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Display information for equation id:math.2687.20 on revision:2687

* Page found: Lösungen der Dirac-Gleichung (freies Teilchen) (eq math.2687.20)

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TeX (original user input):

\begin{align}

& \left( {{\gamma }^{\mu }}{{k}_{\mu }}-m \right)\left( {{\gamma }^{\nu }}{{k}_{\nu }}+m \right)={{\gamma }^{\mu }}{{k}_{\mu }}{{\gamma }^{\nu }}{{k}_{\nu }}-{{m}^{2}}={{E}^{2}}-{{k}^{2}}-{{m}^{2}}=0,\quad  \\

& mit\,{{\left( {{\gamma }^{\mu }} \right)}^{2}}=\pm 1,{{E}^{2}}={{k}^{2}}+{{m}^{2}},\hbar =c=1 \\

\end{align}

TeX (checked):

{\begin{aligned}&\left({{\gamma }^{\mu }}{{k}_{\mu }}-m\right)\left({{\gamma }^{\nu }}{{k}_{\nu }}+m\right)={{\gamma }^{\mu }}{{k}_{\mu }}{{\gamma }^{\nu }}{{k}_{\nu }}-{{m}^{2}}={{E}^{2}}-{{k}^{2}}-{{m}^{2}}=0,\quad \\&mit\,{{\left({{\gamma }^{\mu }}\right)}^{2}}=\pm 1,{{E}^{2}}={{k}^{2}}+{{m}^{2}},\hbar =c=1\\\end{aligned}}

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MathML (2.437 KB / 478 B) :

(γμkμm)(γνkν+m)=γμkμγνkνm2=E2k2m2=0,mit(γμ)2=±1,E2=k2+m2,=c=1
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Identifiers

  • γ
  • μ
  • kμ
  • m
  • γ
  • ν
  • kν
  • m
  • γ
  • μ
  • kμ
  • γ
  • ν
  • kν
  • m
  • E
  • k
  • m
  • m
  • i
  • t
  • γ
  • μ
  • E
  • k
  • m
  • c

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