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Display information for equation id:math.1926.25 on revision:1926

* Page found: Zeitliche Translationsinvarianz (eq math.1926.25)

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TeX (original user input):

\begin{align}
  & I=\frac{\partial \bar{L}}{\partial {{{\dot{q}}}_{f+1}}}=\frac{\partial \bar{L}}{\partial \left( \frac{dt}{d\tau } \right)}=L+\sum\limits_{k=1}^{f}{\frac{\partial L}{\partial {{{\dot{q}}}_{k}}}\left( -\frac{1}{{{\left( \frac{dt}{d\tau } \right)}^{2}}} \right)\frac{d{{q}_{k}}}{d\tau }\frac{dt}{d\tau }} \\
 & =L-\sum\limits_{k=1}^{f}{\left( \frac{\partial L}{\partial \left( {{{\dot{q}}}_{k}} \right)} \right){{{\dot{q}}}_{k}}}=T-V-2T=-(T-V) \\
\end{align}

TeX (checked):

{\begin{aligned}&I={\frac {\partial {\bar {L}}}{\partial {{\dot {q}}_{f+1}}}}={\frac {\partial {\bar {L}}}{\partial \left({\frac {dt}{d\tau }}\right)}}=L+\sum \limits _{k=1}^{f}{{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}\left(-{\frac {1}{{\left({\frac {dt}{d\tau }}\right)}^{2}}}\right){\frac {d{{q}_{k}}}{d\tau }}{\frac {dt}{d\tau }}}\\&=L-\sum \limits _{k=1}^{f}{\left({\frac {\partial L}{\partial \left({{\dot {q}}_{k}}\right)}}\right){{\dot {q}}_{k}}}=T-V-2T=-(T-V)\\\end{aligned}}

LaTeXML (experimentell; verwendet MathML) rendering

MathML (40.712 KB / 4.429 KB) :

I = L ¯ q ˙ f + 1 = L ¯ ( d t d τ ) = L + k = 1 f L q ˙ k ( - 1 ( d t d τ ) 2 ) d q k d τ d t d τ = L - k = 1 f ( L ( q ˙ k ) ) q ˙ k = T - V - 2 T = - ( T - V ) missing-subexpression 𝐼 ¯ 𝐿 subscript ˙ 𝑞 𝑓 1 ¯ 𝐿 𝑑 𝑡 𝑑 𝜏 𝐿 superscript subscript 𝑘 1 𝑓 𝐿 subscript ˙ 𝑞 𝑘 1 superscript 𝑑 𝑡 𝑑 𝜏 2 𝑑 subscript 𝑞 𝑘 𝑑 𝜏 𝑑 𝑡 𝑑 𝜏 missing-subexpression absent 𝐿 superscript subscript 𝑘 1 𝑓 𝐿 subscript ˙ 𝑞 𝑘 subscript ˙ 𝑞 𝑘 𝑇 𝑉 2 𝑇 𝑇 𝑉 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle I=\frac{\partial% \bar{L}}{\partial{{{\dot{q}}}_{f+1}}}=\frac{\partial\bar{L}}{\partial\left(% \frac{dt}{d\tau}\right)}=L+\sum\limits_{k=1}^{f}{\frac{\partial L}{\partial{{{% \dot{q}}}_{k}}}\left(-\frac{1}{{{\left(\frac{dt}{d\tau}\right)}^{2}}}\right)% \frac{d{{q}_{k}}}{d\tau}\frac{dt}{d\tau}}\\ &\displaystyle=L-\sum\limits_{k=1}^{f}{\left(\frac{\partial L}{\partial\left({% {{\dot{q}}}_{k}}\right)}\right){{{\dot{q}}}_{k}}}=T-V-2T=-(T-V)\\ \end{aligned}}}

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MathML (experimentell; keine Bilder) rendering

MathML (4.911 KB / 588 B) :

I=L¯q˙f+1=L¯(dtdτ)=L+k=1fLq˙k(1(dtdτ)2)dqkdτdtdτ=Lk=1f(L(q˙k))q˙k=TV2T=(TV)

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  • I
  • L¯
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  • L¯
  • d
  • t
  • d
  • τ
  • L
  • k
  • f
  • L
  • q˙k
  • d
  • t
  • d
  • τ
  • d
  • qk
  • d
  • τ
  • d
  • t
  • d
  • τ
  • L
  • k
  • f
  • L
  • q˙k
  • q˙k
  • T
  • V
  • T
  • T
  • V

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