I had already watched these lectures back in 2014, and I was stunned by the intuitive approach used by the professor. Now that I come back in 2020 and watch them again to brush up on tensors, I'm still stunned :).
from my Baghdad Iraq, I thank you so much Sir, for making these precious lectures free for us. it really made my reading Electromagnetism more accessible to me.
I'm happy to see so many positive comments about your style of teaching and your love of the subject. I add my voice to those already here. Thank you very much!
This is an amazing series of videos! I've been studying Tensor Calculus since I was 24 due to a strange fascination I had with Einstein's Theory in my youth :) It has been difficult and I've read a myriad of books on the subject... some very technical some very practical, but all of them I always considered partial in some aspect, and none of them gave a full account of the subject by itself. After so much reading and work I managed to understand the subject and gain some insight into it... but I was never satisfied with what I learned. due to the fact that none of the books I've read made the assumptions clear upfront and the underlying themes more clearly visible from the get go. This is the very first time I've seen someone put the knowledge in display in a very clear and intuitive manner... and with really amazing teaching skills as a plus! For that you have my eternal praise and admiration. Thank you very much, and if I ever I'm able financially (different country, rate exchange is not favorable to me atm.), I'll be sure to buy your book... it clearly surpasses any of those I've read in the past :) Best wishes and good luck!
Absolutely beautiful. I am finding these lectures very satisfying and enjoyable. I took college geometry last semester and it was neat to get insights into how geometry is often a "cleaner" way of defining objects. The parabola being defined by declaring a directrix and a focus and then building it from there is a beautiful thing compared to just plopping an equation down and being drilled on the mechanics of the numbers involved. Your insight into the way that the tensor calculus can free us from the drudgery of coordinates to allow us to see the beauty of the forms is an inspiration. Bravo and kudos to you for such a wonderful presentation!
I really like your approach towards geometry first. A lot of my physics books just expect you to take definitions for granted and I hate that. Laying the foundation in geometry helps so much.
+The Misanthropic Nihilist Well, the whole object of tensor calculus is to express geometry in the language of algebra! So it shouldn't be a surprising place to start!
MathTheBeautiful Like, I am probably going to binge watch the whole lecture series. Professor Grinfeld is answering so many questions I had in multivariate calculus that wasn't clearly explained. I like where he uses geometric intuition to build upwards.
Wow, I really love the way you teach! I can see you're concerned with making the students grasp the mathematical concepts. Your videos are being extremely helpful for me! A lot of what I've seen throughout my academic life is starting to make sense. I really want to master this once and for all.
+Akira Uema Seconded. My favorite lectures in grad school were sometimes ones where the board was only filled once in the hour; the time and effort was spent on a single piece of information, and 1 or 2 enlightening examples. Mathematics is the most intense single mental exercise anyone can do.
I am starting to realise now why tensor calculus is so powerful. Studying engineering, so much emphasis is put on a coordinate system, that sometimes its hard to see light at the end of the tunnel when crunching through some elaborate theorem. My God I wish they had explained things in the language of tensor calculus instead.
I got a few lectures further on before I really got the idea of covariant and contravariant bases. I'm enjoying re-watching with better understanding. Lights blinking on frequently - thanks.
I've tried to study tensors several times using different books and articles from rather specialized to relatively popular-written, and I had given up. So far your lectures look really clear and motivative, and they really inspire me to continue studying. Thank you, Pavel! As always, perfect! I don't know if I'll be able to watch all of them, but I sincerely hope so:)
Wow, I always regarded Tensor Analysis as something quite unapproachable with my knowledge, but you, sir are doing a great service to education and to us students. Thank you, sir!
I don't really understand a specific thing though. In the example of the polar coordinates, since we have a basis per point essentially how do we describe a point in that space, there's no basis we can pick because we want to make all the definitions coordinate system independent.
Is it possible to do a video on how you can "discover" a coordinate system based on some description that isn't radial, and how you would describe it? For example, assume that we have coordinates that are sine waves that are stacked on top of each other. How could we describe a coordinate basis in this manner?
James Pedid If the sine waves are vertical shifts of one another, then they do NOT form a coordinate. You cannot take two random, arbitrary shapes, tack them together, and call that a coordinate system. Coordinate systems fundamentally do not work this way.
Given a vector based on the inputs of R, how would you write it in terms of the covariant basis and vice versa? If I am understanding, you defined R as a function that produces a vector based on co-ordinates, and writing a vector in terms of the covariant basis also produces a vector based on co-ordinates. In fact, R is used to define position vectors and the covariant basis is just another way of expressing that. So if the covariant basis is an equivalent way of expressing vectors, there should be some way to transform one notation into the other. I noticed that if you have an arc R=R(Z(t)), then if you take the partial derivative, you naturally get a decomposition in terms of the covariant basis, but it seems a bit strange to define vectors as partial derivatives of the position vector instead of just writing them as the position vector. I have also found that the inputs of R are only equal to the components of the vector in the covariant basis if the Z's are independent of each other and the basis vectors are independent of the Z's. If we write R(Z)=Z^i V_i then take the partial derivative with respect to Z^j and make the previously stated assumptions, then dR/dZ^j=V^j.
I was really having trouble coming to terms with the difference in basis depending on where you "drop" a vector down on a polar coordinate system, but now i get it. The problem that would require such a coordinate system also carries with it baggage. You could imagine the problem for polar to be the speed of a grape spinning around in a circular sink. Of course its speed component in theta is going to be higher the further away it is from the origin (sink center). Please let me know if this thinking is disordered. Your lectures are awesome and really scratch the itch of my engineering outlook towards mathematics.
I'm finding you're classes very clear and well explained. Thank you! Just a quick question on this one. When you define the polar r coordinate geometrically you define it as the distance to an origin. Then you define another arbitrary origin from where you draw the R position vectors. Shouldn't they be the same? This confuses me a bit.
+Pablo Fernández Esteberena I don't think it has to be the same and the extra bit of generality helps in this case. It help highlight the fact that only differences in R, not necessarily R itself, that play the key role.
hello...question ? @ 20:59 when you refer to the function in question as its gradient you switched variables from R to F , are these the same function, if not how are they related ....? ty very much..
let me see if i have figured this out, R is the function of a vector stemming from the origin, witch traces a function F as F ' goes about its business ' lol... ??? so the GRAD of F is the der of F in reference to the coor. sys, acting apon the der of the function R actiong on the coor sys . did I get it swami ????
Hey I've got a question! At 23:00, why did you normalise e_theta with r^2 if its length is just r. A similar scenario came up with the stretched cartesian system and I didn't quite understand that either. Could somebody shed some light on this please?
Michael Mitchell You divide the basis vectors by the length squared, not the length. The basis vectors are not being normalized, the gradient itself is. The idea is that the components scale by a factor A, and the basis by a factor B. Then, to normalize, you must divide by AB. Otherwise, the equation is incorrect because it implies the length of the gradient is coordinate-dependent, which violates the definition of length.
dear prof. grinfeld, one question, how do we see that the covariant/contravariant basis commute (in euclidean space or surfaces later), that is, they are "holonomic basis"? or do they not? thanks!
The problem they're talking about at 18:20 is true whether you normalize or not. That's just a fact of life with polar coordinates. It is not a flaw with this definition of a basis.
thank you for the very helpful lectures. At 17:24 you say "in a lot of Physics books the base vector e_theta is normalized to unity, which in my book is absolute madness". Could some physicist out there make the case for normalizing the base vectors, or why physicists persist in working with normalized base vectors if they add unnecessary complication? Do equations with physical meaning such as Maxwell's equations, give the same value at every point when transformed to different coordinate systems even when the base vectors are not normalized and are changing in size and direction at different points?
I am following along with your course parallel to my normal studies and greatly enjoying it thus far. While working through the book however (I am on chapter 9) I found a persistent difficulty in deriving the covariant basis for different coordinate systems starting from pure geometry. I feel that I am at a loss in doing it for spherical for instance, even though the book technically covers it. It's particularly frustrating because all the rest of the formulae follow quite simply (and elegantly). Is there a trick to get better at it? Particularly I am unable to find the length of the vectors when subtracting them.
Hi Erik, That's a very good comment and a valid point. When it comes to basic curvilinear coordinate systems (Polar, Cylindrical, Spherical) on only needs to understand the derivative of a uniformly rotating vector. Once you have that under your belt, it becomes a matter of applying that insight in various ways. Is that helpful? Thanks, Pavel
Hi Pavel. Thank you for your answer, I apologise for not having replied to your answers on my other questions. It is helpful but I am not really sure where to start? I have no problem differentiating the position vector along r for instance, the problem lies when the "position vector arc" becomes curved. My thought process is currently something like this: 1. Figure out the direction of the vector resulting from taking the difference of two position vectors a certain distance along Z^i. 2. Figure out its length (Problem: if the arc along Z^i is curved, how do I find this length? e.g. Z^3=phi in spherical). 3. Divide by the distance along Z^i (Again, how to determine this?). 4. If not uniform, take the limit.
The only part that I find confusing in trying to connect this with the little bit of differential geometry on manifolds that I learned is that in diff geo, the "covariant basis" dR/dz is really only a basis of the tangent space (since the underlying manifold is not linear.. and hence can't possibly have a basis). Because you are only describing things in Euclidean spaces here, you can "bring back down" the covariant basis to also describe position vectors (as opposed to only velocity vectors). Is this the correct way to think about the covariant basis?
Well, we are not talking about proper manifolds yet. We are just considering the ambient space where dR/dz produces three vectors, i.e. a legitimate basis in the 3D space.
@@MathTheBeautiful Right, but my point was that dR/dz is the Jacobian of the map R, and Jacobians are maps between tangent spaces. So my understanding is that when we say that dR/dz produces "vectors", these vectors are really on the tangent spaces, but because the space is affine, there is a natural isomorphism between the tangent spaces and the space itself, which allows us to say that these "vectors" are also in the ambient space. The reason why I'm being finicky about this is because I confused polar coordinates and rotating frames of reference for a long time. For example, en.wikipedia.org/wiki/Kinematics#Particle_trajectories_in_cylindrical-polar_coordinates uses the term "polar coordinates" to refer to what I would call a rotating euclidean frame of reference. When I finally understood that polar coordinates *induce* rotating reference frames on the tangent spaces, and that the wiki page is using the natural isomorphism between the space and tangent spaces, everything finally made sense. So I just wanted to double check with you that I was interpreting this properly.
@@samlaf92 I think that what you're trying to say is that at each point there exists its own linear space for which the three vectors dR/dz act as a basis. At another point, there is another linear space with a corresponding basis obtained by dR/dz. However, each one of these linear spaces is three-dimensional.
These lectures are excellent. However I was wondering if you could refer me to any extra materials so it would be possible to work out some problems and become better practiced. In other words can I have some homework please.
looking at a vector in polar coordinates and changing only theta by any amount, you can visualize the vector that connects them as a chord on the circle that contains the start of each vector as its center. The vector would then define a sector of that circle and the length of a sector of fixed theta will change with r since the sector length s=r*theta. Draw a diagram and you'll see what I mean since I know reading off a RUclips comment isn't optimal :P
You CAN choose to place your position vector anywhere you want (it doesn't matter because you're interested in its DERIVATIVE, i.e. the DIFFERENCE between the position vector as you vary whatever coordinates, so you can choose ANY "pseudo-origin" (a word I made up) for the position vector, including--if it helps you visualize what's going on better--the center in your polar coordinate system. Sorry, that's a mouthful. Let me restate it: The Tensor Calculus, as I understand it from this video, essentially uses some tricks to make your analysis independent from a chosen coordinate system. One of the tricks is to "scan" the space using a position vector. The starting point of the position vector is arbitrary and won't affect the scan. You could place it to the far bottom left on the black board, but you could also place it at the center of the polar coordinate grid if you wanted to. Either way, its derivative (which is just based on the DIFFERENCE between the position vectors as you vary them w.r.t to coordinates) will be the same, so the resulting scan will be the same. So, to make things easier for us, let's imagine the position vector starting from the center of the polar grid. It would then point outward toward some concentric circle of radius r. Imagine keeping r constant, and only varying theta, and snapping the arrow-side of the position vector to the location of each point in the space that has coordinates (r, theta), where r=constant, while theta is variable. What will the position vector scan trace out? A circle of course. It will trace all points in the space that share the same constant r. During the scan, the difference between position vectors sampled with theta 0.1 radians apart (or some other small delta increment), will be a small segment essentially tangent to the circle. For the same amount of variation in theta, you'd get TWICE as much circle for r=2 than you would for r=1. Therefore, the size of said tangent must be scaled by r. Imagine the two line segments below are actually not straight, but pieces of circles traced out by the position vector (along the polar coordinate system's grid lines). Theta is varying the same at the center, and the position vector is tracing the result, but at different rs. Obviously, Delta(R) will vary more strongly at higher values of r. R |================== r=2 | DELTA( R ) | |----> [the position vector R is sweeping this way] | | |========= r=1 | DELTA( R ) | | | | . (Center of polar grid) Again, the result holds the same even if R is not centered at the center of the polar grid, since Delta(R) will still give the same result.
Great explanation, @@keikitv ! If Pavel explained it this way, it would have saved a whole lot of time for the people watching this video. The key observations here that made it for me were those: 1. The choice of the "pseudo-origin" as you called it (I would call it "external observer") is arbitrary, because: 2. what we are really interested in are only the DIFFERENCES in the position vector R as we tweak one of the coordinates while keeping the remaining ones constant. 3. Although the difference in angle stays the same for all possible endpoints of R, the "speed" of its endpoint may differ the further it is from the centre of rotation, because it sweeps longer arcs in the same "timespan" (we can imagine the change in angle occurring in time, to make the picture more "dynamic", but the key point here is just how does the coordinate change, and how fast is the corresponding change in R, and their RATIO, which should stay the same no matter where the observer is). Way to obfuscate coordinate changes, eh? :q
Bon Bon What are you even saying? He explained it exactly in that way. Those were points he addressed in previous lectures, which by the way, are requisites to this one. Geez
That's a very good question, Taras! And one that took me some time to figure out. I think the answer is that a function is not necessarily something that can be expressed by a formula. A function is any rule which "returns" a unique output for any valid combination of inputs. In this case, the inputs are two numbers and the returned value is a geometric vector! Looking ahead, Tensor Calculus by definition occupies the space of "unspecified functions" since it operates in the framework of arbitrary (i.e. unspecified) coordinate systems. How do you differentiate a function not given by an expression? See this: ruclips.net/video/hiYgYWJEaMk/видео.htmlm12s
Thank you. Did you assume that the origin of the polar system is in the center of the circle when you differentiated R? earlier 4:48 you said that the origin is arbitrary.
Correct. Since we only calculate differences of position vectors, the location of the origin doesn't matter. So might as well use the most convenient one (from some point of view).
1.Why does the covariant basis at each point look like cartesian if no basis is special? 2. When differentiating R(r,theta) w.r.t. theta I think it depends on the origin. For example if the origin is the alternative origin (5:00) dR/dTheta would not be tangent to the circle with the standart origin. 3.17:36 How would you use the covariant basis to define a point? Do you also have to specify the point whose basis you are using?
1. It doesn't look Cartesian, since not both vectors are unit length. Also the overall orientation is different. The basis is **orthogonal** and as a result, polar coordinates are an "orthogonal" coordinate system. 2. Yes, if you choose different polar coordinates, the resulting basis would be different. However, for any given coordinate system, the choice of origin for the position vector is immaterial. 3. Not 100% sure what you're asking. You will have to use a different basis depending on what point you're at.
The covariant basis is a totally mind-blowing local coordinate system. Should call the general polar coordinate system a global system. In the cartesian coordinate system, both local and global are same.
Hi. (At 10:20). In the example, if you increment x by 0.5, could not R - being an arbitrary vector function of x and y - take on ANY value at all? Why did you place the head of the new R vector where you did, on the x axis?
MathTheBeautiful Thanks so much for the instantaneous reply. Having watched all the lectures, it was very frustrating to have a lingering doubt over this basic (and as it turns out, trivial) point. ATB PeterD.
Hi Joe, you can't get this result by manipulating analytical expressions, and that's perhaps the main point here. (Unless of course you introduce a background coordinate system, which would be the wrong thing to do.)
I still can't loose the feeling that there's something wrong with the way things are defined here. Using the same letter for everything doesn't really help that feeling at all, by the way. Even though the form of presentation isn't very mathematical, it should still be coherent and "formalize-able" I will try to sumit up and point to the difficulties I seem to encounter. Consider a 2-dimensional (just to save space and avoid indices) euklidian space and a (continuous etc.) "Abstract-Vector"-valued Curve function K(s) in it, which parameterized by arc length. This means assume that K' (s) * K' (s) = 1. where K' denotes the derrivative of K. Now assume that there is some general coordinate system with coordinates x,y and a covariant basis i(x,y) ; j(x,y) where i and j are abstract-vector-valued functions of the variables x and y. Then the vector-value of K always has a unique representation in the local basis i,j. That representaion is dependant on s and the choice of a local basis. This requires to specify at what coordinates (x,y) the basis vectors i and j are evaluated. It makes sense to evaluate them at the coordinates of the points of K for all values of s. This requires some bijection between the abstract 2-dimensional space and the coordinates for position vectors. These coordinates would then have to be independant of a choice of reference local basis (?). As far as I can tell that bijection is given in the video by writing R(x,y) which denotes an abstract--vector-valued-function of the coordinates that assigns a position vector to every set of coordinates. Then i and j were defined as: i(x,y) := (d/dx) R(x,y) j(x,y) := (d/dy) R(x,y) It is then assumed that K(s) is a position vector for any fixed s (?). As far as I can tell that's it for the definitions explicitly mentioned in the video. I will try to make the following calculations slightly more detailed and add definitions I assume are implicit in the video. Assume that (since R(x,y) is a bijection) there are functions Getx (v) and Gety(v) generating coordinates x and y for any abstract-vector v such that v = R(Getx(v), Gety(v)) . Pluggin in K(s) into this equation instead of v, K(s) = R [ Getx (K(s)) , Gety(K(s)) ] Now, as far as I can tell without any further justification, it is assumed that the functions Getx(K(s)) and Gety(K(s)) can be differentiated with respect to s. I guess this should be provable from the differentiability of K itself somehow, not however without making restrictions on the bijection coming with the coordinate system. I belive the functions used to transform polar coordinates and cartesian coordinates are either not bijective or pose other difficulties which are reasons to use manifolds and charts in this context. Regardless, I will call Getx(K(s)) =: K_x(s) and Gety(K(s)) =: K_y(s) This then probably (?) leads to the equation used at 32:24 still calling the curve K and avoiding indices: K(s) = R(K_x(s) , K_y(s)) By this point I'm almost sure I don't mean the same thing as the video, but allow me to proceed and differentiate: (not supressing the arguments inside the already differentiated functions) T(s) := d/ds [ K(s) ] = (d/dx R)[ K_x(s) , K_y(s) ] * (d/ds K_x)[s] + (d/dy R)[ K_x(s) , K_y(s) ] * (d/ds K_y)[s] = = i(K_x(s) , K_y(s)) * (d/ds K_x)[s] + j(K_x(s) , K_y(s))* (d/ds K_y)[s] the result in my words: Given some value s of curve length, the COMPONENTS of the curve's Tangent vector with respect to the local basis at the position vector that the curve takes (at that particular value of s) are the derrivatives of the (basis-independant) COORDINATES of that same position vector. Is this what is meant?
I think the answer is yes, but I'll make a comment and you'll let me know if it's helpful. I do think you are overcomplicating the formalism a little bit. Perhaps you are envisioning software implementation, which can be a little challenging. In this narrative, when I say "vector", I mean a directed segment - as drawn in the plane or in 3D. It took me some time to adopt this point of view and to unlearn the dominant view that a vector is a pair or a triplet of numbers. A "vector" is a geometric object in the sense of Euclidean geometry. It can be held in one's hand. This approach is not sufficiently formal for a lot of people but I nevertheless invite you to adopt it alongside the formalized approach. My geometric "vectors" can be added/subtracted, and multiplied/divided by numbers. They can therefore be differentiated with respect to parameters, which yields i = dR/dx and j = dR/dy. So I almost never think of the expression "coordinates of R" and there is no such thing as "basis-independent coordinates". Vectors don't have coordinates, coordinates are attributes of points in space. So points on a curve have (curvilinear) coordinates Z^1, Z^2, Z^3. The components of the unit tangent vector equal the derivatives dZ^1/ds, dZ^2/ds, dZ^3/ds.
Yes, that is helpfull. Thank you. Those curvilinear coordinates is exactly what I mean by basis-independant because they do not/can't depend on a choice of local basis. Since the non-affine coordinate systems can't be represented by just one basis I thought in this case saying basis-independant coordinates is fine. Indeed in the last statement I tried to make a distinction between coordinates (as a representation in terms of the coordinate system as a whole) and components with respect to some local basis. You seem to use this terminology as well, judging from the last paragraph. The reason I'm overcomplicating it is not at all a software implementation, rather trying to apply this to a general characterization of a euklidian space and the assumptions that are made about this space. I imagine you could define some sort of coordinate systems for things like matrices with a Frobenius product as well. Whether that's usefull is a different question...
ANSIcode Are you saying the coordinate basis cannot itself depend on the coordinates? That would be nonsensical. There is no theorem or axiom stating this.
22:50 will e1.e1 work with stretch cordinates, as well ? Since we are considering gradient as function of arclength, how can we get sense of direction with arc length alone???
13:48 But how do you know how far is that 0.5 from the starting point and in which direction, if you don't have your unit vector `r` defined yet? :q I think you're putting Descartes before the horse here :P
Bon Bon He has a well-defined unit vector, given by his chalk. Unit vectors are chosen arbitrarily, and they do not have to be part of the basis. Otherwise, all basis would by definition be normalized, and this simply implies a contradiction. If e1•e1 = 1, and I define e2 = (2)e1, then by your line of reasoning, e2•e2 = 1. However, since e2 = (2)e1, I can prove that e2•e2 = (2e1)•(2e1) = (2^2)e1•e1 = 2^2 = 4, and then your definition implies 4 = 1, which is false. Therefore, your definition of unit vectors is invalid, and your understanding of vector magnitudes are defined is incorrect. Vector lengths are basis independent. This is true by the definition of geometric length.
Professor Grinfeld, I have a question about differentiating a parametric (vector) curve whose parameter is arbitrary. If the equation is γ(Z(t)) and it's decomposed with respect to the covariant basis γ^i(Z(t))*Z_i, then how would I differentiate this? I ran into the trouble of not knowing how the covariant basis Z_i depends on the parameter t. Thank you!
+The Flagged Dragon This is one of the places in the subject that requires a great deal of diligence. Whenever you're about to differentiate, you must first write down a precise identity in terms of the independent variables. There's not a blanket recipe I can give. It's just a matter of explaining to yourself what depends on what in your particular problem and then writing it down as in mathematical terms. In your case, that identity would be γ(t) = γ^i(t)*Z_i(Z(γ(t))) It's different from what you had, so you'll have to think about it a little bit. I think I discuss a similar question in the video entitled Velocity Acceleration Jolt, but you'll be much better off doing it on your own. Thanks for the fantastic question.
+MathTheBeautiful Ahhh that makes more sense! Finding the dependence especially with a changing basis was quite challenging. I thought I found a solution but it differs a small amount from differentiating the equation you gave me. Thank you so much for the response!
This is a great lecture series, thank you! I am confused about one thing: at 33:14 or so, you claim that the z's in the partial derivative of R are the coordinates, but the z's in the total derivatives are the parameterized coordinates of the curve. How can two entirely different z's arise from application of the chain rule? There would need to be a dz[coord]/dz[curve] term in the middle for this to be true. Unless I'm missing something, which is entirely possible!
I am confused, tired, or both. For the definition of a covariant basis vector as a partial derivative of the position vector with respect to Z_sup_i, Z_sup_i is a coordinate of the coordinate system. In some coordinate systems, Z_sup_i are also/equally the (contravariant) components of the position vector, right? But in other coordinate systems such as spherical and cylindrical the Z_sup_i coordinates are not the components of the position vector because the angle components of the position vector are zero, right? If so, for spherical and cylindrical, are the Z_sup_i (contravariant) components of some other vector (not position vector) with some special meaning? Thanks.
There are two " Z^i ". One with a bold Z indicating a contravariant basis vector and the other plain indicating the coordinates. There is very in common between these two objects even though they are denoted but very similar symbols. One may argue that the letter Z is overused, but I think I've struck a pretty good balance in terms of using Z. (Some authors denote the position vector by the bold letter Z and also use it for the Levi-Civita symbol!) In all cases, either the font or the indicial signature (or the context) uniquely determines what object we are talking about. It is true that in some coordinate systems (affine, to be specific) the coordinate Z^i equals the contravariant component of the position vector R. However, this is a largely irrelevant fact and is almost never used except in a couple special instances. In fact, is position vector R is usually not decomposed with respect to any basis. Not that there is anything wrong with that, it's just that the need almost never arises. For the most part (although there are some exceptions I'm quite fond of), the role of the position vector is to be differentiated and then to disappear from consideration.
So is it correct that the coordinates themselves (and the tuple of coordinates) are not a variant, not a tensor, not a vector? The components of the position vector are a tensor.
Mongrav Let me break your question down into individual parts. In tensor calculus, you typically bring the n-tuple along for the ride. So Z^i can be seen as both: the i-th coordinate and as the n-tuple of coordinates. So there's no need to say n-tuple of coordinates - you can just say Z^i. I guess I would technically call Z^i a variant. I would never call it a vector in the context of tensor calculus - an n-tuple of numbers and a directed segment are as different as two objects can be. It's generally not a tensor. (It would be in a more narrow sense, if you only considered linear transformations of coordinates. This is done, for instance, when studying rigid body motion.) The components of any invariant vector field with respect to the covariant basis is a contravariant tensor. Since the position vector is an invariant vector field, its components form a tensor.
MathTheBeautiful ok, thanks. I don't have a follow-on question at this point. The following material is just to help me later if I come to a new question. Let's take cylindrical coordinates: r, theta, z; covariant basis: r_hat, r*theta_hat, k_hat. I understand the non-bold z^i as the coordinates (r, theta, z). (I can recover the basis vectors from differentiating the position vector wrto the coordinates, no problem.) For Cartesian coords, the non-bold z^i happened to also be the contravariant components of the position vector. But here in cylindrical coords, z^i are the coordinates but are not the contravariant components of the position vector, which would be (r, 0, z), not (r, theta, z). The theta component is baked into the particular point and the r_hat basis vector for that point. For the cylindrical and spherical coordinate systems, the nonbold z^i coordinates are not the contravariant components of the position vector but relate to a differential excursion from that position.
Mongrav I think what you are saying is correct. The only thing I would like to add is that I have never thought about what the components of the position vector are wrt the covariant basis. That question just doesn't come up that often. One of the few instances where it does come up is rigid body dynamics, but I don't think it's one of your interests. But since you are thinking about it, I'll mention one fact for you regarding those components: their covariant derivative is the Kronecker delta symbol. See if you can show that.
In the 4b lecture video, I got some hint that says "metric tensor is symmetric and as an exercise it is positive definite". So, of the exercise 62, the covariant basis, interpreted as a matrix, is meaning the metric tensor?
Hi Prof. Grinfeld Question 30:59: Why d_R(s) / d_s is the unit vector? I "sort of" understand why it is the tangent vector, found it here www.math.jhu.edu/~jmb/note/curvtgt.pdf and I say "sort of" because the proof uses the dot product, which we still haven't defined at this point in your lectures in this general coordinates. I can not comprehend why the fact that we use s (arc length) makes d_R(s) / d_s not just tangent, but UNIT tangent vector. Thank you
Hello, I'm very much enjoying your series here, and had a question. You define the vector e_r to be the derivative of R(r,theta) with respect to r, where R is the position vector. This makes sense to me. But let P be the point with polar coordinates (1,0). P also has coordinates (1, 2pi) in polar coordinates. So to find the vector at P, what should I do, differentiate R with respect to r at (1,0), or at (1,2pi)? Now of course in polar coordinates we get the same answer either way. But how does this work in general? Do we only deal with coordinate systems where when a point can be expressed in coordinates two ways, the partial derivatives will turn out to be the same no matter which coordinates we choose? Polar, cylindrical, spherical, and affine coordinates all satisfy this. Or can a point have two vectors?
Well, most theory that's developed is "local", meaning this kind of situation is avoided. But yes, if you imagine an overlapping coordinate system, you may get overlapping everything else.
Actually thinking about it, using polar coordinates, it seems to me the partial derivative with respect to r at (1,0) and (-1,pi) will have opposite signs, even though they designate the same point. I guess we just have to exclude negative values of r?
I think when r is negative, the partial derivative with respect to r will point towards rather than away from the origin. After all, if one increases r from -1 in (-1,pi), the magnitude of r decreases, and we go towards the origin. Or we can cheat and convert to Cartesian coordinates, (r cos theta, r sin theta). If we take the partial with respect to r, that is (cos theta, sin theta). At (-1,pi) which has Cartesian coordinates (1,0), the partial derivative in Cartesian coordinates is (-1,0) which is pointing from (1,0) towards the origin.
At around 17:45 you mention in physics they normalize the theta component in polar systems. You then don't say it directly in the next few minutes, but that wouldn't really solve their problem, right? Because even if the coordinates have equal length, their directions would change all the time, so making them have equal length doesn't make them the same anyway.
@@MathTheBeautiful there's one possibility: they make the picture look less messy, because the arrows don't overlap then ;J But yeah, geometrically, it makes no sense at all.
ok, now |e_theta| is r can only be shown if we take the definition that R = r cos theta i + r sin theta j. is that true or can it shown in some other way?
Great question and the answer is an emphatic NO. This can be done (and is done in this video) on pure geometric grounds. Imagining a background coordinate system is the #1 affliction that precludes students from seeing things in a clear and simple way. If you are choosing between quitting smoking and quitting background coordinate systems, please choose the latter.
@@MathTheBeautiful Wasn't it _you_ who claimed several times that the space has its own "inherent" unit? (which would imply having its own "inherent" coordinate system) Isn't the "yellow" coordinate system just another way to sneak in the "background coordinate system"? (even if its choice turns out to be arbitrary in the end, because it doesn't influence the end result)
sir i have a question= i want to find unit tangent to curve y= sin x in the cartesion coordinate system. ans: parameterized form, r(s,sin s) here z1(s)=s and z2(s)=sin s. equation of unit tangent according to your formula is z1'=1 and z2'=cos s. so T=1 i + cos s j. which is wrong eqution bcoz its length is not 1 for all s. where am i going wrong sir?
Ohhhh!! my parameter 's' is not curve length and thats why my tangent is not of unit length!!! I got it. sorry for the stupid question. and thank you for your videos sir. extremely helpful!!
Stupid Question (which I also had when reading the book): Why is e_r equal to dR/dr and why is e_theta equal to dR/dtheta? In other words why are the basis vectors equal to derivatives of the position vector w.r.t the coordinates?
Thanks. So we cannot define basis vectors without derivatives? I always thought of bases as a set of vectors (geometrical or otherwise) that any other vectors can be represented in terms of. I am just not making the connection between these two concepts.
Yes, it's one of the examples of things that come in the opposite order in applied mathematics compared to linear algebra. In linear algebra, basis comes first and coordinates are built from the basis. The inner product comes first and lengths and angles are built from that. In applied mathematics, coordinates come first and the basis is built from the coordinates. Lengths and angles come first and the inner product len1*len2*cos(angle) is built from that.
MathTheBeautiful is the position vector R(r,theta)= ? I guess my question is what is R? is it some arbitrary function or is it just any vector that points to any point?
Why do you use i and j for the unit cartesian coordinate vectors, instead of e_x and e_y like you do for polar coordinate. This seems to acknowledge the cartesian coordinate system an importance that you repeatedly over this series try to deny.
Yes, you are right! Cannot defend my actions, except to say that I try to not be militant about any particular thing. That said, I use i,j,k for all affine coordinates. Furthermore, affine and, therefore, Cartesian coordinates *are* special in many profound ways. For example, their very possibility signals that the space is Euclidean. Cartesian coordinates can be used to show that the RIemann-Christoffel tensor vanishes in all coordinates systems in a Euclidean space. Etc. So, special symbols for special coordinates seems OK. Also, I often use x,y,z for Cartesian coordinates and r,theta,z for cylindrical coordinates.
@@MathTheBeautiful yes, indeed. Sorry, I think my comment did not come across as intended. I realize these days that one of the major impediment in my learning of mathematics decades ago, was the confusion introduced by poorly explained, and poorly enforced/followed notation. I also came to realize the importance of definitions. I certainly did not mean to be negative towards your teaching, which I find both wonderful and very clear.
Hi Radu! It's a valid point. Just pay attention to the indicial signatures - they are all different! That's why I use the same letter. Otherwise, there are lots of letters and lots of indicial signatures.
I had already watched these lectures back in 2014, and I was stunned by the intuitive approach used by the professor. Now that I come back in 2020 and watch them again to brush up on tensors, I'm still stunned :).
I sent this notification to you... watch this again
This material is just blowing my mind. Also,the way you teach is just amazing. Kudos.
from my Baghdad Iraq, I thank you so much Sir, for making these precious lectures free for us. it really made my reading Electromagnetism more accessible to me.
I'm happy to see so many positive comments about your style of teaching and your love of the subject. I add my voice to those already here. Thank you very much!
Thanks, I love how you make this topic so simple and don't overload us with proofs and trivialities. You teach very well!
+U238 This topic *is* simple!
This is an amazing series of videos!
I've been studying Tensor Calculus since I was 24 due to a strange fascination I had with Einstein's Theory in my youth :)
It has been difficult and I've read a myriad of books on the subject... some very technical some very practical, but all of them I always considered partial in some aspect, and none of them gave a full account of the subject by itself.
After so much reading and work I managed to understand the subject and gain some insight into it... but I was never satisfied with what I learned. due to the fact that none of the books I've read made the assumptions clear upfront and the underlying themes more clearly visible from the get go.
This is the very first time I've seen someone put the knowledge in display in a very clear and intuitive manner... and with really amazing teaching skills as a plus! For that you have my eternal praise and admiration.
Thank you very much, and if I ever I'm able financially (different country, rate exchange is not favorable to me atm.), I'll be sure to buy your book... it clearly surpasses any of those I've read in the past :)
Best wishes and good luck!
Thank you for the kind words! I'm glad you found this series. I wish my book was cheaper, but I don't set the price!
The book is on sale now on Amazon.
Absolutely beautiful. I am finding these lectures very satisfying and enjoyable. I took college geometry last semester and it was neat to get insights into how geometry is often a "cleaner" way of defining objects. The parabola being defined by declaring a directrix and a focus and then building it from there is a beautiful thing compared to just plopping an equation down and being drilled on the mechanics of the numbers involved. Your insight into the way that the tensor calculus can free us from the drudgery of coordinates to allow us to see the beauty of the forms is an inspiration. Bravo and kudos to you for such a wonderful presentation!
I really like your approach towards geometry first. A lot of my physics books just expect you to take definitions for granted and I hate that. Laying the foundation in geometry helps so much.
+The Misanthropic Nihilist Well, the whole object of tensor calculus is to express geometry in the language of algebra! So it shouldn't be a surprising place to start!
MathTheBeautiful
Like, I am probably going to binge watch the whole lecture series. Professor Grinfeld is answering so many questions I had in multivariate calculus that wasn't clearly explained. I like where he uses geometric intuition to build upwards.
Wow, I really love the way you teach! I can see you're concerned with making the students grasp the mathematical concepts.
Your videos are being extremely helpful for me! A lot of what I've seen throughout my academic life is starting to make sense. I really want to master this once and for all.
Yes, it's very surprising that Tensor Calculus is not part of a core curriculum. It's my goal to change that.
+Akira Uema Seconded. My favorite lectures in grad school were sometimes ones where the board was only filled once in the hour; the time and effort was spent on a single piece of information, and 1 or 2 enlightening examples. Mathematics is the most intense single mental exercise anyone can do.
I am starting to realise now why tensor calculus is so powerful. Studying engineering, so much emphasis is put on a coordinate system, that sometimes its hard to see light at the end of the tunnel when crunching through some elaborate theorem. My God I wish they had explained things in the language of tensor calculus instead.
I got a few lectures further on before I really got the idea of covariant and contravariant bases. I'm enjoying re-watching with better understanding. Lights blinking on frequently - thanks.
I've tried to study tensors several times using different books and articles from rather specialized to relatively popular-written, and I had given up.
So far your lectures look really clear and motivative, and they really inspire me to continue studying. Thank you, Pavel! As always, perfect!
I don't know if I'll be able to watch all of them, but I sincerely hope so:)
Hey Evgeniy, Keep up the great work. The best place to ask questions is on Lemma: lem.ma
He is the best teacher ever. Well, probably at the same level as Gilbert Strang.
Not quite there, but I appreciate the compliment!
Thanks for sharing these great lectures in your video series, you really have a clear and engaging style of teaching!
Wow, I always regarded Tensor Analysis as something quite unapproachable with my knowledge, but you, sir are doing a great service to education and to us students. Thank you, sir!
Thanks, that means a lot!
The explanations are very clear and insightful. Wonderful !
I think I should work through the whole series because I only watched 5 minutes and I learned more than from any textbook.
I don't really understand a specific thing though. In the example of the polar coordinates, since we have a basis per point essentially how do we describe a point in that space, there's no basis we can pick because we want to make all the definitions coordinate system independent.
thanks for making this videos, I'm genuinely grateful
Thank, Andrew! It means a lot.
Is it possible to do a video on how you can "discover" a coordinate system based on some description that isn't radial, and how you would describe it? For example, assume that we have coordinates that are sine waves that are stacked on top of each other. How could we describe a coordinate basis in this manner?
James Pedid If the sine waves are vertical shifts of one another, then they do NOT form a coordinate. You cannot take two random, arbitrary shapes, tack them together, and call that a coordinate system. Coordinate systems fundamentally do not work this way.
Thank you for this nice video 🙂
As Engineer, Tensor Calculus is very usefull and i like the brilliance in these Theoremes
14:33 "It would be a crime to write it as cos theta ...it is a geometric object ..."
Your teaching is beautiful indeed . I truly enjoy watching your videos . Thank you .
Thank you, Hanna! I'm really glad you're enjoying them!
26:25
Why introducing gamma(Z(s))? Do you actually mean R(Z(s)) - the curve?
18:45 Good discussion about how to deal with the components of the same vector changing at different points of the space.
Given a vector based on the inputs of R, how would you write it in terms of the covariant basis and vice versa?
If I am understanding, you defined R as a function that produces a vector based on co-ordinates, and writing a vector in terms of the covariant basis also produces a vector based on co-ordinates. In fact, R is used to define position vectors and the covariant basis is just another way of expressing that. So if the covariant basis is an equivalent way of expressing vectors, there should be some way to transform one notation into the other.
I noticed that if you have an arc R=R(Z(t)), then if you take the partial derivative, you naturally get a decomposition in terms of the covariant basis, but it seems a bit strange to define vectors as partial derivatives of the position vector instead of just writing them as the position vector.
I have also found that the inputs of R are only equal to the components of the vector in the covariant basis if the Z's are independent of each other and the basis vectors are independent of the Z's. If we write R(Z)=Z^i V_i then take the partial derivative with respect to Z^j and make the previously stated assumptions, then dR/dZ^j=V^j.
I was really having trouble coming to terms with the difference in basis depending on where you "drop" a vector down on a polar coordinate system, but now i get it. The problem that would require such a coordinate system also carries with it baggage. You could imagine the problem for polar to be the speed of a grape spinning around in a circular sink. Of course its speed component in theta is going to be higher the further away it is from the origin (sink center). Please let me know if this thinking is disordered. Your lectures are awesome and really scratch the itch of my engineering outlook towards mathematics.
I'm finding you're classes very clear and well explained. Thank you!
Just a quick question on this one. When you define the polar r coordinate geometrically you define it as the distance to an origin. Then you define another arbitrary origin from where you draw the R position vectors. Shouldn't they be the same? This confuses me a bit.
+Pablo Fernández Esteberena I don't think it has to be the same and the extra bit of generality helps in this case. It help highlight the fact that only differences in R, not necessarily R itself, that play the key role.
+MathTheBeautiful I think I see it now. Thank you for your quick answer!
hello...question ? @ 20:59 when you refer to the function in question as its gradient you switched variables from R to F , are these the same function, if not how are they related ....? ty very much..
let me see if i have figured this out, R is the function of a vector stemming from the origin, witch traces a function F as F ' goes about its business ' lol... ??? so the GRAD of F is the der of F in reference to the coor. sys, acting apon the der of the function R actiong on the coor sys . did I get it swami ????
Hey I've got a question!
At 23:00, why did you normalise e_theta with r^2 if its length is just r. A similar scenario came up with the stretched cartesian system and I didn't quite understand that either. Could somebody shed some light on this please?
From the previous lecture: ruclips.net/video/p75-f0gN5c0/видео.html
Michael Mitchell You divide the basis vectors by the length squared, not the length. The basis vectors are not being normalized, the gradient itself is. The idea is that the components scale by a factor A, and the basis by a factor B. Then, to normalize, you must divide by AB. Otherwise, the equation is incorrect because it implies the length of the gradient is coordinate-dependent, which violates the definition of length.
dear prof. grinfeld, one question, how do we see that the covariant/contravariant basis commute (in euclidean space or surfaces later), that is, they are "holonomic basis"? or do they not? thanks!
This is discussed later in the course.
MathTheBeautiful Your explain is clear, simple and brevity. I have not problems understanding your lectures on tensor calculus.
Fantastic videos. Thanks so much for uploading these!
The problem they're talking about at 18:20 is true whether you normalize or not. That's just a fact of life with polar coordinates. It is not a flaw with this definition of a basis.
thank you for the very helpful lectures. At 17:24 you say "in a lot of Physics books the base vector e_theta is normalized to unity, which in my book is absolute madness". Could some physicist out there make the case for normalizing the base vectors, or why physicists persist in working with normalized base vectors if they add unnecessary complication? Do equations with physical meaning such as Maxwell's equations, give the same value at every point when transformed to different coordinate systems even when the base vectors are not normalized and are changing in size and direction at different points?
at 10:06, did you mean the corresponding difference in the numerator due to the change of dx (the denominator) or am I misunderstanding something?
Yes, that's correct.
Thanks
I am following along with your course parallel to my normal studies and greatly enjoying it thus far. While working through the book however (I am on chapter 9) I found a persistent difficulty in deriving the covariant basis for different coordinate systems starting from pure geometry. I feel that I am at a loss in doing it for spherical for instance, even though the book technically covers it. It's particularly frustrating because all the rest of the formulae follow quite simply (and elegantly). Is there a trick to get better at it? Particularly I am unable to find the length of the vectors when subtracting them.
Hi Erik,
That's a very good comment and a valid point. When it comes to basic curvilinear coordinate systems (Polar, Cylindrical, Spherical) on only needs to understand the derivative of a uniformly rotating vector. Once you have that under your belt, it becomes a matter of applying that insight in various ways. Is that helpful?
Thanks,
Pavel
Hi Pavel. Thank you for your answer, I apologise for not having replied to your answers on my other questions. It is helpful but I am not really sure where to start? I have no problem differentiating the position vector along r for instance, the problem lies when the "position vector arc" becomes curved. My thought process is currently something like this:
1. Figure out the direction of the vector resulting from taking the difference of two position vectors a certain distance along Z^i.
2. Figure out its length (Problem: if the arc along Z^i is curved, how do I find this length? e.g. Z^3=phi in spherical).
3. Divide by the distance along Z^i (Again, how to determine this?).
4. If not uniform, take the limit.
This video is very much important for explaining position vector, should be placed after video 4b.
At 30:28 you said "unit normal", but, if I don't miss the point, you ment "unit tangent". Right?
Yes i was confused with that eithr unit normal or unit tngent.
Yeah. Someone edit the video please.
this is a beautiful video. thank you very much.
Thank you, that means a lot!
All co-ordinate systems are born equal. That's the key insight.
Yes!
But some may be better for spotting certain features of the system due to more symmetry ;J
the insight that leads to relativity
The only part that I find confusing in trying to connect this with the little bit of differential geometry on manifolds that I learned is that in diff geo, the "covariant basis" dR/dz is really only a basis of the tangent space (since the underlying manifold is not linear.. and hence can't possibly have a basis). Because you are only describing things in Euclidean spaces here, you can "bring back down" the covariant basis to also describe position vectors (as opposed to only velocity vectors). Is this the correct way to think about the covariant basis?
Well, we are not talking about proper manifolds yet. We are just considering the ambient space where dR/dz produces three vectors, i.e. a legitimate basis in the 3D space.
@@MathTheBeautiful Right, but my point was that dR/dz is the Jacobian of the map R, and Jacobians are maps between tangent spaces. So my understanding is that when we say that dR/dz produces "vectors", these vectors are really on the tangent spaces, but because the space is affine, there is a natural isomorphism between the tangent spaces and the space itself, which allows us to say that these "vectors" are also in the ambient space.
The reason why I'm being finicky about this is because I confused polar coordinates and rotating frames of reference for a long time. For example, en.wikipedia.org/wiki/Kinematics#Particle_trajectories_in_cylindrical-polar_coordinates uses the term "polar coordinates" to refer to what I would call a rotating euclidean frame of reference. When I finally understood that polar coordinates *induce* rotating reference frames on the tangent spaces, and that the wiki page is using the natural isomorphism between the space and tangent spaces, everything finally made sense. So I just wanted to double check with you that I was interpreting this properly.
@@samlaf92 I think that what you're trying to say is that at each point there exists its own linear space for which the three vectors dR/dz act as a basis. At another point, there is another linear space with a corresponding basis obtained by dR/dz. However, each one of these linear spaces is three-dimensional.
These lectures are excellent. However I was wondering if you could refer me to any extra materials so it would be possible to work out some problems and become better practiced. In other words can I have some homework please.
The textbook amzn.to/1m3Qqdp has many exercises.
Nice classes. Realy good🇷🇸
Why is the length of e_theta = r? Did not follow that.
looking at a vector in polar coordinates and changing only theta by any amount, you can visualize the vector that connects them as a chord on the circle that contains the start of each vector as its center. The vector would then define a sector of that circle and the length of a sector of fixed theta will change with r since the sector length s=r*theta. Draw a diagram and you'll see what I mean since I know reading off a RUclips comment isn't optimal :P
You CAN choose to place your position vector anywhere you want (it doesn't matter because you're interested in its DERIVATIVE, i.e. the DIFFERENCE between the position vector as you vary whatever coordinates, so you can choose ANY "pseudo-origin" (a word I made up) for the position vector, including--if it helps you visualize what's going on better--the center in your polar coordinate system.
Sorry, that's a mouthful. Let me restate it: The Tensor Calculus, as I understand it from this video, essentially uses some tricks to make your analysis independent from a chosen coordinate system. One of the tricks is to "scan" the space using a position vector. The starting point of the position vector is arbitrary and won't affect the scan. You could place it to the far bottom left on the black board, but you could also place it at the center of the polar coordinate grid if you wanted to. Either way, its derivative (which is just based on the DIFFERENCE between the position vectors as you vary them w.r.t to coordinates) will be the same, so the resulting scan will be the same.
So, to make things easier for us, let's imagine the position vector starting from the center of the polar grid. It would then point outward toward some concentric circle of radius r. Imagine keeping r constant, and only varying theta, and snapping the arrow-side of the position vector to the location of each point in the space that has coordinates (r, theta), where r=constant, while theta is variable. What will the position vector scan trace out? A circle of course. It will trace all points in the space that share the same constant r. During the scan, the difference between position vectors sampled with theta 0.1 radians apart (or some other small delta increment), will be a small segment essentially tangent to the circle.
For the same amount of variation in theta, you'd get TWICE as much circle for r=2 than you would for r=1. Therefore, the size of said tangent must be scaled by r.
Imagine the two line segments below are actually not straight, but pieces of circles traced out by the position vector (along the polar coordinate system's grid lines). Theta is varying the same at the center, and the position vector is tracing the result, but at different rs. Obviously, Delta(R) will vary more strongly at higher values of r.
R
|================== r=2
| DELTA( R )
|
|----> [the position vector R is sweeping this way]
|
|
|========= r=1
| DELTA( R )
|
|
|
|
. (Center of polar grid)
Again, the result holds the same even if R is not centered at the center of the polar grid, since Delta(R) will still give the same result.
Great explanation, @@keikitv ! If Pavel explained it this way, it would have saved a whole lot of time for the people watching this video.
The key observations here that made it for me were those:
1. The choice of the "pseudo-origin" as you called it (I would call it "external observer") is arbitrary, because:
2. what we are really interested in are only the DIFFERENCES in the position vector R as we tweak one of the coordinates while keeping the remaining ones constant.
3. Although the difference in angle stays the same for all possible endpoints of R, the "speed" of its endpoint may differ the further it is from the centre of rotation, because it sweeps longer arcs in the same "timespan" (we can imagine the change in angle occurring in time, to make the picture more "dynamic", but the key point here is just how does the coordinate change, and how fast is the corresponding change in R, and their RATIO, which should stay the same no matter where the observer is).
Way to obfuscate coordinate changes, eh? :q
Bon Bon What are you even saying? He explained it exactly in that way. Those were points he addressed in previous lectures, which by the way, are requisites to this one. Geez
Dude @ 19:24 is channeling my thoughts exactly
quick question:
what exactly is this "T" (unit tension?? )
English is just my second language :-/
T is the "unit tangent". It's a vector that's tangential to the curve that has unit length.
ah that makes sense, thanks!
10:40 I don't understand what is R(x, y). You do not specify the fucntion, how do you differentiate?
That's a very good question, Taras! And one that took me some time to figure out. I think the answer is that a function is not necessarily something that can be expressed by a formula. A function is any rule which "returns" a unique output for any valid combination of inputs. In this case, the inputs are two numbers and the returned value is a geometric vector!
Looking ahead, Tensor Calculus by definition occupies the space of "unspecified functions" since it operates in the framework of arbitrary (i.e. unspecified) coordinate systems.
How do you differentiate a function not given by an expression? See this: ruclips.net/video/hiYgYWJEaMk/видео.htmlm12s
Thank you. Did you assume that the origin of the polar system is in the center of the circle when you differentiated R? earlier 4:48 you said that the origin is arbitrary.
Correct. Since we only calculate differences of position vectors, the location of the origin doesn't matter. So might as well use the most convenient one (from some point of view).
1.Why does the covariant basis at each point look like cartesian if no basis is special?
2. When differentiating R(r,theta) w.r.t. theta I think it depends on the origin. For example if the origin is the alternative origin (5:00) dR/dTheta would not be tangent to the circle with the standart origin.
3.17:36 How would you use the covariant basis to define a point? Do you also have to specify the point whose basis you are using?
1. It doesn't look Cartesian, since not both vectors are unit length. Also the overall orientation is different. The basis is **orthogonal** and as a result, polar coordinates are an "orthogonal" coordinate system.
2. Yes, if you choose different polar coordinates, the resulting basis would be different. However, for any given coordinate system, the choice of origin for the position vector is immaterial.
3. Not 100% sure what you're asking. You will have to use a different basis depending on what point you're at.
"my trusted unit"
thanks, i could not understand covariant basis until i saw your video
The covariant basis is a totally mind-blowing local coordinate system. Should call the general polar coordinate system a global system. In the cartesian coordinate system, both local and global are same.
Hi. (At 10:20). In the example, if you increment x by 0.5, could not R - being an arbitrary vector function of x and y - take on ANY value at all? Why did you place the head of the new R vector where you did, on the x axis?
Peter Donnelly Actually R is not an arbitrary vector function. It's a very specific function: the position vector.
MathTheBeautiful Thanks so much for the instantaneous reply. Having watched all the lectures, it was very frustrating to have a lingering doubt over this basic (and as it turns out, trivial) point. ATB PeterD.
How does one take partial of R(r, theta) and get r*e-sub-theta. Understand the drawing, trying to understand it from calculus view point. Help!
Hi Joe, you can't get this result by manipulating analytical expressions, and that's perhaps the main point here. (Unless of course you introduce a background coordinate system, which would be the wrong thing to do.)
Thank you for this videos so much!
I still can't loose the feeling that there's something wrong with the way things are defined here. Using the same letter for everything doesn't really help that feeling at all, by the way. Even though the form of presentation isn't very mathematical, it should still be coherent and "formalize-able" I will try to sumit up and point to the difficulties I seem to encounter.
Consider a 2-dimensional (just to save space and avoid indices) euklidian space and a (continuous etc.) "Abstract-Vector"-valued Curve function K(s) in it, which parameterized by arc length. This means assume that
K' (s) * K' (s) = 1.
where K' denotes the derrivative of K.
Now assume that there is some general coordinate system with coordinates x,y and a covariant basis i(x,y) ; j(x,y) where i and j are abstract-vector-valued functions of the variables x and y. Then the vector-value of K always has a unique representation in the local basis i,j. That representaion is dependant on s and the choice of a local basis. This requires to specify at what coordinates (x,y) the basis vectors i and j are evaluated. It makes sense to evaluate them at the coordinates of the points of K for all values of s. This requires some bijection between the abstract 2-dimensional space and the coordinates for position vectors. These coordinates would then have to be independant of a choice of reference local basis (?). As far as I can tell that bijection is given in the video by writing R(x,y) which denotes an abstract--vector-valued-function of the coordinates that assigns a position vector to every set of coordinates. Then i and j were defined as:
i(x,y) := (d/dx) R(x,y)
j(x,y) := (d/dy) R(x,y)
It is then assumed that K(s) is a position vector for any fixed s (?).
As far as I can tell that's it for the definitions explicitly mentioned in the video.
I will try to make the following calculations slightly more detailed and add definitions I assume are implicit in the video.
Assume that (since R(x,y) is a bijection) there are functions Getx (v) and Gety(v) generating coordinates x and y for any abstract-vector v such that
v = R(Getx(v), Gety(v)) .
Pluggin in K(s) into this equation instead of v,
K(s) = R [ Getx (K(s)) , Gety(K(s)) ]
Now, as far as I can tell without any further justification, it is assumed that the functions Getx(K(s)) and Gety(K(s)) can be differentiated with respect to s.
I guess this should be provable from the differentiability of K itself somehow, not however without making restrictions on the bijection coming with the coordinate system. I belive the functions used to transform polar coordinates and cartesian coordinates are either not bijective or pose other difficulties which are reasons to use manifolds and charts in this context. Regardless,
I will call Getx(K(s)) =: K_x(s) and Gety(K(s)) =: K_y(s)
This then probably (?) leads to the equation used at 32:24 still calling the curve K and avoiding indices:
K(s) = R(K_x(s) , K_y(s))
By this point I'm almost sure I don't mean the same thing as the video, but allow me to proceed and differentiate: (not supressing the arguments inside the already differentiated functions)
T(s) := d/ds [ K(s) ] = (d/dx R)[ K_x(s) , K_y(s) ] * (d/ds K_x)[s] + (d/dy R)[ K_x(s) , K_y(s) ] * (d/ds K_y)[s] =
= i(K_x(s) , K_y(s)) * (d/ds K_x)[s] + j(K_x(s) , K_y(s))* (d/ds K_y)[s]
the result in my words: Given some value s of curve length, the COMPONENTS of the curve's Tangent vector with respect to the local basis at the position vector that the curve takes (at that particular value of s) are the derrivatives of the (basis-independant) COORDINATES of that same position vector.
Is this what is meant?
I think the answer is yes, but I'll make a comment and you'll let me know if it's helpful. I do think you are overcomplicating the formalism a little bit. Perhaps you are envisioning software implementation, which can be a little challenging.
In this narrative, when I say "vector", I mean a directed segment - as drawn in the plane or in 3D. It took me some time to adopt this point of view and to unlearn the dominant view that a vector is a pair or a triplet of numbers. A "vector" is a geometric object in the sense of Euclidean geometry. It can be held in one's hand. This approach is not sufficiently formal for a lot of people but I nevertheless invite you to adopt it alongside the formalized approach.
My geometric "vectors" can be added/subtracted, and multiplied/divided by numbers. They can therefore be differentiated with respect to parameters, which yields i = dR/dx and j = dR/dy. So I almost never think of the expression "coordinates of R" and there is no such thing as "basis-independent coordinates". Vectors don't have coordinates, coordinates are attributes of points in space.
So points on a curve have (curvilinear) coordinates Z^1, Z^2, Z^3. The components of the unit tangent vector equal the derivatives dZ^1/ds, dZ^2/ds, dZ^3/ds.
Yes, that is helpfull. Thank you.
Those curvilinear coordinates is exactly what I mean by basis-independant because they do not/can't depend on a choice of local basis. Since the non-affine coordinate systems can't be represented by just one basis I thought in this case saying basis-independant coordinates is fine. Indeed in the last statement I tried to make a distinction between coordinates (as a representation in terms of the coordinate system as a whole) and components with respect to some local basis. You seem to use this terminology as well, judging from the last paragraph.
The reason I'm overcomplicating it is not at all a software implementation, rather trying to apply this to a general characterization of a euklidian space and the assumptions that are made about this space. I imagine you could define some sort of coordinate systems for things like matrices with a Frobenius product as well. Whether that's usefull is a different question...
ANSIcode Are you saying the coordinate basis cannot itself depend on the coordinates? That would be nonsensical. There is no theorem or axiom stating this.
22:50 will e1.e1 work with stretch cordinates, as well ? Since we are considering gradient as function of arclength, how can we get sense of direction with arc length alone???
Hi Debendra,
I'm not 100% sure what you mean. Can you ask in a slightly different way?
Looking forward to it!
Pavel
debendra gurung Your question makes no sense. We are not considering arclength there
13:48 But how do you know how far is that 0.5 from the starting point and in which direction, if you don't have your unit vector `r` defined yet? :q
I think you're putting Descartes before the horse here :P
Bon Bon He has a well-defined unit vector, given by his chalk. Unit vectors are chosen arbitrarily, and they do not have to be part of the basis. Otherwise, all basis would by definition be normalized, and this simply implies a contradiction. If e1•e1 = 1, and I define e2 = (2)e1, then by your line of reasoning, e2•e2 = 1. However, since e2 = (2)e1, I can prove that e2•e2 = (2e1)•(2e1) = (2^2)e1•e1 = 2^2 = 4, and then your definition implies 4 = 1, which is false. Therefore, your definition of unit vectors is invalid, and your understanding of vector magnitudes are defined is incorrect. Vector lengths are basis independent. This is true by the definition of geometric length.
Professor Grinfeld,
I have a question about differentiating a parametric (vector) curve whose parameter is arbitrary. If the equation is γ(Z(t)) and it's decomposed with respect to the covariant basis γ^i(Z(t))*Z_i, then how would I differentiate this? I ran into the trouble of not knowing how the covariant basis Z_i depends on the parameter t.
Thank you!
+The Flagged Dragon This is one of the places in the subject that requires a great deal of diligence. Whenever you're about to differentiate, you must first write down a precise identity in terms of the independent variables. There's not a blanket recipe I can give. It's just a matter of explaining to yourself what depends on what in your particular problem and then writing it down as in mathematical terms. In your case, that identity would be
γ(t) = γ^i(t)*Z_i(Z(γ(t)))
It's different from what you had, so you'll have to think about it a little bit. I think I discuss a similar question in the video entitled Velocity Acceleration Jolt, but you'll be much better off doing it on your own.
Thanks for the fantastic question.
+MathTheBeautiful Ahhh that makes more sense! Finding the dependence especially with a changing basis was quite challenging. I thought I found a solution but it differs a small amount from differentiating the equation you gave me.
Thank you so much for the response!
+The Flagged Dragon It took me many attempts to get it right. Really requires sorting everything out.
This is a great lecture series, thank you! I am confused about one thing: at 33:14 or so, you claim that the z's in the partial derivative of R are the coordinates, but the z's in the total derivatives are the parameterized coordinates of the curve. How can two entirely different z's arise from application of the chain rule? There would need to be a dz[coord]/dz[curve] term in the middle for this to be true. Unless I'm missing something, which is entirely possible!
R asraster They are not different objects.
I am confused, tired, or both. For the definition of a covariant basis vector as a partial derivative of the position vector with respect to Z_sup_i, Z_sup_i is a coordinate of the coordinate system. In some coordinate systems, Z_sup_i are also/equally the (contravariant) components of the position vector, right? But in other coordinate systems such as spherical and cylindrical the Z_sup_i coordinates are not the components of the position vector because the angle components of the position vector are zero, right? If so, for spherical and cylindrical, are the Z_sup_i (contravariant) components of some other vector (not position vector) with some special meaning? Thanks.
There are two " Z^i ". One with a bold Z indicating a contravariant basis vector and the other plain indicating the coordinates. There is very in common between these two objects even though they are denoted but very similar symbols. One may argue that the letter Z is overused, but I think I've struck a pretty good balance in terms of using Z. (Some authors denote the position vector by the bold letter Z and also use it for the Levi-Civita symbol!) In all cases, either the font or the indicial signature (or the context) uniquely determines what object we are talking about.
It is true that in some coordinate systems (affine, to be specific) the coordinate Z^i equals the contravariant component of the position vector R. However, this is a largely irrelevant fact and is almost never used except in a couple special instances. In fact, is position vector R is usually not decomposed with respect to any basis. Not that there is anything wrong with that, it's just that the need almost never arises. For the most part (although there are some exceptions I'm quite fond of), the role of the position vector is to be differentiated and then to disappear from consideration.
So is it correct that the coordinates themselves (and the tuple of coordinates) are not a variant, not a tensor, not a vector? The components of the position vector are a tensor.
Mongrav Let me break your question down into individual parts.
In tensor calculus, you typically bring the n-tuple along for the ride. So Z^i can be seen as both: the i-th coordinate and as the n-tuple of coordinates. So there's no need to say n-tuple of coordinates - you can just say Z^i.
I guess I would technically call Z^i a variant.
I would never call it a vector in the context of tensor calculus - an n-tuple of numbers and a directed segment are as different as two objects can be.
It's generally not a tensor. (It would be in a more narrow sense, if you only considered linear transformations of coordinates. This is done, for instance, when studying rigid body motion.)
The components of any invariant vector field with respect to the covariant basis is a contravariant tensor. Since the position vector is an invariant vector field, its components form a tensor.
MathTheBeautiful ok, thanks. I don't have a follow-on question at this point. The following material is just to help me later if I come to a new question.
Let's take cylindrical coordinates: r, theta, z; covariant basis: r_hat, r*theta_hat, k_hat. I understand the non-bold z^i as the coordinates (r, theta, z). (I can recover the basis vectors from differentiating the position vector wrto the coordinates, no problem.) For Cartesian coords, the non-bold z^i happened to also be the contravariant components of the position vector. But here in cylindrical coords, z^i are the coordinates but are not the contravariant components of the position vector, which would be (r, 0, z), not (r, theta, z). The theta component is baked into the particular point and the r_hat basis vector for that point. For the cylindrical and spherical coordinate systems, the nonbold z^i coordinates are not the contravariant components of the position vector but relate to a differential excursion from that position.
Mongrav I think what you are saying is correct. The only thing I would like to add is that I have never thought about what the components of the position vector are wrt the covariant basis. That question just doesn't come up that often. One of the few instances where it does come up is rigid body dynamics, but I don't think it's one of your interests. But since you are thinking about it, I'll mention one fact for you regarding those components: their covariant derivative is the Kronecker delta symbol. See if you can show that.
the basis is r=e^2.pi.t
I don''t understand exercise 62. How could it be positive definite?
Later this week there will be a video on this in the Vector Calculus series.
In the 4b lecture video, I got some hint that says "metric tensor is symmetric and as an exercise it is positive definite". So, of the exercise 62, the covariant basis, interpreted as a matrix, is meaning the metric tensor?
Yes! The proof is discussed here: ruclips.net/video/PMkGPc14YfY/видео.html
Thank you so much.
Hi Prof. Grinfeld
Question 30:59: Why d_R(s) / d_s is the unit vector? I "sort of" understand why it is the tangent vector, found it here www.math.jhu.edu/~jmb/note/curvtgt.pdf and I say "sort of" because the proof uses the dot product, which we still haven't defined at this point in your lectures in this general coordinates.
I can not comprehend why the fact that we use s (arc length) makes d_R(s) / d_s not just tangent, but UNIT tangent vector.
Thank you
Take a look at lesson 1 past about minute 34.
That was fantastic!
Thank you, Douglas!
Hello, I'm very much enjoying your series here, and had a question. You define the vector e_r to be the derivative of R(r,theta) with respect to r, where R is the position vector. This makes sense to me. But let P be the point with polar coordinates (1,0). P also has coordinates (1, 2pi) in polar coordinates. So to find the vector at P, what should I do, differentiate R with respect to r at (1,0), or at (1,2pi)?
Now of course in polar coordinates we get the same answer either way. But how does this work in general? Do we only deal with coordinate systems where when a point can be expressed in coordinates two ways, the partial derivatives will turn out to be the same no matter which coordinates we choose? Polar, cylindrical, spherical, and affine coordinates all satisfy this. Or can a point have two vectors?
Well, most theory that's developed is "local", meaning this kind of situation is avoided. But yes, if you imagine an overlapping coordinate system, you may get overlapping everything else.
Actually thinking about it, using polar coordinates, it seems to me the partial derivative with respect to r at (1,0) and (-1,pi) will have opposite signs, even though they designate the same point. I guess we just have to exclude negative values of r?
No problem as long as you assign negative r values to neighboring points.
I think when r is negative, the partial derivative with respect to r will point towards rather than away from the origin. After all, if one increases r from -1 in (-1,pi), the magnitude of r decreases, and we go towards the origin.
Or we can cheat and convert to Cartesian coordinates, (r cos theta, r sin theta). If we take the partial with respect to r, that is (cos theta, sin theta). At (-1,pi) which has Cartesian coordinates (1,0), the partial derivative in Cartesian coordinates is (-1,0) which is pointing from (1,0) towards the origin.
Misty Mouse No, that is incorrect. The partial derivative is not pointing away.
At around 17:45 you mention in physics they normalize the theta component in polar systems. You then don't say it directly in the next few minutes, but that wouldn't really solve their problem, right? Because even if the coordinates have equal length, their directions would change all the time, so making them have equal length doesn't make them the same anyway.
That's right. I'm not really aware of any problems that normalizing these basis vectors solves.
@@MathTheBeautiful there's one possibility: they make the picture look less messy, because the arrows don't overlap then ;J But yeah, geometrically, it makes no sense at all.
Nice teaching thanks.
ok, now |e_theta| is r can only be shown if we take the definition that R = r cos theta i + r sin theta j. is that true or can it shown in some other way?
Great question and the answer is an emphatic NO. This can be done (and is done in this video) on pure geometric grounds.
Imagining a background coordinate system is the #1 affliction that precludes students from seeing things in a clear and simple way. If you are choosing between quitting smoking and quitting background coordinate systems, please choose the latter.
sorry but in the video there was simply a statement that this is true but not demonstrated. perhaps i am wrong.
ok, so just using the definition of the derivative i was able to prove, to myself, that indeed |e_theta| is r.
Fantastic! That's all that matters.
@@MathTheBeautiful Wasn't it _you_ who claimed several times that the space has its own "inherent" unit? (which would imply having its own "inherent" coordinate system)
Isn't the "yellow" coordinate system just another way to sneak in the "background coordinate system"? (even if its choice turns out to be arbitrary in the end, because it doesn't influence the end result)
sir i have a question=
i want to find unit tangent to curve
y= sin x in the cartesion coordinate system.
ans:
parameterized form, r(s,sin s)
here z1(s)=s and z2(s)=sin s.
equation of unit tangent according to your formula is z1'=1 and z2'=cos s.
so T=1 i + cos s j.
which is wrong eqution bcoz its length is not 1 for all s.
where am i going wrong sir?
(z1', z2') is not the unit tangent, it's just a tangent.
sir but you clearly said (T1,T2) or (z1',z2') are the components of the unit tangent with respect to basis e1 and e2.
at time 35.30 in the video.
Ohhhh!! my parameter 's' is not curve length and thats why my tangent is not of unit length!!! I got it. sorry for the stupid question.
and thank you for your videos sir. extremely helpful!!
Exactly! Sorry I didn't respond to your previous comment.
15:27 "Here the coordinate basis will be different at all points"
Stupid Question (which I also had when reading the book):
Why is e_r equal to dR/dr and why is e_theta equal to dR/dtheta? In other words why are the basis vectors equal to derivatives of the position vector w.r.t the coordinates?
That's the *definition*.
Thanks. So we cannot define basis vectors without derivatives? I always thought of bases as a set of vectors (geometrical or otherwise) that any other vectors can be represented in terms of. I am just not making the connection between these two concepts.
Yes, it's one of the examples of things that come in the opposite order in applied mathematics compared to linear algebra. In linear algebra, basis comes first and coordinates are built from the basis. The inner product comes first and lengths and angles are built from that.
In applied mathematics, coordinates come first and the basis is built from the coordinates. Lengths and angles come first and the inner product len1*len2*cos(angle) is built from that.
Thank you!
MathTheBeautiful
is the position vector R(r,theta)= ?
I guess my question is what is R? is it some arbitrary function or is it just any vector that points to any point?
damn, what a great teacher
Why do you use i and j for the unit cartesian coordinate vectors, instead of e_x and e_y like you do for polar coordinate. This seems to acknowledge the cartesian coordinate system an importance that you repeatedly over this series try to deny.
Yes, you are right! Cannot defend my actions, except to say that I try to not be militant about any particular thing.
That said, I use i,j,k for all affine coordinates. Furthermore, affine and, therefore, Cartesian coordinates *are* special in many profound ways. For example, their very possibility signals that the space is Euclidean. Cartesian coordinates can be used to show that the RIemann-Christoffel tensor vanishes in all coordinates systems in a Euclidean space. Etc. So, special symbols for special coordinates seems OK. Also, I often use x,y,z for Cartesian coordinates and r,theta,z for cylindrical coordinates.
very helpful.
36:33 "isn't that so satisfying?"
Fuck yeah it is...
Yeah, and all we did was use the Chain Rule once.
superb!
You da man
14:46
“SIIIIIIIIIMP! Only an absolute SIMP would write the polar coordinates in terms of the Cartesian coordinate system!!!”
Sounds like Matt Damon #goodwillhunting
I would have preferred Robin Williams
So confusing
Hey Leopoldo! Check out the new Vector Calculus series which is a prequel to this series. It will help set a lot of the things up.
Pavel
is that right ? no , that's not right
Wonderful course, where running out of letters is the most incredible excuse for sloppy notation! :D
I think the notation is one of the most satisfying elements of this subject.
@@MathTheBeautiful yes, indeed. Sorry, I think my comment did not come across as intended. I realize these days that one of the major impediment in my learning of mathematics decades ago, was the confusion introduced by poorly explained, and poorly enforced/followed notation. I also came to realize the importance of definitions. I certainly did not mean to be negative towards your teaching, which I find both wonderful and very clear.
@@freddupont3597 Ah, yes, now I agree with everything you're saying!
too many letters that deserved different letters. It's already difficult, why not just use different letters to make it less confusing
Hi Radu! It's a valid point. Just pay attention to the indicial signatures - they are all different! That's why I use the same letter. Otherwise, there are lots of letters and lots of indicial signatures.