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Display information for equation id:math.1128.15 on revision:1128

* Page found: Effektives Potential (eq math.1128.15)

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TeX (original user input): 

\begin{align}
  & \Rightarrow \frac{1}{2}\mu {{\left( \frac{dr}{dt} \right)}^{2}}=E-U \\ 
 & \Leftrightarrow {{\left( \frac{dr}{dt} \right)}^{2}}=\frac{2}{\mu }\left( E-U \right) \\ 
 & \Leftrightarrow dt=\frac{dr}{\sqrt{\frac{2}{\mu }\left( E-U \right)}} \\ 
\end{align}

TeX (checked): 

{\begin{aligned}&\Rightarrow {\frac {1}{2}}\mu {{\left({\frac {dr}{dt}}\right)}^{2}}=E-U\\&\Leftrightarrow {{\left({\frac {dr}{dt}}\right)}^{2}}={\frac {2}{\mu }}\left(E-U\right)\\&\Leftrightarrow dt={\frac {dr}{\sqrt {{\frac {2}{\mu }}\left(E-U\right)}}}\\\end{aligned}}

LaTeXML (experimentell; verwendet MathML) rendering

MathML (20.168 KB / 2.422 KB) :

⇒ 1 2 ⁢ μ ⁢ ( d ⁢ r d ⁢ t ) 2 = E - U ⇔ ( d ⁢ r d ⁢ t ) 2 = 2 μ ⁢ ( E - U ) ⇔ d ⁢ t = d ⁢ r 2 μ ⁢ ( E - U ) missing-subexpression ⇒ absent 1 2 𝜇 superscript 𝑑 𝑟 𝑑 𝑡 2 𝐸 𝑈 missing-subexpression ⇔ absent superscript 𝑑 𝑟 𝑑 𝑡 2 2 𝜇 𝐸 𝑈 missing-subexpression ⇔ absent 𝑑 𝑡 𝑑 𝑟 2 𝜇 𝐸 𝑈 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle\Rightarrow\frac{1}{% 2}\mu{{\left(\frac{dr}{dt}\right)}^{2}}=E-U\\ &\displaystyle\Leftrightarrow{{\left(\frac{dr}{dt}\right)}^{2}}=\frac{2}{\mu}% \left(E-U\right)\\ &\displaystyle\Leftrightarrow dt=\frac{dr}{\sqrt{\frac{2}{\mu}\left(E-U\right)% }}\\ \end{aligned}}}

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MathML (2.457 KB / 417 B) :

⇒12μ(drdt)2=E−U⇔(drdt)2=2μ(E−U)⇔dt=dr2μ(E−U)

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