The Principle of Stationary Action 

Because you have “run out if ideas”: I would like to see more classical dynamics (Hamiltonian + Poisson brackets in particular, collision cross sections, perhaps orbital motion problems), quantum mechanics (computing probabilities, taking various inner products of wave-functions, spin states, total angular momentum, s matrix, etc), thermo (transport coefficients and the Boltzmann equation), and classical electrodynamics (work through a few examples). (I get that you haven’t run out of ideas but I’d love it if you covered these topics eventually).

What does stationary means ?

Sir what does stationary mean.Why the path of least action isn't always chosen. In the video,mentor told that always action is minimum.That means the path of least action is always chosen only wright ?

@@nitink9879 Stationary means a function where the action functional is not immediately changing (i.e. the 'derivative' of the action functional 'with respect to functions' is 'zero'). By solving the Euler-Lagrange equation, we don't necessarily know whether the function we're getting is a path of (locally) least action, most action, or 'saddle' action. We just know that it makes the action functional stationary. That's why it's more accurate to say stationary action instead of least action. For more details on what this explanation means, I encourage you to watch my calculus of variations playlist: ru-vid.com/group/PLdgVBOaXkb9CD8igcUr9Fmn5WXLpE8ZE_

@@FacultyofKhan Okay sir I will watch . Thank you very much.

@@leif1075 you have to realize that the path is in the height-time space meaning that the height changes parabolically with time not distance.

@@humamalsebai what do you mean?? Ifnyoubare standong still and throw a ball straight upwards. It will fall. Ack straight downwards...thsts pretty clear and intuitive..not a parabola..unless you mean in sone spacetime hyperboloc geonetry3..is that what you mean..thats not what he said in the video in any case.

Glad to help! Thank you for the kind words!

I have a bunch of requests for many topics that I plan on covering, but yes, tensor calculus is on the horizon. I hope to get to it once my summer break begins (June 1)!

Nice, thanks!

So what we're integrating is the Lagrangian, which is E_k - U (kinetic - potential). While L has units of energy, it doesn't actually represent an energy quantity itself (i.e. it doesn't represent the total energy). Also, we aren't setting the derivative of energy with respect to time equal to zero anywhere; we're using the Lagrange equation to find the optimal particle path, which is more complex than dL/dt = 0. As for your last statement ("what we are doing is finding the solution to the Lagrangian that makes the change in the action = 0"), that is correct. However, setting dS/dt equal to zero to find the optimal path isn't even possible because S doesn't depend on t (the t gets integrated out when you take the integral of L). I think you might benefit from watching my Euler-Lagrange derivation video; it should clear up your confusion: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-sFqp2lCEvwM.html

It's the second thing you said - he showcases how when given certain initial conditions, the particle will follow a path that corresponds to the Stationary Action Principle - in our case, the y equation with 0.5g in it - and not something else, like the y equation with 0.4g in it.

Technically yes, it can be the local maximum since Euler-Lagrange doesn't discriminate between the types of stationary functions you might end up with. Usually though, we get the local minimum. And thank you for the kind words!

Most books that I've seen usually connect the Principle of Least/Stationary Action with Newton's second law, so my suspicion is the latter. I don't think it pops out of thin air.

Principle of least action does not come out from nowhere. Consider this: T.E.(total energy of a system)=T(kinetic)+U(potential). Now in a conservative force field the total energy of an isolated system shall remain constant. Hence we cannot differentiate it or integrate it. But if we consider the difference between them, T - U, this is a quantity that differs even in a conservative force field, because for every small action dS the particle or system takes, either its T will increase or decrease and so will U. Now for S (the total action) from any point P1 to P2 we would have to add these very small but finite dS and that triggers the word integrate. Setting the integration equal to a local minima or 0 in most cases of Newtonian mechanics we ensure that the "action" that we take is the least. Hence least action principle in a way is a generalization of newtons second law such that it works with any coordinate system. Now the principle of least action can be proved without ever using newtons second law, and what's perhaps even more interesting is for special cases the principle of least action transforms into newtons second law. The proof however would require you to have some knowledge of multivariable calculus and basic calculus. The Euler Lagrange equations can be derived mathematically while finding the shortest distance between two points. We have to take into consideration the infinite paths that a system can take to reach from one point to another and then restrict all those paths such that d/dy(L)-d/dx(d/dy'(L)) equals 0. L here is a functional such that L[y(ē),y'(ē),x] and x is the independent variable, ē is a variable constant that changes distance, y(ē) = y(x) + ēb(x), b(x) is an arbitrary function of x that gives us the direction of the path that is taken.

I have created a resource for the purpose of making Hamilton's stationary action entirely transparent. The connection between Hamilton's stationary action and F=ma is the Work-Energy theorem. As we know, the Work-Energy theorem follows from F=ma, hence the Work-Energy theorem implies F=ma. Hamilton's stationary action and the Work-Energy theorem have the following in common: both express the physics taking place in terms of kinetic and potential energy, and from both F=ma can be recovered. Using the Work-Energy theorem there is a path that starts with F=ma and proceeds to show validity of Hamilton's stationary action. The demonstration is in the form of interactive diagrams; in each diagram the visitor can move one or more sliders, exploring various configurations. localhost/physics/phys256/energy_position_equation.php

actually it gives you joule*second .

Thanks for the question, but I don't think you're right. Pretty much every textbook has kinetic energy in 3-D expressed the way I've written it in my video. You add the kinetic energy in the x-direction (0.5*m*vx^2) to the kinetic energy in the y-direction (0.5*m*vy^2) to the kinetic energy in the z-direction (0.5*m*vz^2) to get the total kinetic energy. Your expression doesn't really follow that intuition/logic, and I also think it's in the wrong units because you take the square root of the entire velocity-square expression. Hope that clears up any confusion!

Faculty of Khan I can see now that you are absolutely correct! I forgot to square everything above. When I do, what I have written above, simplifies to give the same equation which you have presented. It makes sense now. Thank you FoK.

The Lagrangian is important mainly because it serves as a mathematical tool that allows us to determine the equations of motion of complex systems. We can use the Lagrangian to reduce our physics problem to an optimization problem; the fact that the Lagrange equation is closely linked with Newton's 2nd Law allows us to do this. As far as I know, the Lagrangian doesn't explicitly have a physical significance as much as it does a mathematical significance (i.e. it doesn't measure anything notable about the system other than KE - PE; the total energy which is KE + PE has far more physical significance on the other hand). Here's an answer on quora which refers to this: www.quora.com/What-is-the-physical-significance-of-Lagrangian As for the Feynman lecture, I'm not sure which one you're referring to so I wouldn't know how to help you there. Nonetheless, I hope this answers your question!

Lagrange equations are exactly the Euler-Lagrange equations. They are mathematically equivalent, the difference is only conceptually. For example, Lagrange equations involve the time t and the path [x(t), y(t), z(t)] instead of any other arbitrary variable. Euler-lagrane equations are more general in the sense that the variables could be anything of your choice. They don't necessarily have to involve time and a path. In other words, Lagrange equations are just the Euler-Lagrange equations but for mechanics, thus why the variables are time and path. Now how these equations lead to a functional being stationary (extremized), requires a proof from the calculus of variations. You can prove these equations by either approaching it from a mechanical point of view which involves the Lagrangian (L=T-U) and the action (S). Or from a mathematical point of view that involves a function F(y,y',x) and a functional (I). Either ways you have to introduce a variation of the curve y(x) that extremizes the functional. if we assume that y(x) is the curve that makes the functional (I) stationery then the variation is Y(x)= y(x)+kh(x). where Y(x) is the variation, y(x) is the extremal curve, k is a variable that adjust the variation, and h(x) some arbitrary function. After doing some calculus you will end up at the Euler-lagrange equations.

At 2:29

@@chunwaiwong5355 U stands for potential energy, not velocity.

I don't quite understand your question - I'm not making the Lagrangian stationary (I'm not even sure that's a thing); I'm only making the action integral stationary.

It means that the functional S=0, which means that the langrangian must obey the Lagrange equations, so it helps you find the correct function y(t) that will fit the Lagrange equation

I'll refer you to my intro to Calculus of Variations video (one of the prerequisites to this lesson) for an explanation (watch the first 1:45): ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-6HeQc7CSkZs.html If you'd rather not watch it, here's basically what it says: in regular calculus, you might have a function f(x). The point where f(x) is stationary corresponds to the point where df/dx = 0, since when df/dx = 0, the function f(x) is not changing (i.e. it's stationary). A point where the function f(x) isn't changing could be either a local minimum, a local maximum, or a saddle point. The same idea applies to the action integral, which is now a function of functions (a functional), so instead of considering points where the function is stationary, we are now considering functions where the functional is stationary. A function that makes a functional stationary is one that ensures that the 'derivative' of the functional with respect to that function is zero, so we could have a function that makes the functional a local minimum, a local maximum, or a saddle point. The principle of stationary action basically says that the path the particle follows (i.e. the function that the action functional depends on) must make the action stationary. Usually, the particle's path makes the action a local minimum (which is why you'll often hear the alternative term: Principle of Least Action), but sometimes, you can also get saddle points (and very rarely, local maxima). Hope that helps!

AV Drago: I think you mean S' = 0, but yes, that too.

Got it, I should have seen that one first, by the way, thanks for the videos, first-rate explaining on deep content is hard to find on youtube. Keep up the good work!

The first order change of the the action integral is zero.

It usually means that a function "stops" increasing or decreasing.

@@TyronTention Thank you.

@@TyronTention Right in other words the action is constant i.e. doesn't change over that interval right? So in that sense it's constant.