We've had Lagrangian. Lagrangian with cyclic coordinates, we dealt them, they appear, they don't appear. If they appear there's cool things you can do. Can we formalize this approach? That example that I showed you with a particle moving in a plane on a springiness is attached to a spring. Theta didn't appear, and we were able to get final equations of motion that only have R in there. While Theta. did matter, I can solve for Theta. analytically from that integration constant. Can we do this in a more formal way? That's basically if you hear about Routhian dynamics. If you hear of Routhian dynamics to think it's nothing but Lagrangian dynamics where you've gotten rid of all the cyclic coordinates. You will see at the end just jumping ahead the Lagrangian gets truncated with stuff that won't matter. But in terms of this integration constant, new things appear and then you solve it, and you directly jump to what we just had at the end after we back substituted constraint back into the final equations of motion. That's what's going to happen with the Routhian. Routhian is nothing like an Lagrangian, we've already done this substitution of that constraint. How did we do this mathematically? We have n generalized coordinates, cool, but they're not all equal. Some of them are cyclic, some of them are not cyclic. I'm going to order him the first C are going to be my cyclic coordinates, and the second set of them, everything that's non-cyclic, I'm doing to the right just so I get this nice partitioning otherwise it becomes hellacious book-keeping. Who is what and where? I can always do this without any loss in generality. You've just ordered your coordinates in a way that you've got groupings, that's all. Now, my Lagrangian which depends on all these generalized coordinates and coordinate rates, I can break up into ones that depend on cyclic generalized coordinates, non-cyclic generalized coordinates, cyclic generalized coordinate rates, and non-cyclic generalized coordinate rates. Why do these first q_cs vanished? That's when I have a bar there, I just means they don't appear. [inaudible] Exactly. Because their cyclic, they're not supposed to appear in the Lagrangian, so that's why that one can vanish. We know that by definition I shouldn't be there so no magic. The cyclic coordinates if you now go through the classic Lagrangian stuff, you would take the partial with respect to the q_c dots. Q_c dots would appear, but q_c as you just said don't appear, so this partial by definition has to go to zero because cyclic coordinates don't appear in the Lagrangian and the Q_c would be zero as well otherwise it would not be cyclic coordinates. Either there's no other forces acting on it, non-conservative forces acting on the system or those forces are acting orthogonal to this coordinates partial velocities. Either way, the Q for that coordinate must vanish or it's not a cyclic coordinate. That's cool. Then for everything non-cyclic, we can't make those assumptions and you have your classic Lagrangian's equations. Now, because this was zero and this is zero that means this term differentiated is zero, so that term is an integration constant, and just write this in matrix form. So p_c is the conjugate momenta matrix that you have of all the cyclic coordinates. This constant I'm also calling Beta_c that's just an integration constant which you can solve for. P_c, Beta_c is really interchangeable , but it is a constant. Cool. Now, let's look at what happens for natural dynamical systems and I added this to the slide yesterday. Because kinetic energy remember, when we did this matrix form of Lagrangian dynamics we then wrote out a very general thing because if you have a bouncing ball that's moving on a conveyor belts or a train that's moving at a fixed speed, you have a time-dependent part of your kinetic energy and you have part set of state rates based. When you square those, you will have quadratic q dots, you will have linear q dots and term that just depend on time. That's the train speed that is out of your control. That's what a T_2, T_1, and T_0 terms. The T_2 is the one that's always state rate squared like thing, so that's why in natural dynamical system, if you see this in these textbooks what they're talking about is something it doesn't have that time explicit part, though you only have the quadratic measure of your coordinate rates, so you're not on a conveyor belt accelerating through space. Then if you have something like m over 2x. squared and we take a partial with respect to x., you're always going to end up with m times x. or something like that. That's what that form is going to be. I'm not calling it the partial of m, it's just going to be the Lagrangian taking the partial you will have some relationship. Here, I have already times q., now let me go to a slide here. If the Lagrangian's that you have was q. m q., the partial of your Lagrangian with respect to q. would simply be m times q., that's what you would do. Let's see. So that's that one and this is equal to Beta_c, so that means my q. in this case would be the m inverted times Beta_c set, that integration constant. That is that B matrix that I have in that other slide. That's where this one came from. I've just written it out like that. We know we don't have cyclic coordinates in there, so those are all hashed out, and you'd have this relationship between q. and this integration constant. There will be some and if you have c number of cyclic coordinates, it would be a c by c matrix that came out of the algebra that you solved. Why is that important? Well, because we saw when we have that particle moving in a plane with a spring, when I just solved the Lagrangian's equations for R, there was r., there was r double-dot but there was also Theta. appearing in there. Then we use the constraint to get rid of Theta dots and plug it in. Let's do it formally. When you do your Lagrangian's equations, you are going to have q dots actually appearing in there, cyclic coordinate rates can appear in your solutions. But because this has to be constant, I can solve and for a natural dynamical system, they will be just of this form I can solve for the cyclic coordinate rates. Like earlier I solved for Theta. in terms of the integration constant then it was divided by mr squared, I can do this formally in this math and solve for q. and get rid of that so then you still have no cyclic coordinates, only non-cyclic coordinates, and instead of q dots of the cyclic coordinates, you would have this B matrix times integration constant. Of course, you still have non-cyclic coordinate rates, that's like r. we had earlier. Now, if you carry out this math, you can break this up into terms that depend on the non-cyclic coordinates and non-cyclic coordinate rates, but there will also be a term of the Lagrangian that purely depends on this integration constant. Your other coordinates don't appear , and that's the trick. When I do Lagrangian dynamics and I add five to the Lagrangian, will that have any impact on my equations of motion? No, because I'm always doing the partial of the Lagrangian with respect to states, the partial of Lagrangian with respect to rates, so I'm taking taking derivatives. There's all kinds of partials and derivatives. Any constant to the Lagrangian, poof vanishes, and that's where the Routhian comes in. The Routhian essentially it's your Lagrangian that you've rewritten in terms of instead of q. of the cyclics, you've replaced them with this relationship so we have integration constants in there now, and the non-cyclic coordinates and we immediately will get the thing like we had earlier that was just the R double-dot equation that you need. If our Lagrangian is like this and you go through your classic Lagrangian's equations, this part has to vanish as just said because it's just a constant. What dictates this constant? Let's make sure we're on the same page. How do you find this constant? Initial conditions. Initial conditions, exactly. It's not something upfront, "I've got a bouncing ball that's bouncing across the room," you don't know that. It depends on the particular instance of that dynamical system. You'd have to know what's initial position? What's initial velocity? Then you can compute that. For every dynamical system if throw the ball this way or throw the ball this way, I will have a different integration constant but it comes from initial conditions. You can't just do it generally for the system, but just for a particular system take initial stuff compute that and now we're ready direct with the rest of it, so it's not just something arbitrary. But in the end, the Lagrangian dynamics have to look exactly the same as if you put an R, the Routhian in there instead of the Lagrangian because the difference between them is a constant and that doesn't matter. You can see the steps, you could have done what we did earlier in the example, classic Lagrangian solve it and then take the answer and go, "Well, my Theta dots aren't arbitrary, they have to satisfy this constraint, so I can backs off." Here you take the Lagrangian and you then apply that constant directly to the Lagrangian and then you never have to take all these partials, you get to it directly. It's a little bit of a reduction but all should lead to exactly the same state. I just wanted to highlight, if you hear a Routhians, this is all we're talking about.
