Tensor Calculus 6b: The Covariant Derivative

[@wikiemol2]

One of the best part about this video series is the fact that the students are so inquisitive. It really helps to have someone asking so many questions when you can't ask the professor yourself. I think this is why this stands out as one of the best video lectures in mathematics I've seen online.

[@MathTheBeautiful]

Yes, it was a fantastic audience.

[@Yeeren]

Had a class like this during my undergrad. Professor who loved teaching and the subject, and students who are on-the-ball. Many years later I still remember that course fondly, even though the subject wasn't one close to my heart. Side note: i was delighted to discover that the library of the institute where I'm currently studying has a copy of Pavel's textbook, really makes my life easier.

[@mattking8016]

@Brian Babu He teaches at Drexel University in Philadelphia.

[@joeboxter3635]

@@MathTheBeautiful I'm not sure how to express this. But you speak to your students not as students, but rather as future colleague you are helping get to your level. If you say that's not true, then keep doing what you're doing. We are none the wiser. Lol.

[@MrDrvmmer]

As a self-learner of mathematics in general I have obsessively been after elegant and rigorous proofs, I cannot put into words how much I am indebted to you for such a concise derivation of covariant derivative that I have been looking for in a while as part of my study in differential geometry. I have so far read through many textbooks and notes about the definition of the derivative, but all of them lacked the definion. Maybe I looked at wrong sources, who knows. Yesterday I got to watch your series, Tensor Calculus, from the beginning up to this video. Everything is now crystal clear before my eyes! All in all, I do think that contributions to teaching in exactly the same way as you do like many others in order to dispel academic obscurantism must deserve as much respect and value as shown to prominent scientists. Thank you very much!

[@MathTheBeautiful]

Thank you - it's deeply appreciated!

[@pythagorasaurusrex9853]

Great lecture! I learned a LOT! You have the ability to explain this stuff in a very easy and understandable way. And you only use a minimum of necessary formalism, which makes you focus on what is really important. This is how teaching students is meant to work. (P.S. I am a math teacher in Germany and enjoy every single video you do).

[@Tardistimelord]

*A comment for anyone feeling stuck, as I was for a time*: The material in this video is very well supported by the book in terms of exercises and narrative. See in Chapter 8, p.105 up to p.115, where the full covariant derivative is developed. As was said in earlier videos, this material requires practice and calculation. What I've found is that practicing the derivation and use of the Christoffel has been essential to understanding the components of it. An additional set of material that I felt helped was www.physicspages.com/2013/12/22/christoffel-symbols-in-terms-of-the-metric-tensor/ It uses somewhat different notation and, by understanding that notation, followed by redoing it in this more robust and standard form, the ideas emerged more elegantly (at least for me).

[@justforknowledge6367]

n this lesson you have dealt with covariant derivative of tensor. What does it mean? I studied physics, so my question is from a physics point of view, i.e., what does this mean physically? Is the word 'covarient' related to covarient tensors? Is there something called contravarient derivative? What would that mean? I had followed Murray Spiegel's book on Vectors and Tensors, and there are many unanswered questions that remain.

[@MathTheBeautiful]

Well, it measures the rate of change, just like the ordinary derivative does. Except the dependence on the coordinate system is neutralized.

[@justforknowledge6367]

Thank you very much, Sir! Now only this part remain: Is the word 'covarient' related to covarient tensors? Is there something called contravarient derivative? What would that mean?

[@MathTheBeautiful]

Just For Knowledge The contravariant derivative is the covariant derivative with the index raised. See index juggling.

[@justforknowledge6367]

Okay, Sir! Thank you for your response. I will now continue to follow your material with utmost sincerity.

[@massimilianoc2436]

Dear professor, I'm stuck at 2:58 when you say that the vector field V is an invariant. How can be an invariant if V is a vector field? I suppose that the vector field itself is fixed, but the components change according to the basis chosen to describe components. A vector field can be though as the derivative of a scalar field (i.e. the gradient is a vector field), and you showed the gradient is a tensor. So, at least, I suppose vector field V is a tensor, and as you showed in previous lectures the derivative of a tensor is not a tensor. For this reason I don't understand why you say that the partial derivative of the vector field V is still a tensor. I can't figure out how is a tensor. Thanks in advance.

[@MathTheBeautiful]

+Massimiliano C I'm very glad you asked this question. This is an extraordinarily important point to dwell on. I cannot emphasize it enough. The reason you are having a hard time figuring this out is because you have a strong association between a vector and its components with respect to a basis. These are two completely different things. Related? Yes. But the former can exist without the latter. A vector an arrow. It's not a triplet of numbers. it's a physical (or at least geometric) object. You can draw it. You can visualize it. Not only does it not change with a change in coordinates (and is therefore an invariant), it is actually completely unaware of the existence of coordinates. The problem is that when one hears "vector", one immediately thinks of a triplet of number. This is just a bad habit that handcuffs one's imagination and stunts the pursuit of clarity. I rant against this habit throughout my book and these videos. The gradient has two definitions: a geometric one that gives a vector and a coordinate-based one that gives the components of that vector. The vector is an invariant, of course, and its partial derivatives form a tensor - with vector elements! The components of the gradient are a tensor (with scalar elements). Their partial derivatives do not form a tensor. I hope this will help you clarify things.

[@massimilianoc2436]

+MathTheBeautiful thanks for your prompt reply. I've understood the way I need to follow. I can clearly figure out that a vector can be just a geometric object (an arrow with length). What I really need is to fully understand your last sentence. Basically I need to understand why if I think the gradient as geometric vectors, the partial derivative is a tensor, and if I think about it as component based, the partial derivative is not a tensor. At the end, I am calculating the partial derivative of the same "entity". It is clear that I am still missing some foundamental concept about tensors meaning.

[@MathTheBeautiful]

+Massimiliano C Because when you are analyzing the components of a vector alone, you are not dealing with the whole vector. If you only know the components of a vector field, you are not able to reconstruct the vector field itself without knowing what the basis is at every point. So when you evaluate the partial derivative of the components of a vector field you are not quite capturing the intended rate of change since that is ignoring the rate at which the basis is changing at the same time. So you are not really analyzing the same thing in a slightly different manifestation. You're only analyzing "half" of the same thing. That's why you shouldn't expect parallel results.

[@massimilianoc2436]

+MathTheBeautiful Many thanks Professor! Now is clear where and how I need to study more in detail. This course is probably the best resource on the web about this subject, and your reply and clarifications are equally precious. Thanks.

[@deanrubine2955]

The idea of an invariant field is something that's there in space, independent of the coordinates we use to express it. An example of a scalar field S is temperature -- at every point in space there's a value for the temperature. The value of the field is determined by the position in space, S(R), where R is the point in space or the position vector, before any coordinates are introduced. Then when we express it in coordinates we'll get S(z) or S(z') (two different coordinate systems), so two very different functions expressing the same invariant scalar field. We learned the partial derivative of an invariant scalar field S is an order 1 covariant tensor (our first) because ∂S(z')/∂z^i' = J^i'_i ∂S(z)/∂z^i. It's no different for an invariant vector field. What makes it invariant is it's out there in space, before we impose coordinates. The gravitational field (at an instant in time) is an example. It's an acceleration field. At each point in space there's a corresponding acceleration vector. We can place a small test mass at any point in space and observe how it accelerates to measure this field. At each point we get a vector. It's not really any different than the temperature field; we have a value for the field at every point; in one the value happens to be a scalar, in the other it happens to be a vector. They're still both invariant fields, out there in space. Both their partials will be tensors; the partial of the vector field will be a rank 1 covariant tensor with vector elements. We saw this early; the partial of the position vector field is the covariant basis, three vector valued functions of coordinates that transform as a covariant tensor. If V is an invariant vector field (pretend I wrote an arrow over it) ∂V(z')/∂z^i' = J^i'_i ∂V(z)/∂z^i; the proof is exactly the same as for the scalar field.

[@abcddd580]

The covariant derivative has a nice geomteric interpretation when the T^i elements are scalars. In this case of scalar elements, the covariant derivative is a scalar quantity that captures the rate of change of the vector whose components are T^i, with respect to the kth z coordinate, along the ith basis vector. But, what is the corresponding interpretation when the elements of T are vectors? Geomterically, what does the covariant derivative capture then?

[@MathTheBeautiful]

That's a very good question. I think that the beauty of bringing algebra to geometry is that you are able to go beyond direct geometric interpretations. We may be able to come up with a geometric interpretation here (via, say, a dot product with the basis vectors), but, at some point, intermediate elements will cease to have direct geometric interpretations. May it's better to not look for one here and instead enjoy the algebraic elegance and robustness.

[@justforknowledge6367]

Dear Prof. Grinfeld, Please refer to the video "Tensor Calculus Lecture 6b_ The Covariant Derivative - KZbin [360p]" at the time 28:18 to 28:20 (minute:sec). Why do you say what appears as Concovarient tensor? Regards

[@MathTheBeautiful]

It's the definition of the covariant derivative for a covariant tensor. One needs to repeat the entire logic laid out in the video to the case of an covariant tensor.

[@mauriziobaccara]

It seems to me that between what you said at time 3:50 and time 48:57 of this same video there is a contradiction. Probably I misunderstood.

[@MathTheBeautiful]

If I understand your comment correctly, the difference is that in the beginning, it's the ordinary derivative and towards the end, it's the covariant derivative.

[@mauriziobaccara]

yes you are right. What confused me was your sentence "for invariants the ordinary and the covariant derivative coincide" but when you decompose the vector they do not coincide anymore. Thanks.

[@marxman1010]

The start and end are same with derivative of an invariant vector. Very impressive. In the start it shows what the derivative is and shows the product rule of the derivative in the end.

[@kevinz4396]

In exercise 119 (prove 8.13), can I use a metric tensor to change an index of a partial derivative that is “non-vector”?

[@kevinz4396]

Specifically the top part of the partial differential

[@jakewalsh8314]

Dear Professor I have searched your videos and book but can’t find anywhere the idea of the derivative of an invariant being a tensor or more exactly a covariant vector. Can you help? Thanks

[@MathTheBeautiful]

Yes, sure, reach out to me on grinfeld.org.

[@eamon_concannon]

Im trying to do exercise 134 on page116 of your book which is to prove that delta_j T_i is a tensor where T_i is a tensor with VECTOR elements. The second last step involves an equation of the form A_i . Z^i = 0 where A_i is a vector involving covariant derivatives. I need to show A_i = 0 to complete proof. I dont know how i can invoke the linear independence of the Z^i here as we have a dot product. Any feedback would be great, as there are no solutions to this chapter.

[@the_martian_45]

You do a great job at explaining this material. Now, I’m only a junior in high school, so that says something. You are doing (I quote from my history teacher) some damn good teaching.

[@MathTheBeautiful]

Thanks, keep up the great work!

[@the_martian_45]

@@MathTheBeautiful I'm rewatching this playlist now 5 years later as a phd student, and it is just as marveling now as it was when I was in high school.

[@harshitjoshi9184]

Hello sir, I really appreciate the way you took up the topic of tensors. However, with all due respect, I sense a sort of fallacy in the statement when you said the expression at 9:55 is a tensor because it is the component of a vector (tensor). This implication is due to the result that components of a vector represented in covariant basis are contra-variant tensors, however, the result holds only when the vector itself is invariant unlike here. By that logic, the quantity in the brackets at 9:55 should have been contra-variant, but it is mixed tensor. I cannot verify the tensor behavior of the quantity by changing coordinate system. Can anyone please reply with the solution of this?

[@kiuhnmmnhuik2627]

The professor is correct in the sense that the derivation is analogous to the case without the k. The point is that you don't have to look at the expression between parentheses to conclude that it is a tensor. We have the following situation (in LaTeX notation): T_k = S_k^i Z_i (1) where S_k^i is the part in parentheses. We know that both T_k and Z_i are tensors and that they change according to the position of their indices. We know nothing about S_k^i yet. We can rewrite (1) as J^{k'}_k T_{k'} = S_k^i Z_{i'} J^{i'}_i J^k_{m'} J^{k'}_k T_{k'} = S_k^i Z_{i'} J^{i'}_i J^k_{m'} T_{m'} = [S_k^i J^{i'}_i J^k_{m'}] Z_{i'} (2) By comparing (1) with (2) we realize that [S_k^i J^{i'}_i J^k_{m'}] = S_{m'}^{i'} and the two Jacobian matrices tell us that S_m^i transforms like a tensor and the position of the indices are correct. In other words, S_m^i *is* a tensor!

[@harshitjoshi9184]

Kiuhnm Mnhuik Thanks a lot.! You did it very elegantly, now I understand this. You must be a genius. Thanks again😊

[@mikecohen5887]

At 22 you have the partial dotted with the covariant basis and then at 22:40 how do you get the contravariant basis involved?

[@MathTheBeautiful]

In an earlier video, we talked about decomposition by the dot product. It was shown that covariant components (i.e. components w.r.t. a contravariant basis) are obtained by dotting with the covariant basis and contravariant components are obtained by dotting with the contravariant basis.

[@mikecohen5887]

Thanks for your quick answer and also for all these lectures of course and the free pdf of your textbook - amazingly generous and a new business paradigm! Do you and your University and your publisher all have this mission to spread the knowledge?

[@reinerwilhelms-tricarico344]

Cool. Now shouldn't it be possible then to find the covariant derivative by looking for a differentiation that leaves the basis vectors alone, that is, gives zero when applied to the basis, independent on what the basis is? Or, to find the covariant derivative by demanding that the covariant derivative of the metric tensor is zero? It's kind of like reversing the whole story and, instead of starting with these initially somewhat weird Christoffel symbols, ending up with them if one wants to do actual computations of components. Also, what I'm still missing is some nice drawings that really appeal to geometric insight: For instance in a two dimensional arbitrary curvilinear coordinate system: What the Christoffel symbol components exactly do when one goes over a finite distance from one point to another. I vaguely recall that Prof Suesskind gave a nice drawing of that in some lecture on GR: Moving from some point to another point on two different paths, while differentiating a vector field, and then showing the difference one would get when it's done with the usual derivative.

[@David-km2ie]

I am kinda new to tensors but I think you are right. You may define the covariant derivative to a derivative in which the basis are 0. The rest follows like the video in reverse.

[@peterdonnelly5522]

Hi again. At 11:19, why is the definition of the covariant derivative not multiplied by the covariant basis like the equation to its immediate left which 'inspires' it?

[@MathTheBeautiful]

Peter Donnelly By analogy, the Christoffel symbol multiplies the tensor to which the covariant derivative is being applied. (You are right that more than one "analogy" is possible and experience tells us which is right.)

[@MohanAgrawal97]

Around 37:31 , in the equation written on bottom-left side,. can someone tell me how the L.H.S. delta(Z(lower i)) and the first term on R.H.S., del(Z_i vector)/del(Z_k) are any different.

[@rakibullahsazib1268]

At 39:00 you wrote dzi/dzk=dzi/dzk + chritoffel * zm can't we then cancel dzi/dzk of both sides and chritoffel * zm becomes zero. Is there any difference between nbalak(vector(zi)) and d(vector(zi))/dzk?

[@Whizzer]

This is corrected within two minutes. The left hand side was in error.

[@marxman1010]

Invariant is a tensor of order 0. A vector can be an invariant or a variant. Position vector is invariant, but a covariant basis vector is a variant. Furthermore, a V is an invariant vector, but component of that V is a tensor that is variant. At 15:46, derivative of an invariant is a tensor, so the combination of derivative of the vector component and christoffel symbol is a tensor. But why the vector component is a variant while vector itself is an invariant? The covariant derivative operator looks like depending on that a component of an invariant vector is a variant tensor.

[@marxman1010]

Well, components change along with the basis, but the vector does not change.

[@MathTheBeautiful]

If a vector is an invariant, then its components (with respect to the covariant or contravariant basis) are tensors. (If the invariant vector is constant, its components will vary, but their covariant derivative will vanish.)

[@safkanderik7217]

Dear prof, you defferentiate a vector V then u come with an exepression times the covarient basis. the exepression is a tensor (i can agree) but by comparision with the definition u set (compare V^j with T^j) ,it looks like that u took the V^j as a tensor. can u justify it ? Thanks in advanced

[@rkpetry]

From 32:08 the top line equation has the 'uncomfortable' last term that of itself does not say which lower indice 'belongs' to the contraction...it's symmetric, so it doesn't need to say, but my first 'impulse' is to align the m over the j to which it belongs-for beauty sake... A thought: The indice contraction notation might be directly used in high school matrix theory for simple vector proofs and calculations of determinants (not yet differential tensor analysis). Comment: the Christoffel Γ definition has a structure similar to the time-bias in an Earth frame observed by a fast-relative, ≡ (D/c)(V/c), as that is distance times its own derivative velocity. (Prior: An analogy might be a river flowing across a shifting landscape has a cosmic moment.)

[@muhammadumairkhan2681]

Do these concepts of tensor analysis extend to discrete piecewise functions? In particular to 3d meshes?

[@isreasontaboo]

The product rule is (uv)' = u'v + uv' so where does that minus sign suddenly pop out from at 20:05 ?

[@philp4684]

The minus sign is there because the expression on the blackboard is analogous to: uv' = (uv)' - u'v

[@samamafahim6666]

A covariant derivative is the contravariant component of a tensor (which is dV/dZk) with respect to the covariant basis. Is dV/dZk a variant? I am able to prove that it transforms from one system to another according to how tensors transform. We know that the contravariant components of a "vector" with respect to the covariant bases are tensors. But a vector is an invariant, and dV/dZk is a variant.

[@MathTheBeautiful]

Yes, it is a variant.

[@sashastrelnikoff4192]

Other than geometric arguments, is there any way to describe the covariant basis vector? Or is the whole point to use geometric arguments? I would think that when one looks at more complex coordinate systems, or perhaps a 4+ dimensional coordinate system, the geometric arguments become difficult if not impossible. What are the ways around this? Also, do you have any examples of calculating the gradient in multiple coordinate systems and showing that they are equivalent? At this point I would love a concrete example. Thanks a lot!

[@MathTheBeautiful]

Your question is at the very heart of tensor calculus (and linear algebra). Both of these subjects were sculpted precisely to let algebra take over for geometry. You hit the nail on the head (but not hard enough!) when talking about 4D. Geometry works only up to 3D, and is *impossible* (as you suggested) in 4D. Then all we have is the 3D *analogy*. So there is simply no such thing as a covariant basis in 4D. Of course, there are formal books that talk about it, but they are talking about algebraic structures rather than geometric objects. Don't read them, for now. Is that helpful? Yes, I should record an example like that. Or would *you* like to take a crack at it? You'll really enjoy it! For the function take "distance from a given point" and for the coordinate systems use Cartesian, Polar, and "Stretched Cartesian".

[@angelmendez-rivera351]

Sasha Strelnikoff The beauty of tensor calculus is that it works in any number of dimensions. n is arbitrary.

[@jojowasamanwho]

Suppose T^i was not a tensor. Would its covariant derivative be a tensor?

[@MathTheBeautiful]

Likely not.

[@davidatkinson3005]

I'm really enjoying this series (apparently at least a year after most other people, but better late than never). One misunderstanding keeps cropping up for me though, and I'm hoping someone is still looking at this and can help. It seems that all the partial derivatives that are taken with respect to components of the basis use an upper index in the denominator. I thought the upper index signified that these components are for the covariant (lower index) basis. But even when we switch and look at a contravariant basis, as is happening around the 20:00 mark in this video, we still take the partial derivative with respect to an upper index component. Does that mean we're taking the derivative of a contravariant basis vector with respect to the components of a covariant basis vector?

[@fiiIdeferr]

No, we are differentiating both the covariant and the contravariant bases with respect to the coordinates, which are not, in general, the components with respect to any of the above bases; they serve different purpose, since the first identify points in space, the second identify vectors (think at polar coordinates: by (r, theta) you denote a specific point on the plane, not a vector based at it)

[@gotbread2]

You mentioned the "metrinillic" property twice but i cant find any information on that property, are there any resources on that?

[@MathTheBeautiful]

The term "metrinillic" was introduced in my textbook referenced in the description. The term applies to the covariant derivative and means that it "kills" most of the metrics.

[@anantgairola3394]

Dear Professor, So at the end of the lecture you mentioned that the co-variant derivative of an invariant coincides with its partial derivative since the terms with the Christoffel symbol disappear. In this case you gave an example of a vector V (invariant). If we were to however, actually compute that partial derivative of the invariant V, we would have to decompose it with respect to a basis (like you eventually did). But if we were to actually evaluate that expression (the partial derivative), we would end up expanding it and the Christoffel symbol will re-appear, which is in fact how you started the lecture. I know there is no actual question in my comment but did I understand that correctly? Please advise.

[@MathTheBeautiful]

Hi Anant, you do not need to decompose the vector in order to differentiate it. See kzbin.info/www/bejne/npq8mIyNf6qUg80

[@anantgairola3394]

@@MathTheBeautiful , thank you for your prompt response. It makes sense now. This is one of the best lecture series I've seen online; really commendable!! Thank you for all your help.

[@David-km2ie]

How do you proof you have to correct every index. I mean, we used the fact that we corrected 2 times while proving the product rule without proving you have to correct 2 times.

[@David-km2ie]

u know what i mean?

[@MathTheBeautiful]

@@David-km2ie No, I'm not sure what you asking.

[@lawrencemwangi8846]

So in actual sense, if I hav a vector (V) =xi + yj..I get the derivative in terms of x( dv/dx)..the basis i and j also changes?...do we do di/dx and dj/dx...

[@MathTheBeautiful]

+Lawrence Muriuki If the coordinate system is Cartesian (or, more generally, affine), then no. For all other coordinate systems, yes.

[@anwbhr1168]

Have you proven it somewhere that the christoffel is symmetric? I know it must be available at other places..but it would be great hearing it from you. Thanks for these wonderful lectures. And your lectures on Linear algebra too. You are seriously a great teacher.

[@MathTheBeautiful]

I think it's an exercise in the book. In a Euclidean space, it follows from Γ^i_jk = Z^i ⋅ ∂²Z/∂Z^j∂Z^k

[@SashaPersonXYZ]

In which video does he discuss the generalized gradient?

[@SashaPersonXYZ]

+Sashamanxyz As in generalize the concept of 'direction of greatest increase' to arbitrary coordinates systems.

[@MathTheBeautiful]

+Sashamanxyz The fix takes place here: kzbin.info/www/bejne/inScaX6cnqx0hc0h7m21s

[@archishmore6276]

so sir in which case does the covariant derivative becomes 0

[@MathTheBeautiful]

There's much on that in the upcoming videos.

[@JoeHynes284]

holy shit! This video helped something click for me! Thank you!!!!!!!!!

[@mohammedal-haddad2652]

This is not mathematics, this is magic, or something even better.

[@MathTheBeautiful]

Agreed!

[@angelmendez-rivera351]

Justifying the Christoffel symbol is symmetric on its covariant indeces is rather trivial. The symbol can be defined as the contravariant basis Z^k dotted with the partial derivative of the covariant basis with respect to coordinates. The covariant basis, however, is itself the partial derivative of the position vector with respect to coordinates, meaning this term is a second-order derivative of an invariant. Partial derivatives commute, and this implies the symbol is symmetric on the indeces corresponding to the partial derivatives: namely, the covariant indeces.

[@PeterRoditis]

Thank you very much Professor. This exceptional course has to be very good as I am able to work through this course with only a high school background in mathematics. I just need a little help with question 133. Is there a video you were going to make, as there was someone else asking about it. Differentiating the dot product of the contracted indices of T^I. zi leads me to the correct answer but there are a lot of i's

[@mohammedal-haddad2652]

Where in this series it is proven that the christoffel symbol is symmetric with respect to the lower indices? It is so interesting and yet not at all obvious.

[@MathTheBeautiful]

Hi Mohammed, In the new book, see sections 12.1 and 12.3 here: grinfeld.org/books/An-Introduction-To-Tensor-Calculus/Chapter12.html Also see the classic equation (12.50) for the Christoffel symbol in terms of the metric tensor.

[@mohammedal-haddad2652]

@@MathTheBeautiful Thanks!

[@marxman1010]

So derivative of an invariant is a tensor, but derivative of a variant is not a tensor.

[@angelmendez-rivera351]

Radar123 Correct.

[@upscinterviewdristi3190]

You are best lacture

[@rkpetry]

[moved up to include a lot more]

[@realAhmedAbdElGhany]

the intro is scary i thought i got hacked

[@MathTheBeautiful]

Yes, that was not a good choice by me