# Improper use of [nabla operator] in vector analysis

1. Sep 10, 2006

### Staff: Mentor

I just happened across this paper which caught my attention.

http://hdl.handle.net/2027.42/7869

Perhaps this has been discussed here before, but it's new to me and I have learned from several of these books.

What is the correct use or intepration of $\nabla$ and it's use in the 'div' and 'curl'?

What textbooks do contain the correct use?

Apparently "Mathematics of Classical and Quantum Physics", by Byron and Fuller, does give an appropriate treament, since if refers to gradient, div and curl as differential operators, and $\nabla$ is referred to as an operator ei$\partial_i$.

Byron and Fuller explicitly mention that [nabla] is an operator and should not be thought of a vector.

Last edited: Sep 10, 2006
2. Sep 10, 2006

### arildno

Well, you can perfectly well treat $\nabla$ AS IF it is a vector (or at least shares some properties with vectors..), as long as you proceed with some care.
Let's take 2-D polar coordinates as a case example:
The gradient operator may then be written as:
$$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}$$
Let's calculate the divergence of a vector function $\vec{F}=F_{r}\vec{i}_{r}+F_{\theta}\vec{i}_{\theta}$
The divergence can then be written as:
$$\nabla\cdot\vec{F}=\vec{i}_{r}\cdot\frac{\partial\vec{F}}{\partial{r}}+\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}$$
By doing the differentiations (including the differentiations of unit vectors!!) and then perform the dot products yields the correct formula for the divergence of a vector as expressed in polar coordinates.
The same procedure can be done in the 3-D case for the "curl", by taking the cross products at the end of our calculations.

Last edited: Sep 10, 2006
3. Sep 10, 2006

### Staff: Mentor

Tai makes the point that, the divergence is not a dot product because,

$$\nabla\cdot\vec{F}\,\neq\,-\vec{F}\cdot\nabla$$, although I think it is suppose to be

$$\nabla\cdot\vec{F}\,\neq\,\vec{F}\cdot\nabla$$,

in other words, it's non-commutative, which is a property of the dot product.

I remember a discussion of 'dot' vs 'inner' product here somewhere. I have heard them used interchangeably, which apparently is incorrect. It's something like "a square is (always) a rectangle, but a rectangle is not (always) a square."

Last edited: Sep 10, 2006
4. Sep 10, 2006

### arildno

Well, of course Tsai is correct, that is why I said that the gradient operator shares SOME (but not all!) of the properties of vectors.

The (pseudo)-dot products between the gradient operator and a vector (function) is non-commutative and yields in one case a new scalar function and in the other case a new ("scalar") operator.

For computational purposes, however, it doesn't matter much, as long as one computes it in the direction as given.

5. Sep 10, 2006

### neutrino

Astro, did you mean to say Byron and Fuller?

Btw, interesting topic. :)

6. Sep 10, 2006

### Staff: Mentor

Yes, thanks, neutrino. I had Byron White (associate justice of the US Supreme Court in the back of my mind Looks like my mind needs a good cleaning. ).

It is interesting because I've seen some comments in the past, but this is first time I've seen an article about the matter, and I have used several of the textbooks in the past, and still use some as reference. Mathematics is supposed to be 'precise', and I guess it troubles me that in practice this is not always the case.

It also occurs to me that the improper use of nabla is an example of one of many popular misconceptions in mathematics and the sciences. It's a bit like study a language by learning a local dialect and then being misunderstood in a different region where a different dialect of the same language is spoken.

I also wonder if there are other mathematical misconceptions in the sciences, and based on some of the apparent misunderstandings in physics, would it be worthwhile to have a thread/discussion on 'popular misconceptions in mathematics and physics'?

Last edited: Sep 10, 2006
7. Sep 10, 2006

### TD

Correct: the dot product is an example of an inner product, but there exist other inner products.

8. Sep 10, 2006

### fourier jr

off-topic: can anyone guess why it's called nabla? & give one of the alternative names for it when vector analysis was being developed?

Last edited: Sep 10, 2006
9. Sep 10, 2006

### Staff: Mentor

I think I've heard it referred to as "Del".

By the way, here is Tai's other paper - A historical study of vector analysis
http://hdl.handle.net/2027.42/7868

10. Sep 10, 2006

### fourier jr

the name nabla was suggested to peter guthrie tait (doesn't sound familiar to me) by robertson smith because it looked like an assyrian harp. maxwell wrote a letter to tait because he didn't understand what it was & asked what it was called, maybe "atled"?

11. Sep 10, 2006

That would be fantastic. Not to bring Leibniz Notation into the mix (it's been discussed enough), but I find topics such as the improper use of nabla, or why dy/dx is treated as one symbol very interesting.

And by the way... thank you for that PDF on vector analysis. I skimmed through it, and it looks like a fantastic read. Once I get some spare time, I'm going to go through it.

12. Sep 11, 2006

### neutrino

If such a thing is done, it should be made a sticky.
Thanks for the other link as well. I've been meaning to read more on the history of vector analysis.

Last edited: Sep 11, 2006
13. Sep 11, 2006

### Hurkyl

Staff Emeritus
Well, technically speaking, $\nabla$ satisfies the only criterion required to be a vector: it's an element of a relevant vector space. In this case, the space of (scalar-valued) differential operators (on vector functions).

As I understand it, the dot product is not a vector operation per se -- it is an operation on tuples. (It's just that we're usually doing arithmetic with tuples of real numbers) For any adequate pair of "addition" and "multiplication" operations, you have the corresponding dot product between the corresponding tuples.

I think the only real abuse here is calling $\nabla \cdot \vec{f}$ a dot product -- it's only a dot product when you're looking at the coordinate representations in an orthonormal basis. (But it's a common abuse -- just like we call the usual inner product on Euclidean space a "dot product")

14. Sep 11, 2006

### Swapnil

Personally, I think that $$\nabla\circ\vec{F}$$ and $$\nabla\times\vec{F}$$ are extremely bad notations for "divergence" and "curl" because strictly speaking these are only true in cartesian coordinates. For example, if you dot the gradient operator in spherical coodinates with $$\vec{F}$$ you wouldn't get the divengence in spherical coordinates. You would instead have to derive divergence by using the "flux per unit volume" definition. There is no other way to do it.

edit: Warning: Any words with the word "url" like "curl" are going to cause some serious trouble. That's why I put it in quotes. BEWARE!!

Last edited: Sep 11, 2006
15. Sep 11, 2006

### Swapnil

I found one more article by Tai, which actually clearly tells you what are the misconceptions about the Hamilton/nabla/del operator. Plus, he gives a new unified definition for "gradient," "divergence" and "curl." Very Cool!!!!

http://deepblue.lib.umich.edu/handle/2027.42/21026

Last edited: Sep 11, 2006
16. Sep 11, 2006

### neurocomp2003

i thought the summation property had priority over commutative when it comes to the dot product?

17. Sep 19, 2006

### Tomsk

Interesting link, thanks, I have a question though. In equation 6 it says h_i is the metric coefficient, what is that? I can't find anything on it on wikipedia or google, it must just be a small part of how coordinate systems work. Is it something to do with, or related to the Jacobian thing from vector calc?

There's a lot more of this article I don't understand, but I won't ask about it, it's too advanced for me! I got the gist of it though.

Last edited: Sep 19, 2006
18. Sep 19, 2006

### Swapnil

The stuff about metric coefficients isn't too hard. If you understand divergence and curl and know about spherical coordinates and cylindrical coordinates, then you should should be able to understand what metric coefficients are all about.

Metric coefficients usually come up when you are doing coordinate tranformations. Here is a link on this forum that talks about deriving unit vectors in spherical coordinates and it should give you a pretty good idea of what metric coefficients are.

Here an article that I wrote on PlanetMath just an hour ago! Its pretty abstract but short and I am still working on it though. After you finish reading the above discussion, read the following article and see what you can understand.
http://planetmath.org/?method=l2h&from=collab&id=83&op=getobj

Last edited: Sep 19, 2006
19. Sep 20, 2006

### arildno

Complete and utter nonsense. Of course you can!
You just have to do it CORRECTLY, that's all.

Last edited: Sep 20, 2006
20. Sep 20, 2006

### Swapnil

Obviously you have no idea what you are talking about. Let me show you. For example, in cylindrical coordinates the gradient takes the following form:

$$\nabla = \frac{\partial}{\partial\rho}\hat{e}_{\rho} + \frac{1}{\rho}\frac{\partial}{\partial\phi}\hat{e}_{\phi} + \frac{\partial}{\partial z}\hat{e}_z$$

If you dot it with $$\vec{F}$$ then you get:

$$\frac{\partial F_{\rho}}{\partial\rho} + \frac{1}{\rho}\frac{\partial F_{\phi}}{\partial\phi} + \frac{\partial F_z}{\partial z}$$

BUT the CORRECT form of divergence in cylindrical coordinates is:

$$\frac{1}{\rho}\frac{\partial (\rho F_{\rho})}{\partial\rho} + \frac{1}{\rho}\frac{\partial F_{\phi}}{\partial\phi} + \frac{\partial F_z}{\partial z}$$

Hence, you can't just naively DOT the gradient in cylindrical coordinates with the function to get the divergence in cylindrical coordinates.

Last edited: Sep 20, 2006