0

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?

3 Answers3

1

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.

gcc-6.0
  • 148
0

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.

Daniel Underwood
  • 647
  • "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 '18 at 04:46
  • @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. – Daniel Underwood Jul 17 '18 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 '18 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? – Daniel Underwood Jul 18 '18 at 00:46
  • It may also be worth noting for anyone reading that there would also be a boundary term coming in in addition to the $(-1)^n \frac{d^n}{dt^n}$ terms, but I'm assuming those vanish for simplicity. – Daniel Underwood Jul 18 '18 at 00:49
  • Take a functional as an integral of an unknown function. Then the LE formula gives you a differential equation to find the function that takes the functional to the extremum. This equation contains first derivatives. There is no requirement for the function to depend or not depend on the first or higher derivatives. The fact that the higher derivatives are irrelevant comes from the LE equation, not from how you define the Lagrangian. The equation is valid for any Lagrangian. It may just be hard to solve the equation for the Lagrangians written with higher derivatives. – safesphere Jul 18 '18 at 02:06
  • Let us continue this discussion in chat. – Daniel Underwood Jul 18 '18 at 03:11
  • Sorry, I don't chat. I've made my point. It is up to you to decide if it is helpful or not. Good luck! – safesphere Jul 18 '18 at 04:39
  • @danielunderwood ...could you please clarify the relationship between the Euler-Lagrange equation and the function F you noted above in your first equation. Sufficient condition ? necessary condition? sufficient and necessary or equivalent. How is the Euler-Lagrange used ? Is it used to find the function x(t) ? –  Aug 28 '18 at 21:51
  • 1
    @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. – Daniel Underwood Aug 29 '18 at 01:01
  • ahh .good...I also found something useful on YouTube if you are interested "Vid1 " Calculus of Variations Derivation of the Euler-Lagrange Equation and the Beltrami Identity" ...14:56 he seems to indicate Euler-Lagrange is necessary but not sufficient. I need to compare a few examples side by side but am slow ....thank you –  Aug 29 '18 at 15:35
0

  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.

--

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

Community
  • 1
Qmechanic
  • 201,751