Hamilton's Principle (HP), Lagrangian

[Trying2Learn]

TL;DR Summary

Why is it not magic?

I understand the process of the calculus of variations.
I accept that a proper Lagrangian for Dynamics is "Kinetic minus potential" energy.
I understand it is a principle, the same way F=ma is a law (something one cannot prove, but which works)

Still... what do you say to students who say "I get the Calculus of Variations and HP, but it feels like magic."

Do you say "It is because you have been indoctrinated with vectorial dynamics and you have never seen analytical dynamics?"

Even in Feynman's notes, he talks about how he was startled by this and how he reasoned that it was natural and good.

Still... the students say "it seems like magic?"

Without rationalizing why the Lagrangian SHOULD be L = T- V, what would you say to people who say (at first glance) it seems like magic?

 

[Dale]

“Any sufficiently advanced technology is indistinguishable from magic” - Arthur C Clarke

In any case, the job of a physics instructor is to teach the students physics. I don’t think we should be telling them how they should feel about it. A sense of wonder or magic isn’t something to “fix”. Let them feel what they feel, as long as they learn the physics too.

 

[Trying2Learn]

“Any sufficiently advanced technology is indistinguishable from magic” - Arthur C Clarke

In any case, the job of a physics instructor is to teach the students physics. I don’t think we should be telling them how they should feel about it. A sense of wonder or magic isn’t something to “fix”. Let them feel what they feel, as long as they learn the physics too.

Thank you! Beautiful answer!

 

[Cleonis]

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.

 

[Cleonis]

A sense of wonder or magic isn’t something to “fix”.

Well, it depends.

Let me make a comparison:
After rain has passed quite often we see a rainbow. For as long as humans existed the view of a rainbow must have appeared as something magical. It was always attributed to divine intervention.

But then came the time that scholars started to understand the properties of refraction. They could see in their own study how light is refracted at the water-air boundary. Once that understanding was there the scholars had the means to see that rainbows arise from elementary properties of refraction; no divine intervention required.
The aim of the discipline of physics is to try and find whether some phenomenon that presents itself can be understood in terms of elementary properties.

We don't know in advance which things that are encountered lend itself to be understood in terms of elementary properties. Lavoisier and his contemporaries were under the impression that heat is a thing of its own, and they named it 'Caloric'. But then came the time that physicist realized that heat can be understood in terms of kinetic energy of atoms/molecules.

So:
As a physicist you need to keep trying, you need to keep pushing.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 is far from transparent what is happening.

My answer in this thread addresses that:
https://www.physicsforums.com/threads/hamiltons-principle-hp-lagrangian.1044926/#post-6794862

 

[Dale]

The aim of the discipline of physics is to try and find whether some phenomenon that presents itself can be understood in terms of elementary properties.

Sure. And if a student understands that then they have learned the physics. It is not necessary that they also abandon any sense of wonder or magic. When you listen to the great physics teachers, it is clear that they have preserved and even deepened their own emotional reaction to the subject.

As a professional physics teacher I try to communicate my own emotional connection to the topic (excitement and enjoyment) and would never dissuade them from theirs (even negative ones).

 

[Cleonis]

[...] A sense of wonder or magic [...]

I notice that my previous message did not accomplish my intention.

Let me try this example:
Frozen water takes up more volume than the same mass of water in liquid form. That is unusual, and presumably physicist were puzzled by the phenomenon.

(I just found out there are a couple of other substances with that property:
https://www.sciences360.com/index.php/substances-that-expand-when-they-freeze-24357/ )

This sense of puzzlement triggered by a specific observation is independent from a more general sense of awe inspired by things that impress us.

The sense of puzzlement incites the physicist to try and figure it out. With our current understanding of the structure of the water molecule we can account for the phenomenon that as water freezes it expands.My point is: in physics we should never assume from the outset that something is inexplicable.

Hamilton's stationary action can be explained within the realm of Newtonian mechanics. It requires peeling off several layers, but it can be done.

For the first layer:
https://www.physicsforums.com/threads/hamiltons-principle-hp-lagrangian.1044926/#post-6794862

 

[Dale]

in physics we should never assume from the outset that something is inexplicable

And I never suggested the contrary.

 

[Cleonis]

And I never suggested the contrary.

Here is how I came to include that 'in physics we should never' statement.

In your first reply in this thread you gave the quote from Arthur C. Clarke: "Any sufficiently advanced technology is indistinguishable from magic"

It could be that in mentioning that quote you wanted to convey that in your assessment Hamilton's stationary action resides at a level sufficiently advanced that it cannot be accounted for in terms of Newtonian mechanics.For Hamilton's stationary action:
(I use the name 'Hamilton's stationary action' for the scope as introduced by Willliam Rowan Hamilton in 1830. I limit my discussion to Hamilton's stationary action. I make no claims about the more general concept that is referred to by the name 'Principle of stationary action'.)

The usual presentation is to recover F=ma from Hamilton's stationary action.

At this point let me quote physics.stackexchange contributor Kevin Zhou:

[...] in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X)."

The connection between F=ma and Hamilton's stationary action is an instance of that both-directions property. It is possible - without requiring additional assumption - to proceed from F=ma to Hamilton's stationary action.

In the case of the direction from F=ma to Hamilton's stationary action the Work-Energy theorem is the concept that connects the two.

 

[Dale]

It could be that in mentioning that quote you wanted to convey that in your assessment Hamilton's stationary action resides at a level sufficiently advanced that it cannot be accounted for in terms of Newtonian mechanics.

I just meant that the student’s sense of “magic” is perfectly compatible with learning the physics. So we should focus on teaching the physics and not on beating the “magic” out of our students. As I believe I clarified and explained with all of my subsequent posts.

I never stated nor implied that any important physics concept should be left unexplained. Once the student has mastered the physics, if the sense of magic remains, then let it be.

 

[vanhees71]

The direction from "F=ma" to Hamilton's principle simply is that in the cases, where the former is applicable, the latter provides the same equations of motion and provides more refined mathematical tools to investigate them. Particularly the possibility to analyze the symmetries of a mechanical system, given the Lagrangian or the Hamiltonian, leading to Noether's theorem, is a great advantage. In addition you can also treat problems, where a naive "F=ma" approach is pretty difficult (systems with constraints).

From a different point of view, the validity of Hamilton's principle can be understood from the classical limit of quantum mechanics. In the path-integral formalism, for situations where the typical values of the action are large compared to $\hbar$, the saddlepoint approximation is applicable, and then the classical trajectory, defined as the stationary point of the action, and small variations around are what contributes most significantly to the propagator describing the motion of the particle from a fixed initial to a fixed final position.

 

[vanhees71]

I just meant that the student’s sense of “magic” is perfectly compatible with learning the physics. So we should focus on teaching the physics and not on beating the “magic” out of our students.

Well, I think when learning physics you must give up a lot of "magic". E.g., a thunderstorm is not due to the anger of some gods but the result of some electrostatics and discharges, etc. That doesn't mean that there is not a lot of fascination in finding things out in the scientific way and to understand phenomena from "fundamental natural laws", discovered in this way.

 

[Motore]

Well, I think when learning physics you must give up a lot of "magic".

"Magic" in this thread meaning: awe, amazement, peculiarity, not the supernatural kind of magic.
P.S.: That's why the quotation marks around the word magic

 

[vanhees71]

As I said, this other kind of "magic" you don't have to give up.

 

[Cleonis]

[...] Hamilton's principle [...] provides more refined mathematical tools to investigate them.
[...] In addition you can also treat problems, where a naive "F=ma" approach is pretty difficult (systems with constraints).

(I'm replying only now because I revisited only now; my main activity is on physics.stackexchange)

We have that we can derive expression of physics taking place in terms of energy from F=ma

$$ \begin{array}{rcl}
F & = & ma \\
\int_{s_0}^s F \ ds & = & \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \\
-\Delta E_p & = & \Delta E_k
\end{array} $$

(The evaluation that arrives at the above result is in this post:
https://www.physicsforums.com/threads/stationary-point-of-variation-of-action.1009770/#post-6571395 )

We have of course that expressing the physics taking place in terms of interconversion of potential energy and kinetic energy is in many cases more efficient. I will use the expression Energy mechanics to refer in general to that.

In his work Mecanique Analytique (1789) Joseph Louis Lagrange introduced the following two practices:
- Systematic use of generalized coordinates
- Systematic use of Energy mechanics

Both of the above practices were developed decades prior to the introduction of Hamilton's stationary action (1830) by William Rowan Hamilton.

That is: in order to be able to develop the mechanics as introduced by Joseph Louis Lagrange it is not necessary to first formulate Hamilton's stationary action.

(About calculus of variations: Lagrange used calculus of variations to work out cases in Statics, but he didn't extend use of calculus of variations to cases in Dynamics. Maupertuis' action does get mentioned in Mecanique Analytique, with Lagrange stating he didn't think it was particularly relevant.)

About using generalized coordinates:
The concept of using generalized coordinates extends to the concept of force.
See for instance the discussion of generalized force by Richard Fitzpatrick.
So: when some problem involves constraints it can be restated it in terms of generalized force, with no need to involve more than that.

My point is: the 'more refined mathematical tools' that you refer to are already available, in the sense that generalized coordinates and energy mechanics are available anyway, independent of whether or not Hamilton's stationary action is introduced.

 

[vanhees71]

Of course you can use a lot of different principles to formulate classical mechanics. The most important principle is the least-action principle a la Lagrange and Hamilton, because it generalizes to all of the contemporary known physics on a fundamental level.

 

[Cleonis]

Of course you can use a lot of different principles to formulate classical mechanics. [...]

Well, the physics content of the various formulations is the same.

We have that the various formulations of classical mechanics are related by mathematical transformation.

As we know: the transformation between Lagrangian mechanics and Hamiltonian mechanics is Legendre transformation. Legrendre transformation is its own inverse; applying Legendre transformation twice recovers the original function.

The transformation from Newtonian mechanics to Energy Mechanics is integration with respect to the position coordinate:

$$ \begin{array}{rcl}
F & = & ma \\
\int_{s_0}^s F \ ds & = & \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \\
-\Delta E_p & = & \Delta E_k
\end{array} $$

To recover F=ma from Energy Mechanics we do the inverse operation: taking the derivative of kinetic energy with respect to the position coordinate recovers acceleration:

$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma $$

These mathematical transformations change mathematical content only.

Transformation to different mathematical form is analogous to change of perspective. Each mathematical form emphasizes particular characteristics and de-emphasizes others while leaving the object itself unchanged.

The various transformations between different mathematical forms are bi-directional.
Differential form and variational form

Transformation between differential form and variational form is also an invertable transformation.

The derivation of the Euler-Lagrange equation does not involve imposing a restriction on what classes of variational representation de derivation is valid for; the derivation is valid for any variational form.

That is: the very derivation of the Euler-Lagrange equation demonstrates that variational form can always be transformed to differential form.

The fact that variational form can always be transformed to differential form suggests the converse is valid too: that in physics anything that lends itself to being represented in differential equation form can be transformed to representation in variational form.

 

[Trying2Learn]

Well, the physics content of the various formulations is the same.

We have that the various formulations of classical mechanics are related by mathematical transformation.

As we know: the transformation between Lagrangian mechanics and Hamiltonian mechanics is Legendre transformation. Legrendre transformation is its own inverse; applying Legendre transformation twice recovers the original function.

The transformation from Newtonian mechanics to Energy Mechanics is integration with respect to the position coordinate:

$$ \begin{array}{rcl}
F & = & ma \\
\int_{s_0}^s F \ ds & = & \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \\
-\Delta E_p & = & \Delta E_k
\end{array} $$

To recover F=ma from Energy Mechanics we do the inverse operation: taking the derivative of kinetic energy with respect to the position coordinate recovers acceleration:

$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma $$

These mathematical transformations change mathematical content only.

Transformation to different mathematical form is analogous to change of perspective. Each mathematical form emphasizes particular characteristics and de-emphasizes others while leaving the object itself unchanged.

The various transformations between different mathematical forms are bi-directional.
Differential form and variational form

Transformation between differential form and variational form is also an invertable transformation.

The derivation of the Euler-Lagrange equation does not involve imposing a restriction on what classes of variational representation de derivation is valid for; the derivation is valid for any variational form.

That is: the very derivation of the Euler-Lagrange equation demonstrates that variational form can always be transformed to differential form.

The fact that variational form can always be transformed to differential form suggests the converse is valid too: that in physics anything that lends itself to being represented in differential equation form can be transformed to representation in variational form.

Thank you!

 

[strangerep]

The various transformations between different mathematical forms are bi-directional.

Not always. There exist "singular" Lagrangians. (Look up "Dirac Brackets" and his method for handling constrained mechanics. He did a lecture series on this, subsequently published as a book.)

The fact that variational form can always be transformed to differential form suggests the converse is valid too: that in physics anything that lends itself to being represented in differential equation form can be transformed to representation in variational form.

Alas, that is also not true in general. See the "Helmholtz Conditions", which are part of the Inverse problem for Lagrangian mechanics.

 

[apostolosdt]

A lot of natural phenomena look “magical” in a physics class and in fact they are. That’s expected, for we only understand a very tiny part. The opposite “arrogant” attitude is built up from attempting to comprehend Nature purely in mathematical terms.

Take, for instance, the “magical” negative sign in the Lagrangian. Can it be explained in physical terms? Since Feynman’s name already came up, may I remind that’s precisely what he does in Lecture 19, Vol. II, before his mathematical exposition?

 

[vanhees71]

The least-action principle can be understood from quantum theory in Feynman's path-integral formalism, including the "negative sign" in the Lagrangian, $L=T-V$, which comes from integrating out the canonical momenta in the canonical path integral, if the Hamiltonian depends only quadratically on the canonical momenta.

Note that this pattern is not true for the relativistic case, where for free particles
$$L=-m c^2 \sqrt{1-\dot{\vec{x}}^2/c^2} \neq T.$$