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equations:euler_lagrange_equations [2018/03/27 09:05] jakobadmin [Concrete] |
equations:euler_lagrange_equations [2018/04/08 16:13] (current) jakobadmin ↷ Links adapted because of a move operation |
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- | ====== Euler-Lagrange Equations: $\quad \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 $ ====== | + | <WRAP lag>$ \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 $</WRAP> |
+ | ====== Euler-Lagrange Equations ====== | ||
- | //see also [[frameworks:lagrangian_formalism|]]// | + | |
+ | //see also [[formalisms:lagrangian_formalism]]// | ||
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| | ||
<tabbox Concrete> | <tabbox Concrete> | ||
+ | The Euler-Lagrange equation tells us which path is the path with minimal action $S = \int_{t_i}^{t_f} dt L(q,\dot{q})$, where $L(q,\dot{q})$ denotes the [[formalisms:lagrangian_formalism|Lagrangian]]. | ||
- | $$ \text{For particles: } \frac{\partial L}{\partial q_i} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q_i}} = 0 \qquad \text{For fields: } \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 $$ | + | |
+ | $$ \text{For particles: } \frac{\partial L}{\partial q_i} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q_i}} = 0 . $$ | ||
+ | |||
+ | The Euler-Lagrange equation can also be used in a field theory and there it tells us which sequence of field configurations has minimal action. | ||
+ | |||
+ | $$ \text{For fields: } \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 .$$ | ||
+ | |||
+ | The general procedure is that we start with a Lagrangian. The Lagrangian is an object that has to be guessed by making use of symmetry considerations and characterizes the system in question. Then we put the Lagrangian into the Euler-Lagrange equation and this gives us the equations of motion of the system. | ||
---- | ---- | ||
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\begin{eqnarray} | \begin{eqnarray} | ||
0 &=& \delta S = \delta \int_{t_i}^{t_f} dt L(q,\dot{q}) = | 0 &=& \delta S = \delta \int_{t_i}^{t_f} dt L(q,\dot{q}) = | ||
- | \int_{t_i}^{t_f} dt L(q+\delta q,\dot{q}+\delta \dot{q})-S \nolabel \\ | + | \int_{t_i}^{t_f} dt L(q+\delta q,\dot{q}+\delta \dot{q})-S \\ |
&=& \int_{t_i}^{t_f} dtL(q,\dot{q}) + \int_{t_i}^{t_f} dt\bigg(\delta q | &=& \int_{t_i}^{t_f} dtL(q,\dot{q}) + \int_{t_i}^{t_f} dt\bigg(\delta q | ||
{\partial L \over \partial q} + \delta \dot{q} {\partial L \over | {\partial L \over \partial q} + \delta \dot{q} {\partial L \over | ||
- | \partial \dot{q}} \bigg) - S \nolabel \\ | + | \partial \dot{q}} \bigg) - S \\ |
&=& \int_{t_i}^{t_f} dt \bigg(\delta q {\partial L \over \partial q} + | &=& \int_{t_i}^{t_f} dt \bigg(\delta q {\partial L \over \partial q} + | ||
{\partial L \over \partial \dot{q}} {d \over dt} \delta q\bigg) | {\partial L \over \partial \dot{q}} {d \over dt} \delta q\bigg) | ||
- | \nolabel | ||
\end{eqnarray} | \end{eqnarray} | ||
If we now integrate the second term by parts, and take the variation of | If we now integrate the second term by parts, and take the variation of | ||
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q} - \delta q {d \over dt} {\partial L \over \partial \dot{q}} \bigg) = | q} - \delta q {d \over dt} {\partial L \over \partial \dot{q}} \bigg) = | ||
\int_{t_i}^{t_f} dt \delta q \bigg({\partial L \over \partial q} - {d | \int_{t_i}^{t_f} dt \delta q \bigg({\partial L \over \partial q} - {d | ||
- | \over dt} {\partial L \over \partial \dot{q}} \bigg) = 0 \nolabel | + | \over dt} {\partial L \over \partial \dot{q}} \bigg) = 0 |
\end{eqnarray} | \end{eqnarray} | ||
Now, the only way this holds for an arbitrary variation $\delta q$ | Now, the only way this holds for an arbitrary variation $\delta q$ | ||
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\begin{eqnarray} | \begin{eqnarray} | ||
{d \over dt}{\partial L \over \partial \dot{q}} - {\partial L \over | {d \over dt}{\partial L \over \partial \dot{q}} - {\partial L \over | ||
- | \partial q} = 0 \nolabel | + | \partial q} = 0 |
\end{eqnarray} | \end{eqnarray} | ||
+ | This is the Euler-Lagrange equation. | ||
- | ---- | + | Generalized to multiple coordinates $q_i$ ($i=1,\ldots,n$) it reads |
- | ** Euler-Lagrange Expression ** | + | \begin{eqnarray} |
+ | {d \over dt} {\partial L \over \partial \dot{q}_i} - {\partial L \over | ||
+ | \partial q_i} = 0 | ||
+ | \end{eqnarray} | ||
+ | |||
+ | |||
+ | <tabbox Abstract> | ||
Given a Lagrangian function $L:T\mathcal{Q} \longrightarrow \mathbb{R}$, the //Lagrange expression //, denoted as $[]$ is given by: | Given a Lagrangian function $L:T\mathcal{Q} \longrightarrow \mathbb{R}$, the //Lagrange expression //, denoted as $[]$ is given by: | ||
$$ | $$ | ||
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$$ | $$ | ||
- | where $q\in\mathcal{Q}$ and $\dot q$ represents the lift of $q$ to the tangent bundle, i.e $(q, \dot q)\in T\mathcal Q$. Youo can thing $\dot{q}$ as the vector on the point $q$. | + | where $q\in\mathcal{Q}$ and $\dot q$ represents the lift of $q$ to the tangent bundle, i.e $(q, \dot q)\in T\mathcal Q$. You can think of $\dot{q}$ as the vector on the point $q$. |
- | ** Euler-Lagrange Equations** | + | ---- |
- | The form to the Euler-Lagrange equations is | ||
- | $$ | ||
- | [L] = \frac{\partial L}{\partial q} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q}} = 0 | ||
- | $$ | ||
- | |||
- | <tabbox Abstract> | ||
<blockquote> | <blockquote> | ||
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<tabbox Why is it interesting?> | <tabbox Why is it interesting?> | ||
- | The Euler-Lagrange equations are used in the [[frameworks:lagrangian_formalism|Lagrange formalism]] to derive from a given Lagrangian the corresponding equations of motion. | + | The Euler-Lagrange equations are used in the [[formalisms:lagrangian_formalism|Lagrange formalism]] to derive from a given Lagrangian the corresponding equations of motion. |
</tabbox> | </tabbox> | ||