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Display information for equation id:math.1989.35 on revision:1989

* Page found: Hamilton-Jacobische Differenzialgleichung (eq math.1989.35)

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TeX (original user input):

\begin{align}
  & \Rightarrow {{q}_{0}}=\frac{1}{\omega }\sqrt{\frac{2\alpha }{m}}\sin \left( \omega (\beta ) \right) \\
 & 0={{p}_{0}}=\sqrt{2\alpha m}\cos \left( \omega (\beta ) \right) \\
 & \Rightarrow \beta =\frac{\pi }{2\omega }\Rightarrow {{q}_{0}}=\sqrt{\frac{2\alpha }{m{{\omega }^{2}}}} \\
 & \Rightarrow \alpha =\frac{m}{2}{{\omega }^{2}}{{q}_{0}}^{2}=E \\
\end{align}

TeX (checked):

{\begin{aligned}&\Rightarrow {{q}_{0}}={\frac {1}{\omega }}{\sqrt {\frac {2\alpha }{m}}}\sin \left(\omega (\beta )\right)\\&0={{p}_{0}}={\sqrt {2\alpha m}}\cos \left(\omega (\beta )\right)\\&\Rightarrow \beta ={\frac {\pi }{2\omega }}\Rightarrow {{q}_{0}}={\sqrt {\frac {2\alpha }{m{{\omega }^{2}}}}}\\&\Rightarrow \alpha ={\frac {m}{2}}{{\omega }^{2}}{{q}_{0}}^{2}=E\\\end{aligned}}

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MathML (2.802 KB / 494 B) :

q0=1ω2αmsin(ω(β))0=p0=2αmcos(ω(β))β=π2ωq0=2αmω2α=m2ω2q02=E
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Identifiers

  • q0
  • ω
  • α
  • m
  • ω
  • β
  • p0
  • α
  • m
  • ω
  • β
  • β
  • π
  • ω
  • q0
  • α
  • m
  • ω
  • α
  • m
  • ω
  • q0
  • E

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