What is the logic behind Lagrangian mechanics?

In summary, the Euler-Lagrange equations are derived by assuming that the action is minimized or stationary. One defines the Lagrangian for a given system and then takes the time integral of it to obtain the action. This principle is used in various fields, including classical mechanics and electromagnetism, where the Lagrangian approach is a useful alternative to using Newton's laws. However, in more complex systems like General Relativity and Quantum mechanics, the Lagrangian must be guessed or postulated.
  • #1
snoopies622
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I like using the Euler–Lagrange equations to solve simple mechanical systems, but I'm not perfectly clear on the theory behind it. Is it derived by assuming that action is minimized/stationary? Or does one define a system's Lagrangian according to what makes the Euler–Lagrange equations correct, and then define action as the time integral of this Lagrangian? I would like a broad answer that applies not just to classical mechanics but to wherever the Euler–Lagrange equations are used. Thank you!
 
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  • #2
snoopies622 said:
I like using the Euler–Lagrange equations to solve simple mechanical systems, but I'm not perfectly clear on the theory behind it. Is it derived by assuming that action is minimized/stationary?
That's exactly how it is derived.
 
  • #3
How then does one define the action for a given system? I'm afraid there's a circular reasoning here.
 
  • #4
snoopies622 said:
How then does one define the action for a given system? I'm afraid there's a circular reasoning here.
As the time integral of the Lagrangian.
 
  • #5
That's what I mean. If the action is the time integral of the Lagrangian, that implies that one has to know what the Lagrangian is first. Why then, would one start deriving the Euler-Lagrange equation with the physical assumption that the action is minimized without even knowing what the action is?
 
  • #6
So the Lagrangian is the difference between Kinetic and Potential Energy ##L = T - V##. The action is its time integral.

So when you extremize the action you are providing a condition for the path, i.e. out of all possible trajectories only the one with an extremized value for the time integral of ##T - V## occurs.

In many cases the extrema is a minima, so you could view it in that common case as the path that occurs is the one that keeps the time average of the difference between potential and kinetic energy to a minimum.
 
  • #7
snoopies622 said:
Why then, would one start deriving the Euler-Lagrange equation with the physical assumption that the action is minimized without even knowing what the action is?
Are you asking from the historical point of view? Which was first, this chicken or the egg?
 
  • #8
Not historically, just logically. What should i regard as an axiom or first principle, such that from it I can derive the rest? I seem to be arriving at a logical circle. Specifically:

— The Lagrangian of a system is chosen in such a way that it makes the Euler-Lagrange equations work.

— It makes sense to do this if one already assumes that the Euler-Lagrange equations in some way reflect reality.

— To derive the Euler-Lagrange equations, one assumes that the action is stationary.

— In order to apply the principle that the action is stationary, one needs to know what the action is.

— Since the action is the time integral of the Lagrangian, one must therefore first know what the Lagrangian is.

— But the Lagrangian itself is chosen assuming that the Euler-Lagrange equations are correct.

So I see a kind of Escher staircase of reasoning here.
 
  • #9
You know what the kinetic energy is, you reason what the potential energy is. The Least Action principle is then a condition that selects out the path that occurs given the kinetic and potential energies. There's no circularity.
 
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  • #10
Euler-Lagrange equations with L=T-V can be derived from the 2Newton law under certain additinal assumptions.
 
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  • #11
Yes, setting the Lagrangian equal to the difference between kinetic and potential energy and applying the Euler-Lagrange equation yields F=ma. Likewise there is apparently at least one way to define a Lagrangian such that the Euler-Lagrange equations creates Maxwell's equations.

https://quantummechanics.ucsd.edu/ph130a/130_notes/node452.html
In both cases, one starts with the physical law or laws that one wants to arrive at, and chooses the Lagrangian accordingly. Then taking the time (or spacetime) integral of this Lagrangian produces the action, and one can then say that this action is minimized (or extremized, in any case) for the correct dynamic equations of the system.

But to me this seems very different than starting with the minimization/extrem-ation of the action as a physical principle from which one can arrive at the correct physical laws in the first place. That's like taking a shuffled deck of cards, pulling out all the diamonds, placing them on top of the deck, and then saying that "diamonds are always on the top of the deck" is a law of nature.
 
  • #12
snoopies622 said:
But to me this seems very different than starting with the minimization/extrem-ation of the action as a physical principle
nobody forces you to follow Landau and Lifshitz :)
 
  • #13
snoopies622 said:
Yes, setting the Lagrangian equal to the difference between kinetic and potential energy and applying the Euler-Lagrange equation yields F=ma. Likewise there is apparently at least one way to define a Lagrangian such that the Euler-Lagrange equations creates Maxwell's equations.

https://quantummechanics.ucsd.edu/ph130a/130_notes/node452.html
In both cases, one starts with the physical law or laws that one wants to arrive at, and chooses the Lagrangian accordingly. Then taking the time (or spacetime) integral of this Lagrangian produces the action, and one can then say that this action is minimized (or extremized, in any case) for the correct dynamic equations of the system.

But to me this seems very different than starting with the minimization/extrem-ation of the action as a physical principle from which one can arrive at the correct physical laws in the first place. That's like taking a shuffled deck of cards, pulling out all the diamonds, placing them on top of the deck, and then saying that "diamonds are always on the top of the deck" is a law of nature.
In classical mechanics and electromagnetism, the classical laws of nature came first. The Lagrangian could be seen as a useful alternative. It certainly does in many cases make things easier than using Newton's laws directly.

But, in both General Relativity and Quantum mechanics, we do not have Newton's laws; or any obvious laws of nature. And there is no direct generalisation of Newton's laws.

The Lagrangian approach, however, does generalise: although you do have to guess or postulate the Lagrangian or Hamiltonian in each case.

In GR, for example, the principle is extremal proper time, which is a guess of sorts based on the fact that it is true in SR, as an alternative to Newton's first law.

Also, initially Newton's laws were essentially an educated guess. The law of gravitation could have been an inverse cube law, for example.
 
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  • #14
Thanks PeroK, that's very helpful! It reminds me - I have a copy of Dirac's Lectures On Quantum Mechanics from around 1964, in which he talks rather abstractly (at least to me) about what a new physical theory must look like, and he said that it should use the Hamiltonian/Lagrangian approach, because keeping the spacetime length/action invariant is a good unifying premise.
 
  • #15
Specifying the Lagrangian in Lagrange mechanics is essentially on par with specifying how different forces behave in Newtonian mechanics. By itself, Newton's second law ##F = dp/dt## does not tell you anything unless you properly model what ##F## and ##p## are. Physics is guesswork and an attempt to find a description of different systems that is as accurate as possible. Then we have the idea that we can work with guiding principles that have proven useful in the past when setting up a Lagrangian.

wrobel said:
Euler-Lagrange equations with L=T-V can be derived from the 2Newton law under certain additinal assumptions.
And vice versa ... However, Lagrange mechanics does not by itself need this. It is easy to consider cases where the Lagrangian does not take this form (they will just not have a direct equivalent in Newtonian mechanics).

snoopies622 said:
Yes, setting the Lagrangian equal to the difference between kinetic and potential energy and applying the Euler-Lagrange equation yields F=ma. Likewise there is apparently at least one way to define a Lagrangian such that the Euler-Lagrange equations creates Maxwell's equations.

https://quantummechanics.ucsd.edu/ph130a/130_notes/node452.html
In both cases, one starts with the physical law or laws that one wants to arrive at, and chooses the Lagrangian accordingly. Then taking the time (or spacetime) integral of this Lagrangian produces the action, and one can then say that this action is minimized (or extremized, in any case) for the correct dynamic equations of the system.

But to me this seems very different than starting with the minimization/extrem-ation of the action as a physical principle from which one can arrive at the correct physical laws in the first place. That's like taking a shuffled deck of cards, pulling out all the diamonds, placing them on top of the deck, and then saying that "diamonds are always on the top of the deck" is a law of nature.
That is just for cases where you want to show that the Lagrange approach is equivalent to (gives you the same equations of motion as) an already known theory. You could also have gone in the other direction and started with a Lagrangian that defines your model and then shown that it gives you some set of equations of motion. In fact, that is how physics works to a large extent. For example, as a high-energy theorist I like thinking about extensions to the standard model of particle physics. I have some guiding principles (such as gauge invariance) and based on what I am trying to achieve (such as constructing a theory with a new dark matter candidate) I can write down Lagrangians that will accomplish that. It is then up to experiment to test whether or not those models are accurate or not. Often you can put pretty heavy constraints or even rule them out just based on experiments that have already been performed. (Or Weinberg already did it in the 70s ...)
 
  • #16
Orodruin said:
However, Lagrange mechanics does not by itself need this. It is easy to consider cases where the Lagrangian does not take this form (they will just not have a direct equivalent in Newtonian mechanics).
Sure, I just wanted to say that it is a bad idea to deduct Newtonian mechanics from variational principle.
 
  • #17
Thanks all, a follow-up if I may: If action is defined as the time integral of a Lagrangian and this action is at a local extreme, then the Euler-Lagrange equations are true. Does this implication go in the opposite direction as well? That is, given the above definition of action, do the Euler-Lagrange equations imply that the action is stationary? I'm guessing yes but I haven't found a proof of this. (And I regret never having taken a course in the calculus of variations!)
 
  • #18
snoopies622 said:
Thanks all, a follow-up if I may: If action is defined as the time integral of a Lagrangian and this action is at a local extreme, then the Euler-Lagrange equations are true. Does this implication go in the opposite direction as well? That is, given the above definition of action, do the Euler-Lagrange equations imply that the action is stationary? I'm guessing yes but I haven't found a proof of this. (And I regret never having taken a course in the calculus of variations!)

As you can see here, the Euler-Lagrange equation provides only a necessary condition for action to have a weak extremum. Surely it is possible to find a mathematical example when this is not a sufficient condition (this excellent, originally Soviet encyclopedia writes that the condition is necessary), but I will not look for a counter-example.

Your question originally was that the action was defined through the solution of the Euler-Lagrange equation, while the Euler-Lagrange equation was derived so that it was a necessary condition for the action to have a weak extremum, leads to a contradiction. I think you have to think differently here.

Based on Newton's laws, they discovered that the S action function can be calculated using the difference between kinetic and potential energy, and the Lagrange function satisfies the Euler-Lagrange equation. So everything here was built on the Newton laws. Later, as can be seen from the encyclopedia article, it was possible to give the mechanics an entirely different conceptual structure, in which the Newton laws already appear as consequences.

Newton's laws are only reasonable hypotheses. In the new structure of mechanics, the hypotheses are different.
 
  • #19
Thanks Periwinkle, I guess understanding the history of these developments should help me see how all the parts fit together.
 
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  • #20
Apart from satisfying the EL equations, you may have to consider variations on the boundary to make sure that the action is stationary. If you impose boundary conditions, this will not be the case because the variations will be fixed to zero there. However, if you have free boundary conditions, you generally will end up with natural boundary conditions.
 
  • #21
snoopies622 said:
Thanks Periwinkle, I guess understanding the history of these developments should help me see how all the parts fit together.
I also agree with this maximum. There is a fantastic book on this topic, Cornelius Lanczos, The Variational Principles of Mechanics, in several editions. Maybe I'll review this book now to pick up what I've learned before. And the logical pitfalls can be spoken here or on another question.

I remember listening to a lecture on theoretical mechanics at the university before, and I was surprised when the teacher told students about a theorem that he didn't think the problem was clear in theory. (Yet I thought Landau-Lifshitz would answer everything unquestionably.) Of course, I have read in the book of Arnold, V.I., Mathematical Methods of Classical Mechanics ever since, that there is actually an error in Landau-Lifchitz, Classical Mechanics, which is far from insignificant.

In any case, the Cornelius Lanczos book is worth a look. And, of course, the classic Feynman university lecture, which the great Feynman intended to entertain, is definitely worth reading.
 
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  • #22
Well, you are looking for a path that is an extremum between points usually. So, you have some functional of the form ##S = \int^a_b y\{f(x), f'(x);x\} dx## , and you're looking for some path ##f(x)## that is an extremum on this path. Now, if it's an extremum, then if I vary this path, it should make my equation increase *if I have found the right path*.

So, we have to parameterize! I.e ##f(x) \rightarrow f(x,a)## We then say that if ##a = 0##, then ##f(x,a) = f(0,x) + ag(x) ## where g(x) is continuous, and vanishes at our end points. This changes our ##S## as well, so we now write it as ##S \rightarrow S(a) = \int^a_b y\{f(x,a), f'(x,a); x\} dx## Now, the condition that our functional is an extremum is just as you'd expect! A derivative that's set to zero! i.e ##\frac{\partial S}{\partial a}|_{a=0} = 0## This is a necessary condition though.

In other words, in order for this integral to have a "stationary value", S has to be independent of our parameter ##a## for all functions ##g(x)##! This is why we're even able to derive the EL equations, because of that condition. If you haven't worked through it, I suggest you derive them. Your end result will be ##\frac{\partial S}{\partial a} = \int^a_b (\frac{\partial y}{\partial f}-\frac{d}{dx}\frac{\partial y}{\partial f'})g(x) dx## and so here is where our condition comes into play! because ##\frac{\partial S}{\partial a}|_{a=0} = 0## then ##g(x)## must vanish, so the EL fall out: ##\frac{\partial y}{\partial f}-\frac{d}{dx}\frac{\partial y}{\partial f'} = 0## in order for f(x) to be an extremum which, once again, is a necessary condition.

Another mathematical language to look into is the subject of exterior calculus, and you can also apply the logic here with that. You can look into "the Palatini variation" if you want some GR to spice up your life (although it's a lot harder than it looks if you ask me). I'm not sure if this is what you're looking for, but I typed all of this up so I might as well post.
 
  • #23
romsofia said:
In other words, in order for this integral to have a "stationary value", S has to be independent of our parameter ##a## for all functions ##g(x)##! This is why we're even able to derive the EL equations, because of that condition. If you haven't worked through it, I suggest you derive them. Your end result will be ##\frac{\partial S}{\partial a} = \int^a_b (\frac{\partial y}{\partial f}-\frac{d}{dx}\frac{\partial y}{\partial f'})g(x) dx## and so here is where our condition comes into play! because ##\frac{\partial S}{\partial a}|_{a=0} = 0## then ##g(x)## must vanish, so the EL fall out: ##\frac{\partial y}{\partial f}-\frac{d}{dx}\frac{\partial y}{\partial f'} = 0## in order for f(x) to be an extremum which, once again, is a necessary condition.
This is not the correct argumentation. It is not that the variation ##g(x)## must vanish, it is that the action is stationary regardless of what ##g(x)## is that leads to the EL equations. If you had to choose ##g(x) = 0## then there would be no reason to have the EL equations, but the point is that ##g(x)## is arbitrary and therefore the EL equations have to hold.

However, you have also omitted the boundary terms that result from the partial integration, which are important when the function is not fixed at the boundaries (which does happen). The end result, in your notation, should be
$$
\frac{d S}{da} = \int^a_b \left(\frac{\partial y}{\partial f}-\frac{d}{dx}\frac{\partial y}{\partial f'}\right)g(x) dx
+ \left[\frac{\partial y}{\partial f'} g(x)\right]_b^a.
$$
Now, if the function is restricted such that variations at the boundary vanish, then the boundary contributions vanish directly. However, if variations on the boundary are allowed, then ##dS/da = 0## leads you to natural boundary conditions.
 
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  • #24
Orodruin said:
However, you have also omitted the boundary terms that result from the partial integration, which are important when the function is not fixed at the boundaries (which does happen). The end result, in your notation, should be
$$
\frac{d S}{da} = \int^a_b \left(\frac{\partial y}{\partial f}-\frac{d}{dx}\frac{\partial y}{\partial f'}\right)g(x) dx
+ \left[\frac{\partial y}{\partial f'} g(x)\right]_b^a.
$$
I'm so use to the boundary integral going to zero, I didn't think to write it. But, you're right! Should have included it.
 

Related to What is the logic behind Lagrangian mechanics?

1. What is Lagrangian mechanics?

Lagrangian mechanics is a mathematical framework used to describe the motion of objects in classical physics. It is based on the principle of least action, where the motion of an object is determined by minimizing the difference between its kinetic and potential energy.

2. How is Lagrangian mechanics different from Newtonian mechanics?

Lagrangian mechanics differs from Newtonian mechanics in that it uses a more general approach to describe the motion of objects. Instead of using forces and accelerations, it uses a single function, the Lagrangian, to describe the system. This allows for a more elegant and efficient way of solving problems in classical mechanics.

3. What is the role of the Lagrangian in Lagrangian mechanics?

The Lagrangian is a function that encapsulates all the information about a system's kinetic and potential energy. It is used to derive the equations of motion for a system, which can then be solved to determine the behavior of the system over time.

4. What are the advantages of using Lagrangian mechanics?

One advantage of using Lagrangian mechanics is that it allows for a more systematic and elegant approach to solving problems in classical mechanics. It also provides a more general framework that can be applied to a wide range of physical systems, making it a powerful tool for scientists and engineers.

5. Are there any limitations to Lagrangian mechanics?

While Lagrangian mechanics is a powerful tool, it does have some limitations. It is most useful for systems that can be described by a finite number of degrees of freedom. It also does not take into account quantum effects, so it is not applicable to systems on a very small scale.

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