 
<center><math>\mathbf{x^{\mu}}=
 
\begin{bmatrix}
 
\begin{bmatrix}
dx^0  \\
+
x^0  \\
dx^1 \\
+
x^1 \\
dx^2 \\
+
x^2 \\
dx^3
+
x^3
 
\end{bmatrix}
 
\end{bmatrix}
 
</math></center>
 
</math></center>

Revision as of 01:06, 10 July 2017

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


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


Using matrix multiplication, 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]


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

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


[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]