Physics+ Hamiltonian General: UNIZOR.COM - Physics+ 4 All -Hamiltonian

Hi, I'm Zohr. Welcome to UniZohr Education. So, we continue talking about Hamiltonian. Previous lecture was an introduction. And today, I will try to generalize basically whatever the ideas were presented. And that's why it's called general Hamiltonian. Now, this lecture is part of the course called Physics Plus. It's presented on unizohr.com. The website is totally free. There is no subscription needs, no advertisement, uh no sign-in is necessary. Um so, basically, it's all available for free for everybody without any kind of advertisement interruptions, et cetera. Pure knowledge for your consumption. Okay. Now, uh every lecture has uh video and uh text uh presentation. And the text is basically like a textbook. So, you can learn the theory from the notes for this lecture and the video uh in any sequence. And I do recommend you to approach any kind of a lecture from both sides. First, you can um watch the video presentation, then you can uh read the textual part of it on the same page, basically. Or in any other sequence. It's all all for good. Now, back to uh Hamiltonian. Now, very, very brief um repetition of what was in the previous lecture. First, we started with Euler-Lagrange equation, which is the L by uh it's supposed to be a partial derivative. By d q i is equal to derivative. This is a full time derivative of partial derivative by velocity. Then we introduced actually by definition the generalized momentum which is d l by d q i which actually allowed to rewrite basically this since this is now p i and full derivative by time usually we um symbolize with a dot on the front on the top. So the whole thing looks like d l by d q i is equal to p i dot. Right? This is p i the generalized momentum and derivative by time is this dot. Very nice. However, we have introduced one more variable which is generalized momentum but the number of equations is only n. So we have two n variables and um uh only n equations. So we proceeded further in the previous lecture and we found out that considering the Lagrangian is equal to kinetic minus potential energy we have introduced function h which is Hamiltonian which is sum of these two which is full energy of the system and we came to another um set of equations. One equation was like this. And another equation was like this. The only thing is was a minus sign here because it's the minus U. Yeah, T minus U L. So, uh in this case it's Q I derivative. Now, now these are two N equations with two N independent variables Q and P coordinates and momentum. And that basically seems to be a nicer system of equations than the one which is this one. This is the second order differential equation. These are first order. So, we increase the number of variables, but we have decreased the complexity of each equation. Plus it looks much better. Okay. So, that was the last lecture. It heavily dependent on this. How Lagrangian is represented as a as as a difference between kinetic and potential energy. And then we have total energy of the system and then we derived all these equations. So, from this and this we have derived this new system of equations. Fine. But again, we heavily dependent on this. And this is not necessarily true. I mean, it's true for most of the simple mechanical systems, but whenever we want to, you know, use this in much broader um uh, um, uh much much broader number of problems, uh, theories, etc. And and we do, and we will. This is not applicable anymore. Now, let's not get into why it's not applicable and what are those systems where it's not exactly true. But, just trust me that the derivative uh, der- derivation of these equations was extremely helpful in those new cases when this is not true. So, my purpose today is to generalize these things in such a way that it does not depend on the fact that H is actually Hamiltonian is actually a full energy of the system and the Lagrangian is looking as a kinetic minus potential energy. So, I would like to generalize this thing and still come up with something like this where H is something which is related to the Lagrangian, related to um, coordinates and um, momenta and momenta without relying on the specific way how Lagrangian and Hamiltonian look. So, let's assume that we have a general system with some kind of a Lagrangian, but we would still like to to come up with this and express H in terms of Lagrangian, coordinates and momenta. So, that's my my task for today. All right, but how can I basically, you know approach this thing? Well, I don't know the way a function H actually look. All I know that I have to have Lagrangian, I have to have all the coordinates and all the momenta and somehow I have to mix them together to get this function H, which will um with these equations being true, actually. Um and obviously using the fact that Lagrangian and uh uh coordinates and their derivatives satisfy the Euler-Lagrange uh equation. So, that's the general understanding of the whole thing. We we completely put our problem um from another actual angle. We we approach our problem. That consider we have some Lagrangian which has uh the Euler-Lagrange equation satisfied. And the Q and P, where P is defined like this. So, this is it. I'm not really assuming this. This is not given. What's given is this. And I have to prove that this is true for some kind of a function H, which I might must guess, if you speak. Well, guessing is actually kind of easy. And here is why. Um let's basically talk about my assumption that L is equal to T minus U. And try to uh express H which is T plus U in such a way that it does not really depend on T and U, but it depends on Lagrangian and uh coordinates and momenta. So, I have to somehow convert this into a different form, which is not really dependent on T and U. And if I will be able to do this, it would be a good candidate for H function in the general case. And then what I will have to do is I'll just check if the function which I have guessed from from the previous assumptions um if I was able to express H in terms of L, Q, and P and this new H is satisfied these equations, then my task actually is solved. So, I do not present you, "Okay, H is equal to this." and present it as a function. I would like to first have it derived from whatever I know from the some particular case from a simple case and then I will check if it fits the general case. Now, it will fit, obviously. So, how can I do that? Well, it's actually simple. Here it is. How can I express H, which is T plus U in terms of Lagrangian? Well, Lagrangian is T minus U. So, my I can actually put it 2 T minus Lagrangian. Right? 2 T minus T and plus U would be T plus U. That's simple, right? Now, I have to express the kinetic energy in terms of of basically Q and and PI. Well, this is actually not very difficult because again in my very simple formula for kinetic energy, it's uh sum of mass times speed square, right? Now, the speed square is actually uh can be expressed somehow differently. This is MI times VI, which is PI, right? What's the momentum in classical physics? Mass times velocity. times you know, one more velocity. We have a square here. So, what is velocity? Velocity is derivative of uh the coordinate, right? So, that's basically how T looks. Uh I mean, I have to put 1/2 here. Sorry. So, that's my T in classical mechanics. Well, so 2T is uh will basically cancel this two, and my good candidate for H is sum of PI times QI dot minus L. So, this is the formula which depends only on coordinates uh and momenta. Uh well, and derivative of coordinate, which is velocity. And Lagrangian. It does not really depend this formula does not explicitly depend on kinetic or uh potential energy. So, my question is, is this formula good for these equations? So, is H defined as this thing which I just guessed? But it's not not guessed. I used some particular case and came up with a formula which seems to be, you know, satisfy my my uh my my demands. So, it does not depend on energy, it depends only on the Lagrangian, uh coordinates, and momenta. So, I've got this formula. Let me check. I mean, if it fits the bill, if it satisfies these equations, then there is absolutely nothing wrong with saying, "Okay, this is Lagrangian, and we can use it and solve this equation instead of Euler-Lagrange. So, L is satisfied Euler-Lagrange equation, I know that. So, from this formula and these definitions, let me put it on the top. So, that's what's given basically now. H is equal to sum of p i q i dot minus L. So, with this formula, this is given and this is all given. So, Euler-Lagrange Lagrangian, definition of momentum, and this quite abstract formula which I have just invented based on some consideration. Now, this is given, this I have to prove. If I would do this, well, that means that my original problem, which is n equations of second order, can be actually solved with 2n equations with 2n independent variables, but the first order the first order differential equations. Okay. Now, this is now quite straightforward, pure math without any kind of physical intuition, anything like that. I just have to check if these two equations are satisfied. All right. But here I would like to pay some very important I would like you to pay attention to very important consideration. What are my variables right now for H? Well, it This looks like another variable, but we have agreed that momentum and coordinates are functions. Okay? Now, this is not really an independent function. It's the derivative of of QI by by time. However, that's not really um You You see, we have defined PI as such a derivative. But, uh I would like QI to be treated uh QI with a dot, the derivative by time. I would like it to be treated uh as a function of uh independent coordinates, which are time, uh all Qs and all Ps. This is a very important momentum. P and Q, momentum and coordinates, are independent variables, as well as time. And I do not actually uh want to to say that uh derivative of uh coordinate is totally independent on of anything else. I would like to treat it as another function of these arguments, which satisfy certain laws, obviously, but it's supposed to be treated as an independent function. That's very important. It depends of everything. Uh now, in a simple case, for example, of regular um simple mechanical system, we we know that uh momentum is defined as P * V. Well, V is actually Q dot, right? So, they are related in this case. Q dot is a function of momentum. So, that's very important to understand that in this simple case it's just function of momentum. But in more complicated general case, I would like to maintain the possibility that it's actually >> [clears throat] >> a function of everything. And that's the most general case, so to speak. I'm kind of making it a little bit more general just to be sure that I do not miss any any kind of dependencies which might actually be. Because again, I'm abstractly considering two N dimensional space where P and Q are N and N independent coordinates, which means everything else should actually be defined in terms of coordinates, momentum, well, and time. So, it's actually two N plus one dimensional considering the time. So, this is a purely geometrical viewpoint, all right? Okay, so we have basically established this. Now, let's check this. Okay. Uh Let me get some space. I will put D H by D P I should be equal to Q I dot and I will free my space here. So, I have more room. All right. So, I have to check this. All right, so let's start with uh with this. Um so, I have to basically check it for every I. But I will use I for summation here. So, I will check it for I is equal to K in this particular case. So, I will put K here since in this just one particular coordinate. And I will use I for summation. Okay. Fine. So, All right, let's start. DH by DP K equals to Well, let's just look at this one. Well, first of all, this is a sum. Where is dependency on the momentum? Well, first of all, obviously, there is one particular um pair where this is PK and this is QK. And partial derivative by PK of this particular thing is just QK, right? With a dot. Now, but all functions of these all all members of this summation have QI. And QI, as I was saying, with a dot is a function of T all the all the Ps, P1, P2, PN, and all the Q. So, there is a dependency on this as well. So, I have to add sum of PI times derivative of um uh of this by PK which is I have to derive partial derivative of QI by DPK. That would be a full derivative of this sigma. So, this is a product. So, is n different products. One particular component has to be uh used twice for this one, but all of them are supposed to be containing this one. So, it's this times partial derivative of this should be everywhere. But, whenever I'm doing partial derivative of this times this, it's only one because everything else, which is where I is not equal to K, then that that zero out. And minus ZL by ZPK. Um Now, the derivative by DPK is not just itself. Again, L depends on T, P, and Q. Right? So, um Just get a little bit more. L depends on T, Q, and Q dot. That's our original Lagrangian, right? Now, this is dependency on PK is here within the Q within the Q dot. Because there are all different P's here including the PK. But all of them contain this. So, basically I have to put here not just one. I have to put sigma by DQ I with a dot times DQ dot I by DPK. This is the chain rule, right? So, we have many different Q's here, Q1, Q2, Q3, etc. And there is a dependency on P in each one of them. So, that's why I have to differentiate by each QI. This is the sum sum by A by each QI and then from this QI with a dot differentiate by PK. This is the chain rule. Am I right? Let me check. Yes, looks like I'm right. Which is equal to QK with a dot plus >> [sighs] >> sigma PI minus DL by DQ I dot times DQI by DPK. Right? This is DPK and this is DPK. It's a common uh multiplier. So, I put P minus GL by DQI dot in parentheses. Now, what is this? This is the definition basically for PI, so it's zero. Which means the whole thing is equal to Q K. Who is it that? Which is exactly this one. So, we have proven even in this complicated case when uh, when the velocity depends on uh, all the independent variables, all the coordinates, and all the uh, momenta. I still have this thing satisfied. All right, let's check the second one. It's basically the same thing. Once more. >> [snorts] >> DH by DQ K is equal Okay, this is what it is. Now, where is dependency on QK? Here. It depends on all the different Qs. So, we actually have to really do all of them. All right? So, it would be sum PI times DQ I dot by DQ K. All right? Yes. So, P is independent on Q, so it's just a constant multiplier. So, all the dependency on Q K is here. And that's why each one of them should be like this, sum by I. Okay. Now, how about here? Here we have two dependencies. Q depends on Q K and Q prime and Q dot depends on Q K. So, I have to really again do it as a multi-functional derivative. One thing is partial derivative by Q itself. Okay? And another thing, whenever I'm doing uncovering the Q from the Q from the Q dot, would be the whole sigma d l by d Q I dot times d Q I dot by d Q K. Let me check. Yeah, that seems to be right. So, what's here? Well, look. This one and this one have d Q I dot by d Q K as a common. So, it should be equal to Now, this is separate. So, this one minus d l by d Q K plus sigma p I minus d l by d Q I dot times this. Sorry. times this. Same thing here. This is zero. So, I have only this. Okay. Now, what is this? Look at this. Minus This is the same as this. It's Euler-Lagrange equation. Now, this is P I. So, derivative by time from P I is P I with a dot. And that's exactly what should be proven here. So, what we have proven that both equations are satisfied in case we have defined our Lagrangian as as such. And where where where momentum generalized momentum is defined as as this. So, from Euler-Lagrange equation, from definition of momentum, and our intelligently guessed expression for Lagrangian, and thinking basically that P and Q are independent variables, and even the derivative of coordinates velocity is a function of all of these coordinates. This is the most general case. In all these cases if this is given, we have this Hamiltonian satisfying these equations. Now, what I'm actually thinking about if we are talking about equivalency of both approaches, Euler-Lagrange equation or Hamiltonian equation. We have to really prove the backwards as well. So, if the Hamiltonian equations are satisfied where the Hamiltonian is this will the Euler-Lagrange equation be satisfied? Well, that's basically the same thing we did before just I'll just write it differently. So, dH by dQ K as we know it's equal to minus P K dot. Right? We assume this on one hand. On the other hand let's just think what it is. Uh if we will divide if we will differentiate it by by Q K uh this is the derivative of Q1 would be here. So, that's sigma P I times uh Q I dot by QK. This is what it is. Because P is constant independent of coordinate minus uh and here we are also have the same thing basically. Uh dL by d q k and minus sigma d l by d q i times d q i by d q k same thing, right? But now what what happens? If I assume that p i and uh d l by d q, that's the definition. So these two is the same multiplier here, go out. So what we have right now we have that d h by d q k is equal to uh minus d l by d q k and this is as I was saying minus p k dot. So that's what we actually came up with. But what is this? Both of minus can be pluses, right? Now dot is actually d by d t, right? And p k is by definition d l by d q k dot by definition. And what is this? This is the Euler-Lagrange equation. So if we assume that Hamiltonian equations are correct then we basically have derived Euler-Lagrange equation. Which basically now we have both ways. From Euler-Lagrange we derived Hamiltonian equations, from Hamiltonian equation, we all we derived Euler-Lagrange equation. So, these two are equivalent and this one is a generalized definition of Hamiltonian. Hamiltonian as a function is represented from Lagrangian and independent coordinates uh coordinates and uh momentum in this particular formula which we have semi-guessed. I mean, we just derived it in a simple case and maybe decided, okay, maybe it will fit the bill. And it did fit the bill, which we have proven. So, we basically have an expression for Hamiltonian, which is this one. In many In many textbooks, they they just give you this formula and then check that this is actually satisfied the equations. I decided that it's much more kind of natural uh how how did we get this? We just basically somebody guessed it the first time. And how did he guess it? Well, probably the same way I was just trying to explain. They solved it in a simple case, that's what it is, and they have decided, okay, let's check, maybe it works in general case. And so it did. Okay, so now this is a generalized expression for Hamiltonian without any kind of relation to uh the energy or without any relationship to expressed uh to um um uh expression of Lagrangian as a difference between kinetic and potential energy. It's a very, very generalized formula and uh that's what I wanted to derive today. That's it. Now, I do suggest you to read the notes for this lecture. Again, the whole lecture is you go to universidade.com, go to physics plus course, the chapter is Hamiltonian. So, the first lecture was introduction and this is the second lecture in that chapter. That's what That's where it is. Thank you very much and good luck.