### Author Topic: consequence of EL  (Read 1209 times)

#### Jingxuan Zhang

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##### consequence of EL
« on: March 30, 2018, 09:37:25 AM »
From (what I consider to be Euler-Lagrange)
$$\label{1}L_u=(L_{u'})_t$$
how can I derive
$$\label{2}L=u'L_{u'}+C?$$

Or is \eqref{2} even right? are they derived independently? if I integrate both sides of \eqref{1}, what exactly will be on the left?
« Last Edit: March 30, 2018, 09:41:57 AM by Jingxuan Zhang »

#### Victor Ivrii

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##### Re: consequence of EL
« Reply #1 on: March 30, 2018, 12:08:01 PM »
Start from E.-L.:

\frac{d\ }{dt} L_{q'_j}(q,q',t)- L_{q_j}=0,
\tag{A}

consider
\begin{align*}
\frac{d\ }{dt} \Bigl(\sum_j  q'_j L_{q'_j} -L\Bigr)=&
\sum_j  q''_j L_{q'_j} +  \sum_j{\color{blue}{ q'_j\frac{d\ }{dt} L_{q'_j}}} - \frac{d\ }{dt}L \\
=&\sum_j  q''_j L_{q'_j} +  \sum_j {\color{blue}{q'_j L_{q_j}}}-{\color{magenta}{\frac{d\ }{dt}L}}\\
=&\sum_j  q''_j L_{q'_j} +  \sum_j q'_j L_{q_j} -
{\color{magenta}{\Bigl(\sum_j  q''_j L_{q'_j} +  \sum_j q'_j L_{q_j}+L_t\Bigr)}},
\end{align*}
where transition from the first line to the second is due to (A) and from the second to the third due to chain rule.

Therefore

\frac{d\ }{dt} \Bigl(\sum_j  q'_j L_{q'_j} -L\Bigr)=-L_t
\tag{B}

and if $L$ does not depend explicitly on $t$ we have $L_t=0$ and

\frac{d\ }{dt} \Bigl(\sum_j  q'_j L_{q'_j} -L\Bigr)=0.
\tag{C}

Note: $q=(q_1,\ldots, q_n)$ etc
« Last Edit: March 30, 2018, 03:15:59 PM by Victor Ivrii »

#### Jingxuan Zhang

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##### Re: consequence of EL
« Reply #2 on: March 30, 2018, 02:19:45 PM »
Do you mean

\frac{d\ }{dt} L_{q'_j}(q,q',t)- L_{q_j}=0
\tag{A1}

and

\frac{d\ }{dt} \Bigl(\sum_j  q'_j L_{q'_j} -L\Bigr)=0?
\label{C1}

and do you mean $q=(q_j)?$
« Last Edit: March 30, 2018, 02:21:19 PM by Jingxuan Zhang »

#### Victor Ivrii

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##### Re: consequence of EL
« Reply #3 on: March 30, 2018, 03:15:09 PM »
Corrected.

In the Hamiltonian dynamics (beyond scope of this class)

\frac{dq_j}{dt}=H_{p_j}, \quad \frac{dp_j}{dt}=-H_{q_j}
\tag{D}

we have

\frac{d\ }{dt}H = \sum_j   H_{q_j} q'_j +\sum_j   H_{p_j} p'_j +H_t= H_t
\tag{E}

and

\frac{d\ }{dt}H = 0
\tag{F}

as long as $H_t=0$.

Thus,  $\frac{\partial L}{\partial t}=-\frac{\partial H}{\partial t}$, but the left-side is calculated in independent variables $(q,q',t)$ and the right-side is calculated in independent variables $(q,p,t)$; $p_j=L_{q'_j}$  calculated in independent variables $(q,q',t)$ and  $q'_j=H_{p_j}$ calculated in independent variables $(q,p,t)$.
« Last Edit: March 30, 2018, 03:36:26 PM by Victor Ivrii »