Talk:Green's theorem/Archive 1

Archive 1

Stray if

"If it can be shown that if

$\int _{C}L\,dx=\iint _{D}\left(-{\frac {\partial L}{\partial y}}\right)dA\qquad \mathrm {(1)}$

and

$\int _{C}M\,dy=\iint _{D}\left({\frac {\partial M}{\partial x}}\right)\,dA\qquad \mathrm {(2)}$

are true, then Green's theorem is proven." [emphasis added on instance of "if" under discusssion]

The quoted sentences says that:
(it can be shown that if (1) and (2) are true) implies (Green's theorem is proven)
(if (1) and (2) are true) implies (Green's theorem is proven)
"If (1) and (2) are true" doesn't parse correctly.
It should read "If it can be shown that (1) and (2) are true, then Green's theorem is proven." This means that (1) and (2) together imply Green's theorem; this is true. The bold "if" is stray and makes the "sentence" not wrong so much as not a valid sentence. For the record, the statement remains true if the other "if" is pulled instead, but one definitely needs to go. --130.39.152.206 22:29, 19 April 2006 (UTC)

Other form

This is the curl-circulation (tangential) form. Might be nice to have the flux-divergence (normal) version mentioned, too. Revolver 17:48, 27 April 2006 (UTC)

A Burp?

Does anyone else think the burp analogy is a bit silly for an encyclopedia? - Arsian120 05:52, 17 September 2006 (UTC)

A Burp!

Well, in that it implies the discharge of gas from the G.I. tract, the burp is socially inappropriate in some settings. However, it is a pictureseque and useful learning tool. I don't think the analogy fully captures the essence of Green's theorem, but it helps me visualize the content of a volume passing through a region. That's good enough for me to vote to retain the burp analogy. 151.200.241.233 12:30, 3 October 2006 (UTC)

Notation

I'm intrigued by the notation. An integral sign with a little capital letter C at its base. An integral sign with a (possibly directed) circle written over the big curvy S there. The full import of two integral signs written side by side. The full meaning of the partial derivative inside the integral. Ah! The sweet mystery of mathematical notation. All of which is left as an exercise for the reader to grasp. If you a math whiz, the notation is obvious. If you ain't, it ain't. 151.200.241.233 12:30, 3 October 2006 (UTC)

A mistake in the article ?

Quoting:

"In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area."

I just read at http://www.math.gatech.edu/~carlen/2507/notes/vectorCalc/vectorfields.html It says that Greene's theorem is used for calculating Work done over a closed curve, and that the divergence theorem is used for calculating fluxes.

i.e. integral over an area D of curlF (which is the version stated in the article) gives the work done by F when going through the closed path C, and not the total flux.

Read this [1]BlueRaja 16:47, 26 April 2007 (UTC)

Sign convention

You always see Green's theorem written like this:

$\int _{C}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dA$

but I don't think it's obvious to most people that it can also be written like this:

$\int _{C}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial L}{\partial y}}-{\frac {\partial M}{\partial x}}\right)\,dA.$

Is it worth writing both (or at least mentioning this fact)? I can see this symmetry being handy in some manipulations, and frustrating if you don't happen to realize this is the case.

Of course, there's a simple reason why this should be the case: there's absolutely nothing to distinguish $x$ and $y$ on the left hand side, so why should they be distinguished on the right hand side? More explicitly, consider a derivation of Green's theorem from Stokes' theorem:

Let $\omega =Ldx+Mdy$. Stokes' theorem tells us that

$\int _{M}d\omega =\int _{\partial M}\omega$

for any smooth paracompact manifold $M$ with boundary $\partial M$. Computing the exterior derivative of $\omega$, we get

${\begin{array}{rcl}d\omega &=&d(Ldx)+d(Mdy)\\&=&((dL)\wedge dx+L(ddx))+((dM)\wedge dy+M(ddy))\\&=&dL\wedge dx+dM\wedge dy\\&=&({\frac {\partial L}{\partial x}}dx+{\frac {\partial L}{\partial y}}dy)\wedge dx+({\frac {\partial M}{\partial x}}dx+{\frac {\partial M}{\partial y}}dy)\wedge dy\\&=&{\frac {\partial L}{\partial y}}dy\wedge dx+{\frac {\partial M}{\partial x}}dx\wedge dy\\\end{array}}$

and from this point we can group the terms as either

$d\omega =\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)dx\wedge dy$

or

$d\omega =\left({\frac {\partial L}{\partial y}}-{\frac {\partial M}{\partial x}}\right)dy\wedge dx,$

either of which can be considered to be "$dA$." Of course, the convention is to have $dA=dx\wedge dy$, but that's exactly what I'd like to point out in the article: the fact that the sign is one way or another is merely a convention and has no physical significance. Trevorgoodchild (talk) 17:01, 7 March 2008 (UTC)

Do we expect a proof?

Do we expect to see a proof in an encyclopedia? Billlion 14:20, 13 Sep 2004 (UTC)

See WikiProject Mathematics. In this case I think the proof can be ommitted. The thereom is easy to understand and the proof is quite long and technical, so does not contribute to understanding of the theorem. Contrast this with Cantor's diagonal argument where the proof is neccessary for understanding the theorem MathMartin 10:36, 14 Sep 2004 (UTC)

personally I'm Glad the proof is there, "do we expect a proof in an encyclopedia..." of course, this encyclopedia should contain everything possible and more...

I agree. It's nice to have the proof and it has turned useful for me in the past already. --Aphexer 14:52, 16 January 2007 (UTC)

I don't really understand the point of proving the theorem for Type I or II regions, especially since the theorem is true in a much more general context. My main problem is that when I use this theorem I hardly ever have a type 1 or 2 region, I think if you only write a proof for a special case it should be the most important case, which in my opinion this is not.... —Preceding unsigned comment added by 69.207.128.59 (talk) 01:55, 5 December 2008 (UTC)

Area Calculation

Does it make much sense do add this section? What does it say more than what a "normal" integral means? Nijdam (talk) 18:15, 28 February 2010 (UTC)

• Its an important result and perhaps surprising result. Most introductory texts cover it, so does wolfram. It finds practical application in Planimeter. Noodle snacks (talk) 02:03, 1 March 2010 (UTC)

Proof

The article doesn't appear to actually give a proof for the general case, only for simple regions. — Preceding unsigned comment added by 207.239.94.2 (talk) 16:24, 23 August 2012 (UTC)

It is not even a proof for special regions. It proves about 1/2 of the theorem for regions of type 1, and then says a similar thing gets the other half. But absolutely not, since the first 1/2 relied upon x being constant on two of the subcurves, and the second half would rely on y being constant on two of the subcurves, which it is not. This proof is, then, quite incorrect. 32F (talk) 11:52, 5 February 2013 (UTC)

Okay, I fixed this, and also a couple of other things. (The proof talked about "approximating" a general region as a union of regions of a simple type, but it cannot be a mere *approximation*. Also, it cannot be a union either, it is a union of regions whose interiors are pairwise disjoint. Don't know the name for that. Called it "decomposing" the region but if someone knows the correct name that would be better obviously). 32F (talk) 12:36, 5 February 2013 (UTC)

Must have been written by math people...

Not a single mention of curl or circulation. Amazing. Revolver 17:55, 27 April 2006 (UTC)

Green's theorem is just a projection down to the plane of Stokes' theorem. It's trivial conceptually given Stokes' theorem, so explaining it intuitively is pretty much a waste of time. --130.39.152.206 01:26, 30 April 2006 (UTC)

Yup, everything is trivial once you understand it, so there's no point in explaining anything. Spoken like a true math dweeb. Revolver 21:43, 3 May 2006 (UTC)

Hey, anonymous, I seriously hope this isn't your idea of an "intuitive explanation" (taken from Stokes' theorem:

The classical Kelvin-Stokes theorem:

$\int _{\Sigma }\nabla \times \mathbf {F} \cdot d\mathbf {\Sigma } =\int _{\partial \Sigma }\mathbf {F} \cdot d\mathbf {r} ,$

which relates the surface integral of the curl of a vector field over a surface $\Sigma$ in Euclidean 3 space to the line integral of the vector field over its boundary, is a special case of the general Stokes theorem (with n = 2) once we identify a vector field with a 1 form using the metric on Euclidean 3 space. It can be rewritten for the student unacquainted with forms as

$\iint \limits _{\Sigma }\left({\frac {\partial R}{\partial y}}-{\frac {\partial Q}{\partial z}}\right)\,dydz+\left({\frac {\partial P}{\partial z}}-{\frac {\partial R}{\partial x}}\right)\,dzdx+\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\,dxdy=\oint \limits _{\partial \Sigma }P\,dx+Q\,dy+R\,dz$

where P, Q and R are the components of F.

Oh, yeah, clear as mud. It just POPS right out at the reader how the curl is really the "circulation density" at a point, and so all Green's (Stoke's) theorem is saying is that if you "add up" all the "microscopic circulation density" inside a region, you just get the net circulation along the boundary of the region. In other words, it's just like the ordinary FTC you learned in Calc I. But I suppose such an explanation is just too "obvious" for you, and the protracted terms of partial derivatives above must be much more illuminating. Revolver 21:56, 3 May 2006 (UTC)

You might as well hand them Chapter 10 of Baby Rudin and tell them it's obvious what's going on. Revolver 21:57, 3 May 2006 (UTC)

I am the previously anonymous user (130.39.152.206). "Math dweeb" is an ad hominem attack and therefore inappropriate. In general, your tone is not exactly civil discourse; a calm and reasoned approach is usually better for achieving one's objectives, as assorted clichés point out.

I stand by what I said despite your (likely intentional) misreading. Green's theorem is vastly easier to understand as the projection onto the plane of Stokes' theorem, which basically everyone learns two days later, anyway. Perhaps you should help out at Stokes' theorem if you feel it is not explained well. --CalculatinAvatar 19:19, 4 May 2006 (UTC)

I fully agree with the comment that the traditional M-N / P-Q is a very poor exposition of the concept. Such high-handed hermetic mindset is what causes a math book to be put down after one minute instead of 10 minutes. Much of the higher maths has been devised to address physics. As an engineer, I have little patience with obscure tellings from a mathematician. Physics is already complex in itself, another layer of unmotivated maths is a nuisance. — Preceding unsigned comment added by 74.56.95.120 (talk) 15:48, 13 December 2014 (UTC)

Notation Question

Not that it really matters, but I usually see (American mathematics texts) Green's thm stated with functions named P and Q, rather than L and M. In fact, the last section of this article employs these letters. Does anyone object to me going through and switching to P and Q? Chip McShoulder (talk) 21:58, 6 February 2010 (UTC)

Higher maths require being able to go beyond issues of notation.

Green's theorem is supposedly intended to ease evaluation of line integrals. So the traditional way of teaching it boils down to: in order to simply the problem, replace a function by two functions M-N / P-Q. Then evaluate two double integrals.

I am totally dissatisfied with that traditional approach. While it must please mathematicians, it is not intuitive. Would it be so difficult to start with the insight that possessed Mr. Green to come up with something so profound ?

Unfortunately, * teaching * has been replaced by merely * telling *.

http://mathinsight.org/greens_theorem_idea does a far better job * explaining * the concept but is not even listed in the references and other links. Instead a reference is given to Math World, which manages to regurgitate the same hermetic P-Q gospel without bothering with any supporting drawing. — Preceding unsigned comment added by 74.56.95.120 (talk) 15:28, 13 December 2014 (UTC)

Wikipedia is not a textbook, thus it's not our job to teach it. Wolfram Mathworld also isn't the best resource for learning either - it's more a reference or "handbook" like Wikipedia is.--Jasper Deng (talk) 17:14, 13 December 2014 (UTC)