Improper use of [nabla operator] in vector analysis

[Astronuc]

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 e_i$\partial_i$.

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

 

[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.

(I didn't have access to your link, I think)

 

[Astronuc]

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."

 

[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.

 

[neutrino]

Astro, did you mean to say Byron and Fuller?

Btw, interesting topic. :)

 

[Astronuc]

Astro, did you mean to say Byron and Fuller?

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. ).

Btw, interesting topic. :)

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'?

 

[TD]

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."

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

 

[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?

 

[Astronuc]

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

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

 

[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"?

 

[FrogPad]

This thread is awesome.

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'?

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.

 

[neutrino]

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'?

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.

 

[Hurkyl]

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")

 

[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!

 

[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

 

[neurocomp2003]

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

 

[Tomsk]

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

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.

 

[Swapnil]

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.

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.
https://www.physicsforums.com/showthread.php?p=1089868#post1089868

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

 

[arildno]

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.

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

 

[Swapnil]

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

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.

 

[arildno]

No, you don't. You haven't done it correctly, as I said.
You must ALSO remember to differentiate the unit vectors.
We have the relations, for cylindrical coordinates :
$$\frac{\partial\vec{i}_{r}}{\partial\theta}=\vec{i}_{\theta},\frac{\partial\vec{i}_{\theta}}{\partial\theta}=-\vec{i}_{r}$$
And differentiation of unit vectors with respect to either r or z yields zero.

Thus, we have:
$$\nabla\cdot\vec{F}=(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{k}\frac{\partial}{\partial{z}})\cdot(F_{r}\vec{i}_{r}+F_{\theta}\vec{i}_{\theta}+F_{z}\vec{k})=$$
$$\vec{i}_{r}\cdot\frac{\partial\vec{F}}{\partial{r}}+\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}+\vec{k}\cdot\frac{\partial\vec{F}}{\partial{z}}$$
Retaining only those terms yielding non-zero dot products gives:
$$\frac{\partial{F}_{r}}{\partial{r}}+\frac{F_{r}}{r}+\frac{1}{r}\frac{\partial{F}_{\theta}}{\partial\theta}+\frac{\partial{F}_{z}}{\partial{z}}$$
which is the divergence in cylindical coordinates written in an equivalent form to your own.

 

[Swapnil]

$$\nabla\cdot\vec{F}=(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{k}\frac{\partial}{\partial{z}})\cdot(F_{r}\vec{i}_{r}+F_{\theta}\vec{i}_{\theta}+F_{z}\vec{k})=$$
$$\vec{i}_{r}\cdot\frac{\partial\vec{F}}{\partial{r}}+\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}+\vec{k}\cdot\frac{\partial\vec{F}}{\partial{z}}$$

I think you are modifying the definition of a dot product. In your last step above, it doesn't seem like you are "dotting" the above two terms.

 

[arildno]

It is a perfectly standard procedure.
The divergence is the action of an OPERATOR on a VECTOR function yielding a scalar, hence, it isn't any dot product in the usual sense between two VECTORS, even though for calculational purposes, we may ignore that technical detail.

 

[Swapnil]

It is a perfectly standard procedure, your objection is utterly irrelevant.
The divergence is the action of an OPERATOR on a VECTOR function yielding a scalar, hence, it isn't any dot product in the usual sense between two VECTORS, even though for calculational purposes, we may ignore that technical detail.

See, so you did modify the meaning of the dot product. In my previous post I was using the standard meaning of a dot product, one which most students think of when they see $$\nabla\circ\vec{F}$$, where the "product" between $$\frac{\partial}{\partial x_i}$$ and the scalar function $$F_i$$ is treated as taking the partial derivative of that function.

 

[arildno]

Well, YOU modified the meaning of MULTIPLICATION in the first place, since you said:
$$\frac{\partial}{\partial{x}}*F=\frac{\partial{F}}{\partial{x}}$$
I, for one, didn't know that $\frac{\partial}{\partial{x}}$ is a real number..

 

[Swapnil]

Well, I think it is less confusing than completely modifying the definition of the dot product like you did.

Anyways, how would you would you modify the definition of the cross product, $$\nabla\times\vec{F}$$, so that it gives you the right result?

 

[arildno]

Quite simple, in exactly the same way:
Apply the differential operator(s) on the vector function (including diff. of unit vectors) prior to performing the cross product.

As I've said before, it is a perfectly standard procedure found in every decent book on continuum mechanics; even Scaum's Outlines has it.

 

[arildno]

Anyhow, I can't see what you mean by "completely" modifying the meaning.
How would you differentiate $$\frac{\partial\vec{F}}{\partial\theta}$$
where F is given in polar coordinates?
You wouldn't forget to differentiate the unit vectors, would you?

 

[Swapnil]

I wasn't talking about about the fact that you differentiate the unit vectors. My understanding is that the dot product is defined the following way:

$$\vec{A}\circ\vec{B} \equiv \sum_{i=1}^{3} A_i B_i$$
where A and B are vectors in 3 dimension.

Thus, in this context, you are not taking the partial derivative of the function itself but you are taking the partial derivative of its scalar component(s). So if $$A_i = \frac{\partial}{\partial x_i}$$ and $$B_i = F_i\;$$, then you would have $$A_i B_i = \frac{\partial F_i}{\partial x_i}$$ (eventhough I agree that the "product" is not literally a product).

 

[arildno]

Remember that the standard dot product is DISTRIBUTIVE!
(This lies at the very heart of the definition of an inner product, of which the dot product is an example)
That is to say:
$$\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}$$
That property of distributivity in the standard dot product is wholly analogus to splitting up the gradient operator into constituent operators first.
A constituent operator would for example be the "vector"
$$\vec{i}_{r}\frac{\partial}{\partial{r}}$$

 

[Swapnil]

I am not too sure how you are using the distributive law of dot products to justify your claim. Could you please expand upon what you said.

 

[arildno]

Sure, I'll take a simple example using a STANDARD dot product.
Let us look at the 2-D case, but let our vectors be represented in DISTINCT bases:
$$\vec{v}=v_{1}\vec{i}_{1}+v_{2}\vec{i}_{2},\vec{w}=w_{1}\vec{j}_{1}+w_{2}\vec{j}_{2}$$
where the i's are orthogonal unit vectors, likewise the j's, but the i's are not necessarily perpendicular to the j's.
Thus, we have:
$$\vec{v}\cdot\vec{w}=\sum_{m=1}^{2}\sum_{n=1}^{2}(v_{m}\vec{i}_{m}\cdot{w_{n}\vec{j}_{n})$$
Knowing the values of the scalar products between the i's and j's then completes the picture.


In the case of an operator "vector", we just need to know the derivatives of unit vectors as well. Essentially, then, the operator acts as if it were written in a sort of different base than the regular vector (this is not surprising, since, strictly speaking, it isn't a vector in the first place).

 

[Swapnil]

Oh, I see. So going back to your previous post about divergence, you have

$$\nabla\cdot\vec{F}=(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{k}\frac{\partial}{\partial{z}})\cdot (F_{r}\vec{i}_{r}+F_{\theta}\vec{i}_{\theta}+F_{z} \vec{k})$$

and then you using the definition of dot product you gave in the last post we get:

$$\nabla\cdot\vec{F}=\Big(\vec{i}_{r}\frac{\partial}{\partial{r}}\Big)\cdot \vec{F} +\Big(\vec{i}_{\theta}\frac{\partial}{r\partial\theta}\Big)\cdot \vec{F} + \Big(\vec{k}\frac{\partial}{\partial{z}}\Big)\cdot \vec{F}$$

Then you get:

$$\vec{i}_{r}\cdot\frac{\partial\vec{F}}{\partial{r} }+\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}+\vec{k}\cdot\frac{\partial\vec{F}}{\partial{z}}$$

The only "muddy" part is in the last step where you "move" the partial derivatives from the unit vectors to the vector function. Am I correct?

 

[arildno]

The only muddy part is that I switch the "dot" and the differential operator, and in that sense "moves" the diff. operator.
Apart from that switch, we can proceed as usual, for example:
$$\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}=\vec{i}_{\theta}\cdot(\frac{\partial{F}_{r}}{r\partial\theta}\vec{i}_{r}+\frac{F_{r}}{r}\vec{i}_{\theta}+\frac{1}{r}\frac{\partial{F}_{\theta}}{\partial\theta}\vec{i}_{\theta}-\frac{F_{\theta}}{r}\vec{i}_{r}++)=$$
$$\frac{F_{r}}{r}+\frac{1}{r}\frac{\partial{F}_{\theta}}{\partial\theta}$$
the other dot products cancelling by orthogonality relations.

I think that is what you meant by moving the diff. operator?


I would like to emphasize that this manipulative trick has nothing whatsoever to do with a proof of how the curl or div should look like when written in coordinate form.
Rather, it is a calculation device that brings out the correct formulae.

 

[Swapnil]

I meant that you treat the differential operator as a constant and move it to the vector function like the following:
$$\Big(\vec{i}_{r}\frac{\partial} {\partial{r}}\Big)\cdot \vec{F} = \vec{i}_{r}\cdot\Big(\frac{\partial\vec{F}}{\partial r}\Big)$$

But I think that's what you meant too by switching the dot and the differential operator.