Tensor Calculus Lecture 6c: The Covariant Derivative 2

[@xshortguy]

These videos are a treasure trove. I might actually be able to navigate my way through MTW.

[@MathTheBeautiful]

Thank you, it means a lot!

[@jojowasamanwho]

At the end of this discussion it is stated that the Riemann Christoffel tensor does not vanish in a curved non euclidean space. Doesn't that mean that the property of being able to commute covariant derivatives does not hold in such a space?

[@MathTheBeautiful]

Yes, it does.

[@AshrafElDroubi]

So both the position vector and a vector field are invariant of order zero. But we can only decompose the vector field into components wrt basis because we define the position vector before basis and should take it at its own terms?

[@MathTheBeautiful]

Yes!

[@kpmaynard]

Can't believe they missed the omg reference :)

[@rkpetry]

And, the reference that it would be edited out; (Both are still in). p.s. Did you ever notice that 'OMG' can be pronounced 'O-Mega'...?!

[@ashiqmulla147]

6c

[@Sigmath_Bits]

I'm so glad you didn't edit it out You had me laughing, haha xD

[@debendragurung3033]

In which part does he give the Fix on Laplacian. Couple of videos earlier he mentioned about fixing the Laplacian after covariant derivative a

[@rajendrarai8190]

You can prove easily since it is associated with differentiation of covariant basis vector with respect to contravariant elements.

[@Abinasty]

Hi there, you talk about the covariant derivative, but like all the other objects, is there such a thing as the contravariant derivative?

[@orientaldagger6920]

What does Google say?

[@MathTheBeautiful]

Yes, absolutely. It's simply the covariant derivative with a raised index.

[@xaviervangorp4862]

At 31:00 the question of if they commute or not why can't we just say they are dummy indices and swap i and j freely?

[@MathTheBeautiful]

They are live indices, actually!

[@samdietterich2660]

I like his OMG joke.

[@Shad0WmurdeRER]

cringe

[@archishmore6276]

at 22:48 the covariant derivative 1st how does RHS differs from LHS first term

[@MathTheBeautiful]

In the last line? Covariant derivative vs. partial derivative.

[@kapilk1644]

How can the covariant derivative be interpreted geometrically in general? The covariant derivative of the components of a vector gives you the components of the partial derivative of the vector, which I can understand, and the covariant derivative of an invariant is the familiar partial derivative. However, how can I intuitively understand what the covariant derivative of a vector field or a higher order tensor field is?

[@kapilk1644]

What I mean to say is: if you can define a partial derivative geometrically as a rate of change (covariant basis is defined geometrically, for example), is there an similar geometric definition you can make for the covariant derivative?

[@MathTheBeautiful]

Kapil K I'm going to answer this question with a general thought. The way Geometry and 1. Linear Algebra - for things that are straight - and 2. Tensor Calculus - for things that are curved - work together is this: Geometry provides the inspiration, a starting point, and often points in the right direction. Then, the algebraic disciplines take over and advance the Geometric metaphors. Two examples: While R^3 is "equivalent" to the physical space, R^4 is entirely a figment of our imagination which has geometric *analogies* but not *equivalents*. The covariant derivative is another example. Geometry gives us invariants and partial derivatives. Tensors and the covariant derivative are algebraic inventions that extend the geometric ideas and properly connect to them in important special cases. So I, for one, make a point of not asking myself "what's the geometric meaning of a cov. derivative?" I understand the geometric origin and the resulting properties, but its strength is that, while inspired by geometry, it has a life of its own detached from geometry. This perspective is probably not shared by the modern geometric community.

[@kapilk1644]

MathTheBeautiful Out of curiosity, what perspective does the modern geometric community have?

[@owenpeter3]

Has he given out a swatch of lecture notes?

[@archishmore6276]

you teaches the things in very good ways but next year iam going for further studies where I may not be able to use internet .. so can you advice me a book on tensor calculus which which would contain concepts in simplified way you teaches

[@bulbasa9r753]

I haven't bought the book yet, but the lecturer has written a textbook on the subject - there's a link to purchase it in the description.

[@Yeeren]

I have access to the textbook through the library where I study, and it's pretty accessible/ There's even a link here in the video to solutions for the exercises that appear in the book..

[@mongrav1000]

At this point, had your students had any exposure (courses? lectures?) to non-Euclidean geometry in their math program?

[@MathTheBeautiful]

No, I don't believe so.

[@acetabularia326]

At 41.30 So is your father also a mathematician?

[@MathTheBeautiful]

Yes!

[@marxman1010]

Curved space has non-zero Riemann-Christoffel symbol. Astonishing concept.

[@chymoney1]

Cause the derivative of the basis when applying the product rule is 0

[@rkpetry]

1. If one 'ignorantly' includes the Christoffel Γ symbols then these should evaluate ≡ 0, in the case of invariants, having no indices. 2. Something we should review once, is, why the denominator has only contravariants ∂/∂Ƶᵏ. 3. Funny question: As the Christoffel symbol is Greek-gamma Γ, shouldn't the Riemann-Christoffel tensor symbol be Greek-rho Ρ (looking like English-P)?!

[@sepehrs6098]

Answer for 2: that denominator is not a contravariant. It's just a notation indicating all the bases, similar to (i, j, k).

[@lowersaxon]

Did you get from this video alone what “contraction” is?

[@MathTheBeautiful]

Contraction came in an earlier video. Contraction is another word for the kind of summation you encounter in a matrix product.

[@Random-sm5gi]

TENSOR OMGGGG!!!

[@MathTheBeautiful]

Yup, it's that simple!

[@mauriziobaccara]

Riemann-Christoffel tensor vanishes in cartesian or affine coordinate systemthere fore it vanishes in ALL coordinate systems ! NO! Not in spherical coordinate system (curved space).The same logic applies also to Jacobian. They transform a tensor from a coordinate system to ANY another ? NO ! For example not from cartesian to spherical i.e not from flat space to curved space !?!

[@MathTheBeautiful]

You're correct in general! However, this video is about a tensor description of a *Euclidean* space.

[@metalfoot]

81 numbers 🤢