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# PHYS 200: Fundamentals of Physics I

## Lecture 6

## - Law of Conservation of Energy in Higher Dimensions

### Overview

The discussion on the Law of Conservation of Energy continues but is applied in higher dimensions. The notion of a function with two variables is reviewed. Conservative forces are explained and students are taught how to recognize and manufacture them.

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html## Fundamentals of Physics I## PHYS 200 - Lecture 6 - Law of Conservation of Energy in Higher Dimensions## Chapter 1. Calculus Review: Small Changes for Motion in 2D [00:00:00]
Δx and so on. But if ^{3} Δx is tiny this will do; that’s it.But today, we are going to move the whole Work Energy Theorem and the Law of Conservation of Energy to two dimensions. So, when you go to two dimensions, you’ve got to ask yourself, “What am I looking for?” Well, in the end, I’m hoping I will get some relation like U is going to be my new potential energy. Then, if it’s the potential energy and the particle is moving in two dimensions, it’s got to be a function of two variables, _{2}x and y. So, I have to make sure that you guys know enough about functions of more than one variable. So, this is, again, a crash course on reminding you of the main points. There are not that many. By the time I get here I’ll be done.So, how do you visualize the function of two variables? The function of one variable, you know, you plot So, once you’ve got the notion of a function of two variables, if you’re going to do calculus the next thing is what about the derivatives of the function. How does it change? Well, in the old days, it was dependent on So, we’re going to define one derivative, which is defined as follows. You start at the point [Reader note: Partial derivatives will be written as ordinary derivatives to avoid using fonts that may not be universally available. It should be clear from the context that a partial derivative is intended here, since So, this curly Let’s get some practice. So, I’m going to write some function, x and y. You can write any–let’s make it a little more interesting, plus y, or y; that’s some function of ^{2}x and y. So, when I say df/dx, the rule is find out how it varies with x keeping y constant. That means, really treat it like a constant, because number five–What will you do if y was equal to 5? This’ll be 25 and it’s not part of taking derivatives because it’s not changing, and here it’ll be just standing in front doing nothing; it will just take the x derivative, treating y as a constant. You’re supposed to do that here; this is a derivative 3x; so that’s the ^{2}yx derivative. Now, you can take the y part, y, yes, thank you. Then, I can take ^{2}df/dy, so then I look for y changes, this is 2y here, so it’s 2x + 2^{3}yy; so that’s the x derivative and the partial y derivative.Okay, so then you can now take higher derivatives. We know that from calculus and one variable, you can take the derivative of the derivative. So, one thing you can think about is d by dx of the d by dx. That’s what it means. First, take a derivative and take the derivative of the derivative. So, let’s see what I get here. I already took df/dx. I want to take its derivative, right, the derivative of x is 2^{2}x, so I get 6xy. Then, I can take the ^{2}y derivative of the y derivative, d, that’s ^{2}f/dy^{2}d by dy of df/dy, if you take the y derivative there. You guys should keep an eye out for me when I do this. I get this. But now, we have an interesting possibility you didn’t have in one dimension, which is to take the x derivative of the y derivative, so I want to take d by dx of df/dy. That is written as d. Let’s see what I get. So, I want the ^{2}f/dxdyx derivative of the y derivative. So, I go to this guy and take his x derivative, I get 3x from there, so I get 6^{2}x and that’s it. So, make sure that I got the proper ^{2}yx derivative of this, I think that’s fine. Then, I can also take d. That means, take the ^{2}f/dydxy derivative of the x derivative, but here’s the x derivative, take the y derivative of that, put a 2y there, and I get 6x.^{2}ySo, you’re supposed to notice something. If you already know this, you’re not surprised. If you’ve never seen this before, you will notice that the cross derivative, Let’s take a function and let’s ask how much the function changes when I go from some point
Δx.In other words, I am saying the derivative at this location is the derivative at that location, plus the rate of change of the derivative times the change in Δf. But if you’re a little more ambitious but you keep track of the fact the derivative itself is changing, you will keep that term.But another person comes along and says, “You know what, I want to go like this. I want to introduce as my new point, intermediate point, the one here. It had a different value of Δy Δx. If you just do the whole thing in your head, you can see that I’m just exchanging the x and y roles. So, everything that happened with x and y here will come backwards with y and x. But then, the change between these two points is the change between these two points and it doesn’t matter whether I introduce an intermediate point here or an intermediate point there; therefore, these changes have to be equal. This part is of course equal; therefore, you want that part to be equal, Δx Δy is clearly Δy Δx, so the consequence of that is d = ^{2}f/dxdyd and that’s the reason it turns out when you take cross derivatives you get the same answer. It comes from the fact – if you say, where is that result coming from – it comes from the fact, if you start at some point and you go to another point and you ask for the change in the function, the change is accumulating as you move. You can move horizontally and then vertically, or you can move vertically and then horizontally. The change in the function is a change in the function. You’ve got to get the same answer both ways. That’s the reason; that’s the requirement that leads to this requirement.^{2}f/dydxNow, I will not be keeping track of functions to this accuracy in everything we do today. We’ll be keeping the leading powers in ## Chapter 2. Work Done in 2D; Dot Products and Cross Products [00:18:05]So, let’s now go back to our original goal, which was to derive something like the Law of Conservation of Energy in two dimensions instead of one. You remember what I did last time so I will remind you one more time what the trick was. We found out that the change in the kinetic energy of some object is equal to the work done by a force, which was some x. If _{2}F is a function only of x, from the rules of calculus, the integral can be written as a difference of a function at this limit minus that limit, and that was U(x, where _{1}) - U(x_{2})U is that function whose derivative with a minus sign is F. That’s what we did. Then, it’s very simple now to take the U to the left-hand side. Let me see, _{1}K to the right-hand side and _{1}U(x to the left-hand side, to get _{2})K and that’s the conservation of energy._{2} + U_{2} = K_{1} + U_{1}You want to try the same thing in two dimensions; that’s your goal. So, the first question is, “What should I use for the work done?” What expression should I use for the work done in two dimensions, because the force now is a vector; force is not one number, it’s got an I’m going to demand that that’s the power, force times velocity, is what I call the power. And I’m going to look for v has got now ^{2}v _{x}^{2}+ v, because you know whenever you take a vector _{y}^{2}V, then its length is the square root of the x part squared plus the y part squared. Any questions? Okay, so let’s take the rate of change of this, dK/dt. Again, you have to know your calculus. What’s the time derivative of v? The rule from calculus says, first take the derivative of _{x}^{2}v with respect to _{x}^{2}v, which is 2_{x}v, then take the derivative of _{x}v with respect to time. So, you do the same thing for the second term, 2 _{x}v, and that gives me–the 2s cancel, gives me _{y} dv_{y}/dtmdv. This is the rate at which the kinetic energy of a body is changing. So, what’s my next step, yes?_{x} /dtv_{x} + m dv_{y} /dtv_{y}
a, and this is _{x}m times a. Therefore, they are the forces in the _{y}x and y directions, so I write F. So the power, when you go to two dimensions, is not very different from one dimension. In one dimension, you had only one force and you had only one velocity. In two dimensions, you got an _{x}v_{x} + F_{y}v_{y}x and y component for each one, and it becomes this. So, this is what I will define to be the power. When a body is moving and a force is acting on it, the force has two components, the velocity has two components; this combination shall be called “Power.” But now, let’s multiply both sides by Δt and write the change in kinetic energy is equal to F and you guys think about what happens when I multiply _{x}v by _{x} Δt; v = _{x}dx/dt. Multiplying by Δt just gives me the distance traveled in the x direction + F times _{y}dy. So, this is the tiny amount of work done by a force and it generalizes what we had here, work done is force times Δx. In two dimensions, it’s F. It’s not hard to guess that but the beauty is–this combination–Now you can say, you know what, I didn’t have to do all this, I could have always guessed that in two dimensions, when you had an _{x}dx + F_{y}dyx and a y you obviously have to add them. There’s nothing very obvious about it because this combination is now guaranteed to have the property that if I call this the work done. Then, it has the advantage that the work done is in fact the change in kinetic energy.I want to define work so that its effect on kinetic energy is the same as in 1D, namely, work done should be equal to change in kinetic energy. And I engineered that by taking the change in kinetic energy, seeing whatever it came out to, and calling that the work done, I cannot go wrong. But now, if you notice something repeating itself all the time, which is that I had a vector F, I had a vector velocity, which is _{y}I times v times _{x} + Jv. Then, I had a tiny distance moved by the particle, which is _{y}I times dx + J times dy. So, the particle moves from one point to another point. The vector describing its location changes by this tiny vector. It’s just the step in the x direction times I, plus step in the y direction times J. So, what you’re finding is the following combination. The x component of F times the x component of v, plus the y component of F, times the y component of v. Or F component of x times the distance moved in x plus y component of F times the displacement in y.So, we are running into the following combination. We are saying, there seem to be in all these problems two vectors, A. _{y}A, for example, could be the force that I’m talking about. There’s another vector B, which is I times B times _{x} + JB. Right? For example, this guy could be standing in for _{y}F, this could be standing in for v. The combination that seems to appear very naturally is the combination A. It appears too many times so I take it seriously, give that a name. That name will be called a dot product of _{x}B_{x} + A_{y}B_{y}A with B and is written like this. Whenever something appears all the time you give it a name; this is A.B. So, for any two vectors A and B, that will be the definition of A.B. Then, the work done by a force F, that displaces a particle by a tiny vector dr, is F.dr. The particle’s moving in the xy plane. From one instant to the next, it can move from here to there. That little guy is dr; it’s got a little bit horizontal and it’s got a little bit vertical, that’s how you build dr. The force itself is some force which at that point need not point at the direction in which you’re moving. It’s some direction. At each point, the force could have whatever value it likes.So, the dot product, you know, sometimes you learn the dot product as So, once you’ve got the dot product, you’ve got to get a feeling for what it is. The first thing we realize is that if you take a dot product of A, which is the length of the vector _{x}^{2} + A_{y}^{2}A, which you can either denote this way or just write it without an arrow. So A.A is a positive number that measures the length squared of the vector A, likewise B.B. Now, we have to ask ourselves, “What is A.B?” So, somebody know what A.B is? Yep?
x axis, and that makes an angle θ for the _{B}x axis. Now, do you guys agree that the x component of A with the horizontal part of A is the A cos θ? You must’ve seen a lot of examples of that when you did all the force calculation. And _{A}A is equal to the length of _{y}A times sin θ and likewise for _{A}B; I don’t feel like writing it. If you do that, then A.B will be length of A, length of B times (cos θ._{A} cos θ_{B} + sin θ_{A} sin θ_{B})You’ve got to go back to your good old trig and it’ll tell you this A, length of B, cos (θ. Often, people simply say that is _{A} - θ_{B})AB cos θ, where it’s understood that θ is the angle between the two vectors. So, the dot product that you learnt–I don’t know which way it was introduced to you first, but these are two equivalent definitions of the dot product. In one of them, if you’re thinking more in terms of the components of A, a pair of numbers for A and a pair of numbers for B, this definition of the dot product is very nice. If you’re thinking of them as two little arrows pointing in different directions, then the other definition, in which the lengths and the angle between them appear, that’s more natural. But numerically they’re equal. An important property of the dot product, which you can check either way, is the dot product of A with B + C, is dot product of A with B plus dot product of A with C. That just means you can open all the brackets with dot products as you can with ordinary products. Now, that’s a very important property of the dot product. So, maybe I can ask somebody, do you know one property, significant property of the dot product?
A, which certainly doesn’t change on your orientation, or the length of B, or the angle between them. The angle θ will change but the angle with the new _{A}x axis won’t be the same. The angle with the new y axis won’t be the same. Likewise for B, but the angle between the vectors, it’s an invariant property, something intrinsic to the two vectors, doesn’t change, so the dot product is an invariant. This is a very important notion. When you learn relativity, you will find you have one observer saying something, another observer saying something. They will disagree on a lot of things, but there are few things they will all agree on. Those few things will be analog of A.B. So, it’s very good to have this part of it very clear in your head. This part of elementary vector analysis should be clear in your head.Okay, so any questions about this? So, if you want, geometrically, the work done by a force when it moves a body a distance Okay, I’m almost ready for business because, what is my goal? I find out that in a tiny displacement, ## Chapter 3. Conservative and Non-conservative Forces [00:36:16]So, now maybe it’ll be true, just like in one dimension, the integral of this function will be something that depends on the end points. I’m just going to call it F from start to end is really the difference of another function G of the end minus the start, with G as that function whose derivative is F.” Maybe there is going to be some other magic function you knew in two dimensions related to F, in some way, so that this integral is again given by a difference of something there minus something here. If that is the case, then you can rearrange it and get this. But you will find that it’s not meant to be that simple. So, again, has anybody heard rumors about why it may not be that simple? What could go wrong? Okay so, yes?
Well, let me ask you the following question. Another person does this. [Shows a different path from 1 to 2] Do you think that person should do the same amount of work because the force is now integrated on a longer path? So, you see, in one dimension there’s only one way to go from here to there. Just go, right? That way, when you write an integral you write the lower limit and the upper limit and you don’t say any more because it’s only one way in 1D to go from x. In two dimensions, there are thousands of ways to go from one point to another point. You can wander all over the place and you end up here. Therefore, this integral, even if I say the starting point is _{2}r and the ending point is _{1}r, that this is _{2}r and that’s _{1}r, it’s not adequate. What do you think I should attach to this integration? What other information should I give? What more should I specify? Yep?_{2}
xy, okay? I put the powers based on what you guys gave me. So, we picked the force in two dimensions out of the hat. Now, let’s ask, “Is it true for this force that the work done in going from one point to another depends only on the path, or does it depend, I mean, it depends only on the end points or does it depend in detail on how you go from the end points?” You all have to understand, before you copy anything down, where we are going with this. What’s the game plan? So, let me tell you one more time because you can copy this all you want, it will get you nowhere. You should feel that you know where I’m going but the details remain to be shown or you should have an idea what’s happening. If you try to generalize the Work Energy Theorem to two dimensions, this is what happened so far. You found a definition of work which has the property that the work done is the change in kinetic energy. Then, you added up all the changes of kinetic energy and added up all the work done and you said ^{2}K is the integral of _{2} - K_{1}F.dr; that’s guaranteed to be true; that’s just based on Newton’s laws.What is tricky is the second equality that that integral is the difference in a function calculated at one end point minus the other end point. If that was true, if the integral depended only on the end points, then it cannot depend on the path that you take. If it depends on the path, every path you take within the same two end points will give different numbers. So, the answer cannot simply be K + U so that it doesn’t change. So, it’s going to take a very special force for which the answer depends only on the starting and ending point and not on the path. To show you that that’s a special situation, I’m taking a generic situation, namely, a force manufactured by this class, without any prior consultation with me, and I will show you that for that force the answer is going to depend on how you go.So, let’s take that force and let’s find the work done in going from the origin to the point 1,1. So, I’m going to take two paths. One path I’m going to go horizontally till I’m below the point. Then I’m going straight up, okay. So, let’s find the work done when I go this way. So again, you should be thinking all the time. You should say if this guy got struck by lightning can I do anything, or am I just going to say “Well, I don’t know what he was planning to do.” You’ve got to have some idea what I’m going to do. I’m going to integrate dy when you move horizontally. So, when you do the integral, you have F. _{x}dxF happens to be _{x}x, ^{2}y^{3}dxx going from 0 to 1. Now, what do I do with y? You all know how to integrate a function of ^{3}x times dx. What do I do with y? What do you think it means?^{3}
dy, and y goes from 0 to 1, and the y component is xy, where ^{2}dyy goes from 0 to 1. But now, on the entire line that I’m moving up and down, x = 1. Do you see that? x is a constant on the line so you can replace x by 1 and integral of y is ^{2}dyy over 3.The work done is, therefore, 1/3 joules. So, the work done in going first to the right and then to the top is 1/3. You can also go straight up and then horizontally. By a similar trick, I won’t do that now, because I want to show you something else that’s useful for you, I’m going to pick another path which is not just made up of ^{3}x and y segments. Then it’s very easy to do this.So, I’m going to pick another way to go from 0,0 to 1,1, which is on this curve; this is the curve y = x goes through the two points I’m interested in. If you took ^{2}y = 5x, it doesn’t work. But this guy goes through 0,0 and goes through 1,1. And I’m asking you, if I did the work done by the force along that segment, what is the integral of ^{2}F.dr? So, let’s take a tiny portion of that, looks like this, right, that’s dr, it’s got a dx, and it’s got a dy. Now, you notice that as this segment becomes very small dy/dx is a slope; therefore, dy will be dy/dx times dx. In other words, dx and dy are not independent if they’re moving in a particular direction. I hope you understand that. You want to follow a certain curve. If you step to the right by some amount, you’ve got to step vertically by a certain amount so you’re moving on that curve. That’s why, when you calculate the work done, dx and dy are not independent. So, what you really want is F, but for _{x}dx + F_{y}dydy I’m going to use dy/dx times dx. In other words, every segment Δy that you have is related to the Δx you took horizontally so that you stay on that curve. So, everything depends only on dx. But what am I putting inside the integral? Let’s take F, the _{x}x component is x. On this curve ^{2}y^{3}y = x, so you really write ^{2}x, and ^{2}y = x, that is ^{2}x. This ^{6}x [should have said x^{3}^{6}] is really y written in terms of ^{3}x. Then, I have to write F, which is _{y}x times y, which is ^{2}x times ^{4}dy/dx which is 2x. So, I get here, from all of this, (x. Did I make a mistake somewhere? Pardon me?^{8} + 2x^{5})dx
x integral is ^{8}x/9. That gives me 1, 9 because x is going from 0 to 1. The next thing is ^{9}x/7, which is 2 times that. Well, I’m not paying too much attention to this because I know it’s not 1/3, okay? There’s no way this guy’s going to be 1/3, that’s all I care about. So, I’ve shown you that if we took a random force, the work done is dependent on the path. For this force, you cannot define a potential energy whereas in one dimension any force that was not friction allowed you to define a potential energy. In higher dimensions, you just cannot do that; that’s the main point. So, if you’re looking for a conservative, this is called a “conservative force.” It’s a force for which you can define a potential energy. It has the property of the work done in going from ^{7}A to B, or 1 to 2, is independent of how you got from 1 to 2. And the one force that the class generated pretty much randomly is not a conservative force because the work done was path dependent.## Chapter 4. Cross Derivative Test for Potential Energy Equations [00:52:29]So, what we have learned, in fact I’ll keep this portion here, if you’re looking for a conservative force, a force whose answer does not depend on how you went from start to finish, then you have to somehow dream up some force so that if you did this integral the answer does not depend on the path. You realize, that looks really miraculous because we just wrote down an arbitrary force that we all cooked up together with these exponents, and the answer depends on the path. And I guarantee you, if you just arbitrarily write down some force, it won’t work. So, maybe there is no Law of Conservation of Energy in more than one dimension. So, how am I going to search for a force that will do the job? Are there at least some forces for which this will be true? Yes?
So, for example, I’m saying suppose F, which is equal to _{x} = -dU/dx-y, and ^{3}F, which is _{y}-dU/dy would be -3xy. The claim is, if you put ^{2}I times F and _{x}J times F, the answer will not depend on how you go from start to finish. So, let me prove that to you. Here is the proof. The change in the function _{y}U, in the xy plane, you guys remember me telling you, is dU/dx times dx + dU/dy times dy. That’s the whole thing of mathematical preliminaries which was done somewhere, I forgot, here. It says, here, ΔF is dF/dx times Δx + dF/dy times Δy. I’ll apply that to the function U, but who is this? You notice what’s going on here? dU/dx times dx, this is F with a minus sign, that’s _{x},F with the minus sign. Therefore this is equal to _{y}-F.dr. Right? Because we agree, the force that I want to manufacture is related to U in this fashion. Now, if you add all the changes, the right-hand side becomes the integral of F.dr, and the left-hand side becomes the sum of all the ΔUs with a minus sign. Add all the ΔUs with a minus sign, that’ll just become U at 1 -U at 2. It is cooked up so that F.dr is actually at the change in the function U. That’s if you say, what was the trick that you did? I cooked up a force by design so that F.dr was a change in a certain function U. If I add all the F.drs, I’m going to get a change in the function U from start to finish and it’s got to be U._{1} - U_{2}So, I don’t know how else I can say this. Maybe the way to think about it is, why do certain integrals not depend on how you took the path, right? Let me ask you a different question; forget about integrals. You are on top of some hilly mountain. We have a starting point, you have an ending point, okay. I started the starting point, and I walked to the ending point. At every portion of my walk, I keep track of how many feet I’m climbing; that’s like my K. I hope you understand how conservative forces are not impossible to get. In fact, for every function _{1} + U_{1} = K_{2} + U_{2}U of x and y you can think of, you can manufacture a conservative force.So, you may ask the following question. Maybe there are other ways to manufacture the conservative force and you just thought of one, and the answer is “no.” Not only is this a machine that generates conservative forces, my two step algorithm, pick a Finally, you remember that when the class picked a certain force, which I wrote here, I went on a limb and I said I’m going to do the integral of this force along this path and that path and I’m going to get different answers and show you we have a problem. What if the force you had given me was actually a conservative force? Then, I would be embarrassed because then I’ll find, after all the work, this’ll turn out to be again 1/3. So, I have to make sure right away that the force is not conservative. How can you tell? One way to say it is, ask yourself, “Could that be some function
F; therefore, _{y}d over ^{2}Udydx is really d by dy of F. And I want that to be equal to _{x}d by dx of F because that would then be _{y}d over ^{2}Udxdy. In other words, if the force satisfies this condition, the y derivative of F is the _{x}x derivative F. Then, it has the right pedigree to be a conservative force because if a force came from a function _{y}U by taking derivatives, the simple requirement of the cross derivatives are equal for any function U tells me. See, if I take the y derivative of F, I’m taking _{x}d over ^{2}Udxdy. And here, I’m taking d over ^{2}Udydx and they must be equal. So, that’s the diagnostic. If I give you a force and I ask you, “Is it conservative?” you simply take the y derivative of F and the _{x}x derivative of F and if they match you know it’s conservative._{y}So, I can summarize by saying the following thing. In two dimensions, there are indeed many, many forces for which the potential energy can be defined. But every one of them has an ancestor which is simply a function, not a vector, but a scalar function, an ordinary function of y derivative is called F._{y}## Chapter 5. Application to Gravitational Potential Energy [01:03:55]Okay, so let’s take the most popular example is the force of gravity near the surface of the Earth. The force of gravity we know is F; it’s clearly conservative. Then you can ask, “What is the potential _{x}U that led to this?” Well, dU/dy with a minus sign had to be -mg and dU/dx had to be 0. So, the function that will do the job is mgy. You can also have mgy + 96 but we will not add those constants because in the end, in the Law of Conservation of Energy, K, adding a 96 to both sides doesn’t do anything. So this means, when a body’s moving in the gravitational field ½ _{1} +U_{1} = K_{2} + U_{2}mv^{2} + mgy, before, is the same as ½ mv, after. Now, you knew this already when you’re moving up and down the ^{2} + mgyy direction but what I’m telling you is this is true in two dimensions.So, I’ll give you a final example so you guys can go home and think about it. That is a roller coaster. So, here’s the roller coaster, it has a track that looks like this. This is So, if a trolley begins here at the top, what is its total energy? It’s got potential energy equal to the height, it’s got no kinetic energy, so total energy in fact is just its height. And total energy cannot change as the trolley goes up and down. So, you draw a line at that height and call it the total energy. You started this guy off at that energy, that energy must always be the same. What that means is, if you are somewhere here, so that is your potential energy, that is your kinetic energy, it’s very nicely read off from this graph, to reach the same total. So, as you oscillate up and down, you gain and lose kinetic and potential. When you come here your potential energy is almost your whole energy but you’ve got a little bit of kinetic energy. That means, your roller coaster is still moving when it comes here with the remaining kinetic energy. You can have a roller coaster whose energy is like this. This is released from rest here. It is released from rest here. That’s the total energy and this total energy line looks like this. That means, if you released it here, it’ll come down, pick up speed, slow down, pick up speed again, and come to this point, it must stop and turn around because at that point the potential energy is equal to total energy and there is no kinetic energy. That means you’ve stopped, that means you’re turning around, it’ll rattle back and forth. By the way, according to laws of quantum mechanics, it can do something else. Maybe you guys know. You know what else it can do if it starts here? Yes?
dr would be 0, because F and dr are perpendicular. For that reason, you’d drop that and the force of gravity is cooked up so then it becomes U and that’s what we used._{1} - U_{2}Okay, one challenge you guys can go home and do this, take any force in two dimensions,
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