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

Hi, I'm Zor. Welcome to Inor Education. So, we continue talking about pass integrals and today I will uh talk about the properties of pass integrals. Now, this lecture is part of the course called math for actually the part of the course which is called calculus uh which is presented on munisor.com totally free website. You don't need to sign in. Uh no advertisement, no subscription. So everything is free. And there are many other lectures and basically courses. This is a course mass for teens. There is another course mass plus 14. Uh there is a physics for teens, physics plus for teens. There is a relativity course. I mean there are different lectures and all free on unis.com. So today we're talking about properties of pass integrals. Pass integrals or line integrals or curve linear integrals. That's the same thing. Basically I'm using pass integrals because it's more applicable to physics. They like to call it pass integrals where mathematicians usually call it line integrals. In any case, the probably I would say the probably important the most important purpose of pass integrals is in physics. So that's why I decided to use the physics term. Okay. So um we have defined a path integrals as integral of some function along a curve in some n dimensional space. So first of all we have defined a curve in well let me use three-dimensional case. So what is a curve in three-dimensional case from the mathematical standpoint? Um so I mean visually we understand this is a curve um and we're talking about smooth curve without any kind of break breaking or or sharp angles or etc. Smooth curve in a in this case three dimensional case. How did we define it in the previous lecture? We defined it basically as uh three coordinate functions x, y and z which are dependent on some parameter. Parameter can be anything and uh you have some kind of a range of parameters from uh let's say from zero to capital s. This is the lowerase s. Um so let me just put it a bigger one. And uh so the three functions which depend on the parameter which goes from one point to another uh in real numbers. I chose from zero to some kind of s defines a curve with beginning of the curve one h would be uh with s is equal to zero and the ending so this a corresponds s to zero and s equals to capital s corresponds to the end of the curve. So as parameter S goes from zero to capital S, point on the curve moves actually from A to B. And this curve sometimes I call A with a arc or just gamma or something like this. And now we are talking about properties of integrals. So now the first property now all the properties of integrals pass integrals depend on the corresponding properties of regular integrals because we have defined integral along the path of some function. This is basically how I call this pass integrals. It's actually integral from zero to capital s functions function f of x of s y of s z of s ds. So now this is a function of s of the parameter. So this function of three functions and each of them depends on the parameter s and we integrate by s. So this is a regular integral. So the pass integral is defined through s integral through regular integral and all the properties of passive integrals depend on the properties of um regular integrals. And the first one is sum of pass integrals and integrals of sum. So if you have something like this, two functions. This is a sum of two integrals. by gamma. So pass integrals of sum is equal to sum of pass integrals. Why? Again it directly follows from the definition because this integral of um sum of two functions pass integrals is obviously defined as regular integral of u sum of these functions which in turn according to the properties of regular integral it actually uh is equal to sum of integrals and this is the sum of these integrals. So that's the first absolutely trivial property. The second [snorts] which is no less trivial is if you multiply it by some kind of a multiplier. So if you have this times some kind of a constant you put constant outside of the integral. again absolutely trivially follows from the corresponding properties. Now the third property is well again it it follows immediately from the regular integrals but um it's a little bit uh I would say a little bit more entertaining. All right so um here it is. What if I do this integral first? And then I will do integral from B to A. I change the direction of integration. Now and I integrate exactly the same function. So my uh the property which I would like to say is that if you are integrating in opposite direction is basically according to the definition of the of this particular integral that would be integral from from capital s to zero. So this is from zero to s right. So this is from zero to capital s f x of s y of s z of s ds and this is the same function [snorts] y of s z of s ds but integration is an opposite direction from zero to s from s to zero and Again according to the rules of plain integration the integral changes the sign if I will change the limits. Why? Well because the ds would be a different uh every ds would would have a different sign because what is ds? It's infinite decimal increment of our um uh parameter. Now in this case from zero to s we are going this way. So every increment is positive. In this case if this is zero this is s every increment would be negative and that's why the whole um integral change the sign. So that's the third property. Now the fourth property is if we will split if we will split the curve into two curves. Okay. So let's say this is our curve from A to B and we will split it by point M. So this is S of0. This is S of capital S. Now this would be this point would correspond to S of some kind of uh value. Well, let's call it let's call it S M. I shouldn't put S0. It should be S=0. Parameter S is equal to zero and parameter S is equal to capital S and this is S is equal to SM. Well, so what I'm saying is that integral from a function along the whole thing can be split into integral from a to m of f gamma plus integral from m to s uh sorry to b to b f. Now again y exactly the same going to the properties of regular integral. This is integral from 0 to s of f ds. This is integral from 0 to s m f ds and this is integral from s m to capital s f ds and this is a known property of regular integral. So if you're integrating function on some uh segment from A to B then you put some point C in between the whole integral is equal to sum of integral of this plus integral of this. If nothing else then basically uh the area of this is equal to sum of areas of this and this. But it all falls from the corresponding properties of regular integrals. Okay. Now, now we will do something more interesting. The next property is again I started from some references to physics. So this is reference to physics. There are certain um vector functions. It's a vector function and we have defined the um pass integral for vector functions. If you remember integral by gamma d gamma is actually it is actually integral of vector function in three dimensional case it's three components each of them is a function of a point where it was so it's a component fxyz component f_sub_y this is the vector so it's three components in threedimensional case and component of Z of X Y Z. So this is a vector. I put curly brackets and the gamma is actually Z R again vector where our vector is XY Z and this is considered as a scalar dot product. So it's all in the previous lecture. So we have defined pass integral of vector function which has three components and increment along the curve. If r is the radius vector from zero to the point on the curve, it always again has three components x, y, and z. That's the coordinates of the point. So this is a vector and the scalar product as vectors we can do the scalar or dot product is the definition basically of this integral. And now obviously again x is x of s y is y of s and z is zero z of s and that's how the whole scalar product which can be actually expressed as integral of fx * dx + f_sub_y * dy + fz * dz. This is the definition of the scalar product of two vectors. One vector is this. Another vector is this dr is equal to dx dy dz. Three components. This is increment of the vector radius vector to the point of the curve. So basically this is element of the curve. Increment obviously. Yes. Increment of the of the curve. And uh geometrically why is that basically done? Well because if you have some kind of a line where the body is moving and the vector of force is at the angle. So this would be d and this would be f as a vector. Then you have to really to calculate the work we have to calculate projections of this vector on the direction of movement. So this is a piece of a curve right. So and projection is cosine of the angle and this is basically what do that product is. All right. So this is all from the previous lecture. In any case, so this is our integral which obviously can be calculated from zero to s where x and y and z are functions of s. So the whole thing becomes uh some kind of integral of zero to s some function of s d s where function of s is basically this expression where fx has three coordinates each of them depends on s. dx has basically d uh dx is what? It's uh x um dxp ds * ds. So this is the function. This is the derivative of this function. So the whole thing becomes integral of s. Right? If you will substitute this and the corresponding y and z and f would be also as a function of xyz where each x y and z depend on s. It's all becomes a function of s. So we can integrate it uh as a function of s from zero to capital s and that will be the value of this integral. Now in certain cases in certain cases there are certain forces which are very special in physics I'm talking about physics now these special forces and examples are gravity uh electrostatic forces and some others they are called conservative. Now the conservative force forces are um are such forces which are um which have a very important property. If you're moving from this point to that point because conservative forces are driving you now the work which you perform moving from here to here the work by conservative forces will be the same regardless of the path no matter how you move it will be the same. So for example, if you want to um move let's say a stone from the ground to a certain point above it, you can move it straight up or you could move it in this way or in this way and you will still spend exactly the same amount of energy. So that's what work actually is. Now these uh vector functions have a very specific property which I will assume is true right now and I will prove that for these forces for these vector functions which um possess this particular property the pass integral depends only on the beginning and the ending point and does not depend and exactly on trajectory as long as everything is smooth, differentiable etc. And here is the property which I have in mind. Okay. So we are assuming the following vector f of xyz it's a combination of three components. So this is the vector. These are three projections on three axis. Each one of them function of a point. So wherever you have some kind of a curve, this is a point xyz. And yes, we can definitely have x and y and z depending on some parameter. So s parameter goes from uh from zero to capital s. the point xyz moves from a to b. Okay. All right. So now the property which I'm talking about is as follows. Let's assume that there is a scalar function defined exactly on the same curve. And let's assume that f ofx of xyz is equal to partial derivative of this function by x. f_sub_y is equal to partial derivative by y and f_sz is partial derivative by z. Let's just assume this is an assumption. Now from this from this assumption I would like to prove that pass integral of this um vector function f along any pass between a and b this pass or this pass or this pass no matter where it is depends only on the value at point f a and b on the value of function scalar function f. Okay, how can it be proven? Well, actually the proof itself is absolutely trivial. Now what is integral by pass of vector f and I will use dr in this case like I was just explaining it's a scalar product right this is integral of f [music] ofx * dx + f_sub_y * dy + Z * DZ. So this is basically the definition of the pass integral of vector function. Right? Now now we can replace f_sub_x, f_sub_y and f_subz three components of our vector f with partial derivative of scalar function uh lowerase f. So if we'll do that we will have this df by dx df by dy df by dz. Now great. Now what is this? Well, this is a full derivative full differential of function f and if I will use right now the parameterization. So it would be integral from zero to capital S and these are functions of S d I can put it this way that's what it is right so this is a function of s now so the derivative of this function by s is basically the integral which means what? Well, which means that this function since it's a full differential, the integral is equal to f of x, capital S, comma Y, capital S, Z capital minus this is a formula Newton lab x of 0 y of 0 0 0. So this is basically the value of this integral. So as soon as we have defined function vector function f as being with the components equal to partial differentials partial derivatives of scalar function lowerase f as soon as we have defined that I mean as soon as we know that this property does exist immediately from this follows that any pass integral depends only on this function f at points a minus uh sorry b minus a. So this is the x of s y of s z of s and this is x of 0 y of 0 z of 0 0. So this is a very important property. It means no matter how you move from one point to another in let's say some kind of a field which has forces like gravitational field and you're moving against this field then you have to spend certain amount of work certain amount of energy to do this. So no matter how you move you will spend exactly the same amount of energy. Okay. So this is basically a property number five and the property number six immediately follows from here is that if you are integrating function conservative function which has this type of representation along a closed loop. So a is equal to b then the result of that moving would be zero. So if you are moving let's say a stone from this point along any c a long trajectory and back to this point in theory the total amount of energy spent would be zero. Why? Because here you're uh spending some energy and then when you're moving down you are actually working uh you you're helped by by the gravitational field which makes your uh work being negative actually. So you're assuming certain work in this case you're actually assuming certain amount of kinetic energy. So you're spent this energy when you're moving it up and then when you're moving it down you're gaining kinetic energy. So the total amount of energy remains exactly the same which means you spent exactly the same. So that's what the physical kind of sense of the whole thing. Now how to prove this? Well first of all it's obvious from here that the difference is equal to zero. And there is another very cute proof which I would like to say. Let's assume you have some other point here m. [clears throat] Now the from here to here the pass integral is equal to something. Okay let's call it a. Now from the same point a to the same point m along the uh another half of this trajectory you will have to spend exactly the same amount of energy. B is equal to a. Now if you're moving from here to here you're spending a and then you're moving in opposite direction. So it's minus b is equal to a minus a is equal to z. So that's another very trivial proof. That's it. So these are properties of pass integrals and as I was saying in the beginning they are all follow from the corresponding properties of regular integrals. So that's it for today. I would like you to actually read the notes for this lecture. It's on unisur.com. You go to mass4's course is calculus and then you will find um I think it's uh definite integrals part two. You will have this uh pass integrals set of lectures. The first lecture was previous one which defines pass integrals. This is a property and there will be another and as I was saying everything [clears throat] is free and I do recommend you to do both read the lecture notes on this website and um watch the video which uh which I'm just recording right now. So that's it for today. Thank you very much and good luck.