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Display information for equation id:math.1901.8 on revision:1901

* Page found: Eichtransformation der Lagrangefunktion (eq math.1901.8)

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TeX (original user input):

\begin{align}
  & \frac{\partial L}{\partial {{{\dot{q}}}_{k}}}=m{{{\dot{q}}}_{k}}+e{{A}_{k}} \\
 & \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{k}}}=m{{{\ddot{q}}}_{k}}+e\frac{d}{dt}{{A}_{k}}(\bar{q}(t),t) \\
 & \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{k}}}=m{{{\ddot{q}}}_{k}}+e\left( \frac{\partial }{\partial t}{{A}_{k}}+\sum\limits_{l}{\frac{\partial {{A}_{k}}}{\partial {{q}_{l}}}{{{\dot{q}}}_{l}}} \right) \\
 & \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{k}}}=m{{{\ddot{q}}}_{k}}+e\left( \frac{\partial }{\partial t}{{A}_{k}}+\left( \dot{\bar{q}}\cdot \nabla  \right){{A}_{k}} \right) \\
\end{align}

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L q ˙ k = m q ˙ k + e A k d d t L q ˙ k = m q ¨ k + e d d t A k ( q ¯ ( t ) , t ) d d t L q ˙ k = m q ¨ k + e ( t A k + l A k q l q ˙ l ) d d t L q ˙ k = m q ¨ k + e ( t A k + ( q ¯ ˙ ) A k ) missing-subexpression 𝐿 subscript ˙ 𝑞 𝑘 𝑚 subscript ˙ 𝑞 𝑘 𝑒 subscript 𝐴 𝑘 missing-subexpression 𝑑 𝑑 𝑡 𝐿 subscript ˙ 𝑞 𝑘 𝑚 subscript ¨ 𝑞 𝑘 𝑒 𝑑 𝑑 𝑡 subscript 𝐴 𝑘 ¯ 𝑞 𝑡 𝑡 missing-subexpression 𝑑 𝑑 𝑡 𝐿 subscript ˙ 𝑞 𝑘 𝑚 subscript ¨ 𝑞 𝑘 𝑒 𝑡 subscript 𝐴 𝑘 subscript 𝑙 subscript 𝐴 𝑘 subscript 𝑞 𝑙 subscript ˙ 𝑞 𝑙 missing-subexpression 𝑑 𝑑 𝑡 𝐿 subscript ˙ 𝑞 𝑘 𝑚 subscript ¨ 𝑞 𝑘 𝑒 𝑡 subscript 𝐴 𝑘 ˙ ¯ 𝑞 subscript 𝐴 𝑘 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle\frac{\partial L}{% \partial{{{\dot{q}}}_{k}}}=m{{{\dot{q}}}_{k}}+e{{A}_{k}}\\ &\displaystyle\frac{d}{dt}\frac{\partial L}{\partial{{{\dot{q}}}_{k}}}=m{{{% \ddot{q}}}_{k}}+e\frac{d}{dt}{{A}_{k}}(\bar{q}(t),t)\\ &\displaystyle\frac{d}{dt}\frac{\partial L}{\partial{{{\dot{q}}}_{k}}}=m{{{% \ddot{q}}}_{k}}+e\left(\frac{\partial}{\partial t}{{A}_{k}}+\sum\limits_{l}{% \frac{\partial{{A}_{k}}}{\partial{{q}_{l}}}{{{\dot{q}}}_{l}}}\right)\\ &\displaystyle\frac{d}{dt}\frac{\partial L}{\partial{{{\dot{q}}}_{k}}}=m{{{% \ddot{q}}}_{k}}+e\left(\frac{\partial}{\partial t}{{A}_{k}}+\left(\dot{\bar{q}% }\cdot\nabla\right){{A}_{k}}\right)\\ \end{aligned}}}

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Lq˙k=mq˙k+eAkddtLq˙k=mq¨k+eddtAk(q¯(t),t)ddtLq˙k=mq¨k+e(tAk+lAkqlq˙l)ddtLq˙k=mq¨k+e(tAk+(q¯˙)Ak)

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