So, Hammond's law varying action revisited. Let's do an example. Good, we've got time for this. This is kind of interesting. We don't typically use that form. The extended Hamilton's principle is what we use with just equal to zero, because if he argued, dynamical systems tend to be fixed endpoint problems in space. We have initial conditions and vice versa, you could argue have final conditions. It's a completely equivalent kind of a thing. Here, I'm going to show you an approach where we can get the solution to the path of this particle, and this is super simple. It's one dimensional motion. Right? I have a generalized coordinate Q which is basically, my Y vertical altitude coordinate. That's it, and I can find how why must evolve with time, without ever having to find first differential equations, and then analytically solve differential equations. As you can imagine, it doesn't always come for free. You can do it, but you have to pay for it. So, our potential energy MGH easy, and H is the height, and that's my generalized coordinate QQ. So, H is equal to Y, the force is MG minus in the minus three direction. That's just classic constant gravity field, so much kinetic energy is here, potential energy is here, for Hamilton's law, varying actions we needed del t minus del V. There's no other working terms. Right? So, del W was minus del V, which is MG times del Q because Q was H the variation of h is just LH or del que. Here, variation of this is going to be mass times q dot times del que dot classic variations stuff. Taking the partial of this with respect to q dot, which will need is simply going to be of energy, not delta Q is one half m times two times q dot. So, the twos cancel, you're left with them times q dot kind of like linear momentum. Okay, did that look kind of boring? Where are we going with this? Well, this was Hamilton's law varying action in its most general form, and tell W is minus del V here. So, the left hand side, we can take this Dlt and plug it in, and we can take this DLV, and I'll put a minus in front of it and plug that in, and factoring out the mass. So, I get cute dot in here and G times del que. So, I'm good. And on the right hand side, I needed this partial timescale Qs, and there's only one particle. So, its mass times q dot times the variation in Q evaluated at zero, and the final time whatever tea you pick, you know, up to some point time. You can also shift, you know, TF arbitrary. So, that's Hamilton's law varying action would say this must be true. So, how do we solve this without finding differential equations? We make an assumption. We expect a smooth solution for this system. So, you just have to parameter rise all infinity of possible motions into something that's doable, if you know, it's cyclic maybe you want to use sine waves and Kassian waves, or other kind of a function here. I'm using a polynomial. And also over a certain period of time, I can always approximate even a sine curve. Right, with a polynomial series. It just may take infinite terms. Other answers like here, the falling ball problem. What's the solution to the heights? What form does it have? It's just a quadratic equation. Right? So, this polynomial assumption is great. Actually, it will be a finite truncated version of it, versus if it's a spring mass system, it would be an infinite series. But it must have a very particular pattern that you can go, hey, that's nothing but a sine or a cosine. So, here we're just assuming, I don't know what it is, but I'm assuming I can approximate it through a polynomial. As long as it's smooth enough, I can do any polynomial. Any function in terms of this. So therefore, it's derivatives with respect to time derivative of a constant zero derivative of T is one. So, you get a one derivative of T squared, gives you two times T, and you go on, an easy pattern. Right? So, now, what do we do? We needed variations of it. So, the variation of this? The only thing we can vary our these coefficients a zero, a one, a two. They control the shape of the path that results of dropping a particle. Right? And but the variation in paths, and the variation in the path rates must be related. You can't have an arbitrary variation in past. You know, I am drifting to the left yet my velocity keeps getting bigger to the right. That's not possible. Right? So, they have to be correlated. So, you can just different take the variation of this, which would be the L A one plus two times the L A two times T. Again, treating the NRA one A 2, a three infinite series of it as your coefficients that we have to do. Now, this was Hamilton's law varying action for this particular system. Again, I encourage mathematica symbolic manipulators, keep track of all this stuff. But what we can do is you can plug this in, and at initial times as well. So, what happens at t, what happens at t not, because Del que at t not, what is that going to look like up here? What is Del que at Tina, or at zero basically Tina is zero here, delta and not because the variation then if T is zero at initial time, that's going to be it. And the variation at T is just going to be this expression right? So, you do this, and in the end you group all these things that are arbitrary. Were saying, well, I don't know what the motion is, but I can come up with arbitrary path variation. So, I can tweak a not A one A 2 n c to smooth banners, and then I can factor them out. So, what goes inside is all the stuff that relates on a not Della not the other one is the L A one day L A to, and it repeats through the whole pattern. So now, we have to go find a pattern out of this stuff. This is what we just had, and now these must all vanish individually. Right? So, this squiggly bracket term must go to zero. Cool. But within this squiggly bracket term, there's an infinite series. What can you argue about the terms inside that infinite series, as a son or individually from? >> Yeah [INAUDIBLE] >> The coefficients have to go to 0 to pretend T not T is 1, 1 and T two is so it's one that will make it to then two squared is four and you pick a two, and a three. Such that the sum of these happens to go to zero, and there's an infinite series of them. This has to be true for all time. Not Just at a particular time where T is equal to two. This has to be true. Also, if T is equal to four and you will never be able to make that balance, where the sun goes to zero but not the individual terms, because T you can just shift it then an epsilon and you will go up. I'm no longer at that route. Right? So, the only way to make this term goes to zero, for all time is that these terms that depend on time in the serious expansions have to individually go to zero? Well, that's nice because that gives us a series of conditions. The 1st one says, this rounded bracket to a two plus G must be zero. So therefore, a two must be minus 1/2 over G. So, 1/2 times g. This term has to be three times a three must be zero. So, therefore, a three must be zero, and going up through the ladder. If you looked at it and expanded, you will see the pattern that every other hire a term has to be zero. This should jive with your intuition and you said earlier, I know a falling ball problem in a constant gravity field is just a quadratic solution. So, I should only expect up to the A two to be non zero, and then everything should go to zero, and that's what comes out of this condition. So, if you know a three and higher or zero already, even here, that's all going to vanish, and this gives us the same condition that we needed earlier. So, really we come up now with an analytic answer for what this has to be, I don't know what a knot is. I don't know what a one is. I only have a condition on a two, but I do know that a three, a four, a five, a million is going to be zero. So, how do we figure out what a nod and a one is? They are determined once you actually say, I am launching from this height and letting go, right. But I solved for all possible motions regardless of initial conditions. I didn't just solve it for, hey, I'm launching on Tuesday from that location in Vanderburgh. I'm saying, I don't know, this is all infinity of possible answers. So out of it, because it's a two second order, you know, a system that we have, we always need to initial conditions, right? And that's what you're a knot and a ones, they are related to your coordinate rates very trivially here. Right? But you can see how we can use Hamilton's law varying action. I admit, it's a sledgehammer approach to a fly. You know, the easier way to shoo away a fly than swinging a sledgehammer at it. But you can see how this could be useful actually, sometimes when you don't have to deal with the, you know, stuff. So,
