16
$\begingroup$

I learned vector analysis and multivariate calculus about two years ago and right now I need to brush it up once again. So while trying to wrap my head around different terms and concepts in vector analysis, I came to the concepts of vector differentiation, gradient, divergence, curl, Laplacian etc.

The reference that I'm using is very inadequate to give any geometric/physical interpretetions of these (almost) new concepts. I don't have much problem with their formulae and working rules, but I want to look at them in a little more geometric way. For example, the geometric significance of gradient that I got out of my book is the following: if $f:\mathbb{R}^3\to \mathbb{R}$ be a differentiable function then $\nabla f (x,y,z)$ is the vector perpendicular to the level surface $f(x,y,z)=c$ ($c$ constant) at the point $(x,y,z)$.

I would really appreciate if anyone can explain how, in this way, can divergence, curl and Laplacian be interpreted geometrically. (e.g. For a given vector field $\textbf{F}: \mathbb{R}^3\to \mathbb{R}^3$, the relation between $\nabla\times\textbf{F}$ and $F$ and questions like that.) I looked around Google a bit, but couldn't find what I was looking for.

Thanks in advance.

$\endgroup$
4
$\begingroup$

Imagine a volume $V$ (with boundary $\partial V$) centered at a point $p$.

The divergence of $\nabla \cdot F$ of a vector field $F$ can be seen as the limit

$$\nabla \cdot F = \lim_{V \to 0}\frac{1}{V} \oint_{\partial V} F \cdot \hat n \, dS$$

It's not too difficult to geometrically interpret this integral. This is a flux integral--it tells us how much $F$ is normal to the surface elements $\hat n \, dS$. A function with positive divergence must be pointing mostly radially outward from a point--it diverges from that point.

The curl can be constructed in a similar way:

$$\nabla \times F = \lim_{V \to 0} \frac{1}{V} \oint_{\partial V} \hat n \times F \, dS$$

It's probably easiest to picture this in 2d: there, $\partial V$ is a circle and $\hat n$ points radially outward. The curl, then, measures how much $F$ is perpendicular to $\hat n$, or how much it curls around our central point (and if it does curl around, in what direction is it curling?).

Nothing really comes to mind for the Laplacian, but hopefully this helps.

$\endgroup$
  • $\begingroup$ I think the integrals are missing a division by volume and area, respectively. $\endgroup$ – kaba Jan 11 '17 at 15:28
2
$\begingroup$

Here are the ones I remember:

The gradient is as you described it. Also, the gradient points in the direction of "fastest increase" through the field. That gels nicely with the intuition you gave, since it seems intuitive tha the normal to the level curve (which is a region of constant value) would point either upward or downward through other values.

If you imagine a little propeller in the vector field, the curl at that point is the tendency of the propeller to be turned by the flowlines.

The divergence at a point is the tendency of the field to flow outward or inward to that point.

The Laplacian is the one I'm least familiar with, and seems to be the hardest to come up with a picture for. If you interpret it as a combination of the divergence and gradient above, it is something to do with flux of the gradient. As I read about it, it appears to be useful for studying dissolving substances.

$\endgroup$
0
$\begingroup$

Gradient ($\mathbf{\nabla}$)

Imagine you are hiking on a hill. If you can find the steepest way to climb up, you have found the gradient. The hill is a surface and gives you the elevation($h$) as a function of $x$ and $y$.

$s=\mathbf{\nabla}h(x,y)$

$s$ gives you the slope of the tangent on the surface.

Divergence ($\mathbf{\nabla} \cdot$) see also here ,

Imagine you have an empty bucket and you open the tap. The bucket starts filling up with water. Can you imagine where the water is going? Seeing from above the water comes outwards, and you have found the divergence.

$ \nabla \cdot \mathbf{V}(x,y) > 0 $

Positive divergence means that you have a source.

Now let's close the sink and let the bucket rest. No movement of water gives you zero divergence (no source nor sink).

$ \nabla \cdot \mathbf{V}(x,y) = 0$

Finally, drill a hole in the bottom of the bucket, such that the water starts leaving the bucket

$ \nabla \cdot \mathbf{V}(x,y) < 0$

Negative divergence means you have a sink. Let's call $\mathbf{V}$ our velocity field. Btw, I'm not considering compressibility for this case.

Laplacian ($\nabla^2 = \Delta$)

Let's make some tea. Initially you have your teacup with hot water. Now let's put a teabag inside. You will notice that the water starts getting darker, in other words the tea is diffusing.

$h=\nabla^2 C(x,y,z)$

Now your laplacian tells you how homogeneous your mixture is. The more homogeneous the smaller your laplacian becomes. Eventually, without stirring it all your tea will diffuse in your teacup.

Curl ($\mathbf{\nabla} \times$)

Let's stay with our teacup and start stirring it.

$\mathbf{S}=\mathbf{\nabla} \times \mathbf{V} $

The faster you stir your cup, the larger your curl (or vorticity) of your flow becomes.

$\endgroup$
  • $\begingroup$ Unless I'm misunderstanding your analogy, you've got divergence backwards. If water is flowing out of the bucket, then the bucket is a source and likewise if water is flowing into the bucket, then the bucket is a sink. $\endgroup$ – user137731 Nov 23 '16 at 20:58
  • $\begingroup$ @Bye_World The tap acts as a "source" of water and the hole acts as a "sink" $\endgroup$ – ilciavo Nov 23 '16 at 21:14
  • 1
    $\begingroup$ I see. I think a have a better analogy (but feel free to keep yours if you like it better): There's a plane with small valleys and hills and such. It rains. The places where the water collects (the low spots like valleys) are the sinks (because while it's raining water tends to flow into this) and the places where the water runs off (the tops of the hills) are the sources (because while it's raining water tends to flow away from these spots). Anywhere that's flat will be divergence-free. Those will be the places where water doesn't collect much but it also doesn't really run away either. $\endgroup$ – user137731 Nov 23 '16 at 21:18
  • $\begingroup$ @Bye_World I think the divergence and gradient will be zero on valleys and peaks, I like the example of the bucket because I usually have a tap in the class. $\endgroup$ – ilciavo Nov 23 '16 at 21:24
  • $\begingroup$ Given divergence measures the tendency of the water to "spread out" (positive divergence) or "collect" (negative divergence), the peaks and valleys will not have $0$ divergence (unless they happen to be plateaus, but that's not the generic case). But if you like your example better, as I said, feel free to keep it. $\endgroup$ – user137731 Nov 23 '16 at 21:34

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.