Tensor Calculus 1: The Rules of the Game
ฝัง
- เผยแพร่เมื่อ 7 ก.พ. 2014
- This course will eventually continue on Patreon at bit.ly/PavelPatreon
Textbook: bit.ly/ITCYTNew
Errata: bit.ly/ITAErrata
McConnell's classic: bit.ly/MCTensors
Table of Contents of bit.ly/ITCYTNew
Rules of the Game
Coordinate Systems and the Role of Tensor Calculus
Change of Coordinates
The Tensor Description of Euclidean Spaces
The Tensor Property
Elements of Linear Algebra in Tensor Notation
Covariant Differentiation
Determinants and the Levi-Civita Symbol
The Tensor Description of Embedded Surfaces
The Covariant Surface Derivative
Curvature
Embedded Curves
Integration and Gauss’s Theorem
The Foundations of the Calculus of Moving Surfaces
Extension to Arbitrary Tensors
Applications of the Calculus of Moving Surfaces
Index:
Absolute tensor
Affine coordinates
Arc length
Beltrami operator
Bianchi identities
Binormal of a curve
Cartesian coordinates
Christoffel symbol
Codazzi equation
Contraction theorem
Contravaraint metric tensor
Contravariant basis
Contravariant components
Contravariant metric tensor
Coordinate basis
Covariant basis
Covariant derivative
Metrinilic property
Covariant metric tensor
Covariant tensor
Curl
Curvature normal
Curvature tensor
Cuvature of a curve
Cylindrical axis
Cylindrical coordinates
Delta systems
Differentiation of vector fields
Directional derivative
Dirichlet boundary condition
Divergence
Divergence theorem
Dummy index
Einstein summation convention
Einstein tensor
Equation of a geodesic
Euclidean space
Extrinsic curvature tensor
First groundform
Fluid film equations
Frenet formulas
Gauss’s theorem
Gauss’s Theorema Egregium
Gauss-Bonnet theorem
Gauss-Codazzi equation
Gaussian curvature
Genus of a closed surface
Geodesic
Gradient
Index juggling
Inner product matrix
Intrinsic derivative
Invariant
Invariant time derivative
Jolt of a particle
Kronecker symbol
Levi-Civita symbol
Mean curvature
Metric tensor
Metrics
Minimal surface
Normal derivative
Normal velocity
Orientation of a coordinate system
Orientation preserving coordinate change
Relative invariant
Relative tensor
Repeated index
Ricci tensor
Riemann space
Riemann-Christoffel tensor
Scalar
Scalar curvature
Second groundform
Shift tensor
Stokes’ theorem
Surface divergence
Surface Laplacian
Surge of a particle
Tangential coordinate velocity
Tensor property
Theorema Egregium
Third groundform
Thomas formula
Time evolution of integrals
Torsion of a curve
Total curvature
Variant
Vector
Parallelism along a curve
Permutation symbol
Polar coordinates
Position vector
Principal curvatures
Principal normal
Quotient theorem
Radius vector
Rayleigh quotient
Rectilinear coordinates
Vector curvature normal
Vector curvature tensor
Velocity of an interface
Volume element
Voss-Weyl formula
Weingarten’s formula
Applications: Differenital Geometry, Relativity
I could've used this 40 years ago. This is great and the instructor is wonderful. Questions from the students are relevant and enlightening. Now I'm running through the whole series and studying independently. Next step, buy the book.
David Roberts I really want to get the book as well but im kind of broke lol
wrg, ask, teach any by any and any can be perfx
I wanted to get a phenomenal playlist on Tensor Calculus. I immediately thought of you. You teach it like you schooled Ricci and Levita on it. Can't deal with them without your viewpoint! Thanks!
I will get out my old Calculus book and re-learn it. Thank you for this video. Being retired, I likely will never use this, but it will be wonderful to grasp a great idea.
Again, Thank You.
Brilliant lecture. Wish I could convey this as clearly as you do.
You take the time to make sure everything is explained well. Thank you so much. I'm just watching this series for fun and it's been very enjoyable so far.
RIP headphone users....18:00
lmao im deaf now, I read this comment afterwards :(
i read this comment at 17:30
I read this comment thinking that the volume died at 18:00 and went to test it...
I'm wondering why the volume went down and then why it came back up later.
I really love the intuitive approach and pacing of these lectures. I don't have to keep pausing the video, and I can just sit back and enjoy it.
I had been looking for a good class on Tensors as the course was not offered where I went to graduate school. Thank you for posting these videos!
Bless his heart and that of anyone's who makes the effort to talk about tensors.
You make me very happy. I am happy when I understand something and you explain everything so well.
This was great. Much food for thought.Thank you.
I agree with all the comments below. Thank you very much for taking the time and effort of sharing your lessons with us. Much appreciated. Your perspective and insight are abslutely invaluable. I have not words enough to say THANK YOU
Indeed!
very nice lecture and amazing simplicity
Thanks to Tensor Calculus I can finally solve those non-symmetric-center-of-mass problems.
Is there anything Tensor Calculus cannot do!
Thank you for sharing good sir!
Wow, he's good. At 6:00 he addressed directly the misapprehensions I had about what vectors are.
Careful... what he says there is a bit controversial.. and i think he should make it more clear that what he is describing is the way a vector is to be seen in tensor-calculus, or maybe only in his course.
There is definitely the algebraic view of what a vector is, which doesn't require the geometric interpretation of "an arrow".
en.wikipedia.org/wiki/Vector_space
I think both interpretations can yield some insight, and we should not label one as true, and the other as false..
@@TheLeontheking Correct! To make it clear: the definition of vector given here is just for this course. In my Linear Algebra class, I give an entirely different definition of a vector. If I were to teach a class on Machine Learning, I would give a third definition.
so far this has been a review of vector calculus so it's really easy for me. I can't wait until I get to the 2nd ordered tensors! you're awesome professor!
I think this maybe the hard part - looking at something you already know from a different perspective.
MathTheBeautiful
It's awesome really. Calc 3 was hard. You make it so much more simpler.
Beautiful ! Thank you a lot for this amazing video.
At 9:49 - ... so a "capital u" is simply a larger version of a the small letter u ; )
Fantastic series. Respect.
So far I have watched only the introductory 'zeroth' lecture. But shortly thereafter I was on Amazon and purchased the book. I'm looking forward to slowly going through these series of lectures.
Did you reach the end of the book, and understand everything ?
@@joluju2375 I read maybe half of the book, and got a fair bit out of it. Tensor calculus is exponentially more difficult than vector calculus. I never developed a good intuitive feel for covariant vs contravariant derivatives for instance, or the Cristoffel symbol.
@@rontoolsie Ok, useful. Don't give up !
+MathTheBeautiful Part of The Rules of the Game is missing at about 29:35. Would you mind writing a little to describe what's missing?
An EXCELLENT teacher.
Hi, I'm very much enjoying your lectures. So much intuition which I couldn' t develop before. A question, why do you say cartesian coordinates have a prefered scale? Why should there be a stretched or unstretched cartesian coordinates? Like in the example of gradient for stretched cartesian coordinates, we need to use unit basis vectors but the partial derivaive df/dx should remain invariant under stretching of coordinates. Yes the function changes twice as much, but the interval dx should also be twice as large. So the gradient operator for all (stretched or unstretched) Cartesian coordinates will be df/dx i^ +df/dy j^ where i^ is unit vector in x direction.
Is standard sequence enough background for this course? (Not the honors, proof oriented versions of these course btw, just the standard computational skills focused versions)
Yes, that should be enough. I often say that a good prerequisite is being confused about vector calculus. Then Tensor Calculus will help clarify that subject.
the only thing I knew before Tensor Calculus was that I absolutely loved physics and General Reletivity in particular. I got my first real textbook,(Gravitation: by Misner,Thorne and Wheeler) and when I opened it I saw the coolest looking foreign language,(tensors,vectors etc.) and knew I would stop at nothing to learn what those weird looking symbols really meant. That doesn't answer your question precisely but I want to tell you that you don't need to do any kind of strict academic progression. I had to learn these subjects almost backwards, because every time I needed to know why, or where something came from I had to go back and buy a book on linear algebra,or PDE's or calculus II and so on. then go back to the physics, next time I got stuck I would go back again.
this is so helpful thank you so much for everything.
Thank you, I'm glad you found these videos helpful!
He's also doing linear algebra, I'm sure it's great, but I think professor Strang is the best of the best!
Agreed!
Sir, was prof. Strang your professor at MIT?
I like MIT Prof. Arthur Mattuck for Differential Equation....Did Prof. Mattuck teach you at MIT?
You mean that stammering spastic who can't explain the simplest thing and goes on several tangents recursion in one sentence?
Well, he might be better than most university professors, but I wouldn't call it "best of the best". Well, unless, as they say, "In a world of blind people, a one-eyed man can be a king" :q
This guy is just great
18:00 woke up my whole neighborhood
Você aqui, Porto????
now i will buy the book again...after i returned to amazon few years ago.
Great lecture! Many things clicked for me in this lecture. You may want to include a review of the definition of orthogonality before you launch into this. That way the principle of orthogonality is fresh in people's minds. Definition I'm working off of:
"Two vectors *u & v* in R are orthogonal to each other when *u•v* = 0"
from Lay, David. "Linear Algebra and its applications" 4th edition. 2012. Again, thanks you so much for this definition. They don't even offer an undergraduate level tensor calculus at my institution. I'd be pining after tensors, lost in wikipedia if it weren't for your lecture series.
Thanks for the comments.
This is actually *not* the definition I'm using. In vector calculus, the definition is: two vectors are orthogonal if the angle between them is pi/2.
MathTheBeautiful Is this distinction purely to avoid needing to clarify that u and v are non-zero, or something more fundamental?
Titus Pullo More fundamental. The heart of the distinction is that our definition is geometric. It's the same if you define u•v as len(u)len(v)cos(angle). But most people mean u1v1 + u2v2 +u3v3 which, in our setting, is a consequence rather than a definition. (In linear algebra, it's the definition).
MathTheBeautiful Insightful. Thanks very much.
This is a very interesting class. I confess it is hard to try to re-learn vectors as you said on their own terms as opposed to confined to a specific background coordinate system or basis, but I am trying to re-think of them in this way now. But, at 10:10, the dot product is defined, and the definition involves cos(A). Cos(A), as I remember learning, was defined as the x coordinate of the point at angle (A) radians on a unit circle. This requires one to think in terms of an x,y coordinate system. Is there a better way to understand sin(A) and cos(A) outside of the standard x,y coordinate system?
+djttv You are right: it's hard. But it's very much worth it!
Cos is the ratio of the adjacent side to the hypotenuse in a right triangle. (You might enjoy reading this: ualr.edu/lasmoller/trig.html)
That's probably the biggest problem I have with math teachers teaching about vectors: when they say that a vector is "just a column of numbers", totally ignoring their geometry, how they transform, what laws they have to obey in order to be a proper vector, and what's the _gist_ of being a vector at all.
Thank you for this Video!
26:30 It is THUS ALSO a proof that it is true that U' is perpendicular to U, even if U is not of constant length, but when U is at a local maximum ( or minimum or at an inflection,) of its length since that is an interpretation when A' = 0.
At 35:45 are S(t) and R'(t) using the same t, or one of the tees should be given a star?
so a point yes is a sphere and expands or contracts in volume like music, can you sing, Start Again, perfect pitch?
Thanks, Great work
prerequisites for this course? calculus, linear algebra, diff. geometry, topology?
Thanks Professor for this great course!
At 25:51 you are using, applied to vectors, Leibniz rule for derivative of product (en.wikipedia.org/wiki/Product_rule). As we are thinking about vectors only as "object" with length and direction, having 2 operators, namely sum and dot-product, how we are allowed to use this rule?
I mean, we are using a calculus rule applied on vector objects, without any demonstration of validity, aren't we?
What I'm missing, what is wrong with my question?
Thanks in advance.
At 8:50 what have unit vectors got to do with length of vectors. I understood unit vectors only to carry information about direction. When you divide a vector by its magnitude you can get the unit vector corresponding to it's direction but when calculating the unit vector this way doesn't the (arbitrary) distance (in metres, say) above and below the line just cancel and have nothing to do with the unit vector itself. So I'm confused by this part.
Does Euclid have a proposition that amounts to symmetry of the dot product? For vectors u and v as directed line segments one can draw a picture where the dot product u*v is represented as one rectangle, v*u is represented as another rectangle, and these rectangles have the same area.
I'm not convinced that the dot product is about rectangles. It's more about one vector "casting its shadow" upon the other (mathematicians use a fancy word "projection"), to see how much of that vector lies in the direction of the other (this is the `|b|·cos(k)` part in the formula), and then multiply that shadow with the other vector used as a unit (that's where the `|a|` part in the formula comes from). The symmetry comes from the fact that you can as well cast the shadow of the other vector upon the first and express its shadow with the first vector as a unit, and the result will be the same, because the geometry in both cases is the same, the situation is perfectly symmetric. If you're looking for the explanation in Euclid, though, I would go with Pons Asinorum.
@ 25:25 is it dU/dt or dU(t)/dt ... assume U has a vector arrow over it. I'm confused.
Can anyone tell me what kind of math one should be familar with before starting this playlist? So far at Uni I've finished calc 3 and linear algebra. I want to eventually check out this series for when I'm ready.
Give it a shot now and see how it goes. You may be ready!
RiotAtTheTop as a chemical engineering major & physics i can tell u if u’ve taken calc 3 ur ready. Calc 3 covered Surface integrals i suppose & jacobian matrix & change of coordinates. All these concepts really fall under tensor analysis. So ya. Ur ready 100%
Can somebody help me with the exercises in Chapter 2 of the book? I'm having trouble solving the problems regarding directional derivatives. I know what they are, and know how to compute them. I tried looking at the solutions manual by Sulon(easily found in the internet) but didn't make much sense from it.
After 26:03 : Zero, because the dot-product of the two vectors U' and U can only be zero with cos90 (= 0).
Can anyone tell me what textbook is he following? he mention something about the textbook, but I did not see which...little help?
He wrote a book called Introduction to Tensor Analysis and the Calculus of Moving Surfaces (by Grinfeld).
@@NicholasDwork Omg, this man is the author of that book? I had no idea!
MathTheBeautiful
Question about Exercise 13 in this chapter and the solutions that were shared in a different thread. Is the gradient for ex 9 really the line AP, I feel like it should be the line perpindicular to AP that produces the largest change in angle...am I missing something?
Also, a bit more generally, is there an easy way to check?
Meant to say perpendicular to OP
Thank you sir!
Thank you very much for this enlightening introduction to tensor calculus. I am returning to the subject after… 50 years of paying no attention to it, and I must say that your presentation is much more vivid and attractive than the overly formal one I was given during my studies.
The point you make about vectors having to be viewed as "segments with arrows", not couples (or triplets) of real numbers makes perfect sense to me. However, how about elements of the R^n vector space ? It seems that such "vectors" exist only as numbers in the first place. Do you imply somehow that the tensor approach should be limited to objects living in our Euclidian space where "segments" and "arrows" are part of the picture ? Or should one, for that purpose, regard vectors of R^n as mere coordinate representations of "geometrical" vectors in one particular reference frame ?
Hi itube021, thank you for your comment. All ideas certainly continue to work in R^n. But one needs to start somewhere and I start in a Euclidean space. I think I was able to describe it better in the new version of my book grinfeld.org/books/An-Introduction-To-Tensor-Calculus -Pavel
@@MathTheBeautiful The information given, in particular in Section 2.7 of your textbook ("Comparison to the linear algebra approach") addresses my question in a very detailed manner. Thank you for pointing out this reference !
I really love how you take care of addressing seemingly innocuous points which, if left aside as is often done, open the door to misconceptions that will end up undermining the global understanding of the subject by the reader.
Save yourself. Watch in 1.25 speed.
Shiv Akshar Yadavalli how?
By speeding it up lol!?
Click on the cog icon, and select 'speed'.
Life is short and time is precious. You sir are a hero!
2x skips too much.
3:22 "defining them (areas) precisely has proven difficult"
hmm, never heard that before, can you please elaborate??
From 29:27 to 29:28 some part is missing, which is basically the procedure that we derive the results. Even though the idea is that the derivative of a vector is orthogonal to the vector itself, the results (yellow arrows) don't shown that. How is the last arrow orthogonal? Also, where can i find the part that is missing?
Also, all these vectors start from the same point, why? If they are bound to something, a starting point for example, how is this different from the coordinate system?
Paris, Earlier in the lecture it was proved that the derivative was orthogonal to the vector, if and only if, the vector is of constant magnitude (no matter what parameter is used). Here, the magnitude of the position vector R(t) as it goes from t0 to t is not constant so R'(t) is not orthogonal to R(t). It's worth noting that R'(t) is tangent to curve. Without getting into a proof, just picture give R(t) at t=0 a little tiny nudge, h, towards t, then draw the difference vector from the tip of R(t) to R(t+h) and you will see intuitively that it is tangent to the curve and not perpendicular to itself. The magnitude of the this tangent is basically (R(t+h)-R(t))/h and has size. Dividing by h "scaled" it up. I think that was what he probably alluded to in those lost few seconds of the video Hope this helps.
@@mathayes7649 Thanks Mat, this makes sense now. I didn't get the part about the constant magnitude, that's why i got confused. Thanks again for the clarification!!!
Hi Again Prof. Grinfeld.
39:13 to 39:23 You say that if arc length is the parameter than the derivative of the position vector w.r.t arclength is the unit tangent vector. But previously we proved that it is TANGENT if the LENGTH OF THE VECTOR is CONSTANT. HOW IS THE POSITION VECTOR WITH ARCLENGTH AS THE PARAMETER IS CONSTANT? I can not see it . . .It is not as with the circle as we previously did - R(angle)
I would think this way:
| d V(length of path of V) / d (length of path of V) | = | d V(length of path of V) | / | d (length of path of V) | = | d (length of path of V) | / | d (length of path of V) | = | d (length of path of V) / d (length of path of V) | = | 1 | = 1.
Conversely,
| d V(length of path of V) / d (length of path of V) | = | d V(length of path of V) | / | d (length of path of V) | = | d V(length of path of V) | / | d V(length of path of V) | = | d V(length of path of V) / d V(length of path of V) | = | 1 | = 1.
I think the difficulty stems from our habit of always associating motions with time. This is not based on time, but on the length of the path.
I just wish you hadn't taught on a blackboard, it kinda given me goosebumps
24:10 why would the two vectors dotted with each other always be a constant? wouldn't it instead just be |U(t)|^2?
For U(t) in general, |U(t)|^2 will not be constant. But watch again from 23:14 and note that he's talking about those specific cases where U(t) has constant length for any t. For those cases, |U(t)| is constant so |U(t)|^2 is constant and he can write "A" instead of "|U(t)|^2" without losing anything. Either way, the term is going to differentiate to zero when he gets to that step.
AdrianBoyko ah ok, thanks :)
therealjordiano
it was an arbitrary example. he began with suppose we have a vector with constant length. a vector dotted with itself is its length squared, so if its length is constant then the so is the dot product.
imagine a car traveling along a circular path. the parameter would be time and as time passes the vector that points at the position of the car changes direction but its length is always the radius of the circular path.
26:30 Not true if the vector magnitude is parameter dependent. If we are talking about a kinematic trajectory, then if the speed is constant, then the acceleration is orthogonal to the velocity. There can be tangential acceleration though, and in that case, what you said is wrong.
Listen carefully. He says at 26:27 "then the derivative of a constant length vector is ..."
Hey MathTheBeautiful, quick question: after trying to learn this unsuccessfully for years, I finally realized that the gap in my intuition is that I don't see why you can't just make a direct change of variable, eg x=r cos(theta), y=r sin(theta) to get formulas in other coordinate systems. Why the indices, contravariance and covariance? It like it, but in what way exactly does it make things easier. In switching the gradient from Cartesian to polar coordinates why not just make the substitution above? Thank you very much!
The kinds of tasks you are describing are not the purpose of tensor calculus. The purpose of tensor calculus, in part, is to liberate you from having to choose a particular coordinate system in the first place and then suffer from the artifacts imposed by the special features of your choice.
@@MathTheBeautiful Ah, thank you! I believe I caught that in the lecture, but intuitively conflated it with change of coordinates. They say dispelling an unhelpful notion is as good as learning a helpful one :)
@@bcthoburn Tensor Calculus will force you to unlearn a lot of things
At 18:54 you say: V’(α) is a vector orthogonal to V(α) and of the same length. My first thought was: it’s only the same length if α changes at 1 radian/second, then it is a velocity of 1 length unit per second. Then I thought: we are differentiating with respect to α, not w.r.t. time, so the derivative is 1 length unit per radian, so I am OK with it having a value of 1 whatever the rate of change of α w.r.t. time.
But, how can a velocity vector have the “same length” as a position vector? Surely they are in a different vector space and cannot be compared lengthwise? They have the same scalar multiplier times their respective unit vectors, but it does not follow that you must draw them the same length. You give an excellent explanation why this is the case in your video Linear Algebra 2h: What Else Works like Geometric Vectors, from 6:00 to 7:40.
I think your videos are an amazing resource. I am learning so much from them and your style makes learning so easy. My motivation: I have just purchased “Gravitation” by Misner, Thorne and Wheeler, and set myself the goal of understanding it all. I am sure will take a few years, and perhaps I will never reach it. I came across this superb book in Heffers book shop in the seventies, but it was not relevant to my engineering degree. Now I am fortunate to have the time to delve into it.
I learned something new today from this lecture, that is, thinking of vectors as segments in space without any need for a co-ordinate system. This lecture is well presented. You can see the instructor really loves his subject and can probably discuss it all day without recourse to any notes.
One thing doesn't seem to be either quite correct or is not clear to me. At 16:40 he divides by 'h'. But 'h' is not a vector; it's just an index or subscript, I think. Shouldn't he be dividing by V(h), a vector, like so: V(1+h) - V(1) / V(h), not dividing by 'h' the subscript or index value like this: V(1+h) - V(1) / h?
definitely not
only sums and differences of vectors have any meaning implied by the vector space structure. the quotient of between two vectors is left completely undefined, so you are right in that whatever h might be, it isn't a vector
but while h is technically an index, and while V is a function which might take h as its argument in V(h) and that also happens to be a vector-valued function, h does not index V. h is the index of a limit (being taken of the function V), meaning h appears in the subscript of the limit as the value that will tend towards zero
in this limit, h behaves as an infinitesimal scalar and corresponds to the differential term in the denominator of the standard newton definition of the derivative, the form of which superficially resembles a rise over run expression, never a ratio of a form like y2/y1 or V(x)/V(h), but instead some function's slope, of a form like f(x)/x, or V(x+h)-V(x)/h which is your latter expression, or in the limit, the total derivative dV(h)/dh which he wrote here using alpha instead of h
Thank you.
Are there any problem sets that accompany these videos? If they are in the Patreon I would be willing to subscribe. Thanks.
When he says a vector is a length with direction in geometrical terms, I can understand that in 2D and 3D, but I have problem generalizing it to higher dimensions. That definition doesn't easily generalize. But the real difficulty is when I try to imagine this picture in abstract terms. What if instead of R^n we go to a field space of F^n where F is any arbitrary linear field (F, 1, 0 , +, *)? My biggest problem is the definition of length itself. If we define |V|=Sqrt(V*V) where V is a vector in F^n, of course we can easily calculate V*V, as the star operation is defined in our field. But how do we define the Sqrt function? Am I correct to conclude the tensor calculus is only plausible in R^n and cannot be extended to all linear fields?
The definition of a vector depends on the field you're studying. In linear algebra, a vector is any kind of object that can be added to another object of the same kind and multiplied by a scalar. In that context, a directed segment is an example of a vector and is called a "geometric vector".
Here, a directed segment in 2 or 3 dimensions is called a vector. Its length does not require square roots, but only a tape measure. These concepts do not extend beyond 3D.
Tensor calculus on ℝⁿ is covered later in the course in the context of Riemann spaces.
Extensions to other fields are possible. Some aspects are straightforward, others are not. Depends on the problem you are studying.
@@MathTheBeautiful Thank you doctor. I'll continue through your book to understand these concepts. Happy holidays.
At 4:25 , I have a question , how we can define function without coordinate system ?
The lectures are very enlightening ,thank you very much sir.
It's a mapping from physical points to numbers. It won't be an arithmetic expression, which is what we often think of as functions. But it will still be a mapping, which is what we understand functions to be more generally.
As a followup, it is very important to learn to see the world in a coordinate-free way. That will go a long way towards clarify Tensor Calculus which is a study of coordinates.
+Lemma What kind of math should one know before watching this playlist?
Check out the new Vector Calculus series. It's a direct prequel to this series!
25:53 How do we know, though, that they both lie in the same plane of rotation? What this proves is only that they're orthogonal to each other. But it doesn't say in which direction. Heck, it doesn't even say which way the derived one lays on the tangent to the circle of rotation :q How can we obtain this information?
Bon Bon Because the vector can only change within the plane rotation. The difference between the vectors evaluated at any two values of the parameter must lie in the same plane as the two vectors being subtracted.
That's an even more basic fact
ds/ds = 1 might be more intuitive than s'(s) = 1
3-4-18
Both are good.
+Paul Shapshak, PhD
They are only both good for people who already understand what they mean.
A total beginner will have problems understanding "s'(s) = 1" --- heck, I have studied Engineering Physics for 3 years and even I feel uncomfortable with that kind of notation --- whereas "ds/ds = 1" is much more clear.
Laurelindo Yeah, the derivative itself is a quotient so the more "accepted" notation should be the d(variable)/d(variable) notation since it stands for what the derivative ACTUALLY is.
@@Peter_1986 Who would be watching a lecture on tensors that isn't familiar with Lagrange's derivative notation?
@@lafudge2929 even if you are familiar, it's confusing at first glance in a way that the Leibniz notation isn't. Honestly I prefer Leibniz notation in general, as it is more representative of the actual definition of the derivative than the other common notations. dy/dx isn't that far away from lim Δx -> 0 Δy/Δx
Brilliant.
38:00 If anyone else couldn't see why s'(s) =1, I think it's easier to see it as:
(I would use l for length but I don't like sans serif l's, so I'll use w instead)
Let s(w) = w, where w would be the value of the arc length, basically saying "the arc length is the arc length." Then d/dw (s(w)) = d/dw (w) = 1, giving s'(w) = 1.
Then because s'(t) = || R'(t) ||, we have s'(w) = || R'(w) || = 1.
Just using the symbol "s" in the video kept confusing me b/c of its previous use as a function of t instead of as an element in an interval, hopefully that helps anyone else who had that problem.
Edit:
Having gone forward a few more videos, the use of one symbol to refer to both a coordinate and, distinctly, a function is acknowledged as a recurring problem. The notation doesn't seem to have a great way to deal with it, so, for anyone having this same problem I've had, steel yourself and interpret your work carefully.
Thank you
can you add subtitles?
I'm always confused when I have to reparameterize things. If we have arclength = \int ||r'(t)|| dt, and we reparameterize with 2tau=t, do we keep r'(t) as being derivative with respect to t, or does it become with respect to tau?
Because if its wrt tau now, then 2tau=t, 2dtau=dt, and we get arclength = \int ||r'(2tau)|| 2dtau.
So the integral upper bound shrinks by 2, but we multiply inside by 4, and the arclength actually increases... what am I doing wrong?
What is the fuzz at 26:10 about? Isn't it basic math that the tangent (or whatever one like to call the derivative) of a function is orthogonal to a perpendicular line per definition? Therefore it seams to me it has already been assumed that which is to be proven. Or is this just the nature of math in where all theorems are tautologies and some proofs are just easier to see directly than others?
I think that in everything situation you need to sort out for yourself what's definition and what's theorem. The often mentioned example is that in classical geometry Pythagoras' theorem is a theorem but in the coordinate formulation it's a definition.
MathTheBeautiful In the case of Pythagoras' theorem we are talking about an implicit distance function, which is not assumed in non-euclidean geometries but defined. On the other hand the tangent, or derivative, implies a direction and it is hard to me to see if this direction is something you can fiddle around with by a defintion. I.e. to me it seems to be a necessity that the derivative must be orthogonal to the vector itself. But are you saying this is not the case, i.e. am I assuming something I should not?
Edit: Indeed you are saying this is not the case at 22:32.
@38:00 Not sure why s'(s)=1? I get from the integral that s(s)=1s but why s'(s)=1?
Because if f(x)=x then f'(x)=1.
Didn't quite understand the starting equation in prove on 25:00. u(t) x u(t) = A, how this equation is bound to the condition |u(t)| = const? Can somebody explain how the first equation is appeared?
O I've got it. In any point t, function u(t) produces some vector, let's say V. We now about this vector V, that's it length is const, let's say X. So V times V is |V| x |V| cos 0, which is X times X times 1. So X^2 is also constant, let's say A. I need some time to realize this, not so obvious.
Yup, you start with the assertion that your vector `u(t)` has a constant length all the time, which can be expressed as:
u(t) • u(t) = const
because the dot product of a vector with itself is the length of that vector, squared.
Now you differentiate both sides of this equation:
d/dt[ u(t) • u(t) ] = d/dt[ const ]
and since the derivative of a constant is zero, you get:
d/dt[ u(t) • u(t) ] = 0
Now you figure out the derivative of the dot product, which works in a similar way to ordinary multiplication (but DON'T confuse it with multiplication! it's a totally different thing!), which means that you have to use the product rule to work it out, because both factors are functions of `t` and change depending on it. So you get:
d/dt[ u(t) ] • u(t) + u(t) • d/dt[ u(t) ] = 0
Now both terms are the same (the dot product is commutative, so you can swap the factors just fine), so you can group them together:
2 · du(t)/dt • u(t) = 0
Divide both sides by 2, and finally you get:
du(t)/dt • u(t) = 0
which says that the dot product of u(t) and its derivative is zero, and this can only mean one thing: they're orthogonal to each oter ;) (that is, provided the length of `u(t)` doesn't change indeed; because if it does, this dot product will no longer be 0).
Suggestion: when you state that a vector function of alpha is constant length implies that the dot product of that function with itself is equal to a constant, I would reflect back on why that is... that the dot product of a vector with itself is the magnitude squared of the vector. I could follow the logic but didn't follow where the basis came from.
It comes from the definition of the dot product when the angle is 0.
Quick question: I assume you use the euclidean metric. How would the formulae generalize to non euclidean metrics?
Which part of the video are you referring to? (But, basically, the answer your question is central to the entire course.)
Actually my question was about the definition of the arclength. But generally, is there a powerful way of "plugging in" the metric that is general enough. Thank you very much for this videos :D
@26:47 you said, "... As long as the base is fixed...", Why does the base have to be fixed?
I was wondering that too! I don't see why it has to be fixed. Could someone here please clarify?
With due respect- and this may come across as a silly question perhaps- isn't this lecture just vector calculus? I mean, where's the bit which is unique to Tensors (if any such stuff exists at all)? Also, which of your lectures describes the operation of a Tensor on a vector etc?
Tensor calculus IS just vector calculus and differential geometry, but with an emphasis on managing coordinate systems.
One of the advantage of tensor calculus that I emphasize is that it has very few objects (vectors, numbers, variants) and very few operations (addition, multiplication, and maybe contraction). In particular, there's no such thing is "the operation of a Tensor on a vector".
I can understand why the lenght of the vector R' is 1, but I can't see why it's tangent to the curve.
Can you please explain why is that?! Because I know that R' is perpendicular to R only when R has a constant lenght.
I think the explanation starts around 27:30.
@@MathTheBeautiful prerequisites for this course? calculus, linear algebra, diff. geometry, topology?
Stick waving with a ribbon/velocity is a great example of the orthogonal derivative
Yup, except the velocity vector would be in the opposite direction to the ribbon :) But other than that, I agree, it is a wonderful metaphor indeed :)
Edit: Or you may record that experiment on video and then play it back in reverse, then it would work with the ribbon as the velocity vector :)
i put a beat over it and its fire also helps me remember what hes saying
Sounds like you have a bead on it
What year is he teaching? I learned tensors through general relativity and stress strain analysis. This class would have been sooooooooo helpful when I was first interested in that stuff.
Seven Drexel University juniors and two graduate students.
MathTheBeautiful I have to share this with my girl. But I don't know how she'd react. She's a Mathematician, which makes me all kinds of happy. :)
segment with a direction. whats a segment in space, and how can you decide which direction?
Hi Mahir,
Good question! It's very hard to define a primary concept such as "directed segment", but I do my best in this video: th-cam.com/video/Zx6bizDdkRE/w-d-xo.html
Hope it helps!
Pavel
sir please add subtitles also in these videos
Danke!
Is vector calculus a prerequisite to this?
If you referring to my VC playlist then yeah, I would recommend watching it
Subtitles don't work well; they stop after about 15 minutes. Can you intervene to correct them?
Please accept my apologies with regard to the lack of subtitles. I just don't have the bandwidth to do this.
29:27 t* makes its entrance unannounced, to distinguish it from t time. I guess parameter t* is a function of time t.
thanks
Maybe it's a stupid question, but can parameter alpha itself can be multidimensional, or a vector? Then can we take a derivative of a vector with respect to another vector?
Two hours in and still nothing about tensors in a Tensor Calculus lecture? I'm disappoint :P
Professor, would it be too much of a crime to say that the product rule of two geometric vector is vallid because it's possible to correspond every geometric vector with a vector in R^2 (by cartesian, polar or any other coordinate system) and since it works for R^2 it must work for geometric vectors? Is this a vallid proof?
+Family Father Yes, I believe that is a valid and most direct proof.
Thank you very much! This course will certainly help me when I start studying relativity, you're a great teacher.
As so many famous mathematicians, I sort of taught myself calculus. Math is all the same thing, whatever the names may be.
5:25 ish, how many origins are there
Amazing lectures. You should use regular chalk, though hahaha
I trust you in time I will understand why derivative of position vector with respect to arclength is unit tangent vector
What did he mean by "if you can subtract things you can take derivatives of them?" (7:32)
Evaluating a derivative generally involves subtracting two values, dividing by a number and taking the limit as the denominator goes to 0. So if you can subtract things, you're well on you way to being able to take derivatives of them.
Thank you! Does this only apply to values that are in some sense continuous?
Yes. Even differentiable.
Corresponds to Calculus II - 3D Space - Calculus with Vector Functions from Paul's Online Math Notes. tutorial.math.lamar.edu/Classes/CalcII/VectorFcnsCalculus.aspx
37:00 That holds only when s(t) is monotonic (never decreasing).
The microphone is super noisy. Run it through a noise gate.