Timeline for Is the Euler-Lagrange equation a special case of the principle of least action?
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| Aug 29, 2018 at 15:35 | comment | added | user86411 | 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, 2018 at 1:01 | comment | added | Daniel Underwood | @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. | |
| Aug 29, 2018 at 0:57 | history | edited | Daniel Underwood | CC BY-SA 4.0 |
Rework answer
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| Aug 28, 2018 at 21:51 | comment | added | user86411 | @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) ? | |
| Jul 18, 2018 at 4:39 | comment | added | safesphere | 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! | |
| Jul 18, 2018 at 3:11 | comment | added | Daniel Underwood | Let us continue this discussion in chat. | |
| Jul 18, 2018 at 2:06 | comment | added | safesphere | 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. | |
| Jul 18, 2018 at 0:49 | comment | added | Daniel Underwood | 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. | |
| Jul 18, 2018 at 0:46 | comment | added | Daniel Underwood | @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? | |
| Jul 17, 2018 at 21:03 | comment | added | safesphere | 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. | |
| Jul 17, 2018 at 19:11 | comment | added | Daniel Underwood | @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. | |
| Jul 17, 2018 at 4:46 | comment | added | safesphere | "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. | |
| Jul 16, 2018 at 20:46 | history | answered | Daniel Underwood | CC BY-SA 4.0 |