Euler–Lagrange equation

Statement

Let ${\displaystyle (X,L)}$ be a real dynamical system with ${\displaystyle n}$ degrees of freedom. Here ${\displaystyle X}$ is the configuration space and ${\displaystyle L=L(t,\mathit{q}(t),\mathit{v}(t))}$ the Lagrangian, i.e. a smooth real-valued function such that ${\displaystyle \mathit{q}(t)\in X,}$ and ${\displaystyle \mathit{v}(t)=\dot{\mathit{q}}(t)}$ is an ${\displaystyle n}$-dimensional "vector of speed", where ${\displaystyle \dot{\mathit{q}}(t)}$ is the time derivative of ${\displaystyle \mathit{q}(t)}$. (For those familiar with differential geometry, ${\displaystyle X}$ is a smooth manifold, and ${\displaystyle L:{\mathbb{R}}_t\times TX\to \mathbb{R},}$ where ${\displaystyle TX}$ is the tangent bundle of ${\displaystyle X}$).

Let ${\displaystyle \mathcal{P}(a,b,{\mathit{x}}_a,{\mathit{x}}_b)}$ be the set of smooth paths ${\displaystyle \mathit{q}:[a,b]\to X}$ for which ${\displaystyle \mathit{q}(a)={\mathit{x}}_a}$ and ${\displaystyle \mathit{q}(b)={\mathit{x}}_b}$.

The action functional ${\displaystyle S:\mathcal{P}(a,b,{\mathit{x}}_a,{\mathit{x}}_b)\to \mathbb{R}}$ is defined via $${\displaystyle S[\mathit{q}]={\int }_a^{b}L(t,\mathit{q}(t),\dot{\mathit{q}}(t))dt.}$$

A path ${\displaystyle \mathit{q}\in \mathcal{P}(a,b,{\mathit{x}}_a,{\mathit{x}}_b)}$ is a stationary point of ${\displaystyle S}$ if and only if

${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{q}^i}(t,\mathit{q}(t),\dot{\mathit{q}}(t))-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}^i}(t,\mathit{q}(t),\dot{\mathit{q}}(t))=0,i=1,\dots ,n.}$

When we say stationary point, we mean a stationary point of ${\displaystyle S}$ with respect to any small perturbation in ${\displaystyle \mathit{q}}$. See proofs below for more rigorous detail.

Derivation of the one-dimensional Euler–Lagrange equation

The derivation of the one-dimensional Euler–Lagrange equation is one of the classic proofs in mathematics. It relies on the fundamental lemma of calculus of variations.

We wish to find a function ${\displaystyle f}$ which satisfies the boundary conditions ${\displaystyle f(a)=A}$, ${\displaystyle f(b)=B}$, and which extremizes the functional $${\displaystyle J[f]={\int }_a^{b}L(x,f(x),f^{\prime }(x))\mathrm{d}x.}$$

We assume that ${\displaystyle L}$ is twice continuously differentiable. A weaker assumption can be used, but the proof becomes more difficult.

If ${\displaystyle f}$ extremizes the functional subject to the boundary conditions, then any slight perturbation of ${\displaystyle f}$ that preserves the boundary values must either increase ${\displaystyle J}$ (if ${\displaystyle f}$ is a minimizer) or decrease ${\displaystyle J}$ (if ${\displaystyle f}$ is a maximizer).

Let ${\displaystyle f+\epsilon \eta }$ be the result of such a perturbation ${\displaystyle \epsilon \eta }$ of ${\displaystyle f}$, where ${\displaystyle \epsilon }$ is small and ${\displaystyle \eta }$ is a differentiable function satisfying ${\displaystyle \eta (a)=\eta (b)=0}$. Then define $${\displaystyle \mathrm{\Phi }(\epsilon )=J[f+\epsilon \eta ]={\int }_a^{b}L(x,f(x)+\epsilon \eta (x),f^{\prime }(x)+\epsilon {\eta }^{\prime }(x))\mathrm{d}x.}$$

We now wish to calculate the total derivative of ${\displaystyle \mathrm{\Phi }}$ with respect to ε. $${\displaystyle \begin{array}{cc}\frac{\mathrm{d}\mathrm{\Phi }}{\mathrm{d}\epsilon }& =\frac{\mathrm{d}}{\mathrm{d}\epsilon }{\int }_a^{b}L(x,f(x)+\epsilon \eta (x),f^{\prime }(x)+\epsilon {\eta }^{\prime }(x))\mathrm{d}x\\ & ={\int }_a^b\frac{\mathrm{d}}{\mathrm{d}\epsilon }L(x,f(x)+\epsilon \eta (x),f^{\prime }(x)+\epsilon {\eta }^{\prime }(x))\mathrm{d}x\\ & ={\int }_a^b\left[\eta (x)\frac{\mathrm{\partial }L}{\mathrm{\partial }f}(x,f(x)+\epsilon \eta (x),f^{\prime }(x)+\epsilon {\eta }^{\prime }(x))+{\eta }^{\prime }(x)\frac{\mathrm{\partial }L}{\mathrm{\partial }{f}^{\prime }}(x,f(x)+\epsilon \eta (x),f^{\prime }(x)+\epsilon {\eta }^{\prime }(x))\right]\mathrm{d}x.\end{array}}$$

The third line follows from the fact that ${\displaystyle x}$ does not depend on ${\displaystyle \epsilon }$, i.e. ${\displaystyle \frac{\mathrm{d}x}{\mathrm{d}\epsilon }=0}$.

When ${\displaystyle \epsilon =0}$, ${\displaystyle \mathrm{\Phi }}$ has an extremum value, so that $${\displaystyle {\frac{\mathrm{d}\mathrm{\Phi }}{\mathrm{d}\epsilon }|}_{\epsilon =0}={\int }_a^b\left[\eta (x)\frac{\mathrm{\partial }L}{\mathrm{\partial }f}(x,f(x),f^{\prime }(x))+{\eta }^{\prime }(x)\frac{\mathrm{\partial }L}{\mathrm{\partial }{f}^{\prime }}(x,f(x),f^{\prime }(x))\right]\mathrm{d}x=0.}$$

The next step is to use integration by parts on the second term of the integrand, yielding $${\displaystyle {\int }_a^b\left[\frac{\mathrm{\partial }L}{\mathrm{\partial }f}(x,f(x),f^{\prime }(x))-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{\partial }L}{\mathrm{\partial }{f}^{\prime }}(x,f(x),f^{\prime }(x))\right]\eta (x)\mathrm{d}x+{\left[\eta (x)\frac{\mathrm{\partial }L}{\mathrm{\partial }{f}^{\prime }}(x,f(x),f^{\prime }(x))\right]}_a^b=0.}$$

Using the boundary conditions ${\displaystyle \eta (a)=\eta (b)=0}$, $${\displaystyle {\int }_a^b\left[\frac{\mathrm{\partial }L}{\mathrm{\partial }f}(x,f(x),f^{\prime }(x))-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{\partial }L}{\mathrm{\partial }{f}^{\prime }}(x,f(x),f^{\prime }(x))\right]\eta (x)\mathrm{d}x=0.}$$

Applying the fundamental lemma of calculus of variations now yields the Euler–Lagrange equation $${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }f}(x,f(x),f^{\prime }(x))-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{\partial }L}{\mathrm{\partial }{f}^{\prime }}(x,f(x),f^{\prime }(x))=0.}$$

Alternative derivation of the one-dimensional Euler–Lagrange equation

Given a functional $${\displaystyle J={\int }_a^{b}L(t,y(t),y^{\prime }(t))\mathrm{d}t}$$ on ${\displaystyle C^1([a,b])}$ with the boundary conditions ${\displaystyle y(a)=A}$ and ${\displaystyle y(b)=B}$, we proceed by approximating the extremal curve by a polygonal line with ${\displaystyle n}$ segments and passing to the limit as the number of segments grows arbitrarily large.

Divide the interval ${\displaystyle [a,b]}$ into ${\displaystyle n}$ equal segments with endpoints ${\displaystyle t_0=a,t_1,t_2,\dots ,t_n=b}$ and let ${\displaystyle \mathrm{\Delta }t=t_k-t_{k-1}}$. Rather than a smooth function ${\displaystyle y(t)}$ we consider the polygonal line with vertices ${\displaystyle (t_0,y_0),\dots ,(t_n,y_n)}$, where ${\displaystyle y_0=A}$ and ${\displaystyle y_n=B}$. Accordingly, our functional becomes a real function of ${\displaystyle n-1}$ variables given by $${\displaystyle J(y_1,\dots ,y_{n-1})\approx \sum _{k=0}^{n-1}L\left(t_k,y_k,\frac{y_{k+1}-y_k}{\mathrm{\Delta }t}\right)\mathrm{\Delta }t.}$$

Extremals of this new functional defined on the discrete points ${\displaystyle t_0,\dots ,t_n}$ correspond to points where $${\displaystyle \frac{\mathrm{\partial }J(y_1,\dots ,y_n)}{\mathrm{\partial }{y}_m}=0.}$$

Note that change of ${\displaystyle y_m}$ affects L not only at m but also at m-1 for the derivative of the 3rd argument. $${\displaystyle L(\text{3rd argument})\left(\frac{y_{m+1}-(y_m+\mathrm{\Delta }{y}_m)}{\mathrm{\Delta }t}\right)=L\left(\frac{y_{m+1}-y_m}{\mathrm{\Delta }t}\right)-\frac{\mathrm{\partial }L}{\mathrm{\partial }{y}^{\prime }}\frac{\mathrm{\Delta }{y}_m}{\mathrm{\Delta }t}}$$$${\displaystyle L\left(\frac{(y_m+\mathrm{\Delta }{y}_m)-y_{m-1}}{\mathrm{\Delta }t}\right)=L\left(\frac{y_m-y_{m-1}}{\mathrm{\Delta }t}\right)+\frac{\mathrm{\partial }L}{\mathrm{\partial }{y}^{\prime }}\frac{\mathrm{\Delta }{y}_m}{\mathrm{\Delta }t}}$$

Evaluating the partial derivative gives $${\displaystyle \frac{\mathrm{\partial }J}{\mathrm{\partial }{y}_m}=L_y\left(t_m,y_m,\frac{y_{m+1}-y_m}{\mathrm{\Delta }t}\right)\mathrm{\Delta }t+L_{y^{\prime }}\left(t_{m-1},y_{m-1},\frac{y_m-y_{m-1}}{\mathrm{\Delta }t}\right)-L_{y^{\prime }}\left(t_m,y_m,\frac{y_{m+1}-y_m}{\mathrm{\Delta }t}\right).}$$

Dividing the above equation by ${\displaystyle \mathrm{\Delta }t}$ gives $${\displaystyle \frac{\mathrm{\partial }J}{\mathrm{\partial }{y}_m\mathrm{\Delta }t}=L_y\left(t_m,y_m,\frac{y_{m+1}-y_m}{\mathrm{\Delta }t}\right)-\frac{1}{\mathrm{\Delta }t}\left[L_{y^{\prime }}\left(t_m,y_m,\frac{y_{m+1}-y_m}{\mathrm{\Delta }t}\right)-L_{y^{\prime }}\left(t_{m-1},y_{m-1},\frac{y_m-y_{m-1}}{\mathrm{\Delta }t}\right)\right],}$$ and taking the limit as ${\displaystyle \mathrm{\Delta }t\to 0}$ of the right-hand side of this expression yields $${\displaystyle L_y-\frac{\mathrm{d}}{\mathrm{d}t}{L}_{y^{\prime }}=0.}$$

The left hand side of the previous equation is the functional derivative ${\displaystyle \delta J/\delta y}$ of the functional ${\displaystyle J}$. A necessary condition for a differentiable functional to have an extremum on some function is that its functional derivative at that function vanishes, which is granted by the last equation.

Example

A standard example is finding the real-valued function y(x) on the interval [a, b], such that y(a) = c and y(b) = d, for which the path length along the curve traced by y is as short as possible.

${\displaystyle \text{s}={\int }_a^b\sqrt{\mathrm{d}{x}^2+\mathrm{d}{y}^2}={\int }_a^b\sqrt{1+y^{\prime 2}}\mathrm{d}x,}$

the integrand function being $L(x,y,y^{\prime })=\sqrt{1+y^{\prime 2}}$.

The partial derivatives of L are:

${\displaystyle \frac{\mathrm{\partial }L(x,y,y^{\prime })}{\mathrm{\partial }{y}^{\prime }}=\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}\text{and}\frac{\mathrm{\partial }L(x,y,y^{\prime })}{\mathrm{\partial }y}=0.}$

By substituting these into the Euler–Lagrange equation, we obtain

${\displaystyle \begin{array}{cc}\frac{\mathrm{d}}{\mathrm{d}x}\frac{y^{\prime }(x)}{\sqrt{1+(y^{\prime }(x){)}^2}}& =0\\ \frac{y^{\prime }(x)}{\sqrt{1+(y^{\prime }(x){)}^2}}& =C=\text{constant}\\ \Rightarrow y^{\prime }(x)& =\frac{C}{\sqrt{1-C^2}}=:A\\ \Rightarrow y(x)& =Ax+B\end{array}}$

that is, the function must have a constant first derivative, and thus its graph is a straight line.

Canonical momenta and constants of motion

The conjugate momentum p_k for a generalized coordinate q_k is defined by the equation $${\displaystyle p_k\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_k}.}$$

An important special case of the Euler–Lagrange equation occurs when L does not contain a generalized coordinate q_k explicitly, $${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{q}_k}=0\Rightarrow \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_k}=0\Rightarrow \frac{d{p}_k}{dt}=0,}$$ that is, the conjugate momentum is a constant of the motion.

In such cases, the coordinate q_k is called a cyclic coordinate. For example, if we use polar coordinates t, r, θ to describe the planar motion of a particle, and if L does not depend on θ, the conjugate momentum is the conserved angular momentum.

Generalizations

Single function of single variable with higher derivatives

The stationary values of the functional

${\displaystyle I[f]={\int }_{x_0}^{x_1}\mathcal{L}(x,f,f^{\prime },f^{″},\dots ,f^{(k)})\mathrm{d}x;f^{\prime }:=\frac{\mathrm{d}f}{\mathrm{d}x},f^{″}:=\frac{{\mathrm{d}}^{2}f}{\mathrm{d}{x}^2},f^{(k)}:=\frac{{\mathrm{d}}^{k}f}{\mathrm{d}{x}^k}}$

can be obtained from the Euler–Lagrange equation

${\displaystyle \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }f}-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}^{\prime }}\right)+\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}^{″}}\right)-\cdots +(-1{)}^k\frac{{\mathrm{d}}^k}{\mathrm{d}{x}^k}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}^{(k)}}\right)=0}$

under fixed boundary conditions for the function itself as well as for the first ${\displaystyle k-1}$ derivatives (i.e. for all ${\displaystyle f^{(i)},i\in \{0,...,k-1\}}$). The endpoint values of the highest derivative ${\displaystyle f^{(k)}}$ remain flexible.

Several functions of single variable with single derivative

If the problem involves finding several functions (${\displaystyle f_1,f_2,\dots ,f_m}$) of a single independent variable (${\displaystyle x}$) that define an extremum of the functional

${\displaystyle I[f_1,f_2,\dots ,f_m]={\int }_{x_0}^{x_1}\mathcal{L}(x,f_1,f_2,\dots ,f_m,f_1^{\prime },f_2^{\prime },\dots ,f_m^{\prime })\mathrm{d}x;f_i^{\prime }:=\frac{\mathrm{d}{f}_i}{\mathrm{d}x}}$

then the corresponding Euler–Lagrange equations are

${\displaystyle \begin{array}{c}\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_i}-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_i^{\prime }}\right)=0;i=1,2,...,m\end{array}}$

Single function of several variables with single derivative

A multi-dimensional generalization comes from considering a function on n variables. If ${\displaystyle \mathrm{\Omega }}$ is some surface, then

${\displaystyle I[f]={\int }_{\mathrm{\Omega }}\mathcal{L}(x_1,\dots ,x_n,f,f_1,\dots ,f_n)\mathrm{d}\mathbf{x};f_j:=\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_j}}$

is extremized only if f satisfies the partial differential equation

${\displaystyle \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }f}-\sum _{j=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_j}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_j}\right)=0.}$

When n = 2 and functional ${\displaystyle \mathcal{I}}$ is the energy functional, this leads to the soap-film minimal surface problem.

Several functions of several variables with single derivative

If there are several unknown functions to be determined and several variables such that

${\displaystyle I[f_1,f_2,\dots ,f_m]={\int }_{\mathrm{\Omega }}\mathcal{L}(x_1,\dots ,x_n,f_1,\dots ,f_m,f_{1,1},\dots ,f_{1,n},\dots ,f_{m,1},\dots ,f_{m,n})\mathrm{d}\mathbf{x};f_{i,j}:=\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}}$

the system of Euler–Lagrange equations is

${\displaystyle \begin{array}{cc}\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_1}-\sum _{j=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_j}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{1,j}}\right)& =0_1\\ \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_2}-\sum _{j=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_j}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{2,j}}\right)& =0_2\\ ⋮⋮& ⋮\\ \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_m}-\sum _{j=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_j}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{m,j}}\right)& =0_m.\end{array}}$

Single function of two variables with higher derivatives

If there is a single unknown function f to be determined that is dependent on two variables x₁ and x₂ and if the functional depends on higher derivatives of f up to n-th order such that

${\displaystyle \begin{array}{cc}I[f]& ={\int }_{\mathrm{\Omega }}\mathcal{L}(x_1,x_2,f,f_1,f_2,f_{11},f_{12},f_{22},\dots ,f_{22\dots 2})\mathrm{d}\mathbf{x}\\ & f_i:=\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_i},f_{ij}:=\frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_j},\dots \end{array}}$

then the Euler–Lagrange equation is

${\displaystyle \begin{array}{cc}\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }f}& -\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_1}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_1}\right)-\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_2}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_2}\right)+\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{x}_1^2}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{11}}\right)+\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_2}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{12}}\right)+\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{x}_2^2}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{22}}\right)\\ & -\cdots +(-1{)}^n\frac{{\mathrm{\partial }}^n}{\mathrm{\partial }{x}_2^n}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{22\dots 2}}\right)=0\end{array}}$

which can be represented shortly as:

${\displaystyle \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }f}+\sum _{j=1}^n\sum _{{\mu }_1\le \dots \le {\mu }_j}(-1{)}^j\frac{{\mathrm{\partial }}^j}{\mathrm{\partial }{x}_{{\mu }_1}\dots \mathrm{\partial }{x}_{{\mu }_j}}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{{\mu }_1\dots {\mu }_j}}\right)=0}$

wherein ${\displaystyle {\mu }_1\dots {\mu }_j}$ are indices that span the number of variables, that is, here they go from 1 to 2. Here summation over the ${\displaystyle {\mu }_1\dots {\mu }_j}$ indices is only over ${\displaystyle {\mu }_1\le {\mu }_2\le \dots \le {\mu }_j}$ in order to avoid counting the same partial derivative multiple times, for example ${\displaystyle f_{12}=f_{21}}$ appears only once in the previous equation.

Several functions of several variables with higher derivatives

If there are p unknown functions f_i to be determined that are dependent on m variables x₁ ... x_m and if the functional depends on higher derivatives of the f_i up to n-th order such that

${\displaystyle \begin{array}{cc}I[f_1,\dots ,f_p]& ={\int }_{\mathrm{\Omega }}\mathcal{L}(x_1,\dots ,x_m;f_1,\dots ,f_p;f_{1,1},\dots ,f_{p,m};f_{1,11},\dots ,f_{p,mm};\dots ;f_{p,1\dots 1},\dots ,f_{p,m\dots m})\mathrm{d}\mathbf{x}\\ & f_{i,\mu }:=\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_{\mu }},f_{i,{\mu }_1{\mu }_2}:=\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_{{\mu }_1}\mathrm{\partial }{x}_{{\mu }_2}},\dots \end{array}}$

where ${\displaystyle {\mu }_1\dots {\mu }_j}$ are indices that span the number of variables, that is they go from 1 to m. Then the Euler–Lagrange equation is

${\displaystyle \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_i}+\sum _{j=1}^n\sum _{{\mu }_1\le \dots \le {\mu }_j}(-1{)}^j\frac{{\mathrm{\partial }}^j}{\mathrm{\partial }{x}_{{\mu }_1}\dots \mathrm{\partial }{x}_{{\mu }_j}}\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{i,{\mu }_1\dots {\mu }_j}}\right)=0}$

where the summation over the ${\displaystyle {\mu }_1\dots {\mu }_j}$ is avoiding counting the same derivative ${\displaystyle f_{i,{\mu }_1{\mu }_2}=f_{i,{\mu }_2{\mu }_1}}$ several times, just as in the previous subsection. This can be expressed more compactly as

${\displaystyle \sum _{j=0}^n\sum _{{\mu }_1\le \dots \le {\mu }_j}(-1{)}^j{\mathrm{\partial }}_{{\mu }_1\dots {\mu }_j}^j\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{f}_{i,{\mu }_1\dots {\mu }_j}}\right)=0}$

Field theories

Generalization to manifolds

Let ${\displaystyle M}$ be a smooth manifold, and let ${\displaystyle C^{\mathrm{\infty }}([a,b])}$ denote the space of smooth functions ${\displaystyle f:[a,b]\to M}$. Then, for functionals ${\displaystyle S:C^{\mathrm{\infty }}([a,b])\to \mathbb{R}}$ of the form

${\displaystyle S[f]={\int }_a^b(L\circ \dot{f})(t)\mathrm{d}t}$

where ${\displaystyle L:TM\to \mathbb{R}}$ is the Lagrangian, the statement ${\displaystyle \mathrm{d}{S}_f=0}$ is equivalent to the statement that, for all ${\displaystyle t\in [a,b]}$, each coordinate frame trivialization ${\displaystyle (x^i,X^i)}$ of a neighborhood of ${\displaystyle \dot{f}(t)}$ yields the following ${\displaystyle \dim M}$ equations:

${\displaystyle \mathrm{\forall }i:\frac{\mathrm{d}}{\mathrm{d}t}\frac{\mathrm{\partial }L}{\mathrm{\partial }{X}^i}{|}_{\dot{f}(t)}=\frac{\mathrm{\partial }L}{\mathrm{\partial }{x}^i}{|}_{\dot{f}(t)}.}$

Euler–Lagrange equations can also be written in a coordinate-free form as

${\displaystyle {\mathcal{L}}_{\mathrm{\Delta }}{\theta }_L=dL}$

where ${\displaystyle {\theta }_L}$ is the canonical momenta 1-form corresponding to the Lagrangian ${\displaystyle L}$. The vector field generating time translations is denoted by ${\displaystyle \mathrm{\Delta }}$ and the Lie derivative is denoted by ${\displaystyle \mathcal{L}}$. One can use local charts ${\displaystyle (q^{\alpha },{\dot{q}}^{\alpha })}$ in which ${\displaystyle {\theta }_L=\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}^{\alpha }}d{q}^{\alpha }}$ and ${\displaystyle \mathrm{\Delta }:=\frac{d}{dt}={\dot{q}}^{\alpha }\frac{\mathrm{\partial }}{\mathrm{\partial }{q}^{\alpha }}+{\ddot{q}}^{\alpha }\frac{\mathrm{\partial }}{\mathrm{\partial }{\dot{q}}^{\alpha }}}$ and use coordinate expressions for the Lie derivative to see equivalence with coordinate expressions of the Euler Lagrange equation. The coordinate free form is particularly suitable for geometrical interpretation of the Euler Lagrange equations.