Problem with vector field proof

1. Aug 15, 2012

Roidin

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

2. Aug 15, 2012

LCKurtz

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

3. Aug 15, 2012

Roidin

Thanks LCKurtz,

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

4. Aug 15, 2012

LCKurtz

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: Aug 15, 2012
5. Aug 20, 2012

gtfitzpatrick

Hi im doing a similar problem so sooner than start a new thread...
$\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy$?

6. Aug 20, 2012

LCKurtz

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

7. Aug 20, 2012

gtfitzpatrick

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...

8. Aug 20, 2012

LCKurtz

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

I have to go for a couple of hours. I'll check back and see if you have it by then.

Last edited: Aug 20, 2012
9. Aug 20, 2012

gtfitzpatrick

$\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy$ = $\int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0$ but this can only be true if $\oint F.dr = \oint G.dr$ qed?

10. Aug 20, 2012

LCKurtz

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