Tensor Calculus Episode 10 | Is the Affine Connection a Tensor?

[@nab000]

doing a math revision (im 11th grade tho) for a test tomorrow and i have no idea what youre talking about but your voice is really nice and calming :)

[@aarohgokhale8832]

What level are you doing? Cuz I'm pretty sure this comes after multivariable calculus

[@om5621]

@@aarohgokhale8832 Well, he did say he didn't understand what he was talking about...

[@kummer45]

This man gets all the details on the board. HE IS A TEACHER and his explanations are transparent. I have 4 years studying tensor calculus. This man knows what he's doing.

[@AndrewDotsonvideos]

kummer45 :)

[@OayxYT]

I wonder if I can put on my collage application paper, “Subscribed to Andrew Dotson.”

[@AndrewDotsonvideos]

Oayx *harvard wants to know your location*

[@axemenace6637]

For starters, you can spell "college" right.

[@OayxYT]

Maxim Enis I actually misspelled that... thanks for pointing that out.

[@aravindbharathi7801]

Binge-watching the Tensor Calc playlist ahead of my General Relativity finals and the thing that most stuck is that Andrew is super self-conscious about his \mu s

[@frozenmoon998]

You know, the intro gets us straight to the point, even if it is loud - keep it up, A.D.

[@felixbilodeau-chagnon4781]

1:35 "Is it a tensor, or is it what people in the industry call "NOT a tensor"" hahahaha I died

[@silentinferno2382]

I saw the thumb and for a millisecond thought you chopped your hair and decided to go bald

[@AndrewDotsonvideos]

Lol

[@ramajvalia8201]

Clicked link immediately

[@edmund3504]

i don't know what any of this stuff means but it sure looks cool as hell

[@Bank0h]

Took my final A-Level Physics Exam yesterday (UK), and probably wont do much Physics for a while. I don't understand a lot of what goes up on this channel but I enjoy the videos nonetheless, so thanks for the professor memes ( I get those). I definitely have no clue about anything you say in this video but I figured you'd only read recent comments so here it is :).

[@AndrewDotsonvideos]

Thanks a lot! Hope your final went well

[@Bank0h]

@@AndrewDotsonvideos I'll keep watching the vids of course, -but maybe I'll skip the ones on tensor calculus-

[@birupakhyaroychowdhury974]

Excellent explanation...

[@AndrewDotsonvideos]

Thanks!

[@birupakhyaroychowdhury974]

@@AndrewDotsonvideos most welcome...

[@sagnikbhattacharjee3311]

Please can you make a video on Riemann Tensors, Geodesics and curvature related stuff!

[@Braxtoned]

My boy back at it with the T E N S O R S :)

[@joshuapasa4229]

Will you do videos on GR for lvl 100 tensor bois

[@renormalization]

Will you be going over curvature and manifolds after chapter 4 in the textbook?

[@paulboard8221]

that thumbnail looks like a threat

[@federicopagano6590]

I missed your videos buddy im back 2021 😂

[@JaxzanProditor]

Take a shot every time Andrew introduced a new Greek letter index

[@eriklong3594]

Great video man! I saw your channel in the NMSU quantum times recently. Hilarious thing is I am now doing my graduate work at George Mason. Looks like you and I are geographically confused individuals, lol. I can honestly say though that I do miss NMSU and their Physics department. They have some great professors for sure. Any how, good luck with the rest of your studies.

[@AndrewDotsonvideos]

Small world! I actually applied to GMU. I had a skype interview with some really nice professors there from the space-physics (or equivalent) dept.

[@eriklong3594]

@@AndrewDotsonvideos Yea, I recalled that from one of your other videos which I thought was funny. I should say I am impressed with the GMU faculty as well. But NMSU was my first taste of physics so maybe that is why I am partial.

[@mr.sharma125]

The GOAT uploaded

[@sage4670]

@blu berry Greatest Of All Time?

[@49fa75]

@blu berry lol wut

[@49fa75]

@blu berry kek

[@49fa75]

@blu berry f

[@49fa75]

@blu berry nо вы

[@rodrigoaguiar988]

6:40 really surprised me

[@dagisinmines3412]

I've tried to learn this very thing before. I would have said that it is impossible. How come I understood now?? Ty

[@AndrewDotsonvideos]

Dagis Dant :)

[@RockBrentwood]

There's a much more direct way and more satisfying way to address the issue. Tensors are special cases of what are called "natural objects" - which essentially are objects that have well-defined transforms under the action of an infinitesimal coordinate transform x⁰ → x⁰ + Δx⁰, x¹ → x¹ + Δx¹, etc. (I will use x^0 for superscripts x⁰ and x_0 for subscripts x₀). The most fundamental natural objects are the coordinate differential dx^μ and the differential operator ∂_μ ≡ ∂/∂x^μ. All tensors are formed from these as linear combinations with coefficient functions. A covariant rank 1 tensor is an object of the form ω = ω₀ dx⁰ + ω₁ dx¹ + ω₂ dx² + ⋯ = ω_μ dx^μ (using the *summation* *convention* here and below), while a contravariant rank 1 tensor has the form X = X⁰ ∂₀ + X¹ ∂₁ + X² ∂₂ + ⋯ = X^μ ∂_μ. Tensors A ≡ A_μν dx^μ ⊗ dx^ν, B ≡ B_μ^ν dx^μ ⊗ ∂_ν, C ≡ C^μν ∂_μ ⊗ ∂_ν of higher ranks (2,0), (1,1), (0,2) respectively are built up using the tensor product operator ⊗; similarly for tensors of higher ranks still (3,0), (2,1), (1,2), (0,3), etc. And, of course, the coordinate functions φ = φ(x⁰,x¹,x²,⋯) fall into this classification as tensors of rank (0,0). Their transforms are determined by the properties that (a) scalars transform in the expected way via the partial derivative operators Δφ = Δx^μ ∂_μ φ, (b) Δd = dΔ (thus Δdx^μ = d(Δx^μ) = dx^ν ∂_ν(Δx^μ), (c) the Kronecker tensor δ ≡ dx^μ ⊗ ∂_μ is invariant (from which you can derive the transform Δ(∂_μ)), (d) ordinary products and tensor products transform by the product rule, from which the transforms for higher-order tensors are derived. Thus, for instance, the μν component (ΔA)_μν of ΔA works out to be (ΔA)_μν = Δx^ρ ∂_ρ(A_μν) + A_ρν ∂_μ Δx^ρ + A_μρ ∂_ν Δx^ρ. The transform is known in the literature as the Lie Derivative. A natural object is an object that has a Lie Derivative. Tensors do not exhaust the full range of natural objects. The affine connection is also a natural object that can be expressed as the 2nd order differential operator: ∂² ≡ dx^μ ⊗ dx^ν ⊗ (∂_μ ∂_ν - Γ_μν^ρ ∂_ρ). Its transform properties are completely determined by this fact. The component (ΔΓ)_μν^ρ of the transform for ΔΓ is the expression you would expect from a rank (2,1) tensor *plus* an additional term ∂_μ ∂_ν Δx^ρ that arises from the second order differential operator ∂_μ ∂_ν = ∂²/∂x^μ∂x^ν. The connection is called *Affine* because its transform isn't just linear in Γ, but includes that extra term. Technically functions of the form f(x) = mx + b are *not* called linear, but affine, while the term "linear" is reserved for the case b = 0, to functions of the form f(x) = mx. So, the connection would be linear if it transformed as a tensor (linear in Γ) without that extra term.

[@RockBrentwood]

Natural bundles -- which is what natural objects reside in - are discussed in more depth here (and in many other places on the web). ncatlab.org/nlab/show/natural+bundle The term "geometric object" is also used for such objects ... like here www.researchgate.net/publication/202940003_Geometric_objects_on_natural_bundles_and_supermanifolds

[@ProLeopardx1]

I wish I knew this stuff but at the same time I'm glad I don't. I'll stick with EE for now I think 😂

[@hassanb9339]

Thanks for the amazing videos, really helping me understand tensors much better. I just had a quick question. Near the start of the video when you have to carry out the chain rule you state that X^alpha is a function of the x^sigma coordinates and that x^'mu is a function of the x^row coordinates. However, when you went to apply the chain rule for the first term which included X^alpha you used x^row instead of x^sigma and then in the second term you use x^sigma instead. Is this not the wrong way around, or am I misunderstanding how the chain rule works here.

[@Drewbie_T]

hmm. before we start switching unprimed for prime, doesn't term 2 give us the primed affine connection? are we transforming because that would result in us just subtracting from the other side and finding term 1 = 0 or am i missing something

[@saadaqcabdi5865]

@Andrew Dotson bro you said x alpha is a function of sigma and you did the chain rule for x alpha as the function of the density symbol rhu why?

[@dent20111]

Andrew would you say doing all this on a white board and basically teaching the audience as you go along helps you retain and solidify all this information in the memory bank?

[@angelmendez-rivera351]

dent20111 As someone who tutors students and teaches other enthusiasts of my years how to derive stuff, I can say that the answer is yes. I am most certain Andrew feels the same. It's universal feeling tutors and willing teachers tend to have.

[@monisahmedrizvi1697]

Mean while ME:why is Calculus so hard

[@luisgonzalez-qw7bp]

Which one?

[@PhysicsBro-xb8qx]

Cool!

[@fleskimiso]

Bruh I am high school but I guess this is not something I can expect on my exam tomorrow.

[@Styleinmyveins]

did he just say "Chain rule me Andrew Sama"?! XD

[@pancreasman6920]

I think you lost the primes when you rewrote the transformations (5:42)

[@skandys9847]

Ever plan on doing videos on deriving stuff? Like SR or maxwell’s demonic laws?

[@AndrewDotsonvideos]

Skandy S honestly I forgot that I made 1 video out of a two part series on deriving the Lorentz transform until now.

[@PHILLYMEDIC69]

Im gonna ask you to calm down sir

[@nitrozox212]

Me : There are two types of quantities - scalar and vector Andrew : Triggered

[@AndrewDotsonvideos]

Nitro Zox wut bout spinors

[@angelmendez-rivera351]

What about gyrovectors?

[@csalazar797]

Lmao I watch these videos like I didn’t take the algebra physics series 😂😂😂

[@zoltankurti]

But velocity dx^i/dt is always a vector. In the previous video you assumed that x'^i = dx'^i/dx^j x^j which is not correct.

[@AndrewDotsonvideos]

Zoltán Kürti I said the position vector transformed as a vector. That doesn’t mean it’s derivative does. I didn’t assume, I said let’s say we have a vector that transforms this way. Then take the derivative and see if the result manifestly transforms as a tensor.

[@AndrewDotsonvideos]

Those vectors may not live in the same space, so there’s no reason to think the derivative should still preserve the original transformation properties.

[@zoltankurti]

@@AndrewDotsonvideos so in this series position is a vector? If yes, you will never have gammas that are non zero, because you only have linear transformations as coordinate changes. If you want to use polar coordinates and such, you must give up position being a vector. And really, position is not a vector. You can only use that formalism in flat spaces, and I guess a tensor calculus series will deal with nontrivial spaces.

[@zoltankurti]

@@AndrewDotsonvideos x' coordinates are arbitrary continuous and invertible functions of the x coordinates. Still don't get why you wrote that transformation rule.

[@angelmendez-rivera351]

Zoltán Kürti The position vector at one point in time is an invariant. The coordinates transform inversely to how the basis transforms. Therefore, it is incorrect to state that if the coordinate space has polar coordinates, then the vector is not a vector. By definition, whether an object is a vector or not is not dependent on the coordinate space its components are given by. The component array is a contravariant rank 1 tensor, and that is not dependent on the coordinate space. *And really, position is not a vector.* It is. By definition. *You can only use that formalism in flat spaces* No, that is completely false. General relativity can handle position in curved spacetime perfectly fine, and still remains a vector. *x' coordinates are arbitrary continuous and invertible functions of the x coordinates.* So? Nobody cares. It is irrelevant. It has absolutely nothing to do with whether the derivative is a tensor or not. The definition of a tensor is not related to continuity or invertibility. *Still don't get that transformation rule.* That just means you do not understand what a tensor is. That rule is the very definition of a rank 1 contravariant tensor. *But velocity dx^i/dt is always a vector.* No. This is incorrect. Velocity IS a vector. dx^i/dt is not. V = V^i·e_i = d/dt[x^i·e_i]. It can be proven very easily that V^i is not equal to dx^i/dt. By the product rule of derivatives, d/dt[x^i·e_i] = dx^i/dt·e_i + x^i·de_i/dt. x^i·de_i/dt = x^j·de_j/dt, since complete tensor contractions are rank 0, and as such, index-independent. de_j/dt := f^i_j·e_i. Therefore, V^i = dx^i/dt + x^j·f^i_j. Unless f^i_j = 0 for all i or for all j, V^i is not equal to dx^i/dt. One does not even need to consider curvilinear coordinates for this to be the case, although this is completely irrelevant, since curvilinear coordinates are orthonormal, and any orthonormal basis is valid. *In the previous video you assumed that x'^i = dx'^i/dx^j·x^j which is not correct.* It IS correct, by definition. dx'^i/dx^j is the Jacobian matrix, and the Jacobian matrix is defined by this equation. Any two coordinate system are related by the Jacobian matrix, which always has a well-defined matrix inverse. The problem is that you failed to understand that V^i, which is a tensor, is not equal to dx^i/dt, which is not a tensor. TL;DR: Andrew is 100% correct here.

[@HilbertXVI]

LIE DERIVATIVES PLEASE

[@theflaggeddragon9472]

Did you try to use tensor calculus to fix your mic? ;) Might not work on potatoes. JK love you Andrew

[@AndrewDotsonvideos]

The Flagged Dragon yeah I was using my toaster as a mic

[@bigdave6952]

which situation requires chain rule vs which one does not is not all that clear.

[@Mforader1792]

Towlie sorry my phones broken see prior video comment. Lol...my bad.

[@arsenymun2028]

how happy i am that i as a mathematician don't have to deal with this shit

[@monjurulhasannabil1988]

Bro couldn't find his mic, so he took a 1-year break :)

[@angelmedina6384]

[@geoffrygifari4179]

wait.... is that christoffel symbols?

[@AndrewDotsonvideos]

;) Kind of but this is maybe more general. We start calling things Christoffel symbols once we've agreed upon a metric and when we specify the connection we're using is the levi-civita connection.

[@Hon2fun]

You have a huge high school audience do you know why that is? I can't relate to any of these comments and its disappointing because i want to nerd out too.

[@AndrewDotsonvideos]

Probably because I still think like a 12 year old

[@danielcann5554]

Andrew I drowned in all the indices please help

[@AndrewDotsonvideos]

Daniel Cann dude same

[@WhatsAFuzo]

i am now lvl 10 tesor boi.

[@Ryan_Perrin]

You're affine connection ;)

[@AndrewDotsonvideos]

Ryan Perrin ;)

[@thephysicistcuber175]

TL, DW: hell no! Wtf are you even thinking?

[@knight3481]

Answer is No

[@vijaypanchalr3]

What's your Instagram account ??

[@mareksajner8567]

first?

[@mareksajner8567]

:D

[@nathanielweidman8296]

"You're a tau..."

[@ClumpypooCP]

Uhh ok then

[@iWrInstincts]

psh this is elementary. i learned this in 1st grade smh my head

[@silentinferno2382]

I learnt this in first grade shaking my head my head

[@chrisallen9509]

Antimonium Heptadiene r/whoosh

[@iWrInstincts]

@@silentinferno2382 yes i learned this in 1st grade smh my ugh

[@silentinferno2382]

@@chrisallen9509 you're whooshing me? Shaking my head my head.

[@angelmendez-rivera351]

Chris Allen r/wooosh

[@good_boy_13]

bruh

[@archiebenn5707]

bruh

[@silentinferno2382]

bruh

[@AndrewDotsonvideos]

bruh

[@angelmendez-rivera351]

bruh

[@salimdeaibes]

Sorry, math tensor calculus is so much more elegant :p

[@angelmendez-rivera351]

Salim Deaibes Usefulness >> Elegance.

[@drlangattx3dotnet]

Hi Your body sometimes blocks our view of what you are writing.