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$ \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 $

Euler-Lagrange Equations

see also lagrangian_formalism


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.

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 $B$.


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 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.


We consider an arbitrary path $(q,\dot{q})$. If it is the path that minimizes the action, we have \begin{eqnarray} 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 \\ &=& \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}


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 $q$.

With this notation, a globally defined local Lagrangian for fields that are sections of some bundle $E$ over spacetime/worldvolume $\Sigma$ is simply a morphism of the form $$ L : E \longrightarrow \wedge^{p+1}T^\ast \Sigma \,. $$ Unwinding what this means, this is a function that at each point of $\Sigma$ sends the value of field configurations and all their spacetime/worldvolume derivatives at that point to a $(p+1)$-form on $\Sigma$ at that point. It is this pointwise local (in fact: infinitesimally local) dependence that the term local in local Lagrangian refers to. […]

Regarding such $L$ for a moment as just a differential form on $J^\infty_\Sigma(E)$ (= the jet bundle), we may apply the de Rham differential to it. One finds that this uniquely decomposes as a sum of the form \begin{equation} \label{differentialofLagrangian} d L = \mathrm{EL} - d_H (\Theta + d_H(\cdots)) \,, \end{equation} for some $\Theta$ and for $\mathrm{EL}$ pointwise the pullback of a vertical 1-form on $E$; such a differential form is called a source form: $\mathrm{EL} \in \Omega^{p+1,1}_S(E)$. This particular source form is of paramount importance: the equation $$ \underset{v\in \Gamma(V E)}{\forall} j^\infty(\phi)^\ast \iota_v\mathrm{EL} = 0 $$ on sections $\phi \in \Gamma_\Sigma(E)$ is a partial differential equation, and this is called the Euler-Lagrange equation of motion induced by $L$. Differential equations arising this way from a local Lagrangian are called variational.

A little reflection reveals that this is indeed a re-statement of the traditional prescription of obtaining the Euler-Lagrange equations by locally varying the integral over the Lagrangian and then applying partial integration to turn all variation of derivatives (i.e. of jets) of fields into variation of the fields themselves. Here we do not consider this under the integral, and hence the boundary terms arising from the would-be partial integration show up as the contribution $\Theta$.

Why is it interesting?

The Euler-Lagrange equations are used in the Lagrange formalism to derive from a given Lagrangian the corresponding equations of motion.

equations/euler_lagrange_equations.txt · Last modified: 2018/04/08 16:13 by jakobadmin