Brief overview of Lagrangian Mechanics

This is a video about Lagrangian Dynamics which is useful if you are about to take analytical mechanics but might be interesting even if you just like math or physics.
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[@valenfeldmann64]

Corrections: - While many people say "Least action", "Stationary action" is correct. - With the pendulum example, omega should be sqrt(g/l) (and that's what should be inside our sin function also) - The reason we can treat y and y' as independent has to do with the derivation of the Euler-Lagrange equation and is not related to us not knowing the whole path. (If I understand right) If you see anything else, feel free to let me know and I'll add it to the list. I'm really impressed with the comments I've seen!

[@cleon_teunissen]

I concur that it is the mathematical operation of partial differentiation that gives the impression that y and y' are treated as if they are independent (while in fact of course they are not). Let me use the catenary problem as example. The chain is of uniform density along its length. A section of chain with a steep slope has more weight per length-along-the-horizontal-axis than a section of chain with shallow slope. To evaluate the integral of the potential energy with respect to the horizontal axis: at every point multiply the height with the weight-per-unit-of-length-along-the-horizontal-axis. So there are two contributions to the integrand: the height at each point, and the slope at that point, which is y and y'. In order to get a measure of the effect of each of the two contributions you perform partial differentiation. Partial differentiation goes back to the product rule of differentiation. d(f(x)g(x))/dx=f'(x)g(x) + f(x)g'(x) Paraphrasing an earlier comment: Stationary action: the principle of least explanation.

[@probability_density]

This is a subtle point, but the physical path is not necessarily the one that minimizes the action. It is more accurate to say that the physical path is one with stationary action. This just means that for small enough perturbations to the physical path, the action does not change. In fact, the left-hand-side of the Euler-Lagrange equation can be thought of as the "derivative" of the action. The statement of the Euler-Lagrange equation, then, is that the physical paths are the "critical points" of the action.

[@cleon_teunissen]

I concur: stationary action. Also: yes, I regard the Euler-Lagrange expression as a differential operator, analogous to, say, the Laplacian operator. The physical path is the point in variation space such that the derivative of the kinetic energy matches the derivative of the potential energy

[@1ucasvb]

A more subtle point is that "stationary" is the correct term because it includes saddle points of the action. The action is never maximized (see Gray & Taylor "When action is not least")

[@samanthaqiu3416]

however local maxima are unstable under infinitesimal perturbations, but local minima are not. Lagrangian dynamics is not complete under uncertainty in the initial state

[@cleon_teunissen]

@@1ucasvb About that assertion 'is never maximized' by Gray & Taylor. That assertion is subject to a condition that involves narrowing down to a subsection of the trajectory. (Which I think is stated in the article.) With the condition: for a sufficiently small subsection the action is never maximized. That is quite a contrived condition. Also, Gray and Taylor do not quantity that 'sufficiently small'. I Implemented an interactive diagram with a case where the physical trajectory corresponds to a maximum of Hamilton's action. You sweep out variation by moving a slider; the diagram shows the response. There are multiple sliders, they make a range of options available. There is a slider for global variation, and there are sliders that allow more granular adjustment. When the slider for global variation sweep is moved the action is maximum for the physical trajectory.

[@goldenboy.]

I love your honesty here and your pursuit of developing an intuition. I too am trying to develop a deeper understanding of Lagrangian Mechanics, hope to see you there. Thanks for the great video.

[@naturallyinterested7569]

The principle of stationary action actually follows directly from the path integral formulation of quantum mechanics. There, the wave function Psi(x, t) that describes the probability of finding the object at x at time t is proportional to the sum of e^{iS} over all paths x_0 to x, where S is the action. Now imagine a path that doesn't have stationary action. On the quantum scale the change in S might be small, so (thinking of discretizing the integral into an infinite Riemann sum) most of these complex arrows will go into the same direction, resulting in a nonzero value for the wave function, resulting in a nonzero probability of finding the quantum particle at any other location than those its path of stationary action takes. But in the classical limit, big objects have large actions. So if we imagine a path of non-stationary action and look at the sum of arrows, we find that e^{iS} is all over the place, as S varies drastically over any path with non-stationary S. So it is just as likely for any vector to be in one direction or its opposite direction, and the integral for that path averages out to zero, meaning no contribution from that path for almost all paths in space from x_0 to . The only paths that contribute to the probability of finding a macroscopic object at (x, t) are those where S is (almost) stationary so the vectors share direction, and those paths are therefore the only paths on which we could ever observe it.. So macroscopic objects are only likely to be found on paths of (nearly) stationary action, and classical mechanics is then what we observe.

[@MDNQ-ud1ty]

Classical mechanics was derived before QM, not after. QM is simply an extension of CM where observations correspond to hermitian operators in a Hilbert space. [I'm making the point that it does not "follow" from QM except that one can go either way mathematically... but historically it didn't follow which I know you know but it isn't really good to explain a simpler theory with a more complex one derived from it(sort of circular logic)]

[@naturallyinterested7569]

@@MDNQ-ud1tyThat misses the point a bit, and I apologize for wording my comment in a bad way. I was trying to comment on why nature happens to obey the principle of stationary action. The point is that nature doesn't do classical mechanics. Nature only does QM (or whatever its unification with GR would be), that's the only game in town. So while from a historical point of view we might say that this "new theory of quantum mechanics" better be consistent with our old friend classical mechanics, from an objective point of view it's actually classical mechanics that better be consistent with QM in the classical limit, as it is just a simplifying approximation of the true nature of spacetime. So historically QM is derived from CM, but fundamentally CM is derived from QM, to the best of our knowledge.

[@MDNQ-ud1ty]

@@naturallyinterested7569 No, that is wrong. Nature doesn't do CM or QM or any *MODEL* that man creates. These are THEORIES not reality. Nature does nature and man creates models to try an have some way to approximate nature. QM is a probabilistic framework of mechanics and so obviously it will have certain features that make it better at explaining certain things(which probability is good at handling). But QM and CM are no more real than any other physical model. They are mathematical models that have been designed through trial and error to explain the phenomena that we experience. They work well because they use mathematics and because if they didn't work well they wouldn't be used. But just because they work well does not mean they are reality. QM is a generalization of CM and so you can derive CM from it... but this does not mean it is the only generalization of CM. Very likely there are much better models of reality but they simply are too complex for mans mind to comprehend at this point. My only other point is that you do not need to go in to QM to try and understand CM's. The additional complexity is not needed to understand least action and it ultimately will be no better of an answer than what CM says. What you are doing is saying QM is real and CM is an approximation. But in fact, both are approximations. One uses QM when works better, else one uses CM. Least action is basically the conservation of energy law encoded in mathematical form. There is no reason to try to prove it since it will be circular no matter what model it is. The fact is that nature, for all intents and purposes, conserves energy for whatever reason it does... a reason unlikely to ever be known. It will take more than a mathematical theory to prove such a thing. My point is not that you can't transfer the framework in to QM but that it doesn't serve the problem. You are removing the context one step away(in to the more general but probabilistic nature of QM) when it doesn't have to be. It doesn't strengthen the understanding but dilutes it. That is, it adds complexity without adding certainty and may detract it. Likely if one understood the process in QM they would then understand it in CM already so what you are doing is saying "lets complexify the issue to make it simpler to understand". Maybe it would work but it shouldn't be the approach. The approach should always be to keep it as simple as possible. Unless you could prove that the mathematical formulation of Lagrangian mechanics cannot logically express it's inherent meaning AND that QM truly does reflect nature then your just adding complexity through transporting the problem in to another framework. The problem will ultimately still exist. The only way I could accept it as being a possible valid solution is if the person had a much stronger background in QM in which case the language may be much easier for them to comprehend. this is unlikely to ever be the case since learning QM before learning CM is backwards and counterproductive. But maybe there are a few people that skipped learning CM... but if they did they likely also have issues understanding QM.

[@naturallyinterested7569]

@@MDNQ-ud1tyI agree that nature does not do our theories, I should have qualified that more. But I would say that QM is a much better fitted model, describing nature much closer than CM, which has some fundamental disagreements with observation (out of which QM grew historically). It is an observational fact that elementary particles cannot be constrained to the path of stationary action, our observations are clearly contradictory to that assumption. Furthermore, there seems to exist a classical limit at which objects do obey that rule, which is what we first observed, so we kind of have it backwards, as we just happen to live on those scales. And I take those observations as my justification to say that fundamentally the principle of stationary action in CM must be, to the best of our knowledge, a corollary derived from some description of QM like the Feynman path integral formulation, taking those results as fundamental and looking at the classical limit. When the day comes that we find a new model that describes more natural phenomena, that role should be reversed and we should expect QM to be a reduced-order case of some GUT.

[@MDNQ-ud1ty]

@@naturallyinterested7569 Yes, but QM was designed to do that. It is a "wrapper" over CM. If it did worse then it would be like other theories that don't work... no one uses them. As far as using QM as the foundation. At the end of the day it is all mathematics. I don't believe QM is any more of a foundation than anything else. It works because it's been "fitted" to work and because it uses probability theory and such. For me the principle of least action is as straightforward as extrema of a curve. In this case it is not "points" but functions(the points are functions as we are finding the extrema over a function space). I think the only real question one has to ask is "Why does nature conform to our simple mathematics so well?" There is no reason for that. Maybe QM says something about that? Do you know something about it that I don't? BTW, there is nothing in the formulation of the Lagrangian that deals with size and what you are describing isn't fundamental to QM but to functionals. Mathematically speaking we are simply dealing with an integral over a parameterized integrand where the parameters are functions. That integral represents some value. The integrand typically will have some functional form representing some relationships that are analogous to differential equations(in fact, it is more than analogy as they are differential equations). We know the relationships between the various ways to express the function in terms of it's derivatives(just like a DE) but we don't have the "solution"(just like a DE). E.g., f' + cos(t)*f*f' = g(t) This DE in theory has a solution f that makes it true. But we don't know what the function that makes it true is. (note none of this has to do with physics). Now suppose we integrate this over some time interval: int(f' + cos(t)*f*f') = A The integrand still unknown in terms of f just like the DE. But we know abstract mathematical relationships between different parts. Now we can define the integral to be: L(f) = int(f' + cos(t)*f*f') If we have various known functions we substitute in to L we will get various outputs: L(h), L(r), etc. Many will not have the value of A. But there will be a class of functions that do give the value A, let F be this class and f in F. Then L(f) = A. But within this class that do give our value A(F is a level set in the function space) not all are created equal. Some functions in F may be better in terms of how they behave. Now, what the principle of least action says is that the function(s) f in F that minimize the variation IS corresponds to how nature behaves(this is empirical and related to the conservation of energy). So out of all the possible f in F, the one that minimize it's variation gives us the "solution" we want. There are many solutions but only one that corresponds to reality. Minimizing the variation is like taking a derivative to find the extrema(minimizing the variation around a point in some sense). You can't take the derivative of the L(f) wrt time though since it will be zero(it's a constant). You can't take the derivative w.r.t to f since technically that doesn't mean anything. But you can take the variation of it wr.t. to f and set that to 0(that is the analogous idea in function spaces). The idea is simple: Nature doesn't play games: It keeps it simple. It doesn't do a lot of nonsense stuff just to do it. It does the most direct thing it is suppose to do. Out of all the possibilities(all the functions that could be solutions) it will do the most direct one. But what nature really does is what it does. It just so happens that what it does is connected to a relatively simple mathematical idea. Nature doesn't take "all paths" there is only one path in nature. "All paths" are the mathematical constructs of all the mathematical possibilities. If, say, nature did something different, then the mathematics would be more complex. our "least action" would be something else and it would give some other method(maybe not even solvable). Luckily it doesn't do that. Another way to say it is "Nature is efficient". But it isn't just efficient, it's the most efficient that can be done(with some basic constraints such as differentiability).

[@SpencerKraisler]

This was an amazing video, thank you so much for making it!! As a math major, the most intuitive proof for the euler-lagrange equations is writing d'alamberts principle in terms of generalized coordinates. :) The concept of generalized coordinates is a bit tricky, but pretty intuitive after doing a few basic examples. Just means every coordinate is independent of the others (i.e. for a pendalum, the x-position depends on the y-position, so they are not "generalized coordinates", but the angle is (since its a single coordinate). D'alamberts principle is a bit trickier to understand, but really its F=ma but also considering constraint forces. Again, do a few examples (e.g. double pendulum) and it starts making sense! Put the 2 together, and you get the Euler-Langrange equation.

[@t0k4m4k7]

Ah yes, the priciple of the least explained principle in physics

[@learn-with-vince]

valuable intuition

[@ComputationalModelingExpert]

Happy to see another video after your previous banger! 😁 I feel you: I am also often missing intuition on why the objective functionals which we want to minimize look how they look in variational problems. Often the choice of the objective functional seems arbitrary and only after deriving the Euler-Lagrange equations it makes sense ... 🤔 Keep going with the videos!

[@cleon_teunissen]

About the visualization that you start at around 2:00 The type of visualization you implemented there is key: you plot the kinetic energy and the potential energy and you demonstrate how they change as you sweep out variation. At that point you are so close to cracking it! When I implemented my own visualizations I did the following: I mirrored the plot of the potential energy, so that it was a plot of the minus potential energy. Returning to your visualization: At 2:53 you demonstrate what the two plots are like for the physical trajectory. There the plots for the two energies mirror each other; the plots have *matching slope* everywhere (with opposite sign). The matching slope is expression of the work-energy theorem; the rate of change of kinetic energy must match the rate of change of potential energy. I will use the expression 'trial trajectory' for the hypothetical trajectory that is evaluated at any point in variation space. When you sweep out variation: the integral of the kinetic energy has its rate of change (as a function of variation sweep), and the integral of the potential energy has its rate of change (as a function of variation sweep). At the point (in variation space) where the trial trajectory coincides with the physical trajectory the two rates of change (as a function of variation sweep) are matching.

[@stabbysmurf]

This is a great explanation, thanks!

[@mobyzwick6612]

One thing that should be mentioned when talking about Lagrangian mechanics, specifically Lagrange's equations of the second kind is the use of generalized coordinates. This is important because the number of Euler-Lagrange equations equals the number of generalized coordinates you need to describe your system or rather its degrees of freedom. From this you can immediately spot conserved quantities such as generalized momenta, which is when the Lagrangian doesn't depend explicitly on the corresponding generalized coordinate (yet another example where you literally need to differentiate between partial and total derivative !). At 8:57 I'd also like to mention that another way you can arrive at the expression for T is by treating your position vector as a multivariable function of the new coordinates. Then finding the total time derivative of the position vector r just becomes a matter of making use of the multivariable chain rule, setting up r_dot in the new basis and then forming the dot product of r_dot with itself in this new basis to get r_dot^2. At 4:34 you talk about the importance of knowing how to take derivatives. To get a better understanding of derivatives it might be worth taking a look at multivariable calculus. Also bear in mind the need for two constants of integration for a second order differential equation at 8:07.

[@mcwulf25]

Excellent video. I never learned this in my physics course.

[@NicolaTesla28]

this video was very insightful but please slow down your voice and animations a bit to let people a fair chance of think as they are listening to you just like 3b1b 👌👌

[@madhavajith6875]

Why your videos are too good ❤

[@valenfeldmann64]

Thank you

[@aapaavishu]

गुरुजी आपला आवाज आणि शिकवण कानाला आराम देऊन जातात! आभार 🙏🏼 This Indian kid is waiting for your next video.

[@gergelybencsik8626]

A tip for your animation: if you use \left( and ight) instead of simple brackets () in Latex it looks much nicer.

[@yash1152]

7:30 thanks a lot for NOT using the stupid approximation in case of pendulum.

[@chemnobeliumlab1520]

awesome!

[@beamshooter]

but where does the euler lagrange formula come from? would love a video showing the derivation

[@cleon_teunissen]

It is possible to have a transparent understanding of how the stationary action concept connects to Newtonian mechanics. In preparation I must first say some things about calculus of variations. As example I take a case that is often solved with calculus of variations: what is the shape of a hanging chain? For short: the catenery problem. The usual presentation of the catenary problem is with the two points of suspension at equal height, but we can generalize to allowing any ratio of horizontal displacement and vertical displacement. Also, when the two points of suspension are not of equal height then the chain can be so short that it from the lowest of the two suspension points the chain points upward. The catenery problem is scale invariant; let's say you have a particular ratio of horizontal and vertical displacement. If you make that half the size, and you make the chain half the length also, then you get the same shape. This scale invariance is a key factor. When you subdivide a catenary in subsections each subsection is an instance of the catenary problem. This divisibility property has no lower limit; all the way to infinitesimally small subsections it remains valid. The way to solve the catenary problem (using calculus of variations) is to obtain a differential equation that accomodates any ratio of horizontal and vertical displacment. You then impose the demand that that differential equation must be satisfied over the whole domain concurrently. The equation that does that is the Euler-Lagrange equation (when populated with the appropriate Lagrangian). Another crucial aspect of calculus of variations: with the catenary problem what we are looking to find is a curve that is a function of the horizontal coordinate. The variation that is applied, on the other hand, is variation of the vertical coordinate. For emphasis: the variation that is applied is exclusively variation of the vertical coordinate. To find the point where the derivative with respect to variation is zero you differentiate with respect to the applied variation. Since the variation is exclusively variation of the vertial coordinate: setting up to take the derivative with respect to variation boils down to taking the derivative with respect to the vertical coordinate. With the above in place I turn to Hamilton's stationary action. Many authors assume that it is about minimization, but there are also classes of cases such that the true trajectory corresponds to a maximum of Hamilton's action. That is why the physics community has shifted to the name 'stationary action' 'Stationary action' is a compact way of stating: 'the true trajectory is the point in variation space such that the derivative of Hamilton's action is zero'. The derivative doesn't have to be a minimum or a maximum, that is immaterial. The criterion that counts is: that the derivative of Hamilton's action (with respect to variation) is zero. As we know: the true trajectory has the property that the sum of potential energy and kinetic energy is conserved. From that follows the property: at every point along the trajectory the rate of change of kinetic energy *matches* the rate of change of potential energy. It is sufficient to establish that in every infinitesimally small subsection of the trajectory those rates of change are matching. When that matching- rate-of-change demand is concurrently satisfied everywhere then that property propagates out to the integral of Hamilton's action. That is: as you sweep out variation: the derivative of Hamilton's action is zero if and only if on every infinitisimally small subsection the derivative is zero. Hamilton's action consists of two elements: kinetic energy and potential energy. To take the derivative of Hamilton's action means you are evaluating two derivatives: derivative of the kinetic energy and derivative of the potential energy. The true trajectory has the property that those two derivatives are matching. In mechanics we are accustomed to taking time derivatives. In the case of mechanics (Hamilton's stationary action): the derivative that you take is *perpendicular* to the time axis. You take the derivative of Hamilton's action with respect to the *position* coordinate. When you populate the Euler-Lagrange equation with Hamilton's action (T - U), you see that the equation takes the derivative of the potential energy with respect to the position coordinate. And while the operation that is performed on the kinetic energy looks different we know that *dimensionally* it must be the same. Sure enough: it is mathematially the same; in effect the Euler-Lagrange equation takes the derivative of the kinetic energy with respect to the position coordinate. On why in Hamilton's action the potential energy is *subtracted* from the kinetic energy: That has to to with the fact that the differentiation is differentiation with respect to position coordinate,; it's not time differentiation It is when we follow change over time that the potential energy and the kinetic energy are counter-changing (from which it follows that the sum is a conserved quantity.) On the other hand: when you sweep out variation (a hypothentical change) then with respect to that change the potential energy and the kinetic energy are co-changing; to find the point in variation space where the two rates of change match you subtract one from the other.

[@MDNQ-ud1ty]

There is a good book called Calculus of Variations by Weinstock where it talks about some of this. In any case, the idea is that of Hamiltonian's principle which is the idea that the "system" tries to minimize the Lagrangian. Why? Well, you have to realize what the minimization is(it is in the sense of functional variation in the sense of calculus of variations). So the Lagrangian is simply the difference in potential and kinetic energy(so in a system with constant total energy it tracks how the energy flows from from potential into kinetic and vice versa). Hamiltonian says that out of all the ways the system could transfer energy from one to the other it always chooses the "optimal/shortest/quickest" way. E.g., the Energies depend on functions(such as position, velocity, acceleration) and these functions are generally not or only partially known(the point of physics is to determine them). The Lagrangian encodes the "dynamics" of the general system(specified in mathematical relationships corresponding to the system) representing the motion. L(q, q') where the q's are "generalized"(abstracted) variables/functions from before. The point here is that L(q,q') is a functional, that is, it depends on functions(and it doesn't explicitly depend on time) What the integral does is sum over the entire time frame the total change in energy of the system's dynamics. Hamiltonian's principle says that out of all possible dynamics(q and q's) *reality* chooses the q and q' that minimize the integral, or minimizes that change over the entire time. E.g., the will be some q's that will produce some dynamics that is not correct. E.g., think of some physical world which is vastly different than ours, it may work there but it won't make sense in ours. Some q's will produce some dynamics that are correct in some sense in that they will have the same "action" but they will still be wrong because they are not "optimal" in some sense. Out of those q's there will be one that is optimal. E.g., we have a "surface" of from QxQ' but there is only a single curve of (q,q') where q' = (q)'. Out of those (q,q') there is only one that will minimize the *functional* w.r.t to the integral(it won't minimize the integral as the integral is already minimized by all the (q,q')'s that worked). E.g., you can surely imagine that there is some q such that it increases T and V both by the same amount which then cancels in the Lagrangian. But that is not necessarily a valid q. Remember, that there are many q's that give the same q'. E.g., q = ax+b, q' = a. There is an entire space of q's to choose from. But there is at least one that would give the minimal "action". In this case what makes them different is simply their position so there is some point in phase space that represents our space. In some sense there are two parts to the integral. One is the minimization of it(minimizing the action). Some possible dynamics of the system simply are not going to produce a minimal action. But there are many systems that can produce a minimal action... but out of all those there is only one that will be the minimal(well, given certain differentiability constraints) in the sense that it will "*minimize the minimal action*". E.g., think of a system that wastes a lot of energy at the start then somehow compensates for it at the end. Overall, it may produce the same minimal action but it isn't correct because somehow it lost energy and then gained it back. That suggests our system is incomplete(something else going on) or is not what we observe in reality. But there is one system that will behave in a *uniform* way(sorta like a uniform approximation) as to always work to minimize loss of energy(or keep it zero). After all, we only have V and T and we can only transfer one to the other(conservation of energy) so if we are losing or gaining energy this would suggest some other term we didn't write down(some "hidden variable"). So by ensuring we find the "least action" we are essentially ensuring that we have the system that isn't doing anything funky. Also, we expect this to be a local behavior, not a global behavior, because why would the interval of time matter. I.e., the system should be "time invariant" and "linear". So ultimately, then, this results in the Euler-Lagrangian differential equation which tells us how the system has to "flow" to always conserve energy(T - V can change but it can't change arbitrarily). The dual system is the Hamiltonian which is working in coordination and has to essentially Obey similar equations(since it is related to the Lagrangian). The main point is that these ideas express the conservation of energy in them. A universe does not get a free lunch. As something is happening in the universe, as long as all the potential and kinetic energy is accounted for, the total energy will always be constant(the universe is completely isolated from whatever is outside it) and inside it can only flow in a very specific way. Remember this too, Int(T - V) is very specific. The kinetic energy is always positive. This puts a very specific type of constraint on the system. It doesn't allow the integral to take on any values. The potential sets a sort of "lower bound". What this means is that there are many systems with, say, very large kinetic energies that will produce nonsense. E.g., just imagine that V = 0, then one has Int(|T|) and then the minimal solution is when T = 0. Say V = 1, then int(T - 1) in this cause T being 100 would produce a large positive value which clearly is not close to the minimum. T = 1 would give a minimum of 0 but it might not describe our system. E.g., T = 1 + a*cos(wt) might describe our system where a is very small. here the integral would be a*Int(cos(wt)) and one could imagine that the integral happens to be near zero and a

[@cleon_teunissen]

About the suggestion that some form of minimization is involved. There are also classes of cases such that the true trajectory corresponds to a maximum of Hamilton's action. That is why the physics community has shifted to the name 'stationary action'; the true trajectory is the one trajectory in variation space such that the derivative of Hamilton's action is zero. I'm aware: it is sorely tempting to get invested in the minimization idea; there are many instances in physics where we do have minimization at play. In the case of Hamilton's stationary action: I have seen the evidence; the suggestion of some form of minimization at play cannot be salvaged.

[@solarestone]

It minimizes the area according to thermal dynamics

[@user-nv4lx7cl4p]

True

[@jhacklack]

The rough intuition is if you apply the Euler-Lagrange Equation to a general potential (U' = F) and general kinetic energy (T = [mv^2]/2) and work through the algebra, it reduces to F = ma, meaning EL is equivalent to Newton's laws of motion

[@marcossidoruk8033]

That is not an intuition, that is merely a demonstration that Lagrangian mechanics is equivalent to newtonian mechanics.

[@jhacklack]

@@marcossidoruk8033 yes and no. I find it intuitive that the principle of stationary action reduces to newtonian mechanics because of my math background, but i understand it doesnt give you a physical gut feeling explanation that you want from intuition.

[@marcossidoruk8033]

@@jhacklack All intuitions in physics are mathematic, that is not the point. The fact that Lagrangian mechanics is equivalent to newtonian mechanics is obvious due to the fact that they both describe the same thing successfully. The problem is that wether or not they are equivalent it doesn't explain why either of them work. There is a much better intuition that one can get that requires knowledge of Hamiltonian mechanics and modern symplectic geometry, but I couldn't possibly explain it in this comment. I suggest you to look into it, especially if you have a math background.

[@cleon_teunissen]

@@marcossidoruk8033 @jhacklack About the relation between Hamilton's stationary action and Newtonian formulation. So we have that when the general potential and the general kinetic energy are inserted in the Euler-Lagrange equation the result is that F=ma is recovered. Now, we have that in physics relations are usually bi-directional. Example: Lagrangian mechanics and Hamiltonian mechanics are related by Legendre transformation. (In fact, Legrendre transformation is its own inverse: if you apply Legendre transformation twice you get the original function back. Quite the bi-directional relation.) We should expect that it is possible to go in all forward steps from F=ma to Hamilton's stationary action. And indeed that is the case. The forward path proceeds in two stages: 1) Derivation of the work-energy theorem from F=ma 2) Demonstration that in cases where the work-energy theorem holds good Hamilton's stationary action will hold good also. Newtonian formulation and Hamilton's stationary action are on equal footing in the sense that they have the same scope, the scope of Classical Mechanics, and the transformation between them is bi-directional. By contrast: Quantum mechanics has a fundamentally larger scope; with Classical Mechanics being a limiting case of Quantum Mechanics.

[@vaakdemandante8772]

The example @ 2:09 is somewhat nonsensical in that it shows that random paths can take negative values for height, which is somewhat impossible. Besides, the first example was about dropping a ball, and the second one is about throwing the ball up, which is not really mentioned and adds to the confusion. I do however like the video as a whole - nice visuals and clear explanation of the subject. If only this was thought in school that way.

[@EdwardYang-cz1me]

I think I see where you’re coming from about 2:09 but there’s actually no problem here because we can choose any arbitrary height to be 0 and it’s not necessarily where the ground is. In this scenario, there is not even a surface to provide a normal force. You make a good point about the different boundary conditions potentially being confusing though.

[@MissPiggyM976]

Good, but some slides are difficult to see on a mobile phone screen.

[@marcossidoruk8033]

At 8:04 there was a mistake integrating the differential equation. The frequency should be sqrt(g/l) and not sqrt(l/g). The longer the pendulum the lower the frequency and the stronger the gravity the higher the frequency.

[@valenfeldmann64]

Good eye! I'll make sure to add a caption (and comment) with the correct value of omega.

[@frogstud]

3:39 Asking why nature exremizes the action is the same as asking why Newton's laws are obeyed in non relativistic classical mech

[@tokajileo5928]

background music completely unnecessary and extremely annoying

[@unflexian]

nah

[@ollieoniel]

Universe isn't lazy. Lame math.