Question

Why is the study of differential forms coordinate independent if flux and other quantities are represented coordinate-wise? For example the fluz of vector field F ⃗ is represented as F_x dy∧dz+F_y dz∧dx+F_z dx∧dy which clearly doesn't work on coordinates other than cartesian without doing a pullback.

Answer 1

Alright @ivancsc1996 what class/books are you studying with? I'd like to catch up to you and study with you.

Answer 2

This is just one of those questions that stuck with me from my personal studies but I never really bothered with it since I haven't need any differential forms in my coursework (I am majoring in physics). Everything I know from differential form comes from Theodore Shiffrin's Math 3510 (you can easily find it on youtube). All the differential geometry and tensor calculus I've learned comes from Pavel Grinfeld's videos in his youtube channel MathTheBeautiful and his book Tensor Calculus and the Calculus of Moving Surfaces, and Paul Dirac's book The General Theory of Relativity (which is an absolutely amazing book). I learned the essentials (differential, integral and vector calculus) from MIT's Opencourseware (1801 and 1802). My classical mechanics, where I really got the hand of the basic calculus subjects and differential equations, comes from Saveliev's Course on General Physics, Goldstein's Classical Mechanics and Savchenko's problems in physics (the Russians really know their mechanics). The Quantum Mechanics from MIT's 8.04 and Dirac's Principles of Quantum Mechanics (another really good book). I haven't really bothered a lot with linear algebra, apart from what I've learned on the go and from a few of the lectures in Theodore's Math 3500.

Answer 3

Awesome, I have actually watched a handful of Pavel Grinfeld's videos this summer and was planning on buying the book but I got distracted last semester. I'm a chemistry major with a math minor (well, I just graduated) and I'm considering doing computational chemistry for graduate school. I'm just trying to get a good handle on quantum mechanics, clifford algebra, and I'm trying to sneak in a little of other stuff in on the side that I find fun. I think I'm going to go ahead and purchase his book and get to studying right now, I think I've avoided tensors for too long and Pavel has quite a lucid view of tensors which is rare of most subjects at around this level. I'll be sure to read Dirac's book as well, he's pretty much the man when it comes to relativistic quantum mechanics.

Answer 4

Pavel's approach is great since he defines mathematical objects in both a extrinsic and intrinsic manner. Instrinsic approach is necessary since you ultimately want to build a theory on the surface disregarding ambient space but his extrinsic approach is great for building intuition. I hope you do really good on your studies. I think Pavel can answer pretty much any question (he is always up to date with the comments on his videos)!!!!! Hope we stay in touch.

Answer 5

Thanks I appreciate it, especially the tip of him keeping up with comments. I might be able to discuss some things, such as the question being asked here, but since I don't know a bit of the technical wording (what's a pullback? Something with differential forms apparently). But maybe in explaining things to me it helps you realize why.

Flux is independent of your coordinate system as far as I can tell, for instance the amount of fluid flowing through a directed area element should be the same no matter how you decide to look at it, or even a contained charge whether it's inside a sphere or cube should make no difference to the net flux through it. I'm probably not answering your question rigorously enough for what you're looking for though. I know for a while I couldn't believe that Laplace's equation was really coordinate independent, which seems sort of related to your problem as well.

Answer 6

My problem falls back on this. The flux of a vector field is defined as\[Flux = \int\limits_{S}^{}F^{\rightarrow}.dS^{\rightarrow}\]Which is clearly invariant since it is written on geometric quantities (I don't know how to write vectors or dot products). The problem comes when trying to evaluate the integral. In most non-trivial problems this integral has to be transformed into a "arithmetic" integral with the choice of a coordinate system. The whole beauty of tensor calculus is to be able to do this kind of things without actually having to choose a coordinate system. The porblem is that I haven't found much on integration in this regime. For example I guess the circulation of a vector field in tensor notation would be\[\int\limits_{c}^{}F^{\rightarrow} . dr^{\rightarrow}=\int\limits_{c} g_{i j}F^{i}dx^{j}\]but this really doesn't make much sense. In differential forms this problem is solved by introducing the pull back. The integrand becomes \(F^{\rightarrow}.dr^{\rightarrow}= \sigma\) with \(\sigma \in (\mathbb{R}^{n})^*\) which just means sigma is a one form. Then the trajectory c is parametrized as \(g^{\rightarrow}=(x(t),y(t),z(t))\) for some interval \([t_i,t_j]\). Finally the integral is calculated as \[\int\limits_{c} \sigma = \int\limits_{t_i}^{t_j} g^{\rightarrow *}(\sigma)\]The quantity \(g^{\rightarrow *}(\sigma)\) means the pullback of sigma which is basically to plug in \(x \rightarrow x(t)\) and the others to make sigma a function of time. Apart from the fact that in the parametrization a Cartesian coordinate system was already used, when defining the sigma you do it as \(\sigma = F_x dx + F_y dy + F_z dz\) which still resorts to Cartesian coordinates. I don't see anything invariant in it.

Answer 7

Now that I think about it, I just solved my own problem. In tensor notation I said the circulation should be \[\int\limits_{c} F^{\rightarrow}.dr^{\rightarrow} = \int\limits_{c} g_{i j}F^{i}dx^j\]But this is the same as \[\int\limits_{c} F^{\rightarrow}.dr^{\rightarrow} = \int\limits_{c}F^{i}dx_j\]Which is great because \(dx_j\) is covariant which is nothing more than a linear function on vectors which is exactly the same as a one form. Then you just have to make the parameterization in \(x_j\) coordinates and do the pull back. Now I have to think if this would work in curved space being \(x_j\) the coordiates of the manifold and how to generalize this for flux as well and other stuff like that. Thanks dude!!!!

Answer 8

Sorry I made a fluke on the second integral, the subscript on dx should be and i not a j

Answer 9

Differential forms and tensor analysis don't necessarily mean that you can evaluate an integral without choosing a coordinate system. Their utility lies in the fact that relationships expressed in those ways are manifestly independent of the coordinate system that you choose - i.e. that you can choose whichever coordinates you want, and in particular you may choose particularly simple or convenient ones. You still may have to pick them, though, in order to get a quantitative answer.

Answer 10

I'm in chapter 2 of Grinfeld's book making sure I am in complete understanding of what he says every step of the way. I'll probably be asking questions about it on here if you'd be interested in a little practice answering my questions too haha.

Answer 11

@Jemurray3 yes, you are right. The beauty is that you win most of the advantages of coordinate systems without actually having to choose one and therefore retaining geometric info at the cost of still not being able to do concrete calculations. Plus, its just nice to have formulas that will work everywhere :)