Homework Help: Need some more help on curl

1. May 2, 2012

zezima1

I look at the curl as how much it causes a stick to rotate. So like suppose we have the force field F = (y,0,0)

Then we see the curl is nonzero, because the force in the x direction is increasing as we move perpendicular to the direction of the x-axis. Therefore suppose we place a twig with one end fixed a the x-axis and y=0 and the other end at same x but at y=1. Then there would be a net torque on the twig explaining why the curl must be nonzero.

I have always had this "twig-picture" of the curl, but I'm starting to think it's wrong to think of it like that. Because I think that you could find vector fields with zero curl where you could still place twigs and get rotation. As an example look at the dipole of two charges on the picture attached. Definately. If you placed a twig with ends at the two green dots, you would get rotation since the horizontal force gets bigger the closer you get to the two charges. However, the curl still happens to be zero. What is wrong with my arguments here? And what is the curl really a measure of? And how long should you make your twig?

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2. May 2, 2012

tiny-tim

hi zezima1!

any non-uniform field will make things rotate …

it only needs the force to vary along the length

3. May 2, 2012

zezima1

hi Tiny-Tim, I didn't think about that since it was quite some time ago, but yes I should've done that.

But are you then agreeing that the vector field will make it rotate? But what then characterizes an irrotational vector field? What is it that doesn't rotate for a such vector field? I've seen dozens of statements like place a paddlewheel at any point in a vector field with zero curl and it wont rotate. But does that hold for the above on the picture?

4. May 2, 2012

tiny-tim

if a stick is fixed so that it can rotate, then almost any field will make it rotate some way

but if the field has zero curl, then it can't rotate all the way round

a paddlewheel is (approximately) the same all the way round … an "ideal" paddlewheel therefore won't rotate at all in a field with zero curl, and (for non-zero curl) the torque on it will be proportional to the flux of the curl through the wheel

5. May 2, 2012

sweet springs

Hi. Curl is circulation around loop, so not twig, why do not you try loop?
Let us consider the force field that works on small loop. On the loop torque works in your first case , not in your second case.
Regards.

Last edited: May 2, 2012
6. May 2, 2012

zezima1

hmm.. Not sure if I understand this whole paddlewheel thought. We agree that the curl of the field of a dipole is zero right?
Well on the picture I have placed a paddlewheel at a point. Surely that would rotate? - Since the force from the field on the upper half will be less than the force from the field on the lower half.

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7. May 2, 2012

tiny-tim

yes, the force will be less, but the torque will be the same, since the (sin of the) angle between the field line and the spoke of the wheel is greater

8. May 2, 2012

zezima1

Oh wait.. What? We agree that the torques come from the horizontal forces right? Well surely they both push completely perpendicular to the spokes of the paddlewheel, the one pushing on the upper part being lower than the one pushing on the lower part. But I think you are right, that if you really were to sum up all the contributions one every little inch of the wheel you would get zero net torque - because, that's what you are saying right?

9. May 2, 2012

tiny-tim

nooo

10. May 2, 2012

zezima1

umm.. Look at the paddlewheel on the picture. It is situated such that the distance from its center to the two charges is the same. So the most southern point on the paddlewheel will experience a horizontal to the right. Similarly the northern point will experience a horizontal force from the wheel but this will be less since this point is farther away from the charges. Both of them push perpendicular to the spokes (though on the drawing the spokes are rotated a bit).

11. May 2, 2012

tiny-tim

yes, but none of the other spokes are perpendicular to the force