Physics+ Noether Conservation: UNIZOR.COM - Physics+ 4 All - Noether Theorem

Hi, I'm Zor. Welcome to news education. [clears throat] So today we continue talking about neuter theorem. And uh the previous lecture was basically a conclusion of purely theoretical and relatively general case of neoto theorem in um extended uh configuration space. Now this lecture is about application of the general approach to three different um conservation laws. Momentum, angular momentum and energy which uh actually are considered as separate cases in some other lectures in in this particular course uh which basically follow in the menu this one. But I think it's very important to start from something very general and then apply to particular cases. But in any case in this particular um situation you have basically two different approaches. One is from the general to these three and then the next lectures will be like a separate uh proof independent of the general of each of these um conservation laws based on neutral theory. Okay. Now this is uh part of the course called physics plus for all presented on uniserve.com. The website is totally free and um it contains other courses as well. Uh you there is no subscription no uh advertising uh even sign in is optional. So it's all just pure knowledge for your consumption. Now every lecture has video and textual part and the textual part is usually well at least in the physics part of the course um more well precise with more details because during the lecture just for saving time I can skip something. So I do recommend you to basically get acquainted with both the video presentation whatever I'm doing right now and the parallel text which is on the same web page uh and you can go into the details. All right. So let me just very briefly remind what was uh discussed before. So first of all we did talk about um um generalized coordinates which are time and q1 etc q n. So time is well time and Q are um space coordinates in in this generalized uh configuration space which I have actually changed to y0 y1 etc yn to have uh basically unified formula for i with some index and the index can be from 0 to n zero means it's time and y means uh the space coordinates and uh it does actually make sense because it makes all the formulas much more readable. Okay, so next was um the fact that in the um extended configuration space which contains space coordinates and time as n plus first coordinate or zero's coordinate whatever you want to to call it. So in this extended configuration space the motion is represented by a curve which has basically some beginning and and the end. Beginning uh uh would be uh for time one which is equal to uh basically some some value and uh the ending would be t2. But both t and q and basically y are functions of some parameter because the curve in whatever the dimensional space can always be represented as parameterized uh formulas. So every yi there is y i of x where x is parameter which is uh changing from a to b and in particular y0 which is time at a would be t1 and y0 b would be t2. uh same thing every particular coordinate q1 etc qn from a would signify the space coordinate of the beginning of the motion and when x is equal to b so corresponding y of b would be the ending point of motion and motion is a curve in this n plus one dimensional space next there was a um action functional which basically depends on the whole trajectory if I'm not using the index it means basically all indices as a group. So that was uh basically on one hand it's integral from one time to to another and this is log jan of time and all the cubes and and it's derivative. So that's a regular log jan. So um apostrophe means derivative by time gt but at the same time we can do this uh substitution since t is a function of x. So I'm substituting the whole integral as integral by x of l of uh tq prime. These are functions of x and instead of dt I will change since t is a function of x it would be u index in this case by x means the derivative by x * dx because dt is derivative by x * dx and then I basically have to replace everything as a function of uh x right now t is a function of x without problem q is a function of x without problem q prime time which is actually dq by dt uh can be converted into dq by dx divided by dt by dx which is d qx by tx. So if I would change that my under integral I will have the function integral from a to b I put it l nice of basically y and yx dx where y means all the coordinates which is t and q as functions of x and yx is uh d qx by not d just qx is qx by by qx okay derivative by x divided of q divided derivative t now I mean uh obviously there are indices index i index i Okay. And y0 of x is actually dx. Now with all this my whole so this is not a typical drum because right now l is equal to l of t [snorts] q qx / x * tx. this one. So now L is basically function of all Y's and basically uh Y uh with X index derivative by X because we now have here from 0 to one indices which means time and space coordinates and this is also from 0 to one. Maybe I should really put instead of this I will put uh y i comma y i x comma so the whole group. So we have n + one of these coordinates from time to the last space and then n + one of the derivative by x. So now it's a function of x and in this particular terminology we have come up with the newer theorem which we have proven in the previous lecture and I will just put the result here. Okay, the result is now the action function should not change uh with uh motion or which means exactly the same thing. It's derivative by x supposed to be equal to zero. Um now um now um what kind of um condition we have applied to this functional to be action functional to be um invariant. Well we have to model the concept of motion ma mathematically and that we have modeled with transformation of coordinates. So epsilon transformation of coordinates is a mathematical model of motion basically. So that's how physics is modeled using mathematics and transformation of coordinates is basically a function. So if my uh old coordinates were y i where i is from zero to n zero means time and 1 2 3 n means space coordinates is actually changed to y epsilon whatever the epsilon is some kind of function of epsilon where epsilon is infinite decimal um uh variable which signifies just an infinite decimal movement. And if my action function, if I will do this, which is basically a speed of change of my functional, my my action at epsilon is equal to zero. Now if this is equal to zero, it means that my functional is um invariant, doesn't really change. And that's the main condition from which we have derived in the previous lecture the final result which is something which is equal to sum from 0 to n this uh um a function which contains lrjan times t uh derivative of t by x um derivative by yx time zeta i equal constant where zeta i is derivative by epsilon of [snorts] y oops i ep epsilon So this is where my i uh y i's converted as a function of epsilon. So this is basically a change in the corresponding coordinate speed of change of the corresponding coordinates. If y is equal to 0 that's the time. If y if y is equal from 1 to n that's the coordinate. And this is the speed of change if we apply this transformation. Which means if there is a motion by infinite decimal um I know distance if you wish but again distance in extended coordinates. Um so if this is the speed of um change of the coordinates when the motion starts then this is supposed to be um conserved quantity and that's basically the end of the theorem uh neuter theorem. Now these things are called generators by each coordinate corresponding. Okay. Now what I'm going to do today is apply this to certain uh conservation laws based on certain very simple uh motions. So my first motion is let's just consider that some Q k um is changing to q k epsilon which is equal to q k + epsilon. What does it mean? It means that we are just moving along one particular space coordinate. Time is not changing and other not equal to k uh coordinates are not changing. What does it mean? Well, it means that again my Z is equal to D by DQ of Y I of epsilon. Right. So in this case the derivative by I is equal to 1. So the Q I mean zeta K is equal to one and all others are not changing. I said which means there is just for example t epsilon uh is equal to t. So there is no epsilon involved here since this is a constant. So the constant derivation by epsilon is equal to z. So one particular z is equal to one and all others are zero. Okay. So what does it mean for our case? For our case it means that the J which is supposed to be the uh invariant uh conserved quantity by not theorem would be equal to only for i is equal to k to to k this thing is one for others is zero so that's probably means this y k derivative by x so that's my property which is conserved. Okay. Okay. So now if this is the property which is concerned conserved well um then uh what we can do [clears throat] is say that um it's okay since this property is conserved let's just consider what this is was um it's derivative of L by Y derivative of Y K derivative by Y K and then derivative by X that that's that's what it actually means. So So let's do it this way. Derivative by Y K of this thing is constant. That's what it means. Uh, sorry. And and then derivative by derivative by kx sorry is constant. Yeah, that's what it means. Okay. And now considering that we know what um function L is let's just see what what this thing means. Uh so again L l is equal to this is log jan you remember it was t uh y and instead of uh derivative by from y by t we do we did yx by tx * dx. So this is derivative by X from Y and this is well actually I can probably have Q here. So it will be a little bit less abstract and Q here. Okay. Now we are actually now K derivative by Y K. Now yk is the same thing as q k. So that's it's not time it's a space coordinate. We're talking about only space coordinates right now. So I put it by cube. Uh okay. Now we want to differentiate it by qk. All right. Um if this is the function uh which is the product of two functions. So my uh derivative would be derivative by this function time dx plus derivative by this times this. Okay. So let's start with the second one. derivative by this times this. So that would be just L, right? because derivative of TX um uh uh is is is is basically one second sorry no it's not this we are differentiating by by q kx okay so this is I all the different I and this is I and we differentiate by q k x which is here. So only one parameter here. So all other parameters now usually if you have a function of multiple arguments it's private it's a partial derivative by first argument and then from first by who whatever we are differentiating but C is not really dependent on Q K derivative by X Q is also not dependent on its own there is only one particular member here so um we Now so we have to uh partially derivative of L by D let's call it V K where VK is Q K X / T K X because this is actually the derivative by time. Remember we just replaced it with this. So partial derivative of this times um derivative of this by qk which is 1 / t uh tx but I forgot that k doesn't have actually index t is without index only q has indices and multiply by tx tx is just a constant this because it doesn't will depend on this. So this is so this is this dx derivative of t by xx is cancelling and now we have d l by d vk. Now this is something familiar. What is this derivative of the lrjan by by speed actually? What is q k uh x / t uh x? This is d I would say q this is derivative of space coordinate q k by time. So this is derivative by time by by speed by Q. Okay. Which is what? This is the definition of momentum along the case coordinate. So what we have come up with that this thing is constant means this is constant and this is a momentum. So if my mo motion is very uh smooth y k epsilon is equal to y k + epsilon that's what we started from that means it's uh the motion which actually is uh the motion with a constant speed along one particular uh axis of coordinate. Well, obviously we know that in this particular case momentum is conserved. This is the law of conservation of momentum. What I have just done, I have derived the law of conservation of momentum from the newer theorem which we have derived from other properties from the symmetries of the of the space transformation. So symmetry of the space transformation in mathematical language looks like um this type of transformation of coordinates and it results in the um momentum linear momentum conservation. That's it. That's the first conservation law which I wanted to talk about today. Now the second conservation law is exactly the same. It's an angular momentum. Now, how do we um put into coordinates the motion the rot rotation of of the solid body? Well, there are two parameters time and angle of rotation as a function of time, right? There is absolutely no difference between this and this because we are talking about generalized coordinates and in in generalized coordinates language in the language of lranjan and action functional there is no difference what kind of coordinates you're using these are generalized coordinates not necessarily ukidian uh which basically define the position of the uh mechanical system. In this case there are only two coordinates. So n is equal to 1. n + 1 is equal to 2. So we have two coordinates time and uh and angle. And the whole thing actually is exactly the same. I'm not changing anything except I'm calling instead of Q I'm using uh um the the TA angle which means that I can put here DL by D theta prime constant which is a definition in the generalized coordinates. This is a definition of uh angular momentum. So I'm not going to do any kind of a proof. That's that's the proof basically which I did the first time considering the coordinates are generalized. It's exactly applicable to in in this case. So that's the angular momentum and the third one is conservation of energy. This is uh slightly differently because I mean it's good that I have kind of generalized all coordinates t and q space and time as y but now we are talking about separation of these things. Okay. So what happens with energy? Now uh let's say our time is transformed to and all others coordinates do not transform. one. So that makes my generator with an index zero equal to one. That's a derivative by epsilon at epsilon is equal to one to zero and all other generators are zero. Okay. So this is only transformation of time which means we are experimenting today and then we are experimenting in exactly the same position but tomorrow let's say not tomorrow infinite decimal time later let's put it this way and again let's recall our uh function the the not function that J which is J which is equal to sigma from i 0 to n this coordinates i a by x * i is constant preserved. Now in this case only 0 is actually um equal to one and all other indices are zero which means that j is equal to only the first one which is L by Y0 which is TX constant. So derivative of this uh extended lanjan uh by TX is uh constant is conserved. Well, let's now calculate the whole thing again remembering that CL let me is equal to L of um T uh Q and again Q X / t x * t x and here I actually mean all indexed values here group group group arguments. Okay. So now we have to calculate the derivative by tx and again I will use instead of this why is it not working? take a look um I will use V I okay so let's just do the calculation all right now in this case we have both components this is the product this depends on dx and this depends on tx and we have to differentiate product. So it's uh this one by derivative of this plus derivative of this by dx times this one. All right. So it's um derivative of so it's this times derivative of this. So it's L times derivative of this by TX is one right. So L is just a like a multiply plus derivative of this by TX. Now the only thing which [clears throat] well all these coordinates um Q1, Q2, Q3 etc. They're all divided by the same tx. So that should be sum by i from 1 to n uh derivative of l times by this argument which is vi times derivative of u each coordinate which is uh derivative by TX which means this is a multiplier. So it would be uh q ix and derivative of 1 / tx is minus so it's minus would be here d x² right derivative of 1 / tx by tx would be - 1 / tx² times this one. So which means this and by the way what is this derivative by x of qi divided by derivative of x by tx? Well derivative by derivative is derivative of qi by t which is velocity. That's what it is. So our formula is this one. This is constant. Okay fine. So we know this is constant. This is conserved quantity from Y theorem. My question is what is this? Well, obviously in general you can leave it as this [clears throat and cough] assuming that vi is actually q i prime which means time derivative. [snorts] However, in most of the CA on all cases which we are uh considering so all the stationary mechanical systems um uh conservative mechanical systems I have to say all the mechanical systems in all these cases L is equal to T minus U kinetic energy minus potential energy right so Okay. And um what is kinetic energy? Kinetic energy in again all the conservative systems which we are discussing it was actually a quadratic formula of the speeds. Like remember if you have one particular u uh point mass moving with a speed uh v it's u kinetic energy is mv² / 2 right so in all systems when more than one dimension is considered if you consider this v i * vj this is [snorts] i and j from 1 to Yeah. So this is a general homogeneous quadratic form. This is the kinetic energy. The kinetic energy for all these um conservative mechanical [snorts] systems is a homogeneous quadratic form of the uh velocities. Right? Now if this is true then what is this particular expression and this particular expression? Well now we're basically talking about only this that's important. So it's very easy to prove that this part. So again J is equal to L minus sigma D L * Z* VI [snorts] I so there is a theorem that if L is equal to G minus U and T is sigma a i j i j then d l by d v i * v i sigma is equal to [snorts] 2t. Now this is called oilers's theorem. It's very easy to prove and the proof is in the notes for this lecture. It's like three or five lines of code. I just don't want to do it right now. It's very easy to prove. And what's the result of this? The result of this is by the way u is independent of uh the velocity. So basically derivative of L is derivative by by T. So I can put as well T here. Doesn't really matter. Just put T. But D is this. So what's the result? The result is J is equal to L minus uh 2T which is L is this minus T + U right? T minus 2 T is minus T and minus U is which is energy with a negative sign. So this E is a preserved conserved quantity. So energy is conserved quantity if we are changing only the time coordinate from time from um transformation of time which is symmetrical which means preserves the action functional follows the preservation of energy. So the law of conservation of energy depends on certain qualities of our space which means well time space. Um if you are changing the time in a symmetrical way so to speak which means it preserves the action functional then the law of conservation of energy is a consequence logical mathematical consequence of this. Just think about what we have basically demonstrated with these three laws that each law depends on certain symmetry. So my energy is conserved is generator uh by time is equal to one and all the others are zero. My uh linear momentum is only one particular case coordin case generator is equal to one and the rest is zero. And angular momentum is also when only two coordinates time and angle then the corresponding time generator would be zero and angle generator would be one. So all these are just different cases of how the transformation actually is done or how the motion is done. So the motion in time motion in space if they are symmetrical in a way that they preserve the action functional from the new general theorem follows these four particular cases of motion this is a time motion I mean change of the time we are doing the same thing but later let's say or the space if we are doing the same thing but in a different uh location. If this different if we are changing time or we are changing the location in a smooth way in a nice way so that our symmetry preserved our action function or preserved then follow the corresponding conservation laws. Now there are some other lectures in the same um chapter of the course which is dedicated to meter theorem and these lectures three lectures are dedicated to these three laws um momentum linear momentum angular momentum and energy momentum but they are actually derived in a kind of from the beginning without generalizing the newer theorem in completely general case and then going to specific cases. They are just specific cases by themselves and it's also you know certain it presents certain interesting way of doing this but I separately uh defined all these components. I separately proved uh the preservation of linear momentum, angular momentum and energy um in in in some way may be similar form to a generalized but still some specifics for every particular case. Now this lecture is about how to use the generalized approach first and then put into particular cases based on certain values of generators. Okay, that's it. I do suggest you to read the notes for this lecture. These notes are very detailed and uh basically all the calculations are correct. That includes by the way this I was mentioning the similar u uh oil theory. I proved it in a very I think nice way. So I do suggest you to read all these notes. It's very very important. And obviously as everything else [snorts] uh the website again is totally free. I recommend you to look all the courses which are there and there are courses in math for uh math plus physics for includes a lot of different uh uh components different chapters and well this is already a next chapter of [snorts] physics plus which is looking John and then after that I will go to Hamiltonian all the preparations for something in in the future. That's it for today. Thank you very much and good luck.