4-gradient - Revision history

Vanwdani at 18:47, 15 May 2018

2018-05-15T18:47:58Z

2017-07-10T14:39:48Z

4-gradient

2017-07-10T14:33:37Z

4-gradient

2017-07-10T14:32:16Z

4-gradient

2017-07-10T14:31:47Z

4-gradient

2017-07-10T14:21:07Z

4-gradient

2017-07-10T14:17:22Z

4-gradient

2017-07-10T14:16:31Z

4-gradient

2017-07-10T14:13:24Z

4-gradient

2017-07-10T14:11:29Z

4-gradient

@@ Line 1: / Line 1: @@
-<center><math>\textbf{\underline{Navigation}}</math>
+<center><math>\underline{\textbf{Navigation}}</math>
 [[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]]
@@ Line 91: / Line 91: @@
-<center><math>\textbf{\underline{Navigation}}</math>
+<center><math>\underline{\textbf{Navigation}}</math>
 [[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]]

@@ Line 48: / Line 48: @@
 \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}=
@@ Line 72: / Line 78: @@
 \begin{bmatrix}
 \frac{\partial}{\partial x_0}  \\
 \frac{\partial}{\partial x_1}  \\
 \frac{\partial}{\partial x_2}  \\
 \frac{\partial}{\partial x_3}
 \end{bmatrix}=

@@ Line 76: / Line 76: @@
 \frac{\partial}{\partial x_3}
 \end{bmatrix}=
-\frac{\partial^2}{\partial t^2}-\nabla^2
+\frac{\partial^2}{\partial t^2}-\nabla^2\equiv \Box
 </math></center>
 ----

@@ Line 75: / Line 75: @@
 \frac{\partial}{\partial x_2}  \\
 \frac{\partial}{\partial x_3}
-\end{\bmatrix}=
+\end{bmatrix}=
 \frac{\partial^2}{\partial t^2}-\nabla^2
 </math></center>

← Older revision		Revision as of 14:31, 10 July 2017
Line 65:		Line 65:


		+
		+	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
		+	</math></center>
	----		----

← Older revision		Revision as of 14:21, 10 July 2017
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>

← Older revision		Revision as of 14:13, 10 July 2017
Line 38:		Line 38:


−	<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 ]</math></center>	+	<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>

← Older revision		Revision as of 14:11, 10 July 2017
Line 38:		Line 38:


−	<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 ]</math></center>	+	<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 ]</math></center>