Trying2Learn said:
Summary: Why is it not magic?
The reason why it is not magic
can be explained.
Granted: it looks as if it is impenetrable, but if you look at it at just the right angle it becomes transparent.
The thing is: in the usual presentation the Euler-Lagrange equation is derived using integration by parts. The integration by parts does the job, but it has a disadvantage: it does not offer any clue as to
why Hamilton's stationary action holds good.
The following is an important clue:
The Euler-Lagrange equation is a differential equation.
At that point you go:
"Hang on, Hamilton's action is expressed in the form of evaluating an
integral, but the Euler-Lagrange equation is a
differential equation. What happened to the integration?"
There is an interesting combination of properties there: Calculus of Variations is formulated in
integral form, yet it is solved with a
differential equation. Understanding how that combination comes about is key.As we know, it was in the wake of the Brachistochrone challenge that Calculus of Variations was developed. Johann Bernoulli had issued the Brachistochrone challenge, and his older brother Jacob Bernoulli was among the mathematicians who was able to solve it. This means Jacob Bernoulli solved the problem without having Calculus of Variations, nor any precursor of it.
Jacob Bernoulli recognized a particular feature of the brachistochrone problem, and he presented that feature in the form of a lemma:
Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.
(
Acta Eruditorum, May 1697, pp. 211-217)Let me rephrase what Jacob Bernoulli had recognized:
Take the solution to the Brachistochrone problem, and take a arbitrary
subsection of that curve. That subsection is also an instance of solving the brachistochrone problem. This is valid at any scale; down to arbitrarily small scale. So you can treat the problem as a concatenation of arbitrarily small subsections.
This informed Jacob Bernoulli: a differential equation exists that solves the Brachistochrone problem. Of course, it did not provide a tangible clue what that differential equation is. I like to think that knowing that it
must exist gave Jacob Bernoulli the perseverence to carry the problem to the end.The condition of the
derivative of the integral being zero is a remarkably constraining condition.
Jacob's lemma, generalized:
In order for the derivative of ##\int_{x_1}^{x_2}## to be zero: for all individual subsections between ##x_1## and ##x_2##, down to arbitrarily short subsections, all the respective derivatives of the corresponding subsection integrals must be zero concurrently.
So:
The Euler-Lagrange equation can also be derived using
differentiation operations only.
There are a couple of sources where that is done. For example, in their book 'Calculus of Variations' Gelfand and Fomin offer a derivation of that type.
I recommend the derivation of the Euler-Lagrange equation by Preetum Nakkiran:
https://preetum.nakkiran.org/lagrange.html
Preetum Nakkiran uses the Catenary problem as motivating example.
The catenary is a curve. Divide the x-axis in equally spaced intervals. Name the successive x coordinates: ##x_0##, ##x_1##, ##x_2##, ##x_3##, ##x_4##, etc.
For the following triplet: ##x_0##, ##x_1##, ##x_2##
##x_0## and ##x_2## are treated as fixed points, and variation of ##x_1## is applied.
Generalized along the entire length of the curve:
Triplet: ##x_n##, ##x_{n+1}##, ##x_{n+2}##
##x_n## and ##x_{n+2}## are treated as fixed points, and variation of ##x_{n+1}## is applied.
So that is a strategy of
concatenation of a set of equally spaced arbitrarily short subsections.
Hamilton's stationary action
As we know: ##F=ma## can be recovered from Hamilton's stationary action. Hamilton's stationary action looks totally different from ##F=ma##, yet ##F=ma## can be recovered from it. The question is: how does that come about?
We need to find intermediary steps.
I regard the Work-Energy theorem as the Rosetta stone of Hamilton's stationary action.
The derivation of the Work-Energy theorem from ##F=ma## is straightforward, so that is a solid relation.
Hamilton's stationary action and the Work-Energy theorem have the following in common: both express the physics taking place in terms of potential energy and kinetic energy. Hamilton's stationary action and the Work-Energy theorem are closely related; there is simply no room for them
not to be closely related.
In december 2021 I posted an answer with the following:
-Derivation of the Work-Energy theorem from F=ma
-Demonstration that in cases where the Work-Energy theorem holds good Hamilton's stationary action will hold good also.
https://www.physicsforums.com/threads/stationary-point-of-variation-of-action.1009770/#post-6571395
(In fact, that december 2021 post was a reply to a thread started by you.)The relation between Hamilton's stationary action and the Work-Energy theorem hinges on Jacob's Lemma. I cannot emphasize the importance of Jacob's lemma enough.