Is the Euler-Lagrange equation a special case of the principle of least action? I have some confusion after reading a few dozen stackexchange articles of the "principle of least action".

I follow the derivation of the Euler-Lagrange equation , it appears to treat the functional in a generalized geometric and algebraic fashion. But now here I get lost after the derivation.

Is the principle of least action an application of the Euler-Lagrange equation with the functional being the Lagrangian ( i.e. the difference between the kinetic and potential energy ) as applied to some specific case?

up vote 2 down vote accepted

If we have functional defined as $$S({\boldsymbol {q}})=\int _{t_0}^{t_1}L(t,{\boldsymbol {q}}(t),{\boldsymbol {\dot {q}}}(t))\,\mathrm {d} t$$ Then we have a theorem that says that a function $\mathbf{q}$ for which this functional is stationary must satisfy the Euler-Lagrange equations.

Now the principle of least action simply states that physical system must evolve in such a way between times $t_0$ and $t_1$ such that the action (which is functional with the same form as above where $L$ is the Lagrangian and $\mathbf q$ are the generalized coordinates) must be stationary. Now we can use the above theorem to say that $\mathbf q$ will satisfy the Euler-Lagrange equation which gives a more practical way to find such function by solving this equation.

  1. The Euler-Lagrange (EL) expression $\frac{\delta F[\phi]}{\delta \phi^{\alpha}(x)}$ is the functional/variational derivative of a functional $F[\phi]$.

  2. The usual$^1$ principle of least action (which more precisely should be called the principle of stationary action) asks for $\phi$-configurations with vanishing functional derivatives $\frac{\delta S[\phi]}{\delta \phi^{\alpha}(x)}\approx 0$ of the action functional $S[\phi]$, i.e. $\phi$-solutions to the EL equations.

    So to answer OP's title question (v2): The EL equations follow from the principle of stationary action.

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$^1$ Be aware that some authors (e.g. Goldstein) attach a different meaning to the principle of least action, cf. my Phys.SE answer here.

The Euler-Lagrange equations solve a particular variational problem where you want to extremize a functional is of the form

$$ F = \int_a^b f(x(t), \dot{x}(t), t) dt $$

One such example for $f$ is the Lagrangian $L$, in which case $F$ would be the action $S$. From my point of view, this is the principle of least action and it is solved by the Euler-Lagrange equations for a particular choice of the function under the integral.

Something that seems to vary from person to person is what exactly the Euler-Lagrange equations are. Certainly the solution $\delta F = 0$ for the above functional is the Euler-Lagrange equation, but some people define the Euler-Lagrange equations to be the solution to any function under the integral.

One case of particular interest for physics is the case that the Lagrangian may be defined in terms of a Lagrangian density $\mathcal{L}$ as $L = \int \mathcal{L} d^3x$. When this is the case and you vary the action, you get what are essentially equations of motion for a field. You can also take the function under the integral to have multiple space coordinates $x^1, x^2, ...$ to obtain equations of motion for a system of particles. You can also choose to include higher-order derivatives in the function, but I'm not aware of any use of that in physics. All three of these are fairly straightforward once you understand varying a functional and are a decent exercise to try out.

  • "it is possible to think up Lagrangians with higher-order derivatives. In such a case, the EL equations would likely no longer be the solution" - This is incorrect. The EL solution does not require for the functional to have no dependence on higher derivatives. You are confusing the cause and result here. The fact that Lagrangians (and consequently all physical systems) depend only on first derivatives is a result of the least action principle, not a cause for it. – safesphere Jul 17 at 4:46
  • @safesphere I was thinking of Euler-Lagrange equations as just $\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = 0$, which isn't the case if $L$ has higher-order derivatives. I'd imagine that calling the higher-order generalization EL equations as well. I'll edit my answer to clarify that. I'm not sure what you mean by Lagrangian depending only on first order as a result of least action unless that's due to wanting 2nd order EOM. – danielunderwood Jul 17 at 19:11
  • Let me illustrate on a simpler example. Consider a differentiable function f(x). The only condition for its extremum is that the first derivative is zero. It does not matter how the higher derivatives behave. So any condition you put on, say, the second derivative does not change the fact that the extremum is defined only by the first derivative. Same for the Lagrangian. It doesn't matter what it depends on itself. The condition for the extremum of the functional only contain first derivatives. The EL equations hold true for any Lagrangian regardless of which derivatives it depends on. – safesphere Jul 17 at 21:03
  • @safesphere it's true that it would only depend on the first derivative of $f$, just as a Lagrangian with any $q^{(n)}(t)$ where $n$ represents differentiation would have a dependence on $\frac{\partial L}{\partial q^{(n)}}$; however, any such term would have a coefficient of $(-1)^n \frac{d^n}{dt^n}$, changing the EL equations. There are many higher order terms such as $2 \dot{q} \ddot{q} = \frac{d}{dt} \dot{q}^2$ that could be eliminated by the freedom to add a total derivative, but that wouldn't always be the case. Is there something I'm missing here? – danielunderwood Jul 18 at 0:46
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    @Sedumjoy Note that I have reworked my answer a bit. I meant to weeks ago, but missed it. The Euler-Lagrange equations are the equations for which $\delta F = 0$ Where $\delta F = F[f + \delta f] - F[f]$ is the variation of $F$. The $\delta f$ is a slight change in $f$ and is often given by an expresion like $\delta f = \epsilon \eta(t)$ where $\epsilon$ is small and $\eta$ is an arbitrary function. – danielunderwood Aug 29 at 1:01

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