Problem with vector field proof

[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

 

[LCKurtz]

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?

 

[Roidin]

Thanks LCKurtz,

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

 

[LCKurtz]

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?

 

[gtfitzpatrick]

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]

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?

 

[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...

 

[LCKurtz]

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...

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.

 

[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?

 

[LCKurtz]

$\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy$ = $\int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0$ 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$$
\oint_C \vec F\cdot d\vec R -\oint_C\vec G\cdot d\vec R =\oint_C (\vec F-\vec G)\cdot d\vec R=\oint_C\vec H\cdot d\vec R =\ ...$$