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Display information for equation id:math.2126.63 on revision:2126

* Page found: Freie Wellenausbreitung im Vakuum (eq math.2126.63)

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\begin{align}
& {{\delta }_{1}}={{\delta }_{2}}+n\pi \Rightarrow \sin \left( {{\delta }_{2}}-{{\delta }_{1}} \right)=0,\cos \left( {{\delta }_{2}}-{{\delta }_{1}} \right)=\pm 1 \\
& \Rightarrow \frac{{{{\bar{E}}}_{1}}}{{{a}_{1}}}\pm \frac{{{{\bar{E}}}_{2}}}{a2}=0 \\
\end{align}

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δ 1 = δ 2 + n ⁢ π ⇒ sin ⁡ ( δ 2 - δ 1 ) = 0 , cos ⁡ ( δ 2 - δ 1 ) = ± 1 ⇒ E ¯ 1 a 1 ± E ¯ 2 a ⁢ 2 = 0 missing-subexpression formulae-sequence subscript 𝛿 1 subscript 𝛿 2 𝑛 𝜋 ⇒ subscript 𝛿 2 subscript 𝛿 1 0 subscript 𝛿 2 subscript 𝛿 1 plus-or-minus 1 missing-subexpression ⇒ absent plus-or-minus subscript ¯ 𝐸 1 subscript 𝑎 1 subscript ¯ 𝐸 2 𝑎 2 0 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle{{\delta}_{1}}={{% \delta}_{2}}+n\pi\Rightarrow\sin\left({{\delta}_{2}}-{{\delta}_{1}}\right)=0,% \cos\left({{\delta}_{2}}-{{\delta}_{1}}\right)=\pm 1\\ &\displaystyle\Rightarrow\frac{{{{\bar{E}}}_{1}}}{{{a}_{1}}}\pm\frac{{{{\bar{E% }}}_{2}}}{a2}=0\\ \end{aligned}}}

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δ1=δ2+nπ⇒sin⁡(δ2−δ1)=0,cos⁡(δ2−δ1)=±1⇒E¯1a1±E¯2a2=0

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