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--- equations:euler_lagrange_equations [2017/12/21 10:52]
jakobadmin [Researcher]
+++ equations:euler_lagrange_equations [2018/04/08 16:13] (current)
jakobadmin ↷ Links adapted because of a move operation
@@ Line 1: / Line 1: @@
+<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 ======
-<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.
+//see also [[formalisms:lagrangian_formalism]]//
+<tabbox Intuitive>
-<tabbox Layman>
+The basic idea behind the Lagrangian formalism is that nature is guided by a principle of "minimal action". The Euler-Lagrange equations give the path with a minimal amount of "action" that a system follows.
-The basic idea behind the Lagrangian formalism is that nature is guided by a principle of "minimal action". The Euler-Lagrange equations gives the path with a minimal amount of "action" that a system follows.
+In principle, there are many possible paths how some given particle or multiple particles could get from some point $A$ to another point $B$. The Euler-Lagrange equations are used to calculate the correct path that a particle follows between $A$ and $B$.
-In principle there are many possible paths how some given particle or multiple particles could get from some point $A$ to another point $B$. The Euler-Lagrange equations are used to calculate the correct path that a particle follows between $A$ and $B$.
-<tabbox Student>
+<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]].
+$$ \text{For particles: } \frac{\partial L}{\partial q_i} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q_i}} = 0 . $$
+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.
+----
+**Derivation**
+We consider an arbitrary path $(q,\dot{q})$.  If it is the
+path that minimizes the action, we have
+\begin{eqnarray}
+&=& \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  \\
+&=& \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 \dot{q}} \bigg) - S \\
+&=& \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)
+\end{eqnarray}
+If we now integrate the second term by parts, and take the variation of
+$\delta q$ to be $0$ at $t_i$ and $t_f$,
+\begin{eqnarray}
+\delta S = \int_{t_i}^{t_f} dt \bigg(\delta q{\partial L \over \partial
+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
+\over dt} {\partial L \over \partial \dot{q}} \bigg) = 0
+\end{eqnarray}
+Now, the only way this holds for an arbitrary variation $\delta q$
+of the path $(q,\dot{q})$ is when
+\begin{eqnarray}
+{d \over dt}{\partial L \over \partial \dot{q}} - {\partial L \over
+\partial q} = 0
+\end{eqnarray}
+This is the Euler-Lagrange equation.
+Generalized to multiple coordinates $q_i$ ($i=1,\ldots,n$) it reads
+\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:
+$$
+[L] = \frac{\partial L}{\partial q} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q}} =   \frac{\partial L}{\partial q^i} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q}^i}
+$$
+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  $q$.
+----
-<note tip>
-In this section things should be explained by analogy and with pictures and, if necessary, some formulas.
-</note>
-<tabbox Researcher>
 <blockquote>
@@ Line 63: / Line 117: @@
   * Bleecker, D., Gauge Theory and Variational Principles, Addison-Wesley, Reading, MA, 1981.
-<tabbox Examples>
+<tabbox Why is it interesting?>
---> Example1#
+The Euler-Lagrange equations are used in the [[formalisms:lagrangian_formalism|Lagrange formalism]] to derive from a given Lagrangian the corresponding equations of motion.
-<--
---> Example2:#
-<--
-<tabbox History>
 </tabbox>

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