Difference between revisions of "4-gradient"

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<center><math>\nabla_i=\frac{\partial}{\partial r_i}=\partial_i</math></center>
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<center><math>\underline{\textbf{Navigation}}</math>
  
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[[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
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[[Mandelstam_Representation|<math>\vartriangleright </math>]]
  
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</center>
  
<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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=4-gradient=
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From the use of the Minkowski metric, converting between contravariant and covariant
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<center><math>\mathbf x_{\mu} \equiv \eta_{\mu}^{\mu} \mathbf x^{\mu}</math></center>
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Where we have already defined the covariant term,
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<center><math>\mathbf{x_{\mu}}= \begin{bmatrix}
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x_0  & -x_1 & -x_2 & -x_3
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\end{bmatrix}</math></center>
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and the contravariant term
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<center><math>\mathbf{x^{\mu}}=
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\begin{bmatrix}
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x^0  \\
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x^1 \\
 +
x^2 \\
 +
x^3
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\end{bmatrix}
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</math></center>
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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.
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<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 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>\partial^{\mu}=\frac{\partial}{\partial x_{\mu}}</math></center>
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<center><math>\mathbf \partial^\mu \equiv
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\begin{bmatrix}
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\frac{\partial}{\partial x_0}  \\
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\\
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\frac{\partial}{\partial x_1}  \\
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\\
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\frac{\partial}{\partial x_2}  \\
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\\
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\frac{\partial}{\partial x_3}
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\end{bmatrix}=
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\begin{bmatrix}
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\frac{\partial}{\partial t}  \\
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\\
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\frac{\partial}{\partial x}  \\
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\\
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\frac{\partial}{\partial y}  \\
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\\
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\frac{\partial}{\partial z}
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\end{bmatrix}=
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\begin{bmatrix}
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\frac{\partial}{\partial t} \\
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\nabla
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\end{bmatrix}
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</math></center>
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Since it is an operator, the dot product of two partial differentials yields an operator known as the D'Alembert operator.
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<center><math>\partial^{\mu} \partial_{\mu}=
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\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
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\begin{bmatrix}
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\frac{\partial}{\partial x_0}  \\
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\\
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\frac{\partial}{\partial x_1} \\
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\\
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\frac{\partial}{\partial x_2} \\
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\\
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\frac{\partial}{\partial x_3}
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\end{bmatrix}=
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\frac{\partial^2}{\partial t^2}-\nabla^2\equiv \Box
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</math></center>
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----
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<center><math>\underline{\textbf{Navigation}}</math>
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[[Frame_of_Reference_Transformation|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
 +
[[Mandelstam_Representation|<math>\vartriangleright </math>]]
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</center>

Latest revision as of 18:47, 15 May 2018

Navigation_

4-gradient

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


xμημμxμ


Where we have already defined the covariant term,

xμ=[x0x1x2x3]

and the contravariant term

xμ=[x0x1x2x3]


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.

μ=xμ


μ[x0x1x2x3]=[txyz]=[t]


μ=xμ


μ[x0x1x2x3]=[txyz]=[t]


Since it is an operator, the dot product of two partial differentials yields an operator known as the D'Alembert operator.

μμ=[x0x1x2x3][x0x1x2x3]=2t22


Navigation_