Lecture 44 : Knetic Energy and Acceleration in Terms of Generalized Coordinates

Hello and welcome to another lecture session of Classical Motion of a Single Particle. In this lecture, we will simply continue our previous discussion on the Generalized Coordinates and Generalized Velocities. Already we have defined generalized coordinates to be the minimum number of variables which are independent of each other and needed to describe the location of a particle during its motion ok. So, for example, for a single particle of course, you can understand in 3D space the maximum possible coordinates is 3 ok. So, the number of generalized coordinates can be less than or equal to 3 ok. However, if you have a system with more than one particles then of course, you have to calculate accordingly the number of generalized coordinates ok. For every generalized coordinate which we denoted by q i ok, the corresponding. So, we say this is generalized coordinate ok. The corresponding generalized velocity is given by q i dot that is d dt of q i ok generalized velocity. We also mentioned that the word generalized is actually used because this type of concept of generalized coordinates and velocities are used not only in mechanics, but also in different domains of science and social science ok. So, now coming back to this the case of very usual classical mechanical system. So, most of the cases we have actually if you see that these generalized coordinates can have normal position coordinates or actually distance some distance and angular coordinates that means, some angles ok. So, most of the cases unless you really come across a very strange system you can actually have normal coordinates as the generalized coordinates in general they consists of normally normal coordinates signifying some distance and also or actually the position and also the angle subtended by the position vector of the particle with respect to some specified direction ok. So, that is why whenever we are talking about corresponding generalized velocity also for a single particle. So, I just write for a single particle in classical mechanics ok, then the generalized velocities will also be correspondingly usual velocity and angular velocity ok. So, this is the concept we in general use for our incoming formalism ok. Now, you will see that I mean already this part is discussed that the use of generalized coordinates are quite advantageous with respect to the use of normal or usual Cartesian coordinates because in several systems symmetry wise Cartesian system is not the correct one. For example, just locating the position of a particle on a sphere or on a cone I mean Cartesian coordinates can be used, but this is super difficult ok. Then also if I just calculate the distance covered by a particle on a certain sphere ok or on a certain cylinders lateral surface ok, then you will also see that the use of Cartesian coordinate is a little bit complicated. However, according to the symmetry of the problem if we choose from time to time spherical polar coordinate systems, plane polar coordinate systems, cylindrical polar coordinate system many problems can be very easily solved. So, it is not that Cartesian coordinates cannot be or should not be used, the point is that if we want to save time and we want to solve the problem much easier and without avoiding unnecessary problems or unnecessary errors in the calculation we would try to use ok being smarter the generalized coordinates. So, here for example, I will start by saying that I can now say that the position vector of a particle can be written just for a single particle. So, I am now just going for mathematical generality a little bit general in an I would say n dimensional space ok, in an n dimensional space if we have small n number of effective coordinates that means, generalized coordinates ok. We know that n is actually greater than equal to small n ok. Now what is the meaning of generalized coordinate that means, that whenever we are talking about this collection of coordinates we are already taking care of all the holonomic constraints for the motion ok. So, in n dimensional space you will have n Cartesian coordinates always. Now let us say if your system has no holonomic constraint then your Cartesian system will be itself a system of generalized coordinates. So, your n will be then 3 your small n will be then 3 in 3D space ok. Now let us say I am saying that ok the particle is now moving on the surface of a sphere of constant radius then again this will remain 3, but this will now become 2 because there is one constraint ok. Now that is why you see that this n minus small n is nothing but the number of constraints ok. Or in other way you can say that if you just start from the number of Cartesian coordinates and you just subtract the number of holonomic constraints ok, you think that you suppose that corresponding to every holonomic constraint you will be able to reduce the number of coordinates by 1 ok. Then finally, the effective or the necessary coordinates which are independent of each other ok will be small n in number and that is. So, small n is equal to capital N minus k where k is the number of holonomic constraints ok yes. So, I will now say this R will have two representations possible. One is always in Cartesian system another is any other system containing the generalized coordinates I mean two I mean I am saying that two equivalent representations are possible it can have infinite number of representations actually ok. So, in for example, Cartesian system it will be function of x 1 x 2 x n ok. So, I know that there is no t explicitly here, but now we can go a little bit more general we can add I just use a different color t over here. So, we can say t can be actually put here ok. So, for example, I am saying that my system is in general static in a coordinate system Cartesian coordinate system, but somehow suddenly I have an external hidden mechanism by which I am shaking the system with time the systems the particles own dynamics ok has nothing to do with that is with that shaking ok. This is exactly if you now think of forced harmonic oscillator you have this type of a variation when you are talking about a forced harmonic oscillator ok you had a same type of variation for the force, but here what I am talking about a variation directly in the position coordinates ok. So, if you have for example, the vector position of the particle is actually also explicitly changing with time then here you can actually expect a t ok, but now for instance I am simply saying that let me do like that. So, at first I am just talking about simple coordinate system simple Cartesian coordinate system. So, I will have x 1 x 2 up to x capital N number of components ok. In the system of generalized coordinates you will actually have this R to be another function of q 1 q 2 now this will go to q small n and t ok. Here we are actually talking about t ok in case we include t over here as well ok. So, I do not know like you can or you may or you may not ok. So, for the operational generality for the time being I am actually including the t ok. So, I mean why not because it will somehow give us some extra information and which can be helpful for solving a larger class of problems ok. So, if we have this then I know that one possibility of connecting this thing to this thing is by their transformations ok. So, and that is the only way by which you can represent the position vector in Cartesian coordinates and also in terms of the generalized coordinates ok. So, if I can write this then what I will write. So, here one thing is actually somehow necessary to mention that here the supposition is we assume that we know the transformation from x 1 x 2 x n t to q 1 q 2 q small n t. So, that means, that each of the x i's will be represented as a combination of q i's and t ok, but finally, for t this is not needed. So, I will just write for the sake of transformation rule at least I can forget t from here and I can say I can just saying I am just trying to represent the all the Cartesian components in terms of all the generalized coordinates and t ok I can do. Now, if I write this then I know d x i will be equal to summation over J del x i del q j d q j plus del x i del t d t. So, here you see that since x i is one of the capital N Cartesian components. So, this is a function of all possible q's and t here and I am just writing without losing any mathematical generality this expression for the total differential of x i. You can also write I am just writing this in another color the virtual displacement also for x i and you can now be telling me what will be that from the definition and concept of virtual displacement that will over J del x i del q j delta q j ok. Now, what will be that term here counter part of this term equivalent to this term ok there will be no term like this because delta t is 0 ok. So, in a virtual displacement only the generalized coordinates move ok, but not the time. Now, if you have this. So, from this one first of all ok. So, here I keep this I mean expression because that will be useful for us in defining some generalized concepts ok, but here what I can do from this one I can really derive something much more realistic that is x i dot. So, x i dot is simply d x i d t ok. So, here you see that this is one another advantage of using Cartesian coordinate system that is if you just want derivatives of r or r dot you do not want to think of the unit vectors time derivative because unit vectors are fixed in time ok. So, that advantage we are using here and we are simply saying that if you want to just find the x component of velocity it will simply be given by the I mean the sorry the i-th component of the velocity in Cartesian coordinate system it will simply be the dot of the magnitude of the magnitude of the generalized coordinate that will be given by the dot of the i-th generalized coordinate directly without carrying the unit vector. So, if we write that that will simply be given by. So, d dt. So, I can now I mean put this expression here and which will give simply summation over j del x i del q j d q j d t plus del x i del t d t by d t which is one of course, ok. So, you have an expression for the velocity corresponding to the Cartesian velocity corresponding to the position or generalized coordinate x i ok. Now let me ask you a simple question what is the j? So, the sum is done over what? The sum is done over the number of generalized coordinates. So, j will be running from 1 to small n and not 1 to capital N necessarily only when small n and capital N they are same of course, you can write ok otherwise not. So, you have this expression right. So, I just from this one I can actually try to write the most general expression because now you see that for Cartesian system I have this component wise equality with this. So, for example, what I am just trying to communicate is that if you write r vector is equal to x i cap plus y j cap plus z k cap ok. So, r dot vector will be. So, you know that r vectors 1 component number 1 is x ok. I can actually write for the sake of simply and x 3. So, I mean the first component is x 1 and that is why while I am writing r dot it will be x 1 dot i cap plus x 2 dot j cap plus x 3 dot k cap and my r vector dot the first component will be simply x 1 dot. Whereas, in case of plane polar coordinate system for example, you have r is equal to r r cap and you know that what is the for example, so if I now say that this is the first component and there is 0 theta cap plus 0 phi cap of course, this is the definition then r vectors first component is r. So, would that say that the r dot 1 will be simply equal to r dot ok. You can easily check ok you can actually say yes that is true ok, but then the surprise begins you say that ok r vector 2 is 0. Now does it mean that ok I am actually here I should not consider phi cap because I said it is plane polar coordinates ok. Now if it is r 2 is 0 then what will be r dot 2 is it 0 unfortunately no. You know that this will be equal to r theta dot that is the component along theta cap and this is only because that the r cap and theta cap unit vectors are not independent of time ok. They are not 0 d dt of r cap and d dt of theta cap they are not 0 ok. This we have already discussed while discussing the velocity and accelerations in 2D plane polar coordinate systems ok. Now for this reasons we will actually use Cartesian system, but now for the sake of doing some general treatments and derive some general expressions I will somehow go back to writing vector r as it is ok. So, I will now do that. So, first I will write that ok I have the vector r and so, r dot is nothing, but dr vector dt which is dr dt ok. So, here I am writing just the vector. So, this is actually independent of the chosen coordinate, but if you choose the coordinate to be the Cartesian one then even for every component you can write identical relations that is the only advantage q j dot ok. So, now that is what I said that for Cartesian the advantage is you have x 1 or rather for any x i actually ok i running from 1 to 3 for a 3D space dot will be simply equal to d dt ok will be equal to ok summation over. So, I have written something wrong over here that should be del r del t yeah plus this yeah that will be del x i dot del t ok plus summation over j del x i del q j q j dot ok that is the only advantage if you start from Cartesian system. So, you see here in this notation you have something to observe that is r was initially a function of the generalized coordinates and time. So, when you do del del t ok it will not import any q dots because that whether you it will act on a q which gives simply 0 because del q del t is 0 dq dt is not 0 del q del t is 0 because q's q dots and t's all are independent. So, unless there derivatives are always 0 ok. So, again another possibility is that the del del t is acting on where it will give 1. So, no chance of q dots for this one this one has the same problem because r is actually with only q's and t's. So, while acted upon by q j's it will simply give you a function of q and t ok and then you have q j dots. So, this is the term which actually imports q dots in the expression of r dot and that is why finally, one can write r dot is a function of q i q i dot and t ok where all the i's can run from 1 to small n here. So, now, if I know that I would be interested to calculate two things, the first one is the kinetic energy the why kinetic energy I denote the kinetic energy by the symbol T. So, kinetic energy because for our incoming formulation it will be super important. So, T is half m r dot square and actually you can write half m r vector dot dot r vector dot ok. So, if you now know this expressions in terms of the generalized coordinates. So, you can actually find just by knowing the I mean the expression for kinetic energy in the Cartesian system which is nothing, but of course, always half m x dot square plus y dot square plus z dot square in 3D and actually if you want I should actually write or half m x 1 dot square plus x 2 dot square plus x 3 dot square or for an n dimensional case it should be simply small n dimensional case because you I mean sorry capital n dimensional or for a capital n dimensional space you should have this up to x n dot square ok. So, that is the general expression I can write here n dimensional space ok. So, in n dimensional space and for Cartesian system we also know that the system actually constitutes an orthogonal system with constant unit vectors ok. But this may or may not be true for a system of generalized coordinates that is also a very important point to understand the generalized coordinates may or may not create a set of orthogonal directions ok. May be generalized coordinates actually sometimes may not create a usual coordinate system at all. So, I am just trying to say you that if you have this expression and you know the expression of every x i dots in terms of the in terms of time and the generalized coordinates and generalized velocity by using this expression then actually you should be able to write the expression and I am just writing now this expression. So, it will be actually. So, if I have r dot is equal to del r del t plus summation over j del r del q j q j dot then r dot contracted to another r dot will simply be given by del r del t plus summation over j del r del q j q j dot times I mean scalar del r del t plus summation over I am taking another index k del r del q k q k dot ok and I want now t. So, it will be half m times this. So, it will be m by 2 the whole thing and you will have finally, the relation that t will be equal to half summation over j and k m del r del q j del r. So, dot del r del q k q j dot q k dot plus summation over k m del r del q k dot del r del t plus half m del r del t whole square ok. So, this is the most general expression for the kinetic energy in terms of the time generalized coordinates and generalized velocities. Now, you see if your system has somehow the property which is the most usual case you encounter that is the del r del t is 0 that means, there is no explicit time dependence in the position vector ok. So, for example, x is equal. So, for example, r is equal to x i plus y j plus z k or it is equal to r r cap ok in both cases there is no explicit dependence of time ok. So, the in that case we can actually say del r del t will be 0. So, this term will go away this term will go away. So, your t will be. So, I am writing that for del r del t is equal to 0 t is simply m del r q j del r del q k q j dot q k dot ok. So, you see finally, your kinetic energy becomes simply a quadratic function of the generalized velocities ok. So, this is something a very important result we need that actually. So, I will now at the very end just very simply tell you that why this type of situations are very important calculating this transformations between Cartesian systems and the generalized coordinates because sometimes this actually simplifies your calculation. So, for example, let us say I do an exercise that you have x y z system and you are just talking of a sphere and a particle is moving on the surface of the sphere as usual the length I mean the radius is of constant length. So, you can say that the generalized coordinates are theta and phi and I now know the transformation which is given by x is equal to r sin theta cos phi y is equal to r sin theta sin phi z is equal to r cos theta in the whole business this is something very important that we should know this transformation rule we know this. If we know then we can do many things. So, this r this is the radius and this is a constant. So, what we will do we will try to just find the velocities. So, for example, the x component and y component of the velocities in terms of the generalized coordinates theta phi. So, this is again this is nothing, but d dt of x and this is equal to del x del t plus del x del theta theta dot plus del x del phi phi dot phi dot. Now, what is this? This is 0 and del x del theta is nothing, but. So, r cos theta and cos phi theta dot plus del x del phi will be. So, minus r sin r sin theta and there will be sin phi phi dot you can do the same calculations for v y and you will have. So, for v y you will have del y del t plus del y del theta theta dot plus del x del phi phi dot and what you can have you can have r. So, cos theta once again this is 0 r cos theta sin phi theta dot and then plus r sin theta cos phi phi dot ok. So, yes I have this and finally, I have also v z and v z is nothing, but minus r sin theta theta dot. You now do the total kinetic energy which is nothing, but half m by definition v x square plus v y square plus v z square and you write the expression in terms of theta and phi. So, you will see if you do the v x square plus v y square plus v z square you will have. So, theta dot square you can collect and you will have r square cos square theta cos square phi here and you have r square cos square theta sin square phi. So, you will have r square cos square theta then you will also have r square sin square theta. So, finally, you will have r square. So, I can write this r square plus phi dot square. So, I have this term and this term from where I can see this is nothing but r square sin square theta phi dot square. You can do this using the directly the law of transformation and geometry or you can actually do this using this rule of transformation from here directly and using this coordinates. So, you see that you can also do that for any arbitrary generalized coordinate system. Now look, here in this whole mathematical business we did not actually use the very concept of the second system to be a system of generalized coordinate. This is actually true for the transformation from one coordinate system to any other coordinate system. Exactly when the generalized coordinate systems will come into play is in those summations ok. For example, once again J is again 1 to small n, k is equal to 1 to n ok small n ok. Now before just ending this lecture, I will just give you some task to do as usual and I want you to try that. So, r double dot how to find the expression of r double dot in terms of q, q dot and t ok. So, this will be d dt of del r del t plus summation over again J 1 to n ok del r del q J q J dot ok and then what to do that you will actually act this thing ok on this things. So, first you will see that you will go inside. So, first you will you will just see that this del del t and this one can show they of r dot ok, but when this act on this two things then you will have summation over J first d dt acts on del r del q J q J dot plus summation over J d dt q J dot times del r del q J ok. Now this is nothing, but q J double dot ok. So, if you would do everything completely finally, one should I mean derive this expression which I write and I want you to do. So, this is an exercise to do ok that is it should be given by something like double summation on J and K del 2 r del q J del q K q K dot q J dot ok plus summation over J del 2 r del q J del t q J dot plus summation over del r del q J q J double dot of course, summation is on J plus summation over del 2 r del q K del t r vector here ok q K dot on K and finally, plus del 2 r del t 2 ok. Again we can make some simplifications by saying that my r is not explicitly depending on t and that is why this. So, for del r del t equal to 0 I can actually make this term to be 0 ok this term to be 0 because there is a del del t this term to be 0. So, you will have one term ok which is quadratic in generalized velocities and you will have another term which is carrying the double time derivative of the generalized velocities. Now, remember just like from your memory what was the expression for a 2 D coordinate system and one particle is moving under a central force. So, you said that the the system of symmetry is actually r theta. So, what was the acceleration in r direction and theta direction I mean r cap direction and theta cap direction they were at double dot minus r theta dot square and there were terms like 2 r dot theta dot or r theta double dot. So, you will always have some terms which have quadratic terms for example, r dot theta dot like this or you have r theta dot square that is also a quadratic term in dots and also you have terms like this where you have double dot for example, r theta double dot or you have r double dot. So, this type of thing you can do what I suggest that you derive this expression and from this right from this using plane polar coordinate system and spherical polar coordinate system you derive the expression for r for a particles motion moving on the surface of a sphere of constant radius and in 2 D moving on a circle of fixed radius capital R ok. Thank you very much.