Tensor Calculus 14: Gradient explanation + examples 

I often think my monotone is a weakness in my presentation. I've never had a very expressive voice.

eigenchris I’ve been looking into the metric tensor for a while and you’re the only person who explains it well enough for a person who has only taken linear algebra to understand. The dot product of the basis vectors is highly understandable but I haven’t heard anyone else say the metric is just the dot product if the basis.

That's a shame that you haven't seen the easy description of the metric tensor. If you want to learn more, you can check out my Tensor Calc #12 video.

@@eigenchris I have never heard such a profoundly expressive monotone.

@@eigenchris Perhaps in “real life” it is, but at least here on youtube, your voice is very soothing and a little bit sexy ;) By the way: I love your videos!! 💙

The coordinate transformation given by the equations at the top of the screen are not a transformation of basis vectors; rather, they are a transformation of coordinate components. (A hint for this is that the variables do not have the distinctive vector arrow over them) This happens to be contravariant with the basis vector transformation; the basis vectors for d are 1/2 the length of the basis vectors for c. Hope that helps.

@@IronLotus15 I had the same question and your explanation helped me see, I think you are right. I mean, look at the drawn length of the basis vector for a hint, and also follow the video from that point, it makes it more clear. Thanks.

@@nahblue I'm glad I could help you understand!

also, shouldnt it be a single column? if you write it in array form, its a column of columns. it should be column(1 0 0 0)..etc

@@LearnPhysics1231 been a while since I watched this. If you're in 4 dimensions then that's true.

That formula is the definition of the dc^j. We force there to be no cross terms by definition.

just put more emphasis on "sort of kind of like ", and I guess it works out.

I am going to talk all about Christoffel symbols over the next 5-6 videos. The main place they show up is in the covariant derivative. You are correct that we see a set of Christoffel symbols for each covariant/contravariant index. Since scalat fields have are 0-order tensors, they do not show up.

The procedures I use in these videos should work for any coordinate system. Regardless of the coordinate system, the basis vectors along the "q" coordinate are always defined as "∂/∂q". We don't do the extra steps of "normalizing" ∂/∂q to have unit length everywhere, as this would require more work. Instead, we just allow ∂/∂q to be whatever length it needs to be. The θ coordinate follows this logic as well. We just let the θ basis vector be ∂/∂θ, and allow it to be whatever length it needs to be. This "non-standard" in a 1st or 2nd year physics class, but completely standard in tensor calculus classes.

Yes, the definition of the epsilon covectors is derived from the definition of the "e" basis vectors. This means that every choice of "e" basis (cartesian, polar, etc) will get its own set of corresponding epsilon covectors (called the "dual basis"). I cover this in videos 4-6 in this series. When you mention that the vectors are "orthogonal" to the basis vectors... it's kinda hard to really say what that would mean. Normally, covectors are visualized as stacks of hyperplanes. Covector E1 will be in the plane formed by all basis vectors EXCEPT e1 (take all of e2,e3,e4,... depending on the dimension of the space). If the basis vectors are at strange angles, this doesn't give you the traditional "orthogonal" picture you might expect. You might want to look at these two images for clarity: imgur.com/a/6h02lQp

​@@eigenchris This is called a biorthogonal basis pair in mathematics (en.wikipedia.org/wiki/Biorthogonal_system), where the first basis is from the primal space (V) and the second from the dual (V*). It is typically introduced along with the definition of Schauder basis. There is a theorem that states that if: V is isomorphic to V**, {e_i} is a basis for V, {\tilde{e}^j} are elements from V*, and {e_i},{\tilde{e}^j} form a biorthogonal system, then {\tilde{e}^j} is a basis for V*. An orthonormal basis in a Hilbert space provides such a dual basis through Riesz's representation theorem.

I'm a little unsure how to answer this because your description seems to be the definition of a covector, so I can't really give an answer to "why". If a function that maps vectors to scalars is linear, it will behave as you described.

I just meant that del f is a vector field. In this case it's a constant vector field, so that same vector exists at every point in space.

You might want to watch the previous video (#13). The 2 formulas that appear at the beginning are derived in that video.

@@eigenchris I did rewatch it and now I remember. It is even written on my notebook. Thanks.

The Kronecker delta with 2 lower indices is twice-covariant and made of the tensor product of 2 covectors (eps1 x eps1 + eps2 x eps2). The Kronecker delta with 2 upper indices is twice-cotravariant and made of the tensor product of 2 vectors (e1 x e1 + e2 x e2). The Kronecker delta with mixed indices is 1-covariant, 1-contravariant and made of the tensor product of a vector and a covector.

@@eigenchris A beautiful explanation, thank you so very much! May I ask one more question; I see in your about section you recommend a few links (thank you for that, though the exterior calc link is dead currently) and you're most likely asked this over and over so I do apologize, but have you found any particular textbook that explains things as elegantly as you do? Your explanations are intuitive, interesting, and easy. I have yet to find a textbook that begins from where an undergradate DiffGeo course leaves off (for my course we talked about 1st and 2nd FF but never covered manifolds, tensors, differentials, etc so i assume it was just in euclidean space.) Thank you again for your help, amazing explanation as always :)

I updated my channel description with archived versions of those notes. The main page seems to have gone offline. I can't recommend any particular textbook for Tensors. One of the reasons I made these videos was because I found most of the notes online very difficult to understand. I read some of Gravitation by Misner Thorne and Wheeler to learn relativity but it is not the easiest.

@@eigenchris Yea, I've stopped at the E&M chapter and took a break to learn the math behind it and why that darn bell bongs so much haha. I appreciate your help very much, thanks again :)

Do you have a link to where this is? As I said, there is disagreement as to what these words and operators mean. Sometimes people use the same symbol to tall about different things.

Here is the link: physics.stackexchange.com/questions/126740/gradient-is-covariant-or-contravariant

I think it is a disagreement in symbols too. I h have however been unable ti transform the gradient between basis corrrectly; however i think that is due to my bad algebra, your most likely right.

The answer that I would most agree with is the top-rated one by ZachMcDargh. They agree with the definition of the gradient that I use in this video. Other answers use the term "gradient" to mean what I call the "differential". And yes, the components of the differential are covariant. As I mentioned in the video, there is disagreement about terminology. If you have a specific problem, you can try posting a question here, or on a stack exchange website, and give me the link. I can try to help you out with it.

the components partial f over partial x sub i won't transform correctly, I have to us the Jacobian instead of its inverse, and when i do the inverse metric tensor components ( which I have as dx sub i dot dx sub j) won't transform unless I us two inverse Jacobians, making so that del f needs 2 contravariant and one covariant jacobian transforms to work. I must have forgotten to cancel something or I did something else wrong.

I explain this in the tensor calculus 2 video. It's because taking the derivative of a vecotr R in the direction of the c^j coordinate gives us a tangent vector along the c^j coordinate curve, which is the same thing as the basis vector e_j.

Probably in about a week or so.

eigenchris i'm looking forward to it

In the x,y coordinate system, you can expand df = df/dx dx + df/fy dy.