Problem with vector field proof

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SUMMARY

The discussion revolves around proving that the line integrals of two vector fields, F and G, are equal when their difference is the gradient of a scalar function μ(x,y), specifically when F - G = ▽μ. The participants clarify that the curve C is closed and explore the application of Green's Theorem to demonstrate that the integral of the vector field difference over the closed curve results in zero. This leads to the conclusion that ∫ F.dx = ∫ G.dx for all piecewise smooth curves C in the xy-plane.

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Roidin
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Suppose that F and G are vector fields and that F-G = ▽μ for some real-valued function μ(x,y). Prove that

∫ F.dx = ∫ G.dx for all piecewise smooth curves C in the xy-plane

I just need some help in getting started really. Thanks
 
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Roidin said:
Suppose that F and G are vector fields and that F-G = ▽μ for some real-valued function μ(x,y). Prove that

∫ F.dx = ∫ G.dx for all piecewise smooth curves C in the xy-plane

I just need some help in getting started really. Thanks

If ##dx## is a scalar, what does ##\vec F \cdot dx## mean? Have you stated the problem correctly? Is C a closed curve?
 
Thanks LCKurtz,

F.dx is the line integral of the vector field and yes C is closed.
 
Roidin said:
Thanks LCKurtz,

F.dx is the line integral of the vector field and yes C is closed.

That's unusual notation. And since it matters that C is closed, don't you think it would have been good to mention that?

Let's call ##\vec H = \vec F -\vec G##. So say ##\vec H = \langle h_1,h_2\rangle## and you are talking about$$
\oint_C \vec H \cdot d\vec r =\oint_C h_1\, dx + h_2\, dy$$So now if you use ##\vec H(x,y) =\nabla \mu(x,y)## what happens?
 
Last edited:
Hi I am doing a similar problem so sooner than start a new thread...
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy?
 
gtfitzpatrick said:
Hi I am doing a similar problem so sooner than start a new thread...
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy?

Yes. Now do you see how to argue that it is zero?
 
No. in a word.
I thought i might be able to work something out using greens theorem(as the curve is closed) but its not working out...
 
gtfitzpatrick said:
Hi I am doing a similar problem so sooner than start a new thread...
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy?

LCKurtz said:
Yes. Now do you see how to argue that it is zero?

gtfitzpatrick said:
No. in a word.
I thought i might be able to work something out using greens theorem(as the curve is closed) but its not working out...

Green's theorem is what you want. What do you get and why don't you think it works?

[Edit]I have to go for a couple of hours. I'll check back and see if you have it by then.
 
Last edited:
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy = \int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0 but this can only be true if \oint F.dr = \oint G.dr qed?
 
  • #10
gtfitzpatrick said:
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy = \int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0 but this can only be true if \oint F.dr = \oint G.dr qed?

Yes. It would be more clear if the left side had started with$$
\oint_C \vec F\cdot d\vec R -\oint_C\vec G\cdot d\vec R =\oint_C (\vec F-\vec G)\cdot d\vec R=\oint_C\vec H\cdot d\vec R =\ ...$$
 

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