This section contains a number of graphs which were generated by an applet on the realtive velocity approach to understanding the divergecne and curl of vector fields that we will explain here.
As usual, the graph range in the applet is adjustable, but all of the graphs in this subsection are all centered on the origin and cover the range -3 < x < 3 and -3 < y < 3 unless otherwise indicated.
Now, let's start with the visualization of divergecne:
Consider the vector field F(x,y) = (x,y). For this vector field, div(F(x,y))= 2. If we look at a graph of the vector field, we indeed see what looks like an outward flux from the origin: The velocity vectors all point out from there.
However, the divergence is 2 at all points -- how do we see this positive divergence at other points?
The answer is this: Think of F(x,y) as the velocity field of a flowing fluid. Imagine a bunch of tiny particles of something or other flowing along in this stream, and picture yourself on one of them.
One thing of interest to you will be the relative velocity of the other particles with respect to yours -- i.e., how quickly they are headed for a colision with you, or how quickly you are being separated from the rest of the bunch. There relative velocities are of course computed by subtracting the instantaneous velocity at your present location F(x0,y0) from the instantaneous velocity at the other locations; i.e., you form the relative velocity field:
F(x,y) - F(x0,y0)
The first thing to notice is that since we subtracted off a constant vector field ((x0,y0) is fixed!), we didn't change either the divergence or the curl. But we will have changed the graph.
Now the vector field vanishes at (x0,y0), and if the vectors nearby are pointing away from (x0,y0), then the divergence at (x0,y0) is positive. The nearby particles in the stream will all be getting more distant from you -- i.e., diverging.
On the other hand, if the vectors nearby are pointing towards (x0,y0), then the divergence at (x0,y0) is negative. The nearby particles in the stream wil all be getting more distant from you -- i.e., converging.
A third possibility is that some nearby vectors are pointing towards, and some away from (x0,y0). Then it comes down to whether the net tendency is to converge or diverge.
In our example F(x,y) = (x,y), if you form the relative velocity field F(x,y) = (x-x0,y-y0) the effect is simply to shift the picture so that the old origin sits on the new reference point. The relative motion is the same everywhere for this vector filed, and that's why it's divergence is constant.
Here you see the relative velocity field graphed for two different reference points (x0,y0) which are indicated in red:
The vector field that we have just been considering also has constant curl -- it is zero everywhere. That too can be recognized from the picture, but it may help to consider a case where the curl isn't zero before going into that.
First, here is an example of the "third possibility": We take F(x,y) = (2*x,-y). The vector field itself is graphed on the left, and the realtive velocity near the point marked in red is graphed on the right. The arrows going out at theat point are longer than the arrows going in, so this is positive divergence -- as you can easily compute!
Now for the curl. Consider the vector field G(x,y) = (-y,x). For this vector field, curl(G)(x,y)= 2. If we look at a graph of the vector field, we indeed see what looks like counter-clockwise circulation around the origin:
The velocity vectors all swirl around there counter-clockwise. Notice that there was none of this "swirl" in the graph of the vector field in our first example. And indeed, as we pointed out, it has zero curl.
Back to this example, the curl is 2 at all points -- how do we see this positive curl at other points? Again, by refering to the relative velocity field
G(x,y) - G(x0,y0)
As before, notice that since we subtracted off a constant vector field ((x0,y0) is fixed!), we didn't change either the divergence or the curl. But we will have changed the graph.
Now the vector field vanishes at (x0,y0), and if the vectors nearby are swirling counter-clockwise around (x0,y0), then the curl at (x0,y0) is positive.
On the other hand, if the vectors nearby are swirling clockwise around (x0,y0), then the curl at (x0,y0) is negative.
A third possibility is that some nearby vectors are swirling counter-clockwise, and some clockwise, around (x0,y0). Then it comes down to whether the net tendency is clockwise of counter-clockwise.
In our example G(x,y) = (-y,x), if you form the relative velocity field G(x,y) = (-y+y0,x-x0) the effect is simply to shift the picture so that the old origin sits on the new reference point. The relative motion is the same everywhere for this vector filed, and that's why it's curl is constant.
Here you see the relative velocity field graphed for two different reference points (x0,y0) which are indicated in red:
Next here is a vector field with both the divergnce and curl positive:
The vector field is (x-y,x+y). Notice that it has both "swirl" and "outward streaming". And again, since P and Q are linear functions of x and y, the divergence and curl are constant -- both are 2 -- and passing to the relative velocity field just has the effect of shifting the graph. From the realtive standpoint, all points look the same.
Finally let's look at an example in which the divergence and curl are not constant. Consider the case (x2-y,x+y2), which is graphed below.
Now what is going on at, say, the point (1.5,1.5), which is in the middle of the upper right quadrant??
It is hard to say from this picture, but les's "zoom in" on that point, and graph the relative velocity. In the following graph, the cross-hairs are centered on (1.5,1.5), and the graph covers the range 1 < x < 2 and 1 < y < 2.
What you see here is positive divergence and positive curl. Everyone of the nearby vectors points outward, so there is no question about the divergence being positive. As for the curl, notice that all of the closest vectors are rotated a small bit counter-clockwise from pointing straight out, so there is no question about the curl being positive either. And you can check by computation that that is right.