# Nonconservative gradients (not an oxymoron)

1. Jul 15, 2009

### okkvlt

fx,fy=-y,x
curl(fx,fy)=2

its counterclockwise so the gradient is pointing ccw.
the gradient is the direction of max increase, so
then the surface z=f[x,y] whose gradient is -y,x is a spiral/screw whose z position goes up forever as x and y are traced out counterclockwise

so, a nonconservative field IS a gradient of a surface.
why was i lied to?

2. Jul 15, 2009

### Office_Shredder

Staff Emeritus
I think you're going to need to write this out clearer. How can the curl of a vector be a number? How can the gradient of a three variable equation be a two dimensional vector?

Last edited: Jul 15, 2009
3. Jul 16, 2009

### BobbyBear

Hello there! Okay, lets take it step by step and yes, please do be a little clearer so we can understand exactly what you're saying. I think what you're saying is that you're trying to find the surface whose gradient at each point is (-y,x,0)? How did you make these out to be spirals?

So let's see, you've got the vector field:

$$\vec{V} = (-y, x, 0)$$

Ye? And you've calculated its curl, which is:

$$Curl \vec{V} = (0,0,2)$$

so the field is 'rotating' counterclockwise viewed from above (the positive z axis). The curl of your vector field is constant in all of space, and non-zero. Because it is non-zero means that your vector field $$\vec{V} = (-y, x, 0)$$ cannot be a gradient, in other words, there is no scalar field whose gradient is $$\vec{V}$$. That is, there are no surfaces whose gradient at each point is (-y, x, 0). But! this does not mean that there are no surfaces which are orthogonal at each point to $$\vec{V}$$. In fact, in your case, these surfaces DO exist, because the vector field $$\vec{V}$$ is normal to its curl.

4. Jul 16, 2009

### BobbyBear

Just to help visualize, here is the curl of your vector field V, as viewed from above:

http://img31.imageshack.us/img31/4605/curl.gif [Broken]

Last edited by a moderator: May 4, 2017