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Display information for equation id:math.1110.293 on revision:1110

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TeX (original user input): 

\left. \begin{align}
  & {{F}^{\alpha }}{{^{\beta }}_{,}}_{\nu }=\frac{4\pi }{c}{{j}^{\alpha }} \\
 & {{F}^{\alpha }}^{\beta }={{A}^{\alpha ,\beta }}-{{A}^{\beta ,\alpha }} \\
\end{align} \right\}\Rightarrow \square {{A}^{\alpha }}-{{A}^{\beta ,\alpha }}{{_{,}}_{\beta }}=\frac{4\pi }{c}{{j}^{\alpha }}

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F α ν β , = 4 ⁢ π c j α F α β = A α , β - A β , α } ⇒ □ A α - A β , α , β = 4 ⁢ π c j α fragments missing-subexpression fragments superscript 𝐹 𝛼 superscript subscript 𝜈 , 𝛽 4 𝜋 𝑐 superscript 𝑗 𝛼 missing-subexpression superscript superscript 𝐹 𝛼 𝛽 superscript 𝐴 𝛼 𝛽 superscript 𝐴 𝛽 𝛼 } ⇒ □ A 𝛼 A 𝛽 𝛼 , 𝛽 4 𝜋 𝑐 j 𝛼 {\displaystyle{\displaystyle\left.\begin{aligned} &\displaystyle{{F}^{\alpha}}% {{{}^{\beta}}_{,}}_{\nu}=\frac{4\pi}{c}{{j}^{\alpha}}\\ &\displaystyle{{F}^{\alpha}}^{\beta}={{A}^{\alpha,\beta}}-{{A}^{\beta,\alpha}}% \\ \end{aligned}\right\}\Rightarrow\square{{A}^{\alpha}}-{{A}^{\beta,\alpha}}{{{}% _{,}}_{\beta}}=\frac{4\pi}{c}{{j}^{\alpha}}}}

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Fαβ,ν=4πcjαFαβ=Aα,β−Aβ,α}⇒◻Aα−Aβ,α,β=4πcjα

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