Solving Vector Field with Poincare's Lemma

[Ted123]

Homework Statement

[PLAIN]http://img130.imageshack.us/img130/8540/vecx.jpg

The Attempt at a Solution

I've done (i).

First of all Poincare's Lemma says that if the domain U of {\bf F} is simply connected then:

{\bf F} is irrotational \iff {\bf F} is conservative.

So for (ii)(a), does V being simply connected (is it or not?) mean Poincare's Lemma implies there is a potential function for \bf F (since it would mean \bf F is conservative and hence is a gradient)
DESPITE U not being simply connected - it has a hole in the middle at (0,0)?

 

[ystael]

Yes. V is, in fact, simply connected, and the restriction of \nabla\times\mathbf{F} to V is still zero. (The simplest way to prove that V is simply connected is to see that it's star-shaped with respect to any point on the positive x-axis.)

 

[Ted123]

Yes. V is, in fact, simply connected, and the restriction of \nabla\times\mathbf{F} to V is still zero. (The simplest way to prove that V is simply connected is to see that it's star-shaped with respect to any point on the positive x-axis.)

So for (ii)(b), how do I find \phi in S_1 ?

Presumably then I can solve the simultaneous equations x=r\cos\,\phi and y=r\sin\,\phi in terms of x and y (e.g. using \tan^{-1}) and then verify by explicit differentiation that \nabla \phi (x,y) = \mathbf{F} (x,y) . And then do S_2 and S_3 in a similar way.

 

[ystael]

(r, \phi) are just polar coordinates with a particular choice of range for the angular coordinate; what does that tell you the angular coordinate should be on the positive x-axis?

 

[Ted123]

(r, \phi) are just polar coordinates with a particular choice of range for the angular coordinate; what does that tell you the angular coordinate should be on the positive x-axis?

Well for S_1 the angle would be 0 but what is the range of r, is it \pi ?

For S_2 the angle is \frac{\pi}{2} but again what is r, again is it \pi ?

For S_3 the angle is \frac{3\pi}{2}, is r, -\pi ?

So are the angular coordinates of S_1, S_2,S_3, (\pi, 0), (\pi, \pi/2), (-\pi,3\pi/2) respectively?

 

[Ted123]

Well for S_1 the angle would be 0 but what is the range of r, is it \pi ?

For S_2 the angle is \frac{\pi}{2} but again what is r, again is it \pi ?

For S_3 the angle is \frac{3\pi}{2}, is r, -\pi ?

So are the angular coordinates of S_1, S_2,S_3, (\pi, 0), (\pi, \pi/2), (-\pi,3\pi/2) respectively?

\frac{y}{x} = \tan(\phi)

so \phi (x,y) = \tan^{-1} \left(\frac{y}{x}\right)

and \nabla \phi(x,y) = \mathbf{F}

But how do I find \phi in S_2 and S_3? x can be 0 in both of these.

 

[ystael]

Sounds like you need to go back to a trigonometry text and review how polar coordinates work. r is not an angular coordinate; it's a distance (from the origin). And if x = 0, then you already know what the value of \phi should be, depending on whether y > 0 or y < 0.

 

[Ted123]

Sounds like you need to go back to a trigonometry text and review how polar coordinates work. r is not an angular coordinate; it's a distance (from the origin). And if x = 0, then you already know what the value of \phi should be, depending on whether y > 0 or y < 0.

I know r is a distance but I thought seen as \phi\in(-\pi, \pi) that that distance would be those values.

 

[Ted123]

I know r is a distance but I thought seen as \phi\in(-\pi, \pi) that that distance would be those values.

Can anyone see how I do the last part of (ii)(c)? Why can't \phi be extended to x<0 ?

 

[ystael]

Let \varepsilon be small. What is \phi close to at (-1, \varepsilon)? What is \phi close to at (-1, -\varepsilon)? How does this make it difficult to find a value for \phi at (-1, 0)?

 

[Ted123]

I know r is a distance but I thought seen as \phi\in(-\pi, \pi) that that distance would be those values.

\displaystyle \oint = 0 \neq 2\pi