Path Integrals Problems: UNIZOR.COM - Math4Teens - Calculus - Integrals Part 2 - Path Integrals

Duration: 32:38 3529 transcript words EN summary

Video summary

This lecture continues the discussion on path integrals, also known as line or curve linear integrals, by focusing on solving specific problems to illustrate their definitions and properties. The instructor emphasizes that while physicists often prefer the term "path integral," the terminology does not change the mathematical concept. The core of the lesson revolves around conservative vector fields, which are crucial in physics because they represent forces where the work done depends only on the starting and ending points, not the path taken. Consequently, the integral of a conservative field along any closed loop is zero, as the positive and negative contributions along different segments of the path cancel each other out perfectly. The first major problem addressed is determining whether a given two-dimensional vector function is conservative or non-conservative. The lecture establishes a necessary condition for a vector field to be conservative: if a scalar potential function exists such that its partial derivatives match the components of the vector field, then the mixed partial derivatives must be equal. Specifically, the partial derivative of the x-component with respect to y must equal the partial derivative of the y-component with respect to x. If these values are not equal, the function is definitively non-conservative. While equality of these derivatives is a necessary condition, it is also sufficient provided the domain is simply connected, meaning it has no holes or gaps that could disrupt the continuity of the potential function. To demonstrate this concept practically, the video walks through an example involving a vector field with components $xy$ and $x+y$. By calculating the mixed partial derivatives, the instructor shows they are not equal ($x$ versus $1$), proving the field is non-conservative. This theoretical finding is then verified by calculating the path integral along two different routes from the origin to the point $(1, 2)$: a straight line and a two-segment path. The calculations reveal that the integrals yield different values ($11/3$ versus $4$), confirming that for non-conservative fields, the result depends on the specific path taken between two points. The final segment of the lecture applies these principles to a physical scenario involving gravitational force between a sun and a planet. The instructor derives the vector form of Newton's law of gravitation, showing how the force vector points toward the center of mass (the sun). By expressing the force components in terms of coordinates $x$, $y$, and $z$ and identifying the scalar potential function proportional to $1/r$, where $r$ is the distance from the origin, it is proven that the mixed partial derivatives are indeed equal. This confirms that gravity is a conservative force field, meaning the work done by gravity moving a planet depends only on its initial and final distances from the sun, regardless of the orbital path taken.

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Hi, I'm Zor. Welcome to Iner Education. [clears throat] Um, today we will continue talking about pass integrals. Now, the previous two lectures were dedicated to definition of the pass integrals. Pass integrals sometimes is called line integrals or curve linear integrals. Now, in this lecture, we will just try to solve a couple of problems. Well, three problems to be exact. Uh as an illustration of the definition and the properties of the pass integrals. Um the term pass integrals is more preferable for physicists and there are many application in in physics of these pass integrals. So that's why I'm calling them pass integrals but people call them line or curve linear doesn't really matter. Now this lecture is part of the course math proteins presented on unisord.com. Uh more precisely it's a part of the course which is dedicated to calculus and within the calculus there is a chapter uh called uh pass integrals. So this this is how you get into this particular lecture. Now every lecture has video and text part on the same website. So you can watch the video, you can read the text which is basically like a textbook in any order whatever you prefer. But I do suggest to to take both sources of information, the video information, this this type of a lecture and the text. Now the website unisurid.com is totally free. There are other courses like math plus, there is a physics routines, physics plus um and some others. Okay. So let's go to little problems which we have for pass integrals. Now [clears throat] the previous lecture uh contains certain properties including the property of the vector function uh to be conservative. So let me just remind that if you have let's say you have two-dimensional vector function which means for every point xy in cartasian coordinates um the vector is defined and vector means it has x component and y component. So I will use this type of so this is the x component of the vector. This is y component of the vector. Okay. Now um and there was a property of the vector function which we called conservative. Now the conservative vector functions have a uh basically such a definition we can write it as partial derivative of some scalar function. So lowerase f is a scalar function. This is by dx and f_sub_y at xy is equal to df by dy same. So if there is some kind of a scalar function lowerase f uh defined on the same uh domain as the vector function capital f such that partial derivative by x of this scalar function lowerase f is the x component and partial derivative of y is y component of our vector function. Then vector function is called conservative. And one of the properties of conservative function is the integral of this vector function integral by some uh curve on a plane in this particular case. So it's two dimensional case of vector function. I will put it this way as as as a pass integral which uh basically by definition is integral by curve f_sub_x dx + f_sub_y dy. Now this integral this pass integral depends only on a and b points and does not depend on how we move from a to b. So integral by any pass from a to b would have exactly the same value. It depends only on points a and to be more exact depends on the value of function lowerase f at point a and point b and is independent. So this has been proven in the previous lecture. Uh consequently by the way integral by any loop from some point along any path back to the same point would be equal to zero like in this case because this is a positive this is a negative but their value um from a to b would be the same so they negate each other would be zero. So this is all from the previous lecture. So today I would like to have as my first problem how to find out if given vector function given means we have these two functions of two arguments x component and y component. So that's what definition of the vector is vector function. So if we have these two functions of two arguments, how to determine whether this uh is conservative or non-conservative function? It's important in physics by the way. In other word does the lowercase f function of two arguments xy exist such as it's private it's partial derivative by x is equal to first component and partial derivative of uh lowerase f by y is y component of our vector function. I don't know if it exists. Maybe I can find it maybe not. Question is whether it exists or not. Now if I cannot find it just looking at this function and doing something maybe with this function. If I cannot find it does it mean that the function is non-conservative? No. It means I cannot find it because maybe it exists and I'm just not smart enough to do it. So question is how to have some kind of a necessary condition for these functions which assure us at least that it's uh definitely non-conservative. So if this is non-conservative function I would like to definitely know if it's conservative I have to find but if if it's non-conservative and and I can actually find out just looking at these two function that this is non-conservative so there is no need to find anything now if I cannot prove that they are non-conservative then maybe but it does make sense to find out how can I find function lowerase f. So right now I'm looking for a necessary condition for non-conservativism of the function vector function f and this is actually a very easy thing. Now if I will do the following I will do partial derivative of f_sub_x by dy. What is this? Well, assume the function is conservative, which means that this is the representation of f(x). Now, I'm partially uh differentiating by y. So it would be sorry d by dy of df of xy by dx. Now what is the function if I will do partial derivative of second component by x. Now if my function vector function is conservative then the second component can be represented this way and the whole thing is this. Now this is usually it's the second derivative. Yeah, [clears throat] if we assume that the function is conservative. Now then the second derivative the well derivative by y of the x component is this and derivative by x of the second component of vector function is this. Now if the function lowerase f exists then these supposed to be equal because x and y are independent variables. But there is actually a theorem which can be proven that under a very very broad conditions as long f lower lowerase f is smooth enough um these two are equal to each other. I'm not sure I covered it in my previous lectures but it's kind of feeling that this is the right thing to do. So if such a function lowerase f exists then these two derivatives derivative of the first component by y and derivative of the second component by x must be equal. If it's not if if they are not equal to each other then the function vector function capital f is non-conservative obviously. So again if it's conservative then f exists then we can have these two new partial derivatives and we should have the same result. If we are not having the same result then our initial assumption that the function is conservative is wrong. So we have come up with a necessary condition for vector function to be conservative. So the necessary condition is that derivative of the first component uh by y and derivative of second component by x. We're talking right now only about two dimensional case, right? So if they are equal then it's a necessary condition. If they're not equal then we can definitely say the function is not conservative. Now this is a criteria. Okay fine we have come up with this criteria necessary condition. Now is it sufficient condition? So if these are equal to each other, does it mean that there is ex that there exist lowerase function f? Okay, the answer to this question is slightly above the level of this particular lecture. But I can tell you that if the function f ofxy is defined on a domain which does not have it's called simply connected domain no holes and it's a one piece. So we are talking about domain on the um two dimensional plane. So it should not be something like this. So if this is domain now why is it not a good domain? It's not simply connected domain because it has a hole. Now maybe is this this type of domain it's not a good domain not simply connected domain because there are two parts of it. But if it's defined on some simply connected domain then this condition is also a sufficient condition. Okay. So that's my first problem. We have established a necessary condition and in most cases by the way in most nice cases it's also a sufficient condition for the function to be conservative. Now why am I specifying actually the conservative function here? Why I'm talking about conservative because in physics conservative function do play extremely important role and the vector functions are basically the forces which exist in the universe and the force is basically a vector which is defined at certain spots in space. So it's very very practical and very physics oriented. Okay, got that. Now the second problem is basically an illustration of the first. Okay. So let me just have one particular example of function vector function of xy. It has two components. First component is x * y. Second component is x + y. So for any pair XY we have two components of this vector functions X component this is X component and this is Y component projections if you wish uh so if you have something like this this is a point XY now for instance this is the vector f which is defined at this point. So this is projection on the x and this is projection. So this is f_sub_x and this is f_sub_y. So this is equal to x * y and this is equal to x + y whatever the points xy is. So that's my definition of test kind of a function vector function question is it conservative. Okay let's just do our criteria. We take the uh the y uh derivative of the x component. So dxy by dy and it's equal to x y uh we are differentiating by y which means x is a constant. So a constant goes out from the differentiation. Now the second component which is x + y should be differentiated by x. In this case y is just a constant. So uh differentiating of x plus constant by x that's one. They are not equal to each other which means function cannot be conservative. It's a necessary condition. not equal means non-conservative. So that's it very simple but let me exemplify that this particular case when the function is not conservative we are um actually have a difference between pass integral between two points. So if these are two points conservative function gives the same integral no matter which path you take. non-conservative function doesn't have this property. So I would like to have some example of two points A and B and two passes and have two two integrals two pass integrals not equal to each other. Okay. And here is my example. Let's just have two points. One is origin and another is one two. First I will do integral straight from 0 0 to one two and integrate this function. Second I will do this and this. it's different paths which in this case contains two segments and I will integral um both of them and and and and see the sum of these two whether it's equal or not equal for a conservative function I should have exactly the same value well let's check it out [snorts] okay first let's do integration from zero to one two straight. How can we do it? Well, you remember that pass can be defined parametrically. So x would be some kind of a function of t and y should be some kind of a function of t and then everything actually is um uh from from [clears throat] the two dimensional case we are going to integral by t. But in this particular case I will do it x = t and y is = 2t where t is from 0 to 1. So this is basically the uh definition of this path parametrically with t is equal to 0 we have this point 0 0 with t is equal to 1 we have 1 2 and with everything in between we have all these points along this straight line because it's a linear functions right in which case dx is equal to dt dy y is equal to 2dt right now my integral uh let's call it gamma 1 of fxy d gamma is equal to oops integral from okay let's just do one more like um xy dx + x + y dy right remember first component time dx that's basically definition of the pass integral which again we covered in the previous lecture so you have x component time dx and y component * dy. Now since we have expressed everything in terms of parameter t, now it becomes integral from one to from 0 to 1 for t. x is t, y is 2t, and dx is dt plus x is t, y is 2t, dy is uh 2dt 2d. dt equals. So what do we have? Integral from 0 to 1 2 t² + 6 t dt, right? which is 2. Integral of t² is a t cub / 3 + 6 t is t² / 2 from 0 to 1. That's the formula of Newton labs, right? [clears throat] So at zero they're all equal to zero obviously. So from one you have uh 2/3 + 3 3 is 9/3. So it's 11/3. Okay. So that's the value of integral along this particular path. Good. Now uh let's do exactly the same integral but along this path. which contains two segments. So let's do it one by one. So first segment from 0 to 1. Uh x is = 0. No sorry x is equal to t and y is [clears throat] equal to zero. Right? as I'm moving from 0 0 0 to 1 0 uh this is [clears throat] this is how my coordinates are changing. So my integral would be x * y * dx uh which is zero x * y * dx and dx is dt + sum x + y which is t * dy which is also zero. Obviously dy is equal to 0 * dt that's 0. So the whole thing is equal to zero. So integral of this function along this segment is zero. Now let's talk about that segment. X is equal to 1. Right? As I'm moving from here to here, x is always one and y is from 0 to two. So I can put 2 t where t is from 0 to 1. If t is = 0, y = 0, which is this point. And if t is = 1, y = 2. Okay, same thing. Now integral of um xy function xy the product is 2t 1 * 2t * dx but dx is equal to zero because it's a constant plus their sum 1 + 2t times dy which is 2dt D equals to from 0 to 1 integral from 0 to 1 2 + 4 t dt = 2. Integral of 2 is 2t. Integral of 4t is 4 t² / 2. And this is for the mutton wave from 0 to one. So now this is two two. So if t is equal to 1 this is four and if t is = 0 is zero. So the answer is four. So the first integral was 11/3. The second integral is 4. So they are close by the way but not equal to each other and that's the proof that they are not um that the function is not conservative. Okay. And the third problem which I would like to to investigate basically is more physical kind of a thing. So let's talk about gravity. We all know the law of gravitation learned in school that if you have let's say sun which has mass m and some planet which has mass lowerase m the uh force is equal to between them the force of gravitation of this planet towards sun let's put it this way is this and it's directed towards sun. Let's assume that the sun is uh the center of the uh coordinate system in space. R is the distance obviously and uh so F is the force and the force is directed from the planet to the uh uh to sun to the center of uh coordinates. Okay. Now let me call vector R which is X Y and Z which is position of the planet. Well then this R is actually modul of this vector the size basically of this vector. Right now I would like f to be a vector. So I know that the absolute value is of this vector is this one. But I need a vector which means what? Which means it has to have absolute value of this and it should have a direction from the planet to the sun which is opposite to this one. If sun is the center of the coordinates then vector r is from sun to the planet. I want the backwards. So the backwards would be minus r. Right? But I don't want to change the value of absolute value. I don't want to change the modul. So if I would divide this by r, this would be my unit vector which is directed exa exactly as I want from the planet to the sun because of the minus and it has a unit length because I divide it by its own length. So multiplying uh this value of the the modulo of the of the force by this and r is a vector I will get force as a vector which means I can actually do it this way. So and I can put it minus here and times r here. So this is my formula for vector f where r is xyz which means modula r is equal to square of x² + y 2 + z² that's the length of the vector right okay now uh I would like to prove that this is a conservative force How? By using function f ofx is equal to 1 divided by modul of air with maybe multiply by some constant whatever the constant doesn't really matter because if I'm differentiating uh f the constant goes out. So I would like to basically using the constant I would like to deal with all these parameters. But let me first check what is df by dx. Now this is x² + y² + z² in the power of since it's a denominator so it's -2 derivative by x which is what is - 12 * x² + y 2 + z 2 - 3 * differentiating uh in differentiating time 2 x right the derivative by x of this function uh is 2x so this goes out this goes out and what do I F. This is minus 1 / cube - 3/2 minus means it's denominator. Now 1/2 is a square root and three is a cube. Now square root is r. So what I have uh times x sorry x time x. So I have basically an x [clears throat] component of this. So differentiating this function this function differentiating by x gives me x component. Right? This means what? Means f x component is equal to this thing times x. f_sub_y is equal to the same thing time y and fz is equal to this thing time z and I have exactly this and I have a minus. So I'm using this constant now to multiply it and I have basically my function f. So the uh uh scalar function f of x is g * m * m / r uh by by modulo of vector r. So which is basically square root of x + x² + y 2 + z square. So that's it. So that's a proof that my function is conservative. Well, that's it for today. Thank you very much. Read the notes for this lecture and good luck.