# Cauchy,s theorem on real plane?(R^2)

1. Apr 16, 2006

### eljose

could actually the Cauchy theorem be applied to $$R^{2}$$ of course with some restrictions..namely

-The differential form must be invariant under rotation in the plane
-The differential form must come from a gradient $$\nabla{S}=f/(|r|-a)$$

with this using the same techniques by Cuachy we could se the equality:

$$\Int_{C}dq.F(x,y)//(|r|-a)=2\piF_{r}(a)$$

this means somehow that the integral over a curve on RxR is equal to the normal component evaluated at the point r=a where r is the modulus of the vector (x,y,z)

2. Apr 16, 2006

### eljose

Of course i meant the formula:

$$\int_{C}dq.F(x,y)/(|r|-a)=2\pi{F_{r}(a)}$$

where the integral along a curve of a 2-differential form with a singularity at the point whose distance from the origin is equal to "a", is set to be the normal component (invariant under rotations in RxR) of the differential form F..the derivation would be the same..we set a circle of radius $$\epsilon$$ and make this epsilon tend to zero...

3. Apr 16, 2006

### Hurkyl

Staff Emeritus
By the way, you can use that button "Edit" to edit your posts to fix the LaTeX. (You will have to reload the page to get the updated image)

Anyways, you want to do differential geometry! The key theorem you want to use here is the generalized Stoke's theorem. A special case is Green's theorem:

If D is a region of R², and C is the boundary of D, and P and Q are scalar fields with continuous partial derivatives, then:

$$\oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$$

Notice that if the vector field (P, Q) is the gradient of some scalar field, then the right hand side is zero.

The proof idea for what you want to show is similar to the proof of the similar equation in complex analysis, but with Green's theorem being the theorem that you use to justify changing the curve.

Now, I suspect you haven't stated your problem right. Did you really intend for your integrand to be singular along the entire circle of radius a centered on the origin?

Last edited: Apr 16, 2006
4. Apr 17, 2006

### eljose

-No Hurkyl..i perform the integration about two curves in the form:

$$\oint_{C}F.dq-\oint_{\epsilon}{Pdx+Qdy}$$

the first curve encloses the area marked by the curve except a 2-D "ball" of leghnt $$\epsilon{2\pi}$$ in wich the point (x0,y0) is and note that:

$$a^{2}=(x_{0})^{2}+(y_{0})^{2}$$ of course on the contrary to Cauchy,s theorem in wich "a" was a complex number the corresponding value of a must be somehow an scalar, expressed as a distance from the origin of the point (x0,y0) and we must impose that at least the "radial" part of the differential Form F must be invariant under rotations on x-y plane..by the way generalizing to Surface integrals and using divergence theorem (Gauss) we could express the "charge of monopoles" as the result of a singularity of the form $$a/r^{2}$$ for the Magnetic field B (vector)

-a is some kind of distance (note that there are no 2-D real number)

Last edited: Apr 17, 2006
5. Apr 17, 2006

### Hurkyl

Staff Emeritus
If R is a ring in 2-space, then the boundary of R consists (which I'll call C) of the two curves: C_1, the outer boundary, and C_2, the inner boundary. By definition:

$$\oint_C \omega = \oint_{C_1} \omega - \oint_{C_2} \omega$$