Tensor Calculus 2: The Two Conflicting Definitions of the Gradient

[@larrykeese1948]

The teacher in this classroom is as good or better at explanations than Dr,.Suskind. Great Job Sir. I am 70 and decided to continue my understanding of Theoretical Physics. My degrees are in EE. I really do enjoy the lectures. Hope more kids are watching. Larry

[@abhijithrambo]

Massive Respect!!!!

[@bonbonpony]

Yup, he's better in explaining stuff, that's for sure.

[@ChaineYTXF]

Superb comment, sir!

[@ozzyfromspace]

“Kid” here 😅 We’re all learning

[@alemoropv]

Larry: I have your very same enthusiasm thou younger "aged" - 56, still enthusiam for learning good stuff like this! Your statement made my day.

[@carmelpule6954]

The gentleman giving this lecture should go down in history by his statement, " Space stands on its own! Space has its unit vector irrespective of the coordinate system you impose on it." At my age which is now approaching eighty years, I am sitting here enjoying all this and wondering why my teachers, more than sixty years ago did not make that statement , " Space stands on its own!" I shall try and describe what this statement triggered in my mind and what this lecture has done to me. In my younger years my teachers always made me believe that I should stand firm on the axis of a selected coordinate system and while I move along the coordinate axis, I see the curve or the function shaping itself with respect to me while I move along the coordinate system, which were described to me as being solid ground. I always seem to use the time parameter to describe the movement along the coordinate axis so that I can stand in the some outside position watching all the beauty and wonders of all this spectacle through being an impartial outsider standing away from the fixed coordinate system. So here is a little bit of fantasy that has now been generated in my mind. Rather than standing on the coordinate system, watching the function above me, I should float away and ride on the curve or function itself. If the function is a scalar field of say , temperature, pressure, voltage, mass density distribution, etc, roaming on this scalar field say on an equi-potential surface , I would not know my direction unless I have sensors to see the field surface, but I would be totally lost regarding direction if I were to be shortsighted and could not look afar. So to get some bearing regarding direction I would have to forget about the magnitude of the function and go and detect its rate of change and the rate of change will give me a vector to give me an idea what the function is doing, even if I am standing in the dark in the same position. Obviously there would be many rate of changes , but it would be logical and reasonable to chose to walk along the highest rate of change of the function as that would be an absolute reference to the scalar function Having a vector to hold on to will give me a better understanding of the function while I stand on it rather than standing on either of the three co-ordinate system which my teachers always told me about. Now I can imagine myself standing in one location and cannot see any further than my nose, and all I have at hand is the Vector indicating the maximum rate of change of the function . With the space around me existing on its own, but owning its own unit vectors, well, I could now stand on the function and create the coordinate system around me to my own convenience. Hence with a wave of my magic wand, there it is , a space unit vector to which I can now compare the gradient vector in my hand of the scalar field I now ride on. At last I have a point of reference to which I can refer to as I move along any surface of the function, With the gradient vector of the scalar function as reference, that is, the direction derivative of the function, in the direction relative to the unit vector. Irrespective of the direction of the space unit vector, well now, if I had to stand at a point in space, knowing that I stand on a function, it is possible for me to know what path the function would follow , if I use the unit vector and the direction derivative of the function at any point to build on the present position through a space integral. It would be a wonderful experience standing on the function as it forms and even if the unit vectors had to continuously change direction, I could always make an allowance for that before I integrate to find the next point on the function I am walking on. I think that is enough imagination for today, but it has been an education to simply leave the solidity of fixed the coordinate systems I was religiously preached about, especially the cartesian coordinates and ride on a scalar or a vector field holding on to a vector related to the function while comparing it with a unit space vector of convenience. It is new life for me , so free, so free after eighty years of being imprisoned and locked by the coordinates rather than riding freely on the function and me racking my brains about creating directions through derivatives and who cares if the function is fixed in space with the space itself moving around me. I still feel that I can now handle my functions quite admirably by riding on the function! It is fascinating found freedom. Congratulations to the gentleman making these lectures, I hope that I did not misinterpret his teachings. Perhaps, the fairy tales of father Christmas and how Rudolf the red nosed reindeer navigated through the sky was meant to teach us about learning to put more value to knowing the functions and what value have their derivatives, rather than the fixed coordinate systems we were locked on to when we were young.

[@StankyPickle1]

I find your passion to learn inspiring.

[@bonbonpony]

Before going any further, you should ask yourself a question: "What is space, exactly? Is it real?" Because it is not. Space is something that we imagine in our heads when we measure the world around us. If there weren't any interesting objects around us (that is, if we were a point of consciousness in endless emptyness), we couldn't tell any directions nor distances, so there wouldn't be any space whatsoever. In order for us to define space, we need to have at least two distinct things in it - two places that we can distinguish in any way from one another. More philosophically: being an observer is not enough - there has to be something it observes. Then we have two distinct points: the observer (let's say it's the ORIGIN of your space), and the observed object. Then you can connect them with an imaginary straight line, which defines the first DIRECTION of your space. And you can also proclaim that from now on, this piece of the line between you and the other point - is the UNIT of length, or "1". Now if you have some third point somewhat further away from you (let's say in the same direction as the second one), you can try measuring the (as of yet unknown) distance towards it by comparing it with the (already known by acclamation) distance between you and the second point. How many steps of that length you have to make in order to get to the third point? This number will be the distance, and now you know it too. (But only in comparison to the first distance! Without it, it would be still unknown!) In a similar way, you can find some tricks to figure out distances to other points that lie sideways from your original line. Only then your space would have any other directions (that is, it would also have more DIMENSIONS than one). Everytime you can find a new point that doesn't belong to the space you've got so far, it extends it by a new dimension. But everything I described happens only in your head! There is no such things as points, lines or spaces in the real world. There are only objects that you can observe with your senses (or extend your senses with mechanical tools), and then express what you see by points, lines and spaces in your imagination to reflect that. The outside world has no directions, units, numbers, and it doesn't have grid lines stretched all around it. It's only our own way of measuring the world.

[@alexanderkurz3621]

I love what you write ... it is the kind of poetry that is at the origin of new mathematics ...

[@alexanderkurz3621]

What is so good about the lectures is that, while they are rigorous, the formal maths is introduced as it is needed to solve problems. School maths is often doing the opposite. For example, even simple concepts such as functions are found to be confusing. The consequences are devastating. First, only students who are nerdy enough to like formal notation will enjoy doing mathematics. Second, students, even the good ones, come out of school maths believing that the formalism IS the mathematics. Nothing could be more wrong than that. So thanks a lot, Pavel, for being a brilliant guide to the beauty of maths.

[@Blue-ik8ij]

It's pretty good but nothing rigorous tbf

[@DanielMartinez-ss5co]

I followed his lectures and then I bought his book on Tensors, an enjoying task to learn from it, then I could access to General Relativity Theory, I'm 63 years old and grateful to this kind of videos I could comprehend things I never was prepared for. Thanks Sir!

[@rontoolsie]

By far the best series of lectures on tensors on KZbin-which is not to say that there are not worthy alternatives-but these invoke intuition rather than abstract mathematics. I was convinced enough to purchase the text book to add to the ever growing library that a PhD student of physics must accumulate. The closed captions are hilariously inaccurate though-which may be more of an detriment to a non-native English speaker than to someone monolingual in that tongue such as myself.

[@alexandra-stefaniamoloiu2431]

You answered so many of my questions in mathematics in such an elegant way. I feel enlightened.

[@MathTheBeautiful]

I'm glad to hear you say that because that is how I felt when I was learning this subject.

[@joeboxter3635]

@@MathTheBeautiful Yes, but the universe is a zero sum game. She feels enlightened. I feel confused. Lol. Nice lectures and text on tensors.

[@oscarmejia8306]

Thank you very much for these complete lectures, along with the exercises. They will help me achieve my goals. This resource is invaluable.

[@jordanlawlor89]

Thank you for posting these lectures! I would love to sit in on a class with this professor!

[@PenguinMaths]

I really appreciate these videos, they are greatly helpful in understanding the topic. I look forward to seeing the rest of the series

[@markschedler8841]

About halfway through where the partials in x and x prime are discussed, the explanation as to why the partial in each axis doubles seemed a bit confusing. I found it helpful to consider a fixed point where x prime is 0.5x. That way the value of F is unchanged, but in the limit that defines the partial derivative, the denominator is halved thereby doubling the value of the partial derivative in x prime. So far the lectures are great.

[@ankanmann8017]

you are an awesome teacher sir, your explainatory skills are awesome, it really gave a great a insight

[@abajabbajew]

17:00 the doubling of ∂F' is easier to infer from the halving of the Δx' (=0.05) going 'into' the denominator, I think (i.e. there is no appeal to 'approximate' local linearity of F' involved in this rescaling, just the use of an _identical_ local 'arrow' as before) The conclusion that the Gradient F' must maximize along the chosen (and so any) direction seems to come about from being able to recover both F' and n from the form of the dot product (since the chain rule used in differentiation had 'picked up' the components of the direction vector n as scalers) This was an excellent lecture whose gradual development allows the viewer to follow very closely (and I think that's the key) what you are discussing, and motivates each new step. Thank you once again.

[@midnighttutor]

I am enjoying this series of lectures. I already know what tensor calculus is all about and so this is interesting back fill. However, there is a disadvantage to this sort of inductive teaching style to present a new topic. The student does not appreciated the sort of labored induction of each new concept and gets lost or bored. I think in the end it is better to just present the new idea and then at some point go back and put it in perspective after people are already comfortable with it (meaning they have done homework using it). Then they go "ohh yeah now I get the thread". Same thing with the Feynman physics course in 1961. Half the room apparently was graduate students and caltech faculty. They already knew the material so his unique different way of presenting it gave them the 'aha' feeling. The students apparently....not so much.

[@ES-qe1nh]

Love this lecture so much

[@MathTheBeautiful]

Thank you for letting me know - it means a lot!

[@lawrencemwangi8846]

Perfectly explained. Congrats

[@rajanalexander4949]

Legendary teaching style here; thank you!

[@MathTheBeautiful]

Much appreciated! Glad you liked it.

[@johnkechagais7096]

The units change. This transformation from one co-ordinate system to the other is what happens when comparing metric to imperial systems the gradients is exactly the same measured in different units

[@snnwstt]

The unit can belong to the space or to the system of coordinates. Your system can sit on a sheet of plastic that Poincarré plays to stretch, and Poincarré, in his view, will see your unit change, but you, you won't . Same as if you sit on a particule of water of if you sit on the river bank. There are Lagrangian and there are Eulerian coord. Both are possible. Here, the approach is un-usual as the unit belong the to space itself, not to the system of axis.

[@m.s.6449]

Not really, just imagine a slightly different coordinate change where we stretch only X coordinate and keep Y the same - clearly there is no such unit of distance that measures differently in different directions. In the video it just happens that the numerical effect of coordinate change is similar to what numerically happens after a change of units, so it takes some effort to really distinguish those two concepts. But if you take any more interesting coordinate change - for example cartesian to polar coordinates - you will realize that a change of units acts in a completely separate way to a change of coordinates, since with a polar coordinates we have exactly the same units with exactly the same scale, but expressions of geometric formulas in terms of coordinates will change a lot.

[@snnwstt]

@@m.s.6449 Since the vectors of the coord. system are not necessarily "constant" at/for all points, and not only in direction but also, eventually, in "lenght", it seems more appropriate to tag the "unit of measure" to the (local) vectors of the coord. system rather than to the numerical value (component) describing the linear combination having eventually to imply these vectors.

[@johnkechagais7096]

@@m.s.6449 Would that not imply a distortion of space rather than the measurment?

[@yyc3491]

17:10 To understand why ∂F'∂x'=2∂F/∂x, we should distinguish the "coordinate distance and actual distance". The x and x' here are unitless numbers (coordinate distance such as 0.1, 0.2...) while xi and x'i' have the unit of length (actual distance such as 1m, 2m...). An easy way to understand this is just looking at a fixed ∂F before and after the change of the coordination: ∂F = ∂F' This requires x'i' = xi (because the choice of different coordinations do not change the curve of F) ∵ i' = 2i ∴ x' = 1/2x ∴ ∂F‘/∂x' = ∂F/∂x' = ∂F/(1/2∂x) = 2∂F/∂x.

[@djordjekojicic]

Really good explanations. I have bought your book because of these nice lectures and haven't regretted it. I have one question. Where can I find diagrams mentioned in solutions to exercises?

[@elsurexiste]

Nice teaching! By the way, where is this class taking place?

[@MathTheBeautiful]

Drexel University in Philadelphia, PA, USA.

[@Omar.Alamoudi]

One things I noticed and enjoy is your choices of words, you try to be very precise and that works very well for me at least, I really appreciate the effort. I wish I was a student of yours in my undergrad. I have a question, you state at 32:40 the gradient is 2 numbers, which 2 numbers are you referring to? I remember you stated that the gradient is a vector point in the direction of the greatest increase with a magnitude of the rate of change, I don't remember you stating 2 numbers! Please clarify and thanks :)

[@MathTheBeautiful]

That's the two definitions of the gradient. According to one, it's an arrow. According to the other, it's a pair of numbers: the two partial derivatives: dF/dx and dF/dy. This video is about (the difficulties of) reconciling the two.

[@Omar.Alamoudi]

Thanks! I wanted to make sure you were actually referring to resolving them with respect to the Cartesian Coordinate system!

[@bonbonpony]

Yes, you can tell a good teacher by how he/she uses simple words and goes straight to the point instead of verbal diarrhea full of gibberish words.

[@serbanmike]

@33.50 Professor Grinfeld says : "...the space exists on its own. " If there is a causality principle, should we imply that space was actually created at some point in time? Does the space really exist of its own? Would space have any meaning if there was nothing in this space?

[@alan2here]

Re 33:59 The Unit is Unitless :)

[@arjunnarayanan271]

Thanks for the very useful lectures. I have a question. You say at 23:20 that dividing the basis by its 2-norm squared should make the gradient invariant for cartesian and stretched cartesian co-ordinates. I find this a little hard to see. When you work with the (i,j) basis, all vectors are written in terms of (i,j). This means that in the (i,j) basis, i and j are by default normalized. Similarly, when you work in the (i',j') basis, i' and j' basis, i' and j' are by default normalized. I think what you mean in the video is to **represent** i' and j' **in terms of** i and j, and then divide this representation by its 2-norm squared. Is that correct?

[@MathTheBeautiful]

I think you should try to see things differently. One of the goals of this lecture is to change this mentality that you share with most of the viewers. The term "2-norm" especially gives it away. You think of the basis vectors as defining the unit of length. But that's not how I'm inviting you to see it. You should think of the unit of length as existing before the basis is introduced. Then each basis vector has a well-defined length (and we can then divide by the square of its length).

[@arjunnarayanan271]

MathTheBeautiful In that case, | i | does not mean the same thing as what is generally used, right? For instance, how would I represent the vector i in terms of the basis (i,j)? I would represent it as (1,0) -- a linear combination of i and j with these coefficients would produce the necessary vector (which is i). Similarly, how would I represent i' in terms of the basis (i', j')? I would again represent it as (1,0), because a linear combination of i' and j' with these coefficients will produce the necessary vector (which is i'). For the gradient to remain invariant, you need to represent i' *in terms of i and j*. In your particular example, i' represented in terms of (i, j) will be (2, 0). A normalization would then proceed naturally. I think I see what you are trying to explain. But I find it hard to digest the notion of a "norm" until some sort of metric has been introduced in the space. My confusion probably arose because you had not *defined* | i |, and I probably have a different understanding of | i |.

[@MathTheBeautiful]

Arjun Narayanan Yes, the length (pick-up-a-tape-measure-and-use-it) was defined in the previous video "The Rules of the Game". And that's the definition that's used throughout the course and gives rise the metric. This (length --> inner product) is opposite of the way it's done in linear algebra (inner product --> length). That's because our starting point in this subject is the physical space; while in linear algebra, it's the algebraic structures. So here were start with the concept of length and then have to figure out how to express it in terms of the coefficients (which you described correctly) with respect to any given basis (i, j) or (i', j').

[@arjunnarayanan271]

MathTheBeautiful Okay that makes a lot of sense now, thanks! I never thought of using invariants to define the nature of a space, rather than using the space to define invariants!

[@bonbonpony]

This is wrong. Units don't exist without coordinate systems. The `i` and `j` you introduced ARE the units that define that particular coordinate system and only then you can express everything in terms of those units ("measure the unknown world with them", so to speak). There is no "inherent unit" in space. Space does not have grid lines spanning through it - they are only in our imagination.

[@TheAaronDrew]

Great lectures, would be perfect if the audio didn't have so much static

[@user-ij4zr9li8l]

Dear Professor, Can you please tell us when is the next edition of your tensor calculus book expected to become available?

[@MathTheBeautiful]

Available now for free at grinfeld.org. Not a next edition, but a completely new approach.

[@user-ij4zr9li8l]

@@MathTheBeautiful Can I buy it in pdf or printed form (preferably the latter)?

[@davidhand9721]

It seems like the most natural way to take a gradient would, rather than dividing by |i| squared, differentiate over i instead of x. Like a directional derivative. The problem is changing to i' but still using i to measure x.

[@antonellomascarello4698]

2:02 : Geometric description of the gradient

[@Nobody_114]

At 29:05 isn't f'(s) simply equal to F(n1,n2) since dF/ds of x0+n1s = n1 and of y0 + n2s =n2 because x0 and y0 are constants?

[@kingplunger6033]

It always felt wrong to me when doing vector calc exercises, that some coordinates were implicitly expected. Like "prove this relation" and I just thought "why in cartesian and how would that work in general" when x,y,z were used in the solution.

[@MathTheBeautiful]

Same!

[@goosew3266]

The fly example was killer! Can I get this straight? At the part around 6mins, where you defined the gradient vector as: VF = dF/dx i + dF/dy j your "driving point" is that we can't forget F is a Vector, and x and y are Scalers. Right? F is some geometric object, living in space, but it's "real"! x and y are only the numbers on the measuring stick we use to define its position. Also how do we define how to take a derivative of a vector with a scaler? Is it with the old epsilon delta definition?

[@zorro20010]

F is not a vector its a scalor and the gradient of this scalor is a vector

[@gqwang6277]

17:55, This expression in the new coordinates system will be fours times as large. but the coefficients will be two times if all are evaluated into number. So the "four times" imply the tranform of i'=2i . The i'=2i is in geometric sense and somehow is conflicted with the conception that , in geometric sense, the gradient vector should be the same. Maybe the i'=2i should not be applied here.

[@BleachWizz]

wait the unit is defined previously? this part messed me up. If you define a chalk [ck] as your unit, then your choice of coordinate system means that the vectors i , i' j and j' are not unit vectors for any basis basis still when you write a vector in coordinates a.x + b.y, independent of the basis aren't a and b assuming the vectors to be unitary? I'm not being able to explain my doubt since I feel like every time I'm writing I think of an answer. But still how can you freely scale these things as we do on paper like writing random lengths as 1 and still get trouble in this situation? shouldn't you in the end need to be able to extract what exactly is the number 1 for that basis? like different basis have different but comparable definitions for their real number line, or something like that

[@DanTheMan0922]

This type of geometric analysis of multivariable calculus, is this not normally taught in Calc 3 classes? I was taught these concepts in my Calculus 3 class (granted it was an honors course). We defined planes using a dot product definition and lines using a cross product definition. We defined a surface area as the integral of the magnitude of the cross product between the radial and angular partial derivatives of a cylindrical function.

[@snnwstt]

I am not sure about the stretched cartesian system of coord. As example, if (x, y), are in meter and (xi, eta) in feet, we have a STRETCHED cartesian system since (xi, eta) are numerically (roughly) 3 times (x, y), and then the temperature GRADIANT will be (roughly) numerically 3 times smaller in Kelvin/feet than in Kelvin/meter, but exactly the same if I keep the UNITS, not just the numerical figure. Right? So, the gradiant appears different only if I drop the involved UNITS, right?

[@MathTheBeautiful]

This is certainly a valuable observation. However, in my experience, considering units adds a great deal of confusion to this already subtle question. I've been thinking about units for the past twenty years and every time I end up confusing myself more! I would suggest sticking to numbers without units in this subject (at least for the time being).

[@bonbonpony]

But how would you define numbers without defining the unit first? :q Numbers are not "just there" - they are relationships between the "unknown" vectors and the unit you picked. No wonder you get confused about units (I noticed it in this video too) if you don't have this part correct to begin with.

[@atnn4698]

Hi. At 23:00 when we divide by the norm square - It works again "in the linear sense", I mean if the temperature is not twice with the stretched system it does not compute. I guess the question is WHY WE ALWAYS "IN THE LINEAR SENSE ZONE" (sort of speak) THEN THE DELTA GOES TO ZERO (limit) ? Thank You

[@MathTheBeautiful]

Hi Anton, You are asking a question that's at heart of calculus. It is the source of much of its brilliance and much of the frustration learning it. I don't have a short answer, I only have very very long answers. The basic and surprising truth is that the graphs of "all" functions look like straight lines if you zoom in far enough and many questions can be answered by analyzing the slopes of those lines. That's what we are doing here. When having a discussion of this kind, it's customary to say "in the linear sense" or something like that. I hope this helps at least a little bit. Pavel

[@styx4947]

the "skewed coordinates". Are called Affine coordinates

[@thomashwarn]

Stretched cartesian coordinate transformations correspond to a change of units. Since the gradient is a dimensional quantity it is not surprising it changes under such transformations.

[@Hillis360]

Where can I get help for the Exercises 7-12 of chapter 2 from the book. I tried the solutions manual but didn't understand it.

[@MathTheBeautiful]

I'm rewriting the book and in the new version there will be a much more detailed description of these problems. If you reach out to me by email, I'll send you a recent draft.

[@stearin1978]

Why we can't construct a fcn that doesn't obey "dir. derivative=grad F•n"? Or, actually what we demand from the fcn for this definition to be correct? :)

[@samlaf92]

Your geometric/invariant definition of the gradient's magnitude is the "rate of the increase of the function" in the steepest direction. I'm confused when I try to think about what this *really* means. How do you define rate? Doesn't rate require units? There are no units it seems in "coordinate-free" land... so how can we even define an "infinitesimal" step?

[@samlaf92]

You answer this around 34:00: "the space exists without coordinate systems, so the space has its unit regardless of what coordinate system you impose." Still not sure I get it though :( Not to get too philosophical or anything, but does the Universe really come with "units"? Did "God" decide on this unit? For example, what's the "distance" between an atom's nucleus and its orbiting electrons? vs between two solar systems? Clearly these can't be using the same "units"? (one "obeys" quantum laws, the other relativistic laws for example)

[@MathTheBeautiful]

I ponder this question often and units are a deep and confusing concept. However, as far "the" unit of length is concerned, I believe this question has a pretty straightforward answer: the unit of length is the length of an *arbitrarily selected* reference segment. Then all other lengths are specified as multiples of that length.

[@KieranORourke]

@@MathTheBeautiful That's how the ancient Egyptians did it for constructing the Pyramids. Workers were handed a piece of wood and informed, 'this is a Cubit, now follow the plans which are expressed in Cubits'.... So all lengths became dimensionless ratios.

[@nidhinsathyan6904]

I cannot understand ,why you use i/mag(i)^2?.The square at the denominator thing that you put at the time 22:29.

[@bonbonpony]

The `i` in here is not supposed to mean the "imaginary unit". It's just a letter used to denote one of the unit vectors, nothing more. He could as well use `x` and `y` with hats, or `a` and `b` for that matter.

[@naveeenmishra]

the joke at 33:40 and his explanation :D

[@Baffage]

I understand that numerically, the (analytical) gradient of F(x,y) and F'(x',y') does not take on the same value. But why then does dividing by the square of the unit vector suddenly reconcile it with the geometric gradient? When measuring the gradient with a thermometer and a tape measure, how do you know that the unit vector must have length one for the analytical gradient to take the same value? Is it because the tape measure has length one? ( Note: I believe I think about physical space as being the primary and a coordinate system being strapped on afterwards to facilitate calculations. As for gradients, I can hardly imagine one without a unit added. I am an engineer by education, so my math background is regular engineering math. Is the mindset I described the wrong one for mastering this course? )

[@angelmendez-rivera351]

Baffage Because δF'(x', y')/δx = 2·δF(x, y)/δx.

[@raffaelelacatena1203]

I have some troubles about Chap.1 Exercise No.2 : I found dF/dx = 4 dF'/dx' and not 2. Should it be dF/dx=2 dF'/dx instead? (With F' I mean F'(x',y') ) I check also the solution but it not persuade me.

[@MathTheBeautiful]

2 is correct. This shows how confusing this question is and how it requires very careful reason. If you share your work, I will try to point out the mistake.

[@joluju2375]

Hi everybody. At 5:20 our quest is defined. Then we make several attempts, we fail for the general case as stated at 24:47, and by the end of THIS video, we are still unsuccessful, and the quest is reformulated at 35:30. Am I correct ?

[@MathTheBeautiful]

Correct! The actual solution comes much later: kzbin.info/www/bejne/inScaX6cnqx0hc0

[@AshishPatel-yq4xc]

I don't see how you can apply the chain rule to f(s)= F(x0+n1s , y0+n2s) . which then you say is f'(x)n1 and n1 = (n1s)'. So I don't see how you can can apply the product rule to x0 + n1s which is the x parameter in F.

[@MathTheBeautiful]

Let me know if this helps: f'(s) = dF/dx dx(s)/ds + dF/dy dy(s)/ds Now consider dx(s)/ds. Since x(s) = x0 + n1*s - a linear function in s - we have: dx(s)/ds = n1. So: f'(s) = dF/dx n1 + dF/dy n2

[@AshishPatel-yq4xc]

MathTheBeautiful that makes sense now. thanks.

[@wickeddevice5950]

Really nice and well thought out lectures. Well done. I'm a bit worried that the algorithm to find the greatest temperature change won't work. (That's not really germane to the topic at hand. I'm now wondering if finding it is NP hard instead of listening to the lecture :).

[@clfm20]

at 28.12 he says "this works if n is the unit vector" but wouldn't that make n1 and n2 both equal to 1?

[@angelmendez-rivera351]

Balls in the Air Productions No, it would not. If n1 = 1 and n2 = 1, then ||n|| = sqrt[(n1)^2 + (n2)^2] = sqrt(1 + 1) = sqrt(2) > 1. The components themselves are never determined by the magnitude of a vector. That is why direction is needed.

[@ey3796]

I have a doubt regarding the gradient of stretched coordinates. In the case of Cartesian coordinates, let's assume that an increase in x of dx causes an increase dt in the temperature. So gradient= dt/dx i. Now if we take the stretched coordinates, due to linearity, an increase of 2dx will cause 2dt to be the temperature increase. So gradient= dt/dx i*. Where i*=2i. So I only get twice the increase and not 4 times. Can you please point out where I'm wrong?.

[@MathTheBeautiful]

You're right: the component becomes 2x as big. But if you want to get the actual vector, you'll combine it with a basis vector that's also 2x as big. So the ultimate result is 4x as big.

[@solewalk]

shashank balakumar , I share similar ideas as you, but let me add little more to yours. Try to think of the situation when x is zero, and also when x is non-zero.

[@rkpetry]

I'm taking tensor calculus currently (at another university), and I wish to address some issues: A. superscript doesn't work for me: barely-raised middot-level or upper, might; B. the notion of spread, applied only to the x-y axes, is missing the sense of a valuation-dimension-expansion which should be just as valid though its units (physics units) are different, C. the logical tensor goes awry when relativistic twins claim SR alone is the truth to tell the other: when in fact they must include each-other's velocity to know each-other's time-bias ►Trailer for RELATIVISTIC CLOCK-TRIPPING ... Thank you, I like the more mathematical feel herein. Ray.

[@MathTheBeautiful]

At which university are you taking your course?

[@rkpetry]

'Ah-yuh' coordinate transformations show here too: I'm taking the course in a classroom at UH Manoa, Hawaii, with a professor lecturing by video at UU, Utah, except when she visited us a week, over their timezone discontinuity daylight-to-standard, as well as their Fall Break we don't have here... (And, I've expanded my first comment posted above.)

[@md.shafiullahkhan7896]

Raymond K Petry

[@taraspokalchuk7256]

if F prime and x prime are stratched by 2, why doesn't the 2 cancel in dF'/dx'? why is it 2 times bigger than dF/dx?

[@MathTheBeautiful]

Because the coordinates now travel twice as fast through the values of F.

[@h2ogun26]

aren't dF'=2dF dx'=2dx?

[@TheGbk1991]

Is it possible to get the problem sheets to the lecture somewhere?

[@MathTheBeautiful]

The solution set can be found here: bit.ly/ITACMS_Sol_Set_YT

[@dillpickle5065]

dividing by the square of the unit vector norms makes no sense because in their respective coordinate system their norm are 1. what am I missing here?

[@MathTheBeautiful]

Precisely. They're unit vectors for Cartesian coordinates, but in a "stretched Cartesian" coordinate system, the coordinate vectors are no longer unit vectors. And that is what the adjustment is for!

[@dillpickle5065]

MathTheBeautiful I see. The norm must be with respect to the original coordinate system then.

[@dillpickle5065]

MathTheBeautiful Thanks

[@dillpickle5065]

Your reply was so quick!

[@MathTheBeautiful]

NO!!! I'm glad you made this followup comment! In this narrative, Length exists apart from coordinate systems. Length is the sort of thing you can measure by using a tape measure. It is crucial for mastering this material to accept geometric objects on their terms outside of any coordinates. Everything will be endlessly confusing if you keep thinking back to a background coordinate grid.

[@justforknowledge6367]

Dear Prof., You have written grad as [(∂∕∂x)F , (∂∕∂y)F]. Why not (∂∕∂z)F? Because F can be F(x,y,z), some arbitrary function of x,y,z! You said about Cartesian co-ordinates, but not about 2-dim. Anyway, don't have to reply to this. I understand the issue. It is to prove that the Cartesian coordinates is not generalised, and the expressions change if the units are changed. We must remember not to ignore the trivial while probing the deeper aspects of issues. Science is exact, so we need to be exact not to allow doubts to creep in. After following the lecture half way through did I understand what your imperative was. Anyway, this observation does in NO way belittle your work on the paradigm. Great work! Regards

[@MathTheBeautiful]

Yes, it was a 2D example.

[@intellectelite]

Just For Knowledge he said it in the video

[@vitrums]

I'm just curious if my math is right for the problem why ∂F'∕∂x' = 2 ∂F∕∂x holds for stretched coordinates. The biggest part is that ∂y'∕∂x = 0. Here's what I did: [Z' = F'(x'(x,y),y'(x,y)) = / x'(x,y) = x'(x) = 0.5 x and similar for y'... / = F(x,y)] => [∂F∕∂x = ∂Z'∕∂x = (∂Z'∕∂x') (∂x'∕∂x) + (∂Z'∕∂y') (∂y'∕∂x) = 0.5 ∂Z'∕∂x'] or in other words [∂F'∕∂x' = 2 ∂F∕∂x]

[@vitrums]

My trivial calculations above constituted a pure algebraic proof of the phenomenon. Now. Your reasoning why this happens was quote: "Because the coordinates now travel twice as fast through the values of F". And at first I spent minutes faithfully trying to derive the equation out of this magical phrase. Sadly, no magic happened. And I think the reason was quite obvious - this explanation comes from the world of common spoken english, not math. Instead I came up with a geometric explanation. In 1D case (i.e. Z=F(x)) a derivative is nothing but a tangent, which is the ratio of 2 sides of a right triangle: vertical B to horizontal A. When you squeeze everything horizontally, then tg' = B/(0.5 A) = 2 tg. And only then I could finally visualize the surface Z=F(x,y) being squeezed twice to shape Z'. In 1D this will be similar to a soundwave graphic of a mic recording in a computer program, which can visualize your voice. And by dragging a scale slider back and forth one can see the graphic thinner or thicker.

[@MathTheBeautiful]

Your good work confirms that this simple question is actually not at all simple and requires a great deal of thought. Well done! However, the quote ""Because the coordinates now travel twice as fast through the values of F" is a very good quote. It gives a physical/geometric foundation for what's going on and therefore a starting point. If you operate strictly with algebraic expressions, you will get confused whether something is a definition or a consequence. I get frequently get emails explaining why it should be 1/2 or 1 instead of 2 and the error is always in the lack of a starting point and the resulting circuitous logic.

[@vitrums]

Thank you for checking my calculations. Indeed, I understand the awkward part of this problem - it's definitely more tricky, that it seems. Let's assume for now, that my mind might simply lack that vital genius component, which makes this quote to work for the rest of folks out there.

[@viaprenestina3894]

the gradient is a simple concept but after this lesson I am .... puzzled. Complication Bureau of Simple Matters

[@SWiSHRoyal]

How is the book he is referring to called?

[@suluclacrosnet61]

amzn.to/1m3Qqdp

[@GrandGobboBarb]

he has a link in the video description

[@adnanorkhan]

Kit42 Suluclac Rosnet has anyone pdf of this book?

[@michaelvarney.]

Sure... but you can make your own after you purchase the book.

[@hennadiimadan6993]

I went on to check that Helsinki has non-orthogonal street angles in google maps. And it's just not the case. It's not strictlyNorth-South, but neither is Manhattan.

[@mathman1475]

I think they are referring to the streets around the Temppeliaukion Church .

[@viaprenestina3894]

what about a 3-d gradient? shall we normalize to |i|^3

[@viaprenestina3894]

maybe not

[@angelmendez-rivera351]

What?

[@marrytesfu3163]

Is it fair that one's success in school depends on which teacher you get. I wish I had him as my teacher. Why are most professors bad at teaching.

[@ChaineYTXF]

Recruitment techniques are bad. Then again, if you only chose passionate, pedagogic people, you would have far fewer teachers/professors. You wouldn't be able to fill in most schools. But, with the advent of online lectures, things may crystalize in the long term and people will ultimately flow in the direction of .. the gradient of quality. Schools (as in a physical building filled with teachers and students) may be living their last decades, who knows...

[@taraspokalchuk7256]

16:15 if the function stays the same why do you write F'?

[@MathTheBeautiful]

It's a different function. As a more extreme example, if you consider a function F(x,y) = x+2y + y/x in Cartesian coordinates, the same field in polar coordinates would be given by the function G(x,y) = x cos y + 2x sin y + tan y. Very different functions even if they represent the same field.

[@angelmendez-rivera351]

Taras Pokalchuk The values you get from the function are supposed to be the same, but what defines a function is not its output, its how it transforms the input into an output.

[@AbhishekSachans]

33:39: I lost my unit vector 😀😀!

[@allandavis6116]

I don't think ∂F'∕∂x' = 2 ∂F∕∂x is correct, in fact ∂F'∕∂x' = ∂F∕∂x , as ∂F∕∂x = (change in F) / (change in x) and the distance function along the x axis is not changed by the change of coordinates, and F is not changed by a change of coordinates, and the direction of x is not changed by the change of coordinates, so numerator and denominator of ∂F∕∂x are unchanged and = ∂F'∕∂x'

[@MathTheBeautiful]

Hi Allan, you need to draw a careful picture and be very clear, when you say "distance", whether you mean coordinate distance or actual distance. The same change in coordinate variables x and x' corresponds to two different changes in the actual distance, and therefore two different values in the change of the quantity F. It is very important to dwell on this point until all confusion is removed. I hope this is helpful! -Pavel

[@allandavis6116]

@@MathTheBeautiful Wow, this is weird. If I have new basis for the Cartesian plane e1 = (0,0) -> (2, 0) , and e2 = (0, 0) ->(1, 0), then in the new basis the point (1, 0) now has length 2 ????? Yes, thinking ahead, we get the length from the dot product which is calculated using the metric tensor, which will give us the correct length for (0, 1) in the new basis of 2. So, even along a non-normal basis vector the coordinate does not measure length. I hadn't understood that. So expression like dF' / dx' in the e1, e2 basis only makes sense when you use the metric tensor to calculate dx. By Jove, you're right ! Thanks ! ..... But wait, dF'/dx' must be calculated using the metric tensor ????, which would make dF/dx = dF'/dx', or not?

[@MathTheBeautiful]

@@allandavis6116 I think the conclusion is correct but the reasoning is still a little off. By definition, e1 always goes from (0, 0) -> (1,0). But, in the first coordinate system, it represents a segment of length 1 and in the second coordinate system, it represents a segment of length 2. And it's not a matter of using the metric tensor. It's a matter of pulling out a tape measure and measuring it.

[@jolez_4869]

25:23 Suomi mainittu perkele!

[@clfm20]

oh no, I think I get it now...

[@NothingMaster]

The notion of a gradient could even be used to determine one’s realistic approach in the domain for successfully picking up a gorgeous babe at a bar. And heaven knows I have successfully put it to test on several occasions. Or was it my awesome facial hair that did the trick?

[@urinveisinfeksjon]

@30:50 He says "d-f, d-x", but I'm sure he meant to say "del-f, del-x"... Just beeing pedantic...

[@epsilonxyzt]

Too much talk, boring.The sound not good.