You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where $$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$ So that the vector field $$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$ is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is: $$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$ this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is $$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$ In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, asymptotically rigidly with a constant period right near the z-axis. I think you flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used: $$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$ (this is different from the previous thing). That's why you had a difference of opinion.