Mozibur Rahman Ullah
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Nice phrase - I'll have to use it somewhere :-).jtbell said:Classic case of "topic drift."![]()
Nice phrase - I'll have to use it somewhere :-).jtbell said:Classic case of "topic drift."![]()
oh well, it happens all the time :-)Mozibur Rahman Ullah said:Ah, I thought you were a 13 old year boy! A gender assumption on my part!
yes, you certainly do. I think you explained it very, very well.Mozibur Rahman Ullah said:'m glad that what I wrote was taken in by you and understood. Richard Feynman once said that you only understand a concept if you can explain it to a student - so I guess I can say I understand the exterior covariant derivative! ;-).
Ah, so it's not nearly similar to what I thought it was. This makes sense though (intuition aside - I've given up on intuitive for this). I won't ask why a 3D tensor is a parallelepiped because despite my curiosity, the proof might be a bit too much for me, but thanks, this was so, so much better than google :)Mozibur Rahman Ullah said:The second way of thinking about it is that the arrow itself becomes higher dimensional. This is the tensor product. Thus a 3d arrow is the tensor product of 3 vectors u⊗v⊗w. Whereas an ordinary vector has only one way to add, a 3d arrow has three ways to add. That's not so scary when you see it visually - however I haven't figured out how to insert diagrams here yet. A 2d arrow is u⊗v and is visually represented by a parallelogram in the obvious way. However, there is a proviso - you can internally rescale it without changing the tesor. By internal rescaling I mean you can rescale one side, say u, by a factor of a whilst rescaling the other side, v by 1/a. This is because the magnitude of a tensor is its area. For a 3d tensor, which will be a parallelepipid, its magnitude will be its volume. And so on. Another thing to note is that in this geometric picture you can only define tensors upto the dimension of the ambient space. But in fact, it can be defined for arbitrary dimension. This is best shown mathematically.
I see. It's all quite standard then, not like the vector product. I wonder how this is computed with actual values.Mozibur Rahman Ullah said:the multiplication of manifolds follows the Cartesian product of sets. Have you come across this? Basically, a manifold is a set with a smooth structure. Here, the smooth structure is the set of all charts on the manifold. When you multiply two manifolds M, N together you first multiply the two sets together to get M×N and then the smooth structures on M and N determines a smooth structure on M×N. We still use the symbol for the Cartesian product × for the product of manifolds.
Yes, that makes it a bit more intuitive (no proof though... I guess I'll leave that for later)Mozibur Rahman Ullah said:I think you also asked why T(M×N)=TM×TN. The best thing would be to try a few low dimensional cases to see why its true. For example, the plane which is R2=R×R, the torus T2=S1×S1. Recall here that Sn is the n-dimensional sphere, and so S1 is just the circle. Whilst T2 is the torus which is 2d. And the infinite cylinder Cyl2=R×S1.
Haha - thank you, you flatter me. I do take interest in them, I can't help myself. I think it would be even more impressive if I'd be able to limit myself, and understand that I should look at the basics first, rather than just relying on my curiosity to tell me where to go. After all, I know next to nothing about physics. That's why I came here, and I definitely got what I wanted. A set path I can follow. Physics is truly beautiful, I find it hard to not want to learn as much as I can about it.Mozibur Rahman Ullah said:You're welcome. I think it's fantastic that a thirteen year old is taking so much interest in advanced ideas. You've done remarkably well in taking it in. Well done! If you have any further questions on the material, you're welcome to ask them.
Hint taken - though I do agree that this is a case of topic drift... it's a lost cause, sorry! :Djtbell said:Hey y'all... we do have a differential geometry forum here... hint hint.![]()
Schaum's textbooks, just classic.TensorCalculus said:Hello! So I'm a teenage physics enthusiast, who wants to take my knowledge past A-level (or in America I believe this would be high school level) physics.
I've studied multiple textbooks like Young and Freedman's University physics, studied maths from books like mathematical methods for physics and engineering.
I solidified all that by doing lots (like, LOTS) of the practice problems and some Olympiad papers.
I don't really know where to go from here.
I've resorted to surfing the internet, and finding free courses and watching YouTube videos that satisfy my interest, or reading popular science, reading high-school textbooks, and just sitting around.
I'm not entirely sure what books to buy - I fear that if I accidentally skip straight to something too advanced I'll not have strong basics. My teacher told me to just read popular science, but I really enjoy looking at the math behind things and I feel that there are few popular science books (that I've read at least) that satisfy my curiosity.
I do really, really love physics, and want to try and study it further - unfortunately, I'm 13 and have quite a long time till I can study physics in University/College.
Does anyone have any resources they would recommend to me (my interest particularly lies around astronomy and electromagnetism) , or any advice to give?