Tensor Calculus 5a: The Tensor Property

This course will eventually continue on Patreon at bit.ly/PavelPat...
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

[@esreve1]

You want to know what is happiness? This is happiness 29:30

[@MathTheBeautiful]

Oh yes!

[@lawrencemwangi8846]

This is world class..dont believe you gave it for free..Thanks from Kenya

[@gilbertocunha5938]

I just wish you could see the smile in my face throughout all your lectures! It is such a joy to be able to learn this! Thank you so much for making these lectures available for everyone! You are an amazing teacher :)

[@MathTheBeautiful]

Thank you, that means a lot!

[@datamoon]

Good gravy there was a lot of material in this lecture. It took me 3 days intermittently to get all the way through, but things are starting to gel. THANK YOU FOR THIS!

[@MathTheBeautiful]

Glad to hear that!

[@mariobriccetti6462]

Tensor are starting to feel like magic to me -- like when I first learned calculus

[@rfortunati5906]

These lectures are simply amazing. Absolutely fantastic teacher, who can transmit his passion for the subject. Crystal clear explanations. Thank you so much!.

[@barteklangowski]

These lectures are just great. Thank you so much for posting them!

[@Tardistimelord]

If anyone has not bought the book -- do so. It helps hugely with the lectures. This Lecture starts at Ch 6, p.75. I have no financial interest but am just a fan of something that links so beautifully with linear algebra, calculus, calculus of variations etc.

[@MathTheBeautiful]

Thanks Llewelyn, very good points! (Feel free to send the invoice for your services to my email.)

[@Tardistimelord]

You are more than welcome... have a nice day (from NZ).

[@BarriosGroupie]

I feel you've copied the mistake of most books in not emphasizing from the beginning why co-variants, contra-variants are so named. Tullio Levi-Civita did this early on in his book, by defining an invariant multilinear form from vectors and their duals. If by definition this multilinear form is invariant to a coordinate transformation, then the vectors and their duals have to transform the opposite way to keep the multilinear form invariant. Einstein at least defined a contravariant vector from its scalar product with a covector as being an invariant. Mathematics teachers should try and look into the etymology of mathematical terms, because there's a hell of a lot of interesting history behind them that's helpful to a student in learning the subject.

[@vikrantshrimali2287]

yeah. totally agree.learning tensors with the intuition behind covarient and contravarient basis is fun. anyway I like these videos and the style of teaching.

[@angelmendez-rivera351]

Barrios Groupie Teachers do know the etymology of the mathematical terms. They just do not teach it, because there is no point in mentioning covectors in a tensor calculus course when literally nobody is going to understand what they are talking about. Students would not know what a covector is, nor do they know what dual vector spaces are because this is not taught in linear algebra, and very few students take multilinear algebra. Very few institutions even teach multilinear algebra to begin with. And to understand what covectors are, you need to have taken differential geometry and bundle theory, which not many students do.

[@pedsantiago]

@@angelmendez-rivera351 Could you please elaborate briefly on your last sentence: "And to understand what covectors are, you need to have taken differential geometry and bundle theory". I get what you were saying before this (about dual vector spaces and multilinear algebra), but what's the insight that differential geometry and bundle theory (of which I know nothing about) give into the nature of covectors? The way I personally see it, as an interested layman (I'm no more than that), is that the so-called covectors are simply what we used to call "vectors" before we get into Linear Algebra etc, without any further understanding lol. I'd really appreciate any comment.

[@prabhatp654]

@@pedsantiago that's too modest of you, saying any comment.

[@englemanart]

Best introduction to tensor calculus I have ever seen.

[@ekaptsv]

A brilliant lecture, Sir! So many books I have tried to read to get the point what is wrong with the gradient definition. And it is explained so clear here in this lecture, so I finally got it! Thanks a lot!

[@mcpearce]

This lecture series is excellent. Thank you for sharing.

[@Godakuri]

Fixing the gradient might have been one of the most beautiful things I've seen in my life 1:16:12

[@MathTheBeautiful]

I'm glad to hear that and thank you for letting me know.

[@ramyelbehery9264]

This is very good, I found your lectures along with your book are indispensable in my effort learning tensors. My goal is: starting with first order logic statements, describing a process or a situation, arranged in adjacency matrix to identify invariants and consequently get more understanding. It is like what you said in the first lecture, we cannot be all Gauss, but tensor calculus could solve quite complex problems.

[@englemanart]

This is glorious stuff and very well-presented. I am grateful for these videos.

[@aminemohamedaboussalah6169]

You are one of the best Professor I have never seen. Please keep doing what you are already doing :)

[@MathTheBeautiful]

Thank you and yes sir!

[@wagsman9999]

I am learning a ton from these lectures, thanks very much.

[@dlymper]

One of the best lessons I have seen here in YT, the value cannot be measured! A theoretical question: when you prove whether an object is a tensor, you put the indexes in the correct place beforehand and use that property there after... Shouldn't we start by not knowing where the indexes go and finally prove that they should go up/down?

[@MathTheBeautiful]

The short answer is Yes, you're correct. For a longer answer, see paragraph 5 of this section: grinfeld.org/books/An-Introduction-To-Tensor-Calculus/Chapter7.html#Section_7_1

[@MoTheDeliciousPeach]

'The partial derivative of T_i is not a tensor' I thought I knew something about tensors before starting these lectures. When I heard that, I realised I knew nothing ... lots to learn.

[@MathTheBeautiful]

Well, depends on what you call a tensor. If you restrict your attention to linear transformations of coordinates (this would be the case if you only considered Cartesian or affine transformations), then it would be considered a tensor in this narrower sense.

[@Halo_people]

Thank you so much for posting them!

[@kaan_guven]

1:18:46 such a great enthusiasm!!! :)

[@advanstraeten2841]

At 12.22, and I am afraid this is a very silly question;)!: Let's say that we have a F that is a function of q, so F(q). q is a function of p. My intuition would say: F(q)=F(q(p)).So when I would have to do the thing with the Z"s, I would end up with: F(Z'(Z)). I know I am wrong! However, I have been following these lectures for months now, repeating every part, and understanding them a bit more every next instance, but this step I can't get. Also, I am not trying to understand this by the example: I would like to understand why F(Z')=F(Z(Z') is an identity. It doesn't look as an identity to me. What am I missing, what is my error? The lectures are great! My favorite paste time!

[@MathTheBeautiful]

As a warm up, take x = r cosθ and, on the right, plug in how r and θ depend on x and y. What will you get on the right hand side after you simplify?

[@advanstraeten2841]

Well, what about this?! What a fast response: thanks a lot, I will try it tomorrow!

[@gqwang6277]

13：02 The F in the third part of the first equation, is the F of F(x,y) or rather F of F(r, theta)? That is: F() in the second part = F(Z())

[@shawleezhao]

Metric tensor definition starts at 39:09

[@jaeimp]

Was the question @15:50 with the bookmark answer "The answer is coming" explicitly addressed at some specific point?

[@Arcel4]

I still can't figure out how do one knows when to introduce new index... Anybody can help?

[@gqwang6277]

Jacobian is a variant if considering r and θ as function of (x, y) ?

[@snnwstt]

Something seems wrong at 35:14, or 36:00, or 36:32. We can write Z' == Z'(Z), but NOT Z'== Z(Z') as the transition from the second term, last line ( Z i' ), to the third term seems to imply. Right? The fourth term is OK though, with delta R ( Z'(Z) ) / delta (Zi' (Z) ) == delta R / delta Zi times delta Zi / delta Zi'

[@angelmendez-rivera351]

snnwstt But he didn't write Z' = Z(Z'), he wrote F(Z') = F(Z(Z')), which is very different. The latter is correct, the former is not.

[@SalvatoreIndelicato]

can you insert subtitles?

[@rxdszaPhys]

Question: Instead of the attempted 'definition' of the Laplacian, could we define it in the following manner ; 'attempted Laplacian' F = vec{Z}^j dot{ d_j [ (d_i F ) vec{Z}^i ] } The quantity is the [ ] is the invariant gradient of F ill call it G. It is legal to differentiate G since G is invariant thus giving 'vec H_j' i.e. everything in { }. Now vec{Z}^j dot vec H_j is invariant since we showed that the dot product of 2 vectors is invariant -> 'attempted Laplacian' F is invariant. Furthermore, it reduces to the expected laplacian operator in cartesian coordinates. Is this a valid definition of the laplacian? Also, thanks for these lectures they are great!

[@marxman1010]

At 12:40, F could be a vector function or a scalar function. If F is a vector function, the result looks different.

[@sajidhaniff01]

Excellent! Many thanks

[@AshishPatel-yq4xc]

@1:33:18 in this lecture . Trying to get clear why S_{i,j} is not a tensor. The reason you say S_{i,j } is not a tensor is becasue once you apply the jacobian to it on the lower right of the board, you end up with S_{i' ,j' } =S_{i,j } J^{j}_{j' } J^{i}_{i' } + T_{i } J^{i}_{i', j' } . So the term that "stops it" transforming like a tensor is the last bit + T_{i } J^{i}_{i', j' } ? Had it just needed jacobians to transform from one system to another, that would make it a tensor ? So if we had S_{i' ,j' } =S_{i,j } J^{j }_{j' } J^{i}_{i' } that would have been a tensor ?

[@MathTheBeautiful]

Yes.

[@mohammedal-haddad2652]

At 50:15 it is clear that the delta system is a collection of ones and zeros, but in a later lecture, a proof relied at the fact that only the covariant derivative of the delta system is zero not the partial derivative! What am I missing?

[@MathTheBeautiful]

Both derivatives are zero.

[@keeganmcnamara3365]

For the exercise at 56:50 wouldn't it be easier to only represent the first tensor in terms of its covariant components and then use the new Z_k (dot) Z^j to get a delta^j_k from that definition directly above it on the board and contract that with the contravariant metric tensor you got from expressing the first tensor in its covariant components to get the final Z^ij? Is that okay to do or do you absolutely need to change both of the original tensors into their covariant components?

[@keeganmcnamara3365]

+MathTheBeautiful Did I ask this in an incorrect way?

[@MathTheBeautiful]

+Keegan McNamara You asked the question perfectly. Yes, your way is great and is shorter, too!

[@houssemamami4359]

apart from using Z for everything(coordinates, metric tensor, vectors...) which is very confusing, you also don't change the notations when you change from coordinates to another, I mean when you put the transformation's formula of the tensor without putting a ~ on the T, I understand that you mean the same T that is on the right side but with indices i' . another example: if you don't put a ~ above the Z to say "that's the new coordinates" I could understand that those are the old coordinates Z but with different indices i' . I get very confused and I don't understand anymore what is what. :( the lectures are great but I don't understand why you just don't put a simple ~ to avoid confusions.

[@angelmendez-rivera351]

Houssem Amami The prime denotes a different coordinate system. There is no reason for you to be confused, just get used to it.

[@gqwang6277]

There are covariant base ,of a coordinates system and its contravariant base. Is it possible to construct another coordinates system according to the first contravariant base in which it is covarint? So there will be an another set of metric tensors and basis.

[@mohammedal-haddad2652]

This is a rather long lecture but my favourite part is this: Q: The metric sensor was in the definition of the dot product? A: No, it is the consequence of the definition.

[@Ganeshkumar-kc8ur]

Actually I am not able to understand that question can you explain? In the previos lectures itself he mentioned that the metric tensor is an object that arised from the dot product, which represents the geometrical properties, Hemce it is called a metric tensor. Then what is the point in asking whether the metric tensor is in the definition?

[@prabhatp654]

Related to 29:30, do we impose them to be in this particular form (prod of contravariant and covariant tensors) or they come naturally as when we define basis vectors for coordinate systems, there the linear combination emerges naturally.

[@parthsas2085]

i have a question at 15:55 . In the derived result of the gradient of scalar field, can we use the indexing rule from within the partial derivative at the R.H.S ?

[@BArdekani]

Min 3: I would rather see G(r, theta) defined as F( r*cos(theta) , r*sin(theta) ) rather than using the same symbol F to mean two different maps.

[@abcddd580]

So I understand the contravariant components of a vector are themselves a contravariant tensor. Lets say I have a variant V^i and I used it to form the vector V = V^i*z_i out of the covariant basis vectors, z_i. This means V^i now represents componenets of a vector. So, does that mean V^i must be a contravariant tensor? This would imply all first order variants are tensors, which isn't true. Where have I gone wrong? Thanks!

[@MathTheBeautiful]

If you take a first-order variant that is NOT a tensor and contract it with a covariant basis you will get a vector that is NOT invariant. In reverse, for Vⁱ to be a tensor, it must come from an INVARIANT vector.

[@anfarahat]

At some point in the lecture, you mentioned that Z^i is just the coordinate and is not a contravariant tensor. Why would the superscript be helpful then? It seems that dZ^i (the differential of the coordiante) is a contravariant tensor though. Any insights? One more question: is there a method/algorithm by which we can determine the components of the position vector with respect to the covariant basis, without reference to the definition of the position vector in cartesian coordinates? In other words, is there a geometric/coordinate-independent definition of the position vector?

[@MathTheBeautiful]

+anfarahat Two great questions. There are a number of objects with indices that are not tensors, e.g. the Christoffel symbol. Z^i is another. (These objects aren't tensors in general, but they are tensors in a more narrow sense with respect to linear transformations of coordinates.) The placement of the index as superscript will be useful later. The components of the position vector are usually not needed. When you do need them, you can write them as R.(Z-vector)^i. (The interesting question is the rate of change of those components and that question has a very nice answer.)

[@abhishekjoshi8413]

is it necessary to have the book as it is very costly in india your lectures are awesome though

[@debendragurung3033]

1:15:00 comparing the definition to earlier definition.. that gradient in one cordinate system is related to other by a Jacobian. How do we compare that to this expression..

[@BartAlder]

Masterful clarity.

[@MathTheBeautiful]

+Bart Alder My favorite comments are the ones that have words clarity or simplicity.

[@neuzerg]

maybe you could help one point: in order to be a higher order tensor, the variant should firstly be "higher order variant", correct? But how to properly interpret "higher order" variant? For example, could we "visualize" a third order variant as represented by n x n x n (cubic)matrix? and so on?

[@MathTheBeautiful]

Perhaps, yes, but there is no point in visualizing them like that. It works with matrices, because matrices come with a good definition for multiplications but 3d and higher matrices don't. You could think of 3rd order variants as a set of numbers enumerated by three indices. Mathematica displays them like this: x x x x x x x x x x x x x x x x x x This would be a 3x3x2 system.

[@joeboxter3635]

@1:13:00 confused. If the gradient is a covariant, and the bases vectors are covariant. Then why not use the product of two Jacobians to transform to gradient in new bases.

[@MathTheBeautiful]

Great question. You can't use the Jacobian in forming expressions as the Jacobian is not an object specific to a given coordinate system as it characterizes the relationship between two coordinate systems. In other words, it is not a "variant".

[@UnforsakenXII]

I'm trying to see if it's possible to use these tensors rules to show how the cauchy riemann equations are covariant but it doesn't seem to be popping up for me. Do you happen to have any advice? Should I look at two different cases where the coordinates are defined before I look at any generalized coordinate system?

[@MathTheBeautiful]

Can you describe a little more specifically what you're trying to prove? I'd love to help.

[@UnforsakenXII]

I'm currently reading complex variables and applications by Churchill (1960). After we established the conditions needed for a complex function to be differentiable in Cartesian coordinates i.e. the cauchy riemann conditions, they set out a second section that stated that the conditions take on a new form when you derive them in polar coordinates, there was an extra factor of 1/r or r depending on how you write it so I was wondering if this would be a good place to practice some indexing but I'm not exactly sure if it's something that would be good to try. I was trying to see if I can somehow relate both functions u, v from z = u+ iv but without reference to any coordinates. I'll attempt at setting up an identity again. Sorry for the late reply, wasn't expecting such a fast response and thanks!

[@marxman1010]

What is the position vector in covariant basis?

[@mongrav1000]

Around 56:00 there is a proof that the contravariant basis/bases are contravariant tensors. Is this a complete proof? The final step seems to rely on vector relationship A dot B = C dot B ( = Kronecker delta), therefore A=C. But this isn't generally true for vectors, right? Did I miss some inference here? [I am not seeking analytical rigor, just seeing if I missed a step.] thanks.

[@MathTheBeautiful]

That's an excellent point. I use this logic all the time. (A.B = C.B) does *not* imply A=C. However, if (A.B = C.B) for *all* B then it *does* imply A=C. In particular, it's sufficient that (A.B = C.B) for all elements of a basis, which is the argument I'm using here (since the identity holds for all Z_k). (So this nice linear algebra argument got a little bit lost in the tensor notation, which turned this into an insightful exercise on the meaning of tensor relationships.)

[@mongrav1000]

I now see that the final system of equations can only be satisfied by the contravariant basis, in the correct correspondence. I had been looking at a single equation and should have looked at the system of equations.

[@MathTheBeautiful]

Mongrav Exactly - that's how every tensor identity works: it looks like it's saying something about an individual component, but it's actually saying something about all components at once. This is a great advantage when it comes to working with more complicated expressions: all the rules of integration and differentiation continue to work because individual components are ordinary functions. At the same time, the results have the generality of vector identities.

[@marxman1010]

At. 43:15, in the proving of metric tensor, used Zi with arrow. But if Zi is a vector, the commutativity doesn't establish and the it looks something wrong with the proof.

[@MathTheBeautiful]

If I'm not mistaken as to the spot you're referring to, I think that step is based on the property of the dot product that (alpha*a)•b = alpha*(a•b)

[@marxman1010]

Thank you. Vector elements and vector are mixed in the expression.

[@orientaldagger6920]

At 1:09:00 the "attempted" definition of the gradient there is a dF/dZ_i_superscript where Z_i_superscript is the coordinate variable. Is there such a thing as Z_i_subscript? Not Z_i_subscript with an arrow as that is covariant basis...

[@MathTheBeautiful]

No, there isn't such a thing as Zi.

[@orientaldagger6920]

@@MathTheBeautiful Oops I jumped the gun, you actually answered it at 1:42:00

[@isaackulp2885]

When would you change the placement of indecies my multiplying by the Jacobian, and when do you use the metric? What in the notation distinguishes lower/raising indecies from changing coordinate systems?

[@MathTheBeautiful]

You never multiply by a Jacobian. That's just a theoretical thing you consider in the background. A Jacobian is not something you have in a given coordinate system, only when you study transformations between coordinate systems.

[@ForcesOfOdin]

55:30 you establish = delta^i_k = < Z^i, Z_k> where < , > denotes the dot product. Is it really enough to have these inner products equal to determine that J^i_i' Z^i' = Z^i ? I know that = doesn't necessarily mean that x1 = x2 in general.

[@MathTheBeautiful]

+ForcesOfOdin Yes, if instead of the one dot product of x1 and x2 with one "y", you match the dot products of x1 and x2 with every "y" from some basis, which is what I did there.

[@ForcesOfOdin]

Right, of course. x1 != x2 could project onto some of the basis vectors equally but not all, as the difference vector x2-x1 would then be perpendicular to every basis vector Z_k which is possible only if x1-x2 is the zero vector.

[@advanstraeten2841]

Now at the left side of the board Mr Grinfeld writes the identity. He uses the example of cartesian and polar coordinates. That's great, but it is not said what is the primed and the unprimed at the right side of the board.If this would be clear: great! However: I would like to understand- in abstracto, without the examples at the right side of the board, why it is that: F(Z')=F(Z(Z') and not: F(Z')=F(Z'(Z))? It is exactly in reverse of what I would think. I must admit that I have a tendency to mix things up;)! So I have 2 possibilities: except it, or really understand it. On the other hand: I have seen some examples where identities were written- never seen that before- and I was able to understand them. Sorry to be so insistent: my mistake could be one of the examples where one has to make every mistake one can make: I am certainly trying!:)

[@MathTheBeautiful]

It could absolutely be either identity. Both identities are correct.

[@advanstraeten2841]

So: F(Z') is identical with F(Z) as expressed in the primed coordinates as well as with F(Z') as expressed in the unprimed coordinates??? I am certain that I don't understand what the identities mean;)!

[@lemmavideos2163]

Right. That's why you should consider an example. Start with the exercise I suggested in an earlier comment.

[@advanstraeten2841]

Thanks for your suggestion! I have tried. It was in some sense enlightening, but not with regard to understanding the problem that I have with these identities Probably I am wrong, but it feels like working backwards. I always remember the question that Mr. Grinfeld asks his students: "Say it in words!".Great approach which seems very important in the course. I would like to have somebody state these two identities in words. Later on in the lecture there are identities that I can "get". There are also identities written in other parts of the course- I am always rehearsing to pick up more and more of the pieces-that I "get". It is just this one that I can't figure out a priori. Isn't it true that what follows is in fact based on the identities? Also: I remember that Mr. Grinfeld apologizes for a minor error by saying: I was so focused on getting the identities right;)! But if both F(Z')=F(Z(Z') and F(Z')=F(Z'(Z)) are right, why should he have to focus on the order? I am not trying to have a discussion ,or finding contradictions: I am not questioning anything said by this expert. Only the statement that both identities hold is puzzling, and I am well aware that writing the identity is crucial. Anyway, thanks for your comments, and if I am a pain in your bodily parts I apologize for that! Kind regards, Ad

[@MathTheBeautiful]

Hey Ad, would you like to discuss this over skype or such?

[@jackozeehakkjuz]

That was perfect 12:47.

[@c_dorado]

I'm trying to find the derivative of a contravariant tensor T^i. Eventually I need to evaluate (in the non-tensor term) the derivative of the Jacobian J[i', i] with respect to z^j'. Do I need to express that Jacobian as a natural function of the un-primed coordinates J[i', i](z(z')), since it represents a derivative with respect to z^i, and then apply the chain rule?

[@MathTheBeautiful]

+Camilo Dorado García If you need to differentiate the Jacobian with respect to z^j' then you should express it in terms of the primed coordinates. Since J^i'_i naturally arises in terms of the unprimed coordinates: yes, you should substitute z(z') and use the chain rule which results in J^i'_ij J^j_j'

[@theflaggeddragon9472]

At 39:12 you write Z_ij=Z_i (dot) Z_j. Doesn't the dot product output a scalar meaning that Z_ij should be just a number and not a tensor?

[@MathTheBeautiful]

+The Flagged Dragon You are now going through one of the hardest parts of learning this subject: mastering the notation. The equation Z_ij=Z_i (dot) Z_j represents 4 equalities (in 2D): Z_11=Z_1 (dot) Z_1 Z_12=Z_1 (dot) Z_2 Z_21=Z_2 (dot) Z_1 Z_22=Z_2 (dot) Z_2 so Z_ij is a collection of 4 numbers (2x2) described as pairwise dot products of the covariant basis vectors. It is important to, whenever you encounter an identity in indicial notation, to "unwrap" it into the individual identities as I did above.

[@theflaggeddragon9472]

+MathTheBeautiful Thank you so much! I understand it now. Expanding those indicial expressions does help the understanding a lot. Thank you for being so prompt with your responses. Much appreciated.

[@advanstraeten2841]

I am kind of lazy: which one is the unprimed system: the cartesian or the polar? That was unclear to me...

[@angelmendez-rivera351]

ad van straeten Neither. It's an arbitrary coordinate system. Don't be lazy.

[@theodorei.4278]

Amazing lectures, thank you for that Sir, but I think I'm missing something here. At 0:46 you state that Temeperature is function of space and that temperature is invariant, because we could know its value at some point in space without introducing a coordinate system i.e. the value of temperature is the same irrespective of the coordinate system. It seems to me impossible to comprehend that I know the values of a function called "Temperature" without a coordinate system. To prove my point, suppose that someone asks me what is the temperature at a point in space, one needs to tell me where that point is, and in order to tell me where that point is one needs to give me the coordinates of that point and consequently in order to be able to that one needs to define a coordinate system. To cut long story short, I understand that Temperature is invarinat because irrespective of the coordinate system a point has a specific temperature. What I don't understand is how can we know the values of a "temperature" function without first introducing a(n) "initial/primordial" coordinate system.

[@MathTheBeautiful]

One could point to a particular point in space and ask you: what's the temperature at *this* point? And then another point, and then another point. All without involving a coordinate system.

[@theodorei.4278]

Thank you for your prompt response Sir. Yes, but this might lead to umbiguities, couldn't it? I mean that, at first I might erroneously understand that the "this point" might be another point but close enough to it and then by telling again "move a bit to the left or the right" I might get closer to your "this point" and at the end of the day I might locate your point but it would be more probable to not be able to locate that point. So the need of a metric and a coordinate system would naturally arise. Sorry for being so "needy" to introduce a coordinate system but my engineering background leads me to that direction. Obviously I'm still missing something and that is the reason why I turn to you.

[@MathTheBeautiful]

Yes, it's very difficult to give up one's comfortable starting point which, for most people, is some kind of coordinates. However, it's best to learn to operator from different starting points. A good question for you is, if you are not accepting the concept of "this" point, how are you going to assign your coordinate system in the first place (assuming no one has already done it for you)?

[@theodorei.4278]

You blew my mind. I rest my case. So, I should try to think space without any coordinate system at all, but all physical properties layed out there. So all physical/mahematical objects exist without the need of any coordinate system. The only need for introducing one is for making our lives easier to perform calculations? Hope I get the gist of it. Thank you anyway for answering my questions in such a short time and with so good and convincing replies.

[@MathTheBeautiful]

Yes, that's how I see it. Your original point of view is valid as well. Both are valid and both are valuable.

[@BigDmitry]

You define tensor as a variant with special property, but I believe you didn't define what a variant is. Or maybe I missed it? Is it just a vector in a coordinate system?

[@MathTheBeautiful]

It happens at 9:09. A variant is an object that is calculated in each coordinate system according to a recipe that is the same across all coordinate systems (and is generic in a sense that it doesn't refer to any specific properties of a given coordinate system). The components of a vector is one example.

[@BigDmitry]

I see, thank you. And thanks for the great series, it's very insightful!

[@MathTheBeautiful]

Glad you're enjoying it!

[@anantgairola3394]

Dear Professor Grinfeld, So if I understand correctly, an invariant can never have an index? We can think of either the position vector R as an example or any arbitrary invariant U, formed by the contraction of one co-variant and one contra-variant tensor [U=S^(i)T_(i)]. Is that correct? Thank you once again for all your lectures and help.

[@MathTheBeautiful]

Correct. It's uncommon to decompose the position vector R. But for any other geometric vector, your observation is correct. Also, the Kronecker delta has indices and the exact same values in all coordinate systems. However, it is not (technically) considered an invariant but, rather, a second-order tensor.

[@anantgairola3394]

@@MathTheBeautiful, thank you for the clarification, Professor Grinfeld. On that note, I have a follow up question. I can understand why the kronecker delta could (arguably) be considered an invariant. It has the same values in all coordinate system. From that logic, I was under the impression that the permutation symbol would also be an invariant since it is just a collection of 1s, -1s and 0s and will have the same value in all coordinate systems. However, that all changed after I watched your lecture on relative tensors. I used to think if an object/entity has the same value in all coordinate system then it is regarded as an invariant. So isn't the permutation symbol an invariant? And aren't all invariants also tensors? Thank you once again for your time and effort.

[@MathTheBeautiful]

@@anantgairola3394 Sure it's an invariant, just not in the sense used in tensor calculus. In tensor calculus, invariant means no indices, and having indices means that you have to contract with the Jacobian in order to be classified as anything.

[@anantgairola3394]

@@MathTheBeautiful, thank you once again for your prompt reply, professor. This lecture series has been a blessing. Regards, Anant

[@AshishPatel-yq4xc]

the amount of juggling and using i for the live index but also as a unit vector is totally confusing. Could follow up to lecture 4 now confused with the juggling required and indirection with the various tensors.

[@MathTheBeautiful]

The unit vector i should have a little arrow above the i. The index i is smaller and is placed next to the letter denoting the variant. Does that help clarify the confusion a little bit?

[@AshishPatel-yq4xc]

MathTheBeautiful Yep, that helps. Just need to get used to keeping it straight in my head depending on the context.

[@BarriosGroupie]

34:00 where is it shown that the partial derivative of an invariant is a tensor?

[@MathTheBeautiful]

Starts at around 10:30.

[@mikedelmonaco6193]

during the proof at 1:00:00 why was m necessary? couldn't you have just said k for both of them? I did it that way and it worked out, but is that ok?

[@MathTheBeautiful]

I don't think this is OK. Consider this extreme example: what is your interpretation of a_k b_k c^k d^k? How are you going to pair up the k's?

[@mikedelmonaco6193]

I just did this: rewrite Z^i as Z_k Z^ki now I have Z_k Z^ki *(dot) Z^j this simplifies to Z^ki 𝛿^j_k (using what was proved earlier) which is just Z^ij which is the contravariant metric tensor I now see how it would have been a problem if I rewrote both vectors and only used k. Is what I did valid though? Thank you for the reply by the way

[@MathTheBeautiful]

Yes, what you did is perfect.

[@straightedgesoldierx6111]

MathTheBeautiful So a tensor in general is an object that can be represented as: T^{i'}=T^{i}J_{i}^{i'} correct?

[@straightedgesoldierx6111]

Of course there are the covariant ones and covariant-contravariant ones as well.

[@MathTheBeautiful]

StraightEdgeSoldier[X] It's an object that *transforms* from one coordinate system to another according to that rule.

[@straightedgesoldierx6111]

Ah ok thank you very much.

[@rkpetry]

I agree, one should write F'... (I just wrote and deleted a lengthy comment that amounted to repeating something you'd already said-because I missed the F' you didn't write).... Alternatively, mathematics has special symbols for definitively-equal: ≜, ≡ ?

[@MathTheBeautiful]

I think that ≡ is a very useful symbol to denote an identity compared to an algebraic equation. It's probably not a bad idea to use ≡ when you are about to differentiate both sides of an equality.

[@rewtnode]

Very confusing to use the symbol Z for the metric tensor. It gets lost in all the other Z's. Especially with all the i' and j' , which have the problem that they need a tiny dot and a tiny dash. Makes me cringe. You know there are more letters in the alphabet. Doesn't have to be all Z's. ;-). Still, apart from the notation, this is a great lecture because he doesn't leave anything out and explains very well! Thanks.

[@kevinz4396]

50:20.. equal to 1 or greater than or equal to 1?

[@T4l0nITA]

equal to one

[@jacanchaplais8083]

This is all fantastically done, and I have bought and am enjoying your accompanying book; however, I do not feel that there has been much justification for the metric tensor, as defined by the inner product of two basis vectors, as a raising or lowering operator. It feels that there aren't enough axioms to support this, the contravariant metric tensor as the matrix inverse of the covariant metric tensor and the definition of the contravariant basis. Could you illuminate me as to what I am not seeing/missing?

[@MathTheBeautiful]

What in particular doesn't make sense. Do you feel that these definitions lead to a contradiction of some sort?

[@psharmacgk]

I think he means he's missing the motivation for using the metric tensor as a way of raising and lowering indices, I also don't really see why that works. Contravariant bases vs covariant bases feel rather vague to me by this point in the playlist, I was hoping they'd be explained in more detail later on.

[@jacanchaplais8083]

MathTheBeautiful​ well, ultimately my question is "why does Z^i • Z^j = raising operator?", or the similar expression for the lowering operator. Wherever I've tried to find an answer, the logic seems somewhat circular. P.S. I do not have an issue with the contravariant and covariant bases.

[@rkpetry]

I'm going over the middle section one more time- The notion perception that coordinate basis Z is derived from R... I think you're about to prove that R is irrelevant to Z... Otherwise I'd want R-and-Z in some compound symbol {Ȓ,rZ,...}

[@MathTheBeautiful]

I wouldn't say irrelevant, but once R gives birth to Z_i, it largely disappears from discussions.

[@mohamedmoussa9635]

Wait, isn't Z_i the covariant component?? You say around ~40:00-41:00 that it doesnt mean anything.

[@MathTheBeautiful]

The covariant component of what vector?

[@mohamedmoussa9635]

MathTheBeautiful OK. I need to get my thoughts in order. My initial answer was the position vector. But that can't be right, R(Z^i) just maps to some vector R. R is just a vector defined from some reference point, independent of the origin. However, we can describe R itself in terms of covariant R_i and contravariant components R^i, yes?

[@MathTheBeautiful]

Mohamed Moussa Your guess is good. Z_i is sometimes used for that, but not in these videos and not in my book. If it were, then your original comment would have been right on. Yes, I would probably use R_i and R^i. Interestingly, the need for the components of the position vector arises very rarely (and never in this course).

[@mohamedmoussa9635]

MathTheBeautiful I suppose thats because we mostly deal with changes in things. Thanks for your input.

[@MathTheBeautiful]

Mohamed Moussa Yes.

[@benceracsko6586]

I enjoyed the previous lectures as they seemed to be a great and quite descriptive way to introduce this all, but this lecture disappointed me. You keep referring to vectors as coordinate-independent objects, which are true, but vectors ARE tensors and tensors themselves are coordinate-free objects. Their most exact mathematical definition introduces them as homogenous multilinear functionals acting on vectors and their duals and there is not a single damn reference to any sort of coordinate system at all. You are treating tensors as if their components were the tensors themselves, but that is not true. I get it that in engineering and physics, we ARE constrained to using coordinates, so it is important to be able to calculate with their components, but by not outright screaming this fact into the audience's ears, they are really not going to get the most important property of tensors, namely that they are COMPLETELY AND UTTERLY coordinate-free objects. Also, this is more of a minor thing, but especially since you were saying you do not want to make references to a "background" cartesian system, you cannot really define the components of metric tensors as "dotting" the basis vectors because you cannot "dot" them in the first place. It is also quite... weird, to "dot" vectors with their duals. In fact, the whole dual space thingy was pretty much skimmed through. What we have here is technically working and your students will be able to calculate stuffs, but mathematically speaking, incorrect. Anyways I mean no offense to you as you seem to be a great teacher and I am a student myself, but differential geometry is the area I want to specialize in and these sort of things really bother me...

[@angelmendez-rivera351]

Bence Racskó *I enjoyed the previous lectures as they seemed to be a great and quite descriptive way to introduce this all, but this lecture disappointed me.* I'm going to see just how much attention you truly paid to these lectures you say you enjoyed. *You keep referring to vectors as coordinate-independent objects, which are true, but vectors ARE tensors and tensors themselves are coordinate-free objects.* No, this is incorrect. A vector is a contraction of two-tensors, and such a contraction is an order 0 tensor, but being an order 0 tensor is literally the definition of being coordinate-free. Coordinate-free and coordinate-independent are completely synonymous. Vectors are not order 1 tensors. Also, it is "is true," not "are true," since we are speaking about a single sentence, not multiple. Also, to be strictly rigorous, most vectors are not tensors, because the array of their components may not be a tensor. For example, there are many triplet of real numbers which satisfy the axioms of a vector space, so by definition, they are vectors, but the array of its components may not transform via a Jacobian, hence not being a tensor, and as such, the vector is not a tensor either, and it is not coordinate-free. This happens with polar vectors in physics, and other such objects. As such, your understanding of what a vector is happens to be wrong. Vectors are not defined by tensor properties. They are completely unrelated to tensors. Vectors are defined by being elements of some vector space. *Their most exact mathematical definition introduces them as homogeneous multilinear functionals acting on vectors and their duals and there is not a single damn reference to any sort of coordinate system at all.* That is total bullshit. The definition that two objects are tensors if their components transform via a Jacobian when the coordinates of which the components are functions change, this is definition is exact. Nothing about it is approximate or vague. Also, it is completely equivalent to the definition you talk about. They are both first-order logic propositions which have identical truth tables. As such, they mean literally the same thing. Neither is more "correct," "exact," or "rigorous," than the other. To say otherwise is complete stupidity on your part. *You are treating tensors as if their components were the tensors themselves.* No, he is not. He has literally talked about these components as being entries in some array. This array is what is being considered a tensor of order n, not the components themselves nor the object that is being decomposed into those components, although such an object would also be a tensor. This is also why he specified in the definition that an object is a tensor *if its components* transform via a Jacobian. He never said anything about the decomposed object itself transforming. In fact, he has stated the opposite. You clearly have not been paying attention to his lectures. *they are really not going to get the most important property of tensors, namely that they are completely and utterly coordinate-free objects.* This is by no means the most important property of tensors, because 1. it is not the definition of tensors, 2. some classes of pseudotensors also have this property. However, even if it were the most important, he has stated this multiple times throughout the lecture as well as previous ones. The fact that you are too damn arrogant does not change this. *you cannot really define the components as metric tensors as "dotting" the basis vectors because you cannot "dot" them in the first place.* Yes, you can. The definition of the dot product of two vectors is geometric and coordinate-free. It does not require the existence of a background coordinate system. All it requires is that you know the angle, which is well-defined without coordinates, and the length of the vectors, which is again well-defined without coordinates. The length of a vector only depends on the metric function defined on the space - which does not require coordinates - and a unit - which is arbitrary and also does not require coordinates. *It is also quite... weird, to "dot" vectors with their duals.* How so? The length of these covectors is equally as well-defined. You can superpose a vector space with its dual via a direct sum, and now the dot product between a covector and vector is well-defined. Again, completely free of coordinates. *the whole dual space thingy was pretty much skimmed through.* Because it is not necessary for students to understand tensor calculus. *but mathematically speaking, incorrect.* You talk a lot for someone who 1. doesn't know the definition of a vector 2. didn't pay attention to the lectures. *but differential geometry is the area I want to specialize in and these sort of things really bother me...* Of course they bother you. You have your definitions all wrong. It's no wonder. At this rate, it'll take you decades to be able to grasp differential geometry with a sufficiently deep understanding.

[@angelmendez-rivera351]

Hugh Jones Can you explain to me how on Earth his approach is "half-done"?

[@taraspokalchuk7256]

1:26:40

[@MathTheBeautiful]

Are you referring to the kick?

[@taraspokalchuk7256]

just a bookmark for me@@MathTheBeautiful

[@taraspokalchuk7256]

just a bookmark for me@@MathTheBeautiful

[@federicopagano6590]

Who is watching this in covid time

[@nomadr1349]

I'm watching through this playlist and notice, that it gets more difficult to follow. So I stopped and gave it a thought - you actually never explained/defined what covariance/contrvariance actually is. All you did is to say that a certain quantity is a covariant Jacobian, and from that poin on you started juggling these terms never stopping to explain the physical meaning. And without all you tell afterwards is a mumbo-jumbo.

[@MathTheBeautiful]

I believe the terms "covariant and contravariant" are defined in this lecture. Then learning these concepts, it's best not to thing about their physical meaning. It's more important to focus on their algebraic meaning, at least for now.

[@nomadr1349]

No they are not. I actually had to stop watching and refer to wikipedia to grasp, only after that I could follow further. Besides, I think it is generally never good to learn something mechanically, without knowing the reason. If I know the reason I can do the reasoning (and your task as a teacher is to show how) and derive the rule. You propose to learn the rule first. Sorry, I cannot learn if I do not see the reason. You started so nicely with geometric problems, deriding particular algebraic descriptions, but drifted right into those later. Well, tensors are very real physical objects too, they are NOT a mathematical construct, you should have mentioned this in that first lecture. But in your lectures so far they are still a mistery. Keep watching nonetheless.

[@MathTheBeautiful]

Well, in my mind, I was speaking to a student like you; one, who wants to understand the inner workings and logic of the subject. But it's understandable if you don't see it this way. Perhaps it's because these lectures were being delivered to students who were reading the textbook at the same time, so the lectures are slightly out of order compared to the textbook as I was reacting to my students' questions.

[@advanstraeten2841]

You know nomadr 1349, it is always great to have the opportunity to follow lessons by a great teacher like Pavel Grinfeld. It is still greater that this teacher responds so quickly to all the questions in much detail.And off course you can comment with regard to content and so on. I have noticed that Mr. Grinfeld is very open and encourages discussions. Then: why would you be so rude? If a student would behave like you I would tell him to find him self another teacher! You don't have to agree, but at least you could try to be polite! If you haven't that in you, you will find that there is no place for you in any environment where people discuss scientific matters. As my mother said: don't bite the hand that feeds you;)!

[@lightfreak999]

At 1:13:12 I stood up and started clapping. This was _mind-blowing_.

[@MathTheBeautiful]

+Eiad The only mind-blowing thing is not this is not part of every undergraduate curriculum.

[@lightfreak999]

MathTheBeautiful It's truly a shame. I'm currently an undergraduate student in physics and math, and I really appreciate this lecture series! The topic is extremely fluid and feels like a very natural extension of vector calculus.

[@littlewhitebutterflies4586]

At the beginning...it should've been z prime i, not z i prime

[@kiuhnmmnhuik2627]

He did it on purpose. In a previous video he says that the two are equivalent. I think it's just a way to avoid putting parentheses around Z' (needed to keep ' and i separated to avoid confusion). edit: There's another reason. If we were to use (Z')^i we'd also need to use another index in case i was already used. By writing Z^{i'} we save some letters and know that that Z refers to a different basis than Z^i.

[@littlewhitebutterflies4586]

@@kiuhnmmnhuik2627 Can you show me which video he says that? Because from my knowledge I don't think they're the same thing. The name of the index does not matter (calling the index i or i' doesn't make a difference), so Z^i and Z^i' are the same thing. However, the name of the vector basis DOES matter, because Z and Z' denote two different coordinate systems. So Z^i' gives you the basis vectors of coordinate system Z, whereas Z'^i gives you the basis vectors of coordinate system Z'

[@kiuhnmmnhuik2627]

​@@littlewhitebutterflies4586 I don't remember the video, but all his derivations in this video use the same convention. Note how the Jacobian is J with one i-like index and one i'-like index. He always uses indices without ' for one basis and indices with ' for the other basis. Go to @25:30 for example. He writes (I'll use latex notation): U = S_i T^i U' = S_{i'} T^{i'} = S_i J_{i'}^i T^j J_j^{j'} = ... If S_i T^i and S_{i'} T^{i'} were the same thing, his derivation would have no sense. S_i and S_{i'} are different but U = U' because S_i and T^i transform differently and so the two Jacobians cancel one another. Also note how he says that S' is different from S even though he writes S_{i'}. This means that S_{'i} = (S')^k for some available index k.

[@littlewhitebutterflies4586]

@@kiuhnmmnhuik2627 Okay I was going through my General Relativity book and I believe I found the answer. It is merely a notational convention to set Z^i' = Z'^i. Although usually it does not matter what you call the index, it is a notational convention to say that i' does not equal i, and that i' are the indicies that will give you the components of Z' in coordinate system Z. So there we go, I guess we were both right haha

[@kiuhnmmnhuik2627]

@@littlewhitebutterflies4586 I heard (I'm not a physicist) that the MTW "Gravitation" book uses the same convention.

[@NeedsEvidence]

One of the students snuffles all the time -- really, really disgusting.

[@taraspokalchuk7256]

I think the crucial insight to understand the derivation at 11:04 that some people (@advanstraeten2841 inculding me) had a problem with is that the functions F(Z) and F(Z') are different functions in terms of how they act on the coordinates. The notation hardly reflects this.

[@MathTheBeautiful]

Exactly!

[@catdanceable]

quite an adventure in symbolic logic

[@SalvatoreIndelicato]

can you insert subtitles?

[@MathTheBeautiful]

Hopefully some day - sorry about that!