| Thread Closed |
Differential forms. Why? |
Share Thread | Thread Tools |
| Oct28-03, 08:35 AM | #1 |
|
|
Differential forms. Why?
I was reading lethe's thread on differential forms and suddenly it dawned on me that I had no idea what differential forms were for, or why the process was developed.
Do they replace vector calculus, or are they a more powerful form of linear algebra or what? For me it is much easier to study something if I know why I'm studying it, and what the benefits are. Any enlightenment would be greatly appreciated. BTW I'm just an interested layperson who studies physics simply because I find it fascinating. |
| Oct28-03, 10:40 AM | #2 |
|
|
For the familar three dimensional topics, differential forms are just an alternative to vectors and tensors, and it's easy to go back and forth between them. Even in this simple case, the diff forms have an advantage, in that you don't have to specify a specific basis to use them. This shows up in the absence of superscripts and subscripts, which can get annoying when you get up to tensors.
Then as you increase your dimensionality diff forms become more and more convenient. Physical laws take on a simple and symmetric shape when expressed in them. You learn these mathematical subjects in order to have more freedom of choice in solving problems and proving theorems. If I may bring in an analogy from school math, it's a little like having both rational fractions and decimals. Sometimes one better expresses your thought and sometimes the other. You find out some things about numbers from fractions (least common denominator) and other things from decimals (repeating and non repeating expansions <-> rational and irrational numbers). |
| Oct28-03, 09:44 PM | #3 |
|
|
Thanks selfAdjoint. I just wish there were some way to start out with a simple example but I guess
you can't do that until you've developed the mechanism. |
| Oct29-03, 10:15 AM | #4 |
|
|
Differential forms. Why?math-sters are apt to have some simple examples you can study before developing all the machinery it is fair to expect that Lethe or selfAdjoint have simple of examples of diff forms in the vest pocket of the wizzard suit try just asking: "could you give me a simple example of one?" but be prepared for it to be TOO simple. when an example is really elementary it might be so simple that you cant tell from it why something is interesting well, if the space you are studying is not a moebius strip or a double donut or something but simply the real line (very vanilla) then a "zero-form" is just a real-valued function defined on the real line that is a zero-level differential form, something like 3x, or sin x, or you know what I mean---how much simpler can you get. there is a way of jacking up the level, but you should start with something with more dimensions like 3D euclidean space R3 instead of just the real line if you stick around and keep asking, someone here will give a non-trivial example of a 2-form or a 3-form and they may say things like "change of coordinates" and "Jacobian" something to ask about is the MOTIVE behind math-constructions. why did they bother to invent this? in the case of diff forms it may have had something to do with wanting to be able to write calculus stuff without coordinates, write calculus on the "plane itself" rather than the plane looking like graph paper, with sub-and-super-scripts one and two for x and y. Wanting to do calculus on the "donut itself" or the "saddle surface itself" rather than having to project graph paper all over these things before you could get started. diff forms can give you a surface element or a volume element that you can integrate with and have flux integrals and charge density integrals without ever having to write indices one and two and three or have chosen coordinates x and y and z. Watch out its a trick! The indices (subandsuperscripts) are really there in the background, like cockroaches in an old apartment building, and they are always ready to come out. But with the slick diff form notation you can not see them and you can forget about them. unless you temporarily need them. write Lethe a PM and say "pray tell me about diff forms, Sir!" He probably will. He is a grad student somewhere and they like to practice their explanatory arts. |
| Dec5-03, 02:27 AM | #5 |
|
|
I'm taking an intro vector calculus course and our professor explained to us differential forms and their interesting properties very briefly over a couple lectures. I don't know how much math you have, but from my VERY limited experience, I can tell you some things.
A 0 form, as marcus explained, is a function. A 1 form is the gradient of a function. A 2 form is the curl of a function. And a 3 form is the divergence of a function. Instead of expressing these by [tex]f>\nabla>curl>div[/tex] they simply get expressed in their differential equivalent, which is simply [tex]d[/tex]. This differential depends on what form you're operating on; it is convenient to write everything this way. This one operator does all of the things the corresponding vector operations do. I should say that [tex]d[/tex] is the operator to go between the different forms. In vector notation, it requires three operators to do this. For example, Gauss' Theorem in vector form is: [tex]\int_{\partial E} \vec{F}\cdot d\vec{S}=\int_E{div\vec{F}dV}[/tex] Where [tex]\partial E[/tex] is the boundary of E and has some properties I won't go into. The anolagous expression using differential forms is: [tex]\int_{\partial E} \alpha=\int_E{d\alpha}[/tex] At least, this is my very very limited understanding of what differential forms is. The professor skipped over explaining manifelds and stuff, so the only thing I really know is that it looks nicer. Heh. And I could(and probably am) horribly horribly wrong on some point. And as I can now see, there already is a thread on what differential forms is. You were asking why. Now I feel like a moron for spending 30 minutes on this post. |
| Jan21-04, 03:36 PM | #6 |
|
|
I've been off this board for some time but I see that Lethe's 'differential forms' thread has mutated from the original, 'concise but chatty' sequence of lectures into more of a discussion between equals - which is out of my depth. Perhaps it would be useful to start, as I did on 'that other board', a thread for discussion of Lethe's material, keeping his original thread for the lecture notes themselves (subject, of course, to occasional revision).
I'm still on the steep part of this particular learning curve and Lethe (and others) have kindly answered some of my very basic questions about diff forms in curved space v. vector calculus in flat space. I have more but I'm not sure where to post them. Perhaps here? |
| Jan21-04, 07:05 PM | #7 |
|
|
the exterior derivative of a 0-form is a 1-form whose components make up the traditional gradient. however, not every 1-form is a gradient of some function. this is very important. similarly, not every 2-form is a curl, although every curl can be thought of as a 2-form which is an exterior derivative of a 1-form. similarly, not every 3-form is a divergence, even though every divergence is a 3-form |
| Jan21-04, 07:09 PM | #8 |
|
|
actually, despite appearances, i think my differential forms thread never reached as far as it did on the other board. no one ever did the exercises i dropped into the text, but occasionally people dropped in to argue semantics, or dropped in to say "i haven t read this, but would like to some day". people requested that i not post all the entries at once, so i never finished posting, since involvement was rather low. the thing sort of never got to maturity. i guess there was more involvement here than at sciforums, but at least there, the thread was a continuous unbroken lecture, instead of a discussion. |
| Jan22-04, 02:08 PM | #9 |
|
|
O.K - my own motivation: I teach mech engineering, specifically solid (occasionally, fluid) mechanics at Glasgow Uni (check my profile if you want to know where that is). The machinery of 'modern' differential geometry (i.e <100 years old) has been used in engineering for some time but only in the past few years have the terms 'differential form', 'Lie derivative', 'pullback', etc appeared in our textbooks. So have we been using all the right tools without knowing what we were doing? Obviously, yes, because we kept getting useful answers - like Dirac!
At any rate, I'm hoping to help it creep faster (O.K, ignore the wee mech joke!) and equip my students for the next couple of decades (I should live that long!). Richard Feynmann said that if you can't teach a subject to a First Year Student, you don't really understand it. So my first task is to teach myself. Here's my collection of useful websites: http://www.mech.gla.ac.uk/BB/viewtopic.phtml?t=62 Feel free to browse. At any rate, I still have difficulty in transiting (is that a real word?) from flat space to curved space. Hence some elementary questions which, with Lethe's and others' forebearance, I'll post here. Later! |
| Jan22-04, 03:45 PM | #10 |
|
|
|
| Jan22-04, 06:26 PM | #11 |
|
|
you know, my differential forms thread really is clogged up with more off-topic posts than on-topic ones. should i start the thread over? what do you think, Ron.
|
| Jan23-04, 05:51 AM | #12 |
|
|
So 'Dummies' question no 1: The traditional picture of an infinitesimal length dx is a little arrow, hence a vector. I'll take on board that an infinitesimal length dx is really a 1-form but I'm having problems picturing it as other than a little arrow; certainly not as a level surface (as in MTW). I _can_ picture grad(F) as level surfaces of a function F and hence as a 1-form. However, can't grad(F) take dx as argument to get the number of surfaces pierced (again following MTW)? That brings it back to being a vector? I'm going round in circles. Any way out? Cheers, Ron Yep - Glasgow Uni is in Glasgow but Heriot-Watt Uni isn't in Heriot-Watt. It was more a 'Where the **** is Glasgow?' thing (there are, AFAIK, about a dozen 'Glasgows' worldwide). |
| Jan23-04, 10:54 AM | #13 |
|
|
which is not to say that the pictures are a bad thing. probably, they are a good thing, and having such a picture probably gives you a much better intuition about these objects. but anyway... ok. you definitely do have to stop thinking of [itex]dx^\mu[/itex] as an arrow. if you are looking for an arrow, use [itex]\mathbf{\partial_\mu}[/itex], which is a tangent vector. and if you can picture [itex]df[/itex] as the level functions of [itex]f[/itex], then [itex]dx^\mu[/itex] should be no problem: [itex]dx^\mu[/itex] is just [itex]df[/itex] for the case [itex]f=x^\mu[/itex], a coordinate function. its level surfaces stack along the direction of the coordinate, instead of some arbitrary direction (for some arbitrary [itex]f[/itex]) of course, it is clear from your question that you know this already. anyway, to answer your question: No, [itex]df[/itex] cannot take [itex]dx^\mu[/itex] as an argument. [itex]df[/itex] is a dual vector, as is [itex]dx^\mu[/itex], which means you feed them vectors, like [itex]\mathbf{\partial_\mu}[/itex] if i do feed such a vector to [itex]df[/itex], i get this: [tex]df(\mathbf{\partial_\mu})=(\partial_\nu f)dx^\nu(\mathbf{\partial_\mu})=\partial_\nu f\delta^\nu_\mu=\partial_\mu f[/tex] perhaps you really did mean [itex]\mathbf{\nabla}f[/itex] to be a vector? if so, then there is a way you can give it [itex]dx^\mu[/itex] as an argument. you know, sometimes in trivia games or whatever, they ask the question "what do you call someone who lives in Glasgow?" Answer: "Glaswegian". weird. and um, you look like Sting in that picture. |
| Jan28-04, 01:49 AM | #14 |
|
|
I've had my difficulties with this, but fundamentally it's
not too hard. In the small limit, there are only derivatives and the geometry is necessarily linear: a vector space with its dual. Any point of a manifold, by definition must have a neighbourhood which can be mapped to an n-dimensional coordinate system, and any such system gives us a vector space in the limit. There are two primary types of vector: tangent vectors and gradients or cotangent vectors; we say they comprise the tangent space and cotangent space at the point, respectively. Suppose for example our coordinates are [itex] (x,y,z) [/itex]. Basis vectors of the tangent space are unit vectors along the coordinate axes; basis vectors of the cotangent space are gradients of linear functions whose level surfaces are parallel to the coordinate planes. E.g. the unit vector [itex] \partial_x_1[/itex] has components (1,0,0). It is the tangent to the parametric curve [itex] (t,0,0)[/itex], or any other curve [itex] (t,0,0) + O(>1) in t[/itex]. Because we have a vector space, curves can be added to make [itex] (v_1, v_2, v_3) t + O(>1)[/itex], and they all have the same tangent, [itex] (v_1, v_2, v_3)[/itex]. The unit covector [itex] dx_1[/itex] has components [itex] (1,0,0)[/itex]. It is just the gradient of the function [itex] x[/itex], or any other function [itex] x + O(>1) in (x,y,z)[/itex]. Functions can be added, and so can gradients because differentiation is linear, so for example the function [itex] f(x,y,z) = a x + b y + c z + O(>1)[/itex] has gradient components [itex] (a,b,c)[/itex]. The components of a gradient are directional derivatives along the axes. Let's play this out in full. The derivative of f in the x direction is [itex] Limit(t->0) ( (a x + b y + c z)[/itex] where [itex] (x,y,z) = (t, 0, 0) )/t [/itex]= [itex] Limit(t->0) ( a t )/t =[/itex] [itex]a [/itex] Another way of stating this, obviously, is [itex] \partial f/\partial x = (1, 0, 0) \cdot (a, b, c)[/itex] Note that because all functions look like linear functions in the small limit, the working above always comes out as a scalar product [itex] (a, b, c) \cdot (v_1, v_2, v_3)[/itex] between a gradient vector and a tangent vector. This is the sense of the MTW picture of "counting the number of level surfaces pierced by a curve". Let's get this into standard terminology now. The basis cotangent vectors are the gradients of the coordinate functions: dx, dy, dz, or [itex] dx^\mu (\mu = 1,2,3).[/itex] Any gradient value of a function is a linear combination of these, e.g. [itex] df = a dx + b dy + c dz[/itex] The basis tangent vectors are tangents to curves aligned with the coordinate axes: [itex] \partial_x, \partial_y, \partial_z[/itex], or [itex] \partial_{x^\mu}[/itex]. Any tangent vector is a linear combination of these, e.g. [itex] v = v_1 \partial_x + v_2 \partial_y + v_3 \partial_z.[/itex] Components of gradients are partial derivatives of the coordinates. Partial derivatives are directional derivatives along the coordinate axes. Directional derivatives are scalar products of tangents with gradients. Thus for example [itex] \partial_x f / \partial_x =[/itex] [ trad. notation ] [itex] < \partial_x, df > =[/itex] [ scalar product ] [itex] a[/itex] It follows from the expansion of df and the scalar product, that [itex] < \partial_{x^\mu}, dx^\nu > = \delta_\mu^\nu[/itex] Fundamentally, that's ALL there is to it. Tangents and gradients are dual, and their scalar products are partial derivatives. But there is an extra refinement of the notation, which can be very helpful in some cases. Since a tangent vector can operate on a function, by making a scalar product with its gradient, so as to calculate a directional derivative, we can identify the tangent with just that directional derivative operation; i.e. [itex] v(f) = < v, df >[/itex] This is also written [itex] \partial_v f = < v, df >[/itex] I don't care for this notation, because it overloads the \partial symbol, which we already used for the basis tangent vectors, [itex] \partial_{x^\mu} [/itex]. It would be OK to write [itex] \partial_x (f) = < \partial_x, df >[/itex] but to be consistent, we should write the second form as [itex] \partial_\partial_x f = < \partial_x, df>[/itex] It's probably not a crippling confusion, but if I happen to see [itex] \partial_u[/itex] somewhere, I need to check whether u is a point-function or a tangent, in order to interpret whether [itex] \partial_u[/itex] is a basis vector or a derivative in the basis direction. Just while I'm whinging, and this is purely pedagogical, I've wasted crucial hours trying to understand how a directional derivative operator can be something of dual type to a gradient. How did the scalar product work? Should I be applying one to the other? Major angst, and completely misconceived. (1) Tangent vectors are dual to gradients. (2) Directional derivatives are applied to functions. Yes I know that tangent vectors and d.d. operators are isomorphic; nevertheless, they are to be distinguished. At least in teaching! Apologies: This probably reads way too assertively. I'm just putting down statements as explicitly as I can, so I don't lose track. Appreciations: Excellent explanations, lethe. Please keep it up. I'm very glad to have discovered the site, things are just at my level. Jonathan |
| Feb2-04, 08:19 PM | #15 |
|
Recognitions:
|
|
| Feb3-04, 10:04 AM | #16 |
|
|
|
| Feb3-04, 02:25 PM | #17 |
|
Recognitions:
|
I've noticed that the number of pages is not consistent. For instance, sometimes, when I go to your diff. forms thread, it is 8 pages long. Lately, it has been 3 (much longer) pages long. You could use the number of the post, but they are not explicitly numbered, so that wouldn't be very useful for your viewers. Furthermore, that would probably be quite a burden for you to keep updating the TOC, what with people like me deleting posts and stuff. That's what I think. |
| Thread Closed |
| Thread Tools | |
Similar Threads for: Differential forms. Why?
|
||||
| Thread | Forum | Replies | ||
| Differential Forms? | Differential Geometry | 19 | ||
| differential forms and degree | Differential Geometry | 8 | ||
| differential forms | Differential Geometry | 8 | ||
| Differential forms in Mechanical Eng. | Mechanical Engineering | 5 | ||
| intro to differential forms | General Math | 0 | ||