Vector field plot

Can someone show me what

\frac{sin^2\theta}{r^2} \underline{e_r}

looks like in the plane z = 0? I tried plotting it on maple but couldn't figure it out.

Does it go clockwise around the origin with the arrows decreasing in magnitude as you go further and further away from the origin?

Reply 1

generalebriety

Firstly, draw yourself some lines of constant r (i.e. circles about the origin). Then the arrows on each circle all point in the direction of r increasing / decreasing (so outwards from / inwards to the origin) and have magnitude between 0 and +1/r^2 (so they're big if you're close to the origin and small if you're far away); this magnitude is minimum (zero) at 0 and pi, and is maximum (1/r^2) at pi/2 and 3pi/2, and acts like a cosine wave in between (remember that sin^2 theta is just a simple transformation of cos 2(theta)).

Reply 2

Barny

generalebriety

"Then the arrows on each circle all point in the direction of r increasing / decreasing (so outwards from / inwards to the origin)"

I don't quite get that? So I draw a series of concentric circles and then the arrows go away from the origin? ie. not along the lines of the circle? Also in the plane z = 0 isn't theta constant? Bloody hell I'm terrible at sketching stuff.

Reply 3

generalebriety

Barny

\mathbf{e}_r

is the unit vector pointing along the radius (outwards from the origin);

\mathbf{e}_\theta

is the unit vector pointing anticlockwise and at right angles to

\mathbf{e}_r

;

\mathbf{e}_z

is the unit vector pointing up. So here the only vector is e_r, so everything must be in the direction of e_r (or the 'opposite' direction). In the plane z = 0, z is constant and zero - hence the equation z = 0. :p:

Theta can't be constant, because in order to have a plane, we need to have two variables - and if z is constant, the only two left are r and theta.

Reply 4

Barny

But isn't er just giving us the position vector? To find the direction of the vector field you need to differentiate which gives you ephi? This is in SPC not cylindricals by the way. Also doesn't the fact that the curl is not equal to zero imply some sort of rotational element to the vector field?

Reply 5

generalebriety

Position vector of what? I'm pretty confident that if you're asked to sketch a vector field A(r, theta, phi), you're asked to draw arrows at various points whose lengths represent the magnitude of A there and whose directions represent the direction of A there. Of course, I'm prepared to be told otherwise, because this really isn't my speciality. Finding the rate at which the vector field changes is done by differentiation - I can't really see why you'd differentiate it (or which variable you'd differentiate with respect to) to find its direction.

Whether it's in sphericals or cylindricals doesn't really matter - in the plane z=0, the two are equivalent (and equivalent to plane polars) anyway. In either sphericals or cylindricals, the curl is perpendicular to the plane z=0 (try it - in sphericals it's in the direction e_phi, and in cylindricals it's in the direction e_z, which of course in the plane z=0 are in the same direction), which to my mind implies that the component of the curl in the plane z=0 is zero, and hence that there's no rotational element in the plane, even though there might be along some other surface.