Talk:Del

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[edit] page title

Is "del" really the most common name of the operator? I have heard the names "nabla", "grad" and "gradient" before, but never "del". Maybe the page should be renamed to "Nabla" and just mention "del" as an alternative name for the operator? -- Jochen

The formal operator was certainly called "del" in all of my mathematics and physics courses, which also mentioned the (infrequently used) name "nabla operator". The operations it is typically used to construct are called "div", "grad" and "curl". "Del" is one syllable, "nabla" two, so when you have to speak a lot of equations, it's more natural to use "del". It's also easier to say "del squared" than "laplacian of". -- Karada 11:17, 7 Nov 2004 (UTC)
Asking Google for "del operator" gives 1620 hits, asking for "nabla operator" gives 2380. This would indicate that "nabla" is more common. -- J.Voss 13:50, 7 Nov 2004 (UTC)
But, then, most people say "del", not "the del operator," so it's not clear that this is meant. As below, the google fight for "del+math" vs. "nabla+math" shows del+math on top. So I'm not willing to say either is definitive proof. 24.59.199.228 09:40, 23 November 2006 (UTC)
Just to add to this, the del symbol itself isn't the same as the "grad" operation; you only get the grad operation by "multiplying" a scalar field by del. -- Karada 11:21, 7 Nov 2004 (UTC)

Perhaps it could also be notes that sometimes the Δ symbol is used for \nabla^2. And even the tex editor you use here for math editing uses nabla instead of del. --Jaap 20:00, 18 Apr 2005 (UTC)

It's always been del to me, from calculus to physics at umass. also: phi is 'fee'

See http://www.googlefight.com/index.php?lang=en_GB&word1=del+math&word2=nabla+math

But compare http://www.googlefight.com/index.php?lang=en_GB&word1=%22nabla+operator%22&word2=%22del+operator%22 --Jochen 10:09, 29 September 2005 (UTC)
I am also in favour of a move of del to nabla. I think "del" is a manifestation of loss of culture (so we should fight against this). There is no sense imho of calling this "del". The symbol is a reversed delta, OK, but then call it "led" or "atled". (Besides the fact that many other things would deserve the WP "Del" page much more than this one...) — MFH:Talk 15:13, 16 March 2006 (UTC)
Wikipedia isn't a forum for the preservation of culture. Nabla is the symbol, del is the operator, that makes sense, and the rest is nothing to fret over. 24.59.199.228 09:40, 23 November 2006 (UTC)
PS: if you add Ostrogradsky, it's 27 against 4 - as I said, a problem of culture... (I better don't mention the result for Ostrogradski... ;-) — MFH:Talk 15:44, 16 March 2006 (UTC)

[edit] Coordinates

Is it best to expand on how to convert del to different coordinate systems here or in the other coordinate systems? This is done very poorly at them moment, from new unit vectors (and relationship to i,j,k), why you get (1/r).(d/d theta) e.t.c. --rex_the_first 18:30, 05 Nov 2005 (GMT)

Yeah, sorry about that; I know it's not incredibly obvious. Sadly, there are easy ways to do it (h-factors), but they look way too handwavy for wikipedia, and there are rigorous ways to do it (actual coordinate substitution), but they look way too imposing and deep for Wikipedia. Unless people have proofs that I haven't seen, I don't think anything like that is coming to wikipedia. 24.59.199.228 09:40, 23 November 2006 (UTC)

[edit] So what is ∇?

The page defines ∇ beautifully, without actually saying what it is/does. In other words, unless you're a mathematician, the page is meaningless.

There must be a real life example that will clarify what ∇ does. I believe it could be applied to the surface of an irregular hill, in which case, what do ∇, d/dx, etc, represent? Do the d/dx, etc represent the slopes in three different coordinates, in which case ∇ is what? The direction in which a ball will roll? Something else? --Iantresman 13:28, 13 June 2006 (UTC)

Del is an operator. It does different things depending on what you ask it to do. Gradients of potential energy fields are forces. The divergence of the electric field is related to the charge density at a point. Et cetera.

In the final paragraph of the section on Gradient, we are suddenly confronted with an example of vector fields(?) being directly multiplied with the \nabla operator (rather than multiplying with the results of \nabla). This could be done more gently, at least by explaining what the \vec u represents, references to dot and cross products over vector fields, and the definition/representation of \nabla as a vector field in itself. --Pgagge 12:57, 10 June 2007 (UTC)

[edit] Confusing?

The article says:

Since del does not really have a direction, this is hardly expectable

How does del 'not really have a direction'? It's a product of basis vectors, right, so isn't it just another vector? -- Ornette 10:39, 22 June 2006 (UTC)

Thanks, I cut out that part. Oleg Alexandrov (talk) 23:30, 22 June 2006 (UTC)
I added what I believe this was talking about to the end of the article. Del *doesn't* have a direction -- not in the same way that k does. kx has the same direction as ky, but ∇x and ∇y are orthogonal. Real directions don't act that way under scalar multiplication. 24.59.199.228 09:40, 23 November 2006 (UTC)


[edit] Etymology

I think it would be nice to know where the name "del" came from. (My guess is likely because it looks like a delta upside-down...perhaps?)

[edit] error in identity 4. ?

I guess there is, as I write this, since I don't see an equality sign in 4. Correction please? Thanks

Identities (3) and (4) did not actually make any formal sense and were eliminated in my most recent edit. 24.59.199.228 09:26, 23 November 2006 (UTC)


[edit] Additional formulas

Please include something like this http://planetmath.org/encyclopedia/GradientOfVector.html into the page. I hate having to look all over the internet when i do my math =)

[edit] Product rules with del

In analogy to the one shown in the gradient section, I added some product rules in that same and other sections. Do you think it breaks the flow too much? Perhaps it would be better to move the examples to a separate section on product rules. −Woodstone 16:20, 3 April 2007 (UTC)

[edit] Gradient operator not a vector?

Commenting on the last part of this article:

"Central to these distinctions is the fact that del is not a vector — it is a vector operator. Whereas a vector is an object with both a precise numerical magnitude and direction, del doesn't have a precise value for either until it is allowed to operate on something."

Is this true considering that differential operations are by definition linear. Central to this question is whether or not we can create a vector space of differential operators. Based on the stringent requirements that define a vector space, we can indeed come up with a set of differential operators that define a vector space. Note that magnitude and multiplicative commutation are not requirements. If this is the case, then we can create a linear combination of differential operators that span a basis and can therefore create any arbitrary differential vector operator that can be classified as a vector. Then the next question is, can we have vector in euclidean space with differential elements and still classify it as a vector? It would seem that even in this case, the elements in the gradient operator can still define a vector space (maybe a subspace in euclidean space) which then suggests that the gradient operator is indeed a vector.

Anyone with a strong mathematics background who can comment on this? —Preceding unsigned comment added by 128.193.140.64 (talk) 21:54, 17 October 2007 (UTC)

[edit] Einstein Notation

If I remember correctly, in summation notation you can't have the dummy as a subscript in both terms. Maybe \partial_i e^i is more appropriate? Pwsnafu (talk) 02:59, 11 April 2008 (UTC)