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I really tried thousands of things before deciding to ask here. Searched all over the internet for an answer, but failed to find it.

Let's get started with the Fundamental Theorem of Line Integrals. Take the following statement (found in many places, for example, in this Math.SE question):

Suppose $C$ is a smooth curve given by $r(t)$, $a \leq t \leq b$ and suppose $\nabla f$ is continuous on $C$. Then $$\int_{C}\nabla f\cdot dr = f(r(b)) - f(r(a)).$$

Unfortunately, there must be some mistake here. This theorem clearly implies that if C is a closed curve, the integral must be zero. I will show a counterexample soon. But first, I would like to clarify that I already understand "part" of the mistake: I know the previous statement must have forgot some condition, like some sort of differentiability and such. So this is my question, what are EXACTLY the requirements? At first I expected this to be an easy google-search question, but it turned out that no one seemed to care enough with that kind of details. I am interested in those details, though.

Very well, take a look at this MIT video, where it's shown that the following (famous) vector field is NOT conservative:

$$\vec{F}(x,y) = \dfrac{-y\hat{i} + x\hat{j}}{x^2 + y^2}$$

Very well. I understand that. In fact, if you take $C$ to be the counter-clockwise unit circle around the origin, the integral will be equal to $2\pi$, not zero.

The problem is, $\vec{F}$ can be written as a gradient field!! Look:

$$ f = -arctan(x/y) \implies \vec{F} = \nabla f$$

It is also true that $C$ is a smooth curve (circle) and $\nabla f$ is continuous on $C$.

This shows that the quoted statement for the Fundamental Theorem of Line Integrals must be wrong.

As I said before, I suspect the quoted statement forgot some conditions. I want to know what are those conditions, and more importantly, why those conditions are needed! I even looked at a proof for the theorem, but couldn't detect any steps that needed extra conditions.

All this also raised another question: is it really true that Gradient Fields and Conservative Fields are the same thing (as stated by this Math.SE question)? If they are indeed the same thing, do you agree that Gradient Field was a poor name choice (given that I showed a field, written as a gradient, that is not conservative)? If they are not the same thing (which would disagree with the linked question), then what's the difference? (Meta-parentheses: I hope this part is not considered a duplicate, given that I believe the other question to be lacking - if I'm wrong and it is indeed a duplicate, please help me explaining what exactly should I have done instead).

SUMMARY

1. What is wrong with the quoted Fundamental Theorem of Line Integrals? (what conditions are missing?)

2. Why are those extra conditions needed?

3. With all this in mind, is it really true that "gradient field" is the same thing as "conservative field"? (Please note I am aware of this question but I believe it to be incomplete, therefore I believe this part is not a duplicate. Please read the whole question for details on this. I just don't want a duplicate tag here.)

Thank you all for your time.

EDIT: After searching even more I finally found out a bit of information that helped - this Wikipedia page, but sadly all my questions still stand. It seems that the missing condition is the need of being simply connected, but how exactly does that go in the Fundamental Theorem of Line Integrals? Actually, my textbook (by James Stewart) doesn't mention anything regarding simply-connectedness on the Fundamental Theorem of Line Integrals. There is no region, just a curve!

EDIT 2: I just found this question and this question, answered by Shuhao Cao. Although the major portion of the explanations use deeper knowledge that I am not familiar with, I was able to find out that

the equivalence between a vector field being conservative, its rotation being zero, it being the gradient of a scalar potential and its path integral being path-independent only holds in simply connected domains

(as said by joriki in a comment).

I believe that but didn't understand the reason yet - probably the reason is the lack of knowledge to understand Shuhao Cao's explanation.

Nevertheless, my questions 1 and 2 still stand. Also, if someone can provide a simpler explanation than Shuhao Cao's ones, that would be great.

Community
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Pedro A
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    Excellent question, formatting, and writing. +1 – Zach466920 May 02 '15 at 00:40
  • Thank you Sir @Zach! I always try my best to be precise in my questions, to avoid wasting time of both sides. – Pedro A May 02 '15 at 00:49
  • Stewart certainly does discuss simply connectedness, although you may not have gotten that far at this point. See here. – Viktor Vaughn May 02 '15 at 01:02
  • @SpamIAm Actually I already got there, but Stewart just says "simply-connectedness is needed" instead of explaining the true reasons; also, I have failed to understand why such a condition doesn't show up in the Fundamental Theorem of Line Integrals. But you are right, my choice of words wasn't the best one in that part. I'll fix that. – Pedro A May 02 '15 at 01:53
  • I suggest reading this. I also strongly suggest reading the other material on this site. – Zach466920 May 02 '15 at 14:17

2 Answers2

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The usual statement of the FT of line integrals is correct. You just make sure to understand the statement "$\nabla f$ is continuous on $C$." As you notice for your example $\vec{F}$, you can compute $f(x,y) = - \mathrm{arctan} \left( \frac{x}{y} \right)$. The problem is that this function is not defined when $y = 0$. That is, $\nabla f$ is not continuous on any curve that crosses the $x$-axis. In particular, it is not continuous on any circle $C$ around the origin. In other words, the statement of the theorem implicitly assumes that $f$ is well-defined on the curve.

The crux of the issue is that when you show that $\nabla f = \vec{F}$, you only need to take derivatives. This is a local (i.e. works only for nearby points) calculation. Computing line integrals involve global calculations since the endpoints need not be very close. The extra condition that the domain of $\nabla f$ (or $\vec{F}$) be simply-connected just ensures that $f$ is globally well-defined (up to adding constants). Otherwise, we need to worry about where the curve lies.

user197427
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  • Unfortunately I think you got confused with $f$ and $\nabla f$ in your answer. I agree that $f$ is not defined when $y = 0$, but that means $f$ is not continuous when $y = 0$, while what matters for the theorem is the continuity of $\nabla f$, which is continuous everywhere but at $(x,y) = (0,0)$ (which does not belong to $C$). Therefore I still believe $\nabla f$ is continuous on $C$, and my question still stands. – Pedro A May 05 '15 at 02:50
  • $\nabla f$ is not continuous on $y=0$ since $f$ is not even defined there. If you want, you can see this since the definition of the partial derivatives at a specific point requires the function to be defined at such points. I'm guessing that you are using the chain rule to compute $\nabla f$, but this is also invalid since the inside term $\frac{x}{y}$ is not differentiable (or even defined) when $y=0$. – user197427 May 05 '15 at 05:33
  • I see now. Thank you very much. The key problem was to notice that even though $\nabla f$ seemed to be defined everywhere but the origin, it was actually not defined in a lot more points. I will upvote and accept now. – Pedro A May 11 '15 at 11:07
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I can point out the mistake here. This theorem only applies if $F$ is conservative. For instance, Gravity is conservative, but friction is not (because the force is dependent on the path you take). If this is true we can say, $$\nabla F=G$$ And that the derivative of the composition $F$ and $r(t)$ is... $${{dF} \over {dt}}= \nabla F(r(t)) \cdot r'(t)=G(r(t)) \cdot r'(t)$$ This fortunately was discovered to be the integrals for a line integral... $$\int_C \nabla F \cdot dr= \int_C G(r) \cdot dr=\int_a^b G(r(t)) \cdot r'(t) \ dt=\int_a^b {{d F} \over {dt}} \ dt=F(b)-F(a)$$ What's important to remember is that the function must be conservative. You can check this intuitively or you can verify by checking the curl.

Pedro A
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Zach466920
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  • I believe you meant $\nabla F$ is conservative, right? Also there is a typo in your second equation. But you didn't address my true question - you said it must be conservative, without saying why. Also, checking the curl is not enough to decide if the vector field is conservative. I want to know what change has to be done in my quoted statement for the Fundamental Theorem, and why. – Pedro A May 05 '15 at 02:45
  • @Hamsteriffic actually I do address your question, the field must be conservative because its the only field that allows for the proof shown above to work. And yes, checking the curl is enough. If its not zero, the field, must not be conservative. One of the things that has to be changed in your theorem statement, is the condition that the field is conservative. – Zach466920 May 05 '15 at 14:10
  • @Hamsteriffic P.S. you can edit the question instead of vaguely saying it has a typo somewhere... – Zach466920 May 05 '15 at 14:11
  • Thanks and sorry my late; I didn't notice I could edit other's answers - I felt strange doing that, but I fixed the typos. Sorry about that. – Pedro A May 11 '15 at 11:05