Talk:Laplace operator/Archive 1

Rewriting this article
There should really be mention of the Laplace operator on non-Euclidean spaces, e.g. -y^2(f_xx + f_yy) on the upper half-plane.
 * Yes, the Laplacian should be discussed in a differentiable manifold. Its definition reads $$\Delta:=(\mathrm{d}+\mathrm{d}^*)^2$$. This will be a MAJOR rewrite. I'm inclined to undertake this, unless there is resistance.MarSch 16:24, 18 Mar 2005 (UTC)


 * Be bold! However, it is not clear to me what exactly you want to rewrite. Do you just want to add to this article one or more sections about &Delta; on manifolds? I think it is good to keep at least the first half of the article the way it is, meaning more elementary, and then get into more complicated issues in sections below, or even in a new article if that new stuff becomes really a lot. Anyway, give it a try, and let us see how it goes. Oleg Alexandrov 17:13, 18 Mar 2005 (UTC)


 * Ugh. I completely disagree with the form of the recent edits by User:MarSch. As a geometer, I like the fact that the full abstract definition has been added, but it should appear later in the article, after a simpler high-school/college-level definition.


 * Please keep in mind why people come to Wikipedia in the first place: to learn something new, to refresh thier memory, to look up a forgotten formula. There is nothing worse that one can do to a reader than to overwhelm them with abstractions they don't understand. For example, any chemist, who may have had a few semesters of quantum, would be lost in this article as it currently stands. Ditto for any structural engineer, or electronics engineer. These are people who would use wikipedia, and frankly, they outnumber the geometers by a hundred to one.  The article should cater to that level of understanding first, and then, only later, turn to the more abstract definitions.  As an example of where this works, see the definition of the discrete laplace operator, which appears at the end of the article, not at the beginining.  linas 02:02, 27 Mar 2005 (UTC)


 * I don't like how the article is turning out either. But, this content would be perfect in an article called Laplacian on manifolds. Oleg Alexandrov 02:48, 27 Mar 2005 (UTC)

I don't like the way it's written now. Certainly it should mention the general versions, but it should begin by saying that in the most concrete cases it's the sum of the second partial derivatives with respect to etc. etc. -- something that everyone (except those who don't know what partial derivatives are) will know what it says. Michael Hardy 03:33, 27 Mar 2005 (UTC)

I agree with you that the article needs to be way better, but I don't think the way to do this is by excluding math, instead we need to add motivation and an informal explanation. I've added some headings which I think are necessary and wrote some flavor sentences, which I hope are close to the truth. Please also see Wikipedia_talk:WikiProject_Mathematics MarSch 12:01, 28 Mar 2005 (UTC)


 * Hi MarSch, no one is trying to exclude math here, nor are we trying to be argumentative jerks. Let me make a simple concrete suggestion: Pull up the version of this article from 6 March, and cut and paste that version above the "formal definition" parts you just added.  Thus, the college sophmore will see the traditional $$\frac{\partial^2}{\partial x^2}+  \frac{\partial^2}{\partial y^2}+ etc.$$ that they expect to see. (This is also what most non-math-major adults expect to see as well, i.e.engineers, etc.). That gives them the warm fuzzies that they are in the right place, reading the right article.  Then, as they read further down, they get to your formal definition, and lo, the scales fall from their eyes, and they see the "true" definition.  That's all; let the article progress from the simple, concrete, pedestrian definition, to the formal, abstract, exact definition.  Hook the reader in with something they can immediately identify with, and then take them on the journey, hoping they'll follow.  linas 05:23, 29 Mar 2005 (UTC)

Ok, I have incorporated all relevant material from the 6 march version and I think it is starting to look decent :) MarSch 14:35, 30 Mar 2005 (UTC)


 * I have a suggestion regarding the intro:
 * $$ \Delta = \nabla^2 $$
 * being right in the introduction looks odd since, it is two names for the same operator and one wonders why it's there, it deserves some exposition. And this being an encyclopedia, I'm sure the $$\Delta$$ appears more often as the difference symbol.  Please provide some context for this use of $$ \Delta $$ .  Consider me simple minded but, I've never had a use for this shortcut for $$ \nabla^2 $$.  And, I really don't want to have to grade papers from sophomores pretending sophistication or whatever, with this notation used. This same shortcut is used in the article on the Laplace's equation but, then not in the Helmholtz equation.  I'd rather not see it used at all but, I'd like to know what others think.


 * The D'Alembert operator has it's own article why not just link to it as it's not so much of a mathematical extension to the Laplacian but, is a physics specific one?


 * Otherwise, I think, this article is well organized from basic to generalized and then specialized and interesting. My thanks to the authors.Kckid 23:34, 25 April 2006 (UTC)


 * I agree with Kckid's comments about the use of the $$\nabla^2$$ instead of $$\Delta$$. I feel the $$\nabla^2$$ is more widely used today than the $$\Delta$$ for the Laplacian. Nearly all the physics articles on Wikipedia are using $$\nabla^2$$ over $$\Delta$$. Also for consistency, the Mathworld article cited as a reference uses $$\nabla^2$$ notation exclusively and only mentions the alternative $$\Delta$$.--Nkrupans 16:30, 7 June 2006 (UTC)

Averages, sign
I think this statement is reversed: "the value of the Laplacian of a function is simply that amount by which a point in space is greater or lesser than the value of the averages of that point's neighbors." it should be something like: "the value of the Laplacian of a function is simply that amount by which the averages of that point's neighbors is greater or lesser than that of the value of a point in space." for instance y=x^2, the Laplacian is 2, while clearly at x=0 the value of the function is less than of the average...


 * I changed it. Think either way means the same thing, but it's (of course) better not to be confusing. (You can edit articles directly, as well as this discussion page, by the way.) &#922;&#963;&#965;&#960; Cyp  17:21, 6 Oct 2004 (UTC)
 * 2? What about 4x^2? lysdexia 13:48, 24 Oct 2004 (UTC)

Which phi
I'm wondering if it's better to use \phi or \varphi:


 * $$ \Delta t

= {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial t \over \partial r} \right) + {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial t \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 t \over \partial \phi^2}. $$

or


 * $$ \Delta t

= {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial t \over \partial r} \right) + {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial t \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 t \over \partial \varphi^2}. $$

or maybe we should write:


 * $$ \Delta t(r, \theta, \varphi)

= {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial \over \partial r} t \left( r, \theta, \varphi \right) \right) + {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial \over \partial \theta} t \left( r, \theta, \varphi \right) \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 \over \partial \varphi^2} t \left( r, \theta, \varphi \right). $$

--Jacobolus 05:26, 6 Mar 2005 (UTC)


 * I would say \varphi looks better. But then you could also consider putting \vartheta instead of \theta. Ultimately it of course does not matter as long as one is consistent. Oleg Alexandrov 05:28, 6 Mar 2005 (UTC)


 * Actually, some other pages I visited use \phi and \theta. Maybe we should keep things this way for consistency. Oleg Alexandrov 05:32, 6 Mar 2005 (UTC)

Hey. Wow that was a quick response. I think that the norm in mathematics is to use the $$\varphi$$ symbol instead of $$\phi$$ I think this looks better. Maybe we should change this across multiple articles?

--Jacobolus 05:45, 6 Mar 2005 (UTC)


 * You might need to change lots and lots of pages which have to do with spherical and cyllindrical coordinates. I would say you should not rush and ask for more opinions. You might also consider posting this on Wikipedia talk:WikiProject Mathematics where the mathematicians hang around. Oleg Alexandrov 05:51, 6 Mar 2005 (UTC)


 * Either way, take a look at Nabla in cylindrical and spherical coordinates and whatever leads from there. Oleg Alexandrov 05:51, 6 Mar 2005 (UTC)

Still disagree with recent changes
In spite of Marsch's recent edits, this article still does not look good. Instead of the original nice essay, we now have an incompletely written article, with lots of formulas, and which boldly proclaims that the Laplace operator in Rn is nothing but a particular case of the real thing, which is the Laplace operator on manifolds.

I don't know what Marsch's math biases are, but the Laplace operator in the flat space is very improtant in its own right, and needs an article about it on Wikipedia. This article, in its present form, does not do a good job. What was before, worked well.

I suggest reverting this to the March 6 version. The part about the Laplacian on manifolds, instead of being the center of the article, can become a generalizations section, or even better, a separate article. Oleg Alexandrov 23:58, 1 Apr 2005 (UTC)


 * The 6 march version is not a "nice essay". It contains little more than the definition and one property. Furthermore it contains redundant data; I don't think it is usefull to include the Laplacian in 2D in the standard basis, when there is already the 3D version and the general flat version in any dimension. It further contains some gibberish about approximating a second derivative. There should be an article about that and it should contain a much better description. I don't see that any information was lost. Also I don't read in the article any bias (bold or otherwise) against the flat version. If you have something to say about the flat version that is interesting/important then you should do it, it will probably have a nice generalization :)
 * I oppose reverting MarSch 18:28, 2 Apr 2005 (UTC)


 * You wrote in the article:
 * The earliest form of the Laplacian is due to Pierre-Simon Laplace. who introduced it when studying . At the time it was simply a partial differential operator. Thus it worked on flat space. When this was realized, attempts where made to generalize it to arbitrary smooth spaces.


 * This is why I wrote that for you it matters more what happens on manifolds than what happens in the flat space. Also, the thing either needs to be filled in, or deleted. Let us make sure the articles have some nice finish. Oleg Alexandrov 18:57, 2 Apr 2005 (UTC)


 * Now, are you sure that it was Laplace himself who introduced the operator? Oleg Alexandrov 18:59, 2 Apr 2005 (UTC)


 * No, I am far from sure. This was pure speculation on my part, as explained above, to provide some structure to be filled in and not merely some usefull headings. I was hoping someone would do something with them, so please change them as you see fit. BTW Are you sure this wasn't an April's Fools joke? -MarSch 14:18, 4 Apr 2005 (UTC)


 * I will do some changes today. By the way, it is good to indeed stick to things we really know about. It is good to hope that somebody else will come and correct that, but from my experience here, that seldom happens, and then misinformation can persist. I will also put some of that numerical stuff back about discretizing the Laplacian. It seems you are a theoretical guy. I am more applied, and the discretization of the Laplacian is a good thing to have. Oleg Alexandrov 15:34, 4 Apr 2005 (UTC)

The proof thing
Hi MarSch. If you want to keep the proof, and in general, the differential geometry part, I think you would need to do some of the following:

(a) explain what d is

(b) explain what d* is

(c) Explain your summation notation (I think you use the Einstein convention, but that is not said)

(d) Explain the * operator which you use as some kind of multiplication.

(e) explain what epsilon is or give a pointer

(f) Explain what $$\partial^i$$ is as opposed to $$\partial_i$$

(g) Explain what vol is in this context

The way the proof stands now, is several long lines of formulas, and with underfined notation. Also, if not too hard for you, could you please drop here a couple of lines explaining what the usefullness of that proof is, in this context. Thanks. Oleg Alexandrov 00:20, 13 Apr 2005 (UTC)

Hi Marsch, I agree with Oleg here. The proofs are useful, but they should go into a distinct article possibly titled proof of linearity of laplacian or something like that. And yes, you must explain the notation, link to things like Hodge star since otherwise a lot of readers will assume its "some kind of multiplication". (User:Linas forgot to sign).


 * Well, the proof could even stay in this article, if it is properly explained. Oleg Alexandrov 15:16, 13 Apr 2005 (UTC)


 * My concern is that the article will become too long. Is the proof going to be something of general value, or is it "merely" establishing the correctness of a formula? Does it illustrate an important concept, or is it a mechanical manipulation of little pedagogical interest? linas 15:36, 13 Apr 2005 (UTC)


 * This notation should be explained in an article about exterior calculus somewhere, in one place. Not scattered or repeated on each page using it. That should cover (a) to (g). THe proof is just a few lines that don't hurt anybody and that some might find very interesting. I learned a lot doing this stuff. Removing the proof does not eliminate the notation, since it is used in the def and also the derivation of laplacian in general coordinates. And yes it is also a valuable reference. Don't downplay the importance of that. What is the value of any formula that just falls out of the sky and can be edited be anyone? Also proofs should be directly besides what they prove to prevent drifting apart. MarSch 13:27, 14 Apr 2005 (UTC)


 * You have valid points. As long as you find a way (by referring to some other pages if you wish, or any other way) to explain all the notation, it should be fine. But, I do not agree leaving the proof here with the notation not explained in any way. Oleg Alexandrov 16:23, 14 Apr 2005 (UTC)


 * And you are right, you used this notation in other places too in this article besides the proof. Even more reasons to have the notation disambiguated. Oleg Alexandrov 17:47, 14 Apr 2005 (UTC)

I've added pointers for most of your points except (c) and (f). It's much better already. (f) should be in metric tensor. (c) yes summation convention and increasing multi-index. Kinda tricky to explain/state. -MarSch 15:29, 18 Apr 2005 (UTC)

I'm having problems with
 * and d* is the codifferential defined by d* = *d*,

simply because its hard to see that in one case, the star is a superscript, and in the other its not. This just looks hopelessly confusing to the casual reader. How about an alternate notation, such as $$\overline{d}$$ or d+ or something like that? Maybe a five-pointed star?


 * It seems clear enough to me, however I am willing to change it. One solution would be to enforce latex rendering to make the distinction clearer. For an alternate notation we could use &delta; which is also in use. Will check cochain complex. MarSch 11:45, 19 Apr 2005 (UTC)

Here's another example of the problem with the superscript-star notation: The definition you give here seems to be in notational collision with that in De Rham cohomology. I think you really want to get rid of the superscript-star notation and write $$\Delta=d^+d+dd^+=d*d*+*d*d$$ just to make it extra clear and unambiguous. Note that this is also the same as the notation used in the article Hodge star. Please do *not* change the notation used in those two articles; they're fine.


 * I do not see what is the collision. They use only *d* in the same way as do I. MarSch 11:45, 19 Apr 2005 (UTC)

And, while we're on the topic: I think its is more correct to point out that this operator really should be called the Laplace-Beltrami operator, and that some people call it the Hodge-Laplace operator, and that the word "Laplacian" is usually restricted to that thing defined on Euclidean space.


 * My book didn't mention this, But they seem to use this terminology in Hodge star. I will create some redirects for those names. Please feel free to include this info. MarSch 11:45, 19 Apr 2005 (UTC)

For the summation convention, just link to einstein summation convention in the article. Explain that's for the italic i. For the summation convention on the capital roman J, make sure you link the &epsilon; to Levi-Civita symbol and call it the totally anti-symmetric symbol, and that may be enough to explain the summation.

For point (f), just take the time out and just plain say $$\partial^i=g^{ij}\partial_j$$. Its OK to be wordy here, we're not trying to cram this on a postage stamp for some exam. linas 22:29, 18 Apr 2005 (UTC)

Adding factoids, theorems of general utility
Marsch, could you be persuaded to resurect some of the text from the version of 6 march 2005, dealing with the limit/discrete case? I think it was useful to note that the second derivative (in eucliden space) was f(x+h)-2f(x)+f(x-h) since that leads directly to the topic of digital image filtering with laplacian fileters. Then right after this, it should state that when &Delta f=0 then the average of f over a sphere equals the value in the center, which is I think eye-opening about why the Laplacian is interesting in the first place. (Is this called "Poisson's theorem', maybe??) It might also be nice to add blurbs mentioning heat equation, diffusion equation, schreodinger equation, and poisson equation, since these are the main equations that 90% of readership will be encountering. Also handy to mention that solutions to poissons equation are harmonic functions. linas 22:55, 18 Apr 2005 (UTC)


 * I agree with Linas that the discrete case should be brought back. And it should be before the differential geometry part, as it still refers to the flat laplacian. The discrete Laplacian is very important in Numerical Analysis.


 * Linas, can I challenge you to work on this article (both explaining the differential geometry notation, and the discrete thing, and what not)? I think MarSch will not mind us adding new stuff. We can then all work together to integrate things and make this a good article. Oleg Alexandrov 23:12, 18 Apr 2005 (UTC)


 * Ah, well, yes; but I thought I'd give Marsch first crack. I know that if I spend two hours editing and then Marsch re-writes half of it, I'll just end up getting frustrated.  But if the muse comes... I was thinking of working on some special functions today, which came up in an interesting problem recently.linas 23:28, 18 Apr 2005 (UTC)


 * The stuff you mention sounds interesting. But this is not an article about numerical approximation methods, so I think you should point to one such for your f(x+h)-2f(x)+f(x-h) formula. Then if you could relate this to digital image filtering etc.that would be great. Some of the equations you mention are already in the intro, but the others should definitely also be there. It would probably be best if you first spend two hours adding this stuff, after which I can rewrite half ;) But I certainly don't expect to have to do that. -MarSch 12:00, 19 Apr 2005 (UTC)


 * Well, a blurb about the numerical laplacian will certainly not hurt. After all, this is not an article about differential geometry either. :) Oleg Alexandrov 15:19, 19 Apr 2005 (UTC)


 * OK, I'll give it a shot real soon now. But please do not make threats about re-writing half of it before you've seen the result. linas 05:51, 25 Apr 2005 (UTC)


 * Sorry. That was supposed to be a joke. I thought it was clear, but once again: I do not intend to rewrite anything just yet. MarSch 13:22, 27 Apr 2005 (UTC)


 * Good luck Linas. I think you are doing MarSch a great favor by tackling the unfinished business of explaining the notation. By the way, if you get a nice long article, one idea would be to fork it into a Laplace-Beltrami operator article. But that's all up to you. Oleg Alexandrov 15:14, 25 Apr 2005 (UTC)


 * PS If MarSch does not agree with something, I am sure he will first ask here rather than start rewriting. :) Oleg Alexandrov 15:15, 25 Apr 2005 (UTC)

Wow, linas. I think you have done a great job with this article. -MarSch 13:48, 27 Apr 2005 (UTC)

Remark about Dalambertian
Hi Linas. Good job. Just a remark. I would think that the
 * $$\square = {\partial^2 \over \partial x^2 } + {\partial^2 \over \partial y^2 } + {\partial^2 \over \partial z^2 } - \frac {1}{c^2}{\partial^2 \over \partial t^2 }$$

which I think is called Dalambertian is not the Laplacian. Certainly not in R^4 with the usual Euclidean structure. I think this goes a bit off topic, what do you think? Oleg Alexandrov 18:18, 26 Apr 2005 (UTC)


 * Thanks, I'd forgotten that it has a name. The D'Alambertian is the Laplacian for flat Minkowski spacetime; don't think its off-topic, it gives a flavour for what non-euclidean geometry is like. It is recovered directly from the formula for the Laplace-Beltrami by plugging in the flat minkowski metric.


 * I still plan on adding something very breif on discreteness; I admit the article is now quite long; don't want to split it up, am afraid it will loose continuity. linas 14:32, 27 Apr 2005 (UTC)


 * That's what I said above, that the D'Alambertian is not the Laplacian in the Euclidean space. So, I am not sure it belongs in the definition part, where one talks about the Euclidean Laplacian. It more likely belongs in the section which actually talks about Laplace-Beltrami, which is below. Oleg Alexandrov 16:55, 27 Apr 2005 (UTC)


 * But I will leave this up to you. My idea for this article was Part 1: Euclidean Laplacian, and Part 2: Differential geometry Laplacian. But since you are doing all the work anyway, you can decide yourself which way to go. :) Oleg Alexandrov 17:03, 27 Apr 2005 (UTC)


 * I'd prefer to leave in in the initial section, as a "teaser" for what is to come. Again, college sophmore engineering majors will see the 3D euclidean stuff in class, and have a 50/50 chance of seeing the dalambertian as well, or at least the 4d wave equation (electromagnetism, EE, integrated circuit design etc.). However, these same students will be lost almost immediately in the next section, as it promptly whips into differential forms, and I assume they aren't teaching differential forms to these students. So I'm looking at the dalembertian as a carrot "you too can understand the deeper meaning of the dalembertian, if you apply yourself a bit and wade through the next section" its a sample of what lies ahead, presented in simple terms. linas 22:54, 27 Apr 2005 (UTC)

d and -div adjoints

 * $$\int_M df(X) = \int_M X(f) = \int_M \mathcal{L}_X(f) = \int_M (-1)^?*\mathcal{L}_X(*f) = $$

$$\int_M (-1)^?*(f\mathcal{L}_X\omega + \omega X(f)) = \int_M (-1)^?*(f\mathrm{div}(X) \omega + \omega X(f)) = $$ $$\int_M (-1)^?(f\mathrm{div}(X) + X(f)) =? - \int_M f \mathrm{div} X \; $$

I thought this is all legal. If ? is odd it would mean we are missing a factor 2: strange. Otherwise it would mean that the last term is zero and break the claimed adjointness: very strange. What am I missing? MarSch 14:23, 27 Apr 2005 (UTC)


 * Hmm, not to be glib, but maybe a factor of &omega; ? I'm late to work, I'm rushing, but I'll think about this a bit. I'm thinking that the equation should have been


 * $$\int_M (\omega f\mathrm{div}(X) + \omega X(f)) = 0$$


 * and I wrote the integral without the "implicit" &omega;. linas 14:46, 27 Apr 2005 (UTC)

Yes, without that the formula makes no sense. My new calculation shows you are right for manifolds without boundary.
 * $$\int_M (f\mathrm{div}(X) + X(f)) \omega = \int_M (f\mathcal{L}_X + \mathcal{L}_X(f)) \omega = \int_M \mathcal{L}_X f\omega = \int_M \mathrm{d} \iota_X f\omega = \int_{\partial M} \iota_X f\omega$$

I think it is better to include this short calculation. -MarSch 15:27, 27 Apr 2005 (UTC)


 * I'd prefer to leave the short calculation out; I think it makes the article harder to read without conveying much info. The reader has to burn brain cells to undertand the formula, and then when gets to the end, and probably realizes "oh I don't care, anyway". What we really need to do is to establish some sort of wikipedia house style/house convention for supplying proofs or at least details for formulas like this, without cluttering the main article. That way, the casual reader is not impeded by dense formulas, whereas someone who is worried about detail and correctness can check explore more deeply.  I will create an experimental page to do this shortly; lets see "how it feels", and if such a format could work out.


 * I agree in leaving the calculation out. This elementary article is gradually becoming a differential geometry monster. :) Oleg Alexandrov 23:34, 27 Apr 2005 (UTC)


 * BTW, the surface does not have to be boundary-less; the only requirement is for the support for f to be compact: if support is compact, then you can find an open cover on whose boundary f vanishes, and so the integral on boundry is zero. I really want to keep this adjoint-ness property really really simple, I really just want to say "div is just the transpose of d" and make the point that these two things are really almost exactly the same thing, just kind-of reflected.linas 23:26, 27 Apr 2005 (UTC)


 * BTW, I have to mention this, just in case: I suspect you might be bothered by the idea of making a true statement that holds for all X and f with compact support, and then wildly jumping to the conclusion that this somehow holds for the operators div and d themselves. If this troubles you, you'd be justified, there's subtle stuff going on, and Sobolev spaces are part of what you'd want to resolve this. But that is way deeper than we want to go here.  I just want to say d and div are 'the same thing, sort-of' without diving into something else. linas 23:26, 27 Apr 2005 (UTC)


 * Doesn't this duality hold more generally for d and &delta;? -MarSch 15:23, 5 May 2005 (UTC)

partials
I notice that although there is still the def
 * $$\partial_i= \frac {\partial}{\partial x^i}$$

it is not used immediately following. Not even in the Laplace-Beltrami section. I would prefer to at least use it there exclusively. If you want to keep it before that then we should move this def down. -MarSch 15:41, 27 Apr 2005 (UTC)
 * Hmm, I'd rather stick to partial/partial x notation, just to keep things concrete, and not force the reader to make irrelevent abstractions. The reader who already knows that partial_i = partial/parital x^i is not going to be impeded by the more verbose notation.  On the other hand, there will be readers who will suspect that maybe partial_i is maybe the same thing as  partial/parital x^i, but they're not sure. Even when we explain that "hey duude partial_i = partial/parital x^i" they'll still be thinking there's some conspiracy, some depth or theorem there that they don't understand. To ask them to juggle that uncertainty at the same time they're juggling the other uncertainties is a bit too much for an encyclopedia article.  I know that math books and teachers love to force thier students to juggle as much as possible ... that's how you get good. But I think that's bad style for an encyclopedia or reference work. linas 23:37, 27 Apr 2005 (UTC)
 * I would much prefer the simpler notation of $$\partial_i$$ especially when the formulas start to get more complicated. It is easier to see whether the index is up or down. Also this is standard notation. I especially dislike the use of $$\frac {\partial}{\partial x_i}$$. I don't think I've seen this before. I really think we should use the simpler notation starting from heading "Laplace-Beltrami operator". -MarSch 13:08, 2 May 2005 (UTC)


 * I saw the $$\frac {\partial}{\partial x_i}$$ used a lot though. I think all of us have different biases depending on our background. But I don't care either way. My bigger problem, MarSch, is that you did not yet explain the $$\partial^i$$ notation. Hope you find time to do it sometime. :) Oleg Alexandrov 14:56, 2 May 2005 (UTC)

Agree with Linas's changes
The article looks much cleaner now without the proofs (the proofs were not very well written anyway). Oleg Alexandrov 17:34, 18 May 2005 (UTC)


 * well, I don't like it. But since this is not particular to this article will discuss somewhere else. -MarSch 11:12, 19 May 2005 (UTC)

Elliptic?
Is the Laplacian an elliptic operator or not? In the introduction, it says that it is, but then in the article it is stated that it can be defined on non-Euclidean spaces, like the d'Alembertian. --Joke137 23:07, 18 May 2005 (UTC)


 * Well, see the discussion above, at Talk:Laplace operator. You have a good point. Let us see what Linas has to say. Oleg Alexandrov 23:29, 18 May 2005 (UTC)

Ahh, that'll teach me not to read the whole talk page. It's a problem of semantics more than anything, but one I worry about. Is the operator $$\nabla^2$$ on a curved space with a pseudo-Riemannian (i.e. $$(-,+,\dots,+)$$) metric properly called the Laplacian, d'Alembertian, both, or neither? --Joke137 01:03, 19 May 2005 (UTC)


 * In 4D, it would be D'Alembert operator; in any other set of dimensions, it would be laplacian. Its a hyperbolic operator when its not elliptic. linas 05:56, 6 Jun 2005 (UTC)

Split this article?
should this article be split into two or three, one for Laplace-Bletrami, one for Laplace-deRHAM ?linas 05:56, 6 Jun 2005 (UTC)


 * I would not mind. And the Laplace-Bletrami article could use the terminology Laplace-Bletrami right from the beginning, without calling it Laplacian till the end.


 * By the way Linas, I did some changes yesterday. I think they were more cosmetic than anything, but feel free to revert what you don't like. Oleg Alexandrov 15:26, 6 Jun 2005 (UTC)


 * I don't see the purpose of splitting other than destroying context. --MarSch 14:13, 17 Jun 2005 (UTC)


 * No split, not now. The article is near the limit of acceptable length for an article. If it got significantly longer, it should be split. Its OK for now.  I am distracted with other topics. linas 15:09, 17 Jun 2005 (UTC)

I see Linas's point. Actually, what bugs me more is that in this article there is no clear distinction between the laplacian and its generalizations. I am not that familiar with the literature, so I want to ask you people, is the laplacian just the ellyptic operator which is the sum of unmixed partial derivatives, or one can use this name for the more general operators on manifolds where this operator might not even be ellyptic? In particular, does the following sentence from the introductory paragraph:


 * (or a hyperbolic operator when defined on pseudo-Riemannian manifolds)

apply to the laplacian or to its generalizations? Oleg Alexandrov 05:06, 18 Jun 2005 (UTC)


 * In the book "Geometry of Physics", by Herbert Franklin, from which I got this stuff the Laplacian on Euclidean space and its generalization to manifolds and forms is called simply Laplacian. I was unaware of the other names until linas mentioned them, but I do not think this is a good criterion for judging whether this should be one article or many. --MarSch 12:41, 19 Jun 2005 (UTC)

You are of course right about this being a bad criterion for splitting the article, and I did not mean it to sound so. But I do know that the laplacian on surfaces is called the Laplace-Beltrami operator, and not simply the Laplacian. Oleg Alexandrov 15:41, 19 Jun 2005 (UTC)

The introduction
Patrick and Linas: thanks. I also felt that something about the Laplace equation needed to be said, but I trully think the original place was inappropriate. It is much better now. Oleg Alexandrov 04:16, 20 August 2005 (UTC)

Vector Laplacian
Should we not also add the definition of the "vector Laplacian" (grad div - curl curl), as opposed to the scalar form, and be sure to mention this ambiguity at the beginning of the article? Zeroparallax 09:51, 5 February 2006 (UTC)
 * Would be a good idea to add it, in a new section. I don't know if it would be a good idea to mention that in the intro. The intro does not mention even the Laplace-Beltrami and Laplace-DeRham operators, which are generalizations to sufraces, and thus, simpler than the vector laplacian. Or maybe, since this article is already big, might be nice to create a Vector Laplacian article, and link to it from this article. Comments? Oleg Alexandrov (talk) 16:23, 5 February 2006 (UTC)


 * I agree, a link would be good and it would provide an indexable link to the Vector Laplacian as an article. I can only find the vector Laplacian in Vector operator and it is not expository, just notational. But, then I think the Laplace-Beltrami and Laplace-DeRham operators could use a home of their own, too. Not that I'm an advocate for brevity in math. Kckid 23:50, 25 April 2006 (UTC)


 * I agree that the omission of the vector Laplacian is a major flaw. I suggest that a separate article be created, and that an italic note be placed under the title of the existing article reading, "This is about the Laplace operator as applied to scalar fields.  For the analogous vector operator, see [name of article]" Hyperdeath 17:32, 5 July 2006 (UTC)

This article is an abomination.
I had the misfortune to stumble upon this article. This article really shows the ugly side of Wikipedia. No organization, bizarre choices, too many authors. I feel bad for my beloved Laplace operator. Will some responsible mathematician please rewrite this whole article from scratch? --anon (User:152.3.25.60 on 17 feb 2006)


 * Hmm, what, specifically, is wrong with the article? Can you provide a detailed criticism? I don't see whatever it is that you disagree with. linas 00:12, 18 February 2006 (UTC)


 * Yes, constructive suggestions could be helpful; ranting is not. Oleg Alexandrov (talk) 02:23, 18 February 2006 (UTC)


 * Okay, I apologize. After perusing other math articles, I see that this one is not really so awful.  I guess I had unrealistic expectations.  Here are some criticisms of what is in the article, mixed with suggestions about what should be in the article, in roughly decreasing order of importance/conviction: (User:152.3.25.60 on 20 feb 2006)

1. The d'Alembertian and the other references to pseudo-Riemannian geometry should be relegated to a footnote at most. Think of it this way. Is the wave equation a "special case" of Laplace's equation? Absolutley not. Ellipticity is at the heart of what the Laplacian is all about. Formally describing two concepts by the same formula does not necessarily relate those concepts in a meaningful way.

2. There are only two references. One is MTW (a GR book), and the other is Jost (a geometric analysis book). Huh? Let's see an elementary calculus textbook and an introductory PDE textbook.

3. The two properties listed under Laplace-deRham have nothing to do with Laplace-deRham. The second one was already given earlier under Identities, except now the notation is worse.

4. The Laplacian should be first given in rectangular coordinates in 2 and 3 dimensions, WITH the function f, since this is easiest to digest for undergrads. Writing div(grad f) might be a nice addition to \nabla\cdot\nabla. (While the \nabla^2 notation is terrible, people do use it sometimes.)

5. The Laplacian is a second-order LINEAR elliptic operator. This basic description should be featured. (Indeed, the Laplacian is the quintessential elliptic operator, but this statement is too fuzzy.) In fact, there should be a separate short section on operator (functional analytic) properties of the Laplace-Beltrami operator, starting with a link to an article that describes the functional analytic properties of elliptic operators generally.

6. A second expression for the Laplacian in spherical coordinates is unnecessary. I'd rather see the 2d polar expression (again, for undergrads who don't immediately see that the cylindrical expression already tells us).

7. Laplace's equation is the Euler-Lagrange equation for the "energy" functional. This more or less explains why the Laplacian shows up in most physical and geometric applications. IMHO this would be a better "derivation" of the Laplace-Beltrami operator. In this case, we might define the divergence as the formal adjoint of the gradient. Which brings me to...

8. In this context (Riemannian geometry), I think it makes more sense to say that div is adjoint to grad, and that \delta is adjoint to d, rather than saying that div is adjoint to d.

9. I might be wrong, but I think that the "Laplace-deRham" operator is more frequently called the "Hodge Laplacian" or simply Laplace or Laplace-Beltrami when there's no risk of confusion. The section should be called labelled Hodge Laplacian, and there should be a prominent link to the Hodge Theorem, which by the way, should really be in a separate article from "Hodge Theory." In order to keep this section short and self-contained, the codifferential might be defined as the formal adjoint of d. At the top of the article, it is claimed that the Laplacian "is at the core of Hodge theory and the results of de Rham cohomology," but the relation to de Rham IS the Hodge Theorem, so the mention of de Rham is unnecessary.

10. There is a very good chance that someone looking at this article is actually interested in aspects of Laplace's/Poisson's equation. A prominent link should be presented given right away. (Unfortunately, those articles don't offer much information.)

11. The whole page should be called the Laplacian rather than the Laplace operator, since the former is more commonly used.

12. In general, one should either assume that the reader already knows the basic language of Riemannian geometry or does not. The current treatment defines some things but leaves other definitions to other articles.

13. In the "See also" section, Christoffel does not belong. Riemannian geometry perhaps does belong. Laplace's equation and Poisson's equation definitely belong. I don't even know what non-orthogonal analysis is.

14. If we are already going nuts, the rough Laplacian *might* be a nice addition at the very very end.


 * Agree, these are nice comments, and should be acted upon. Properly speaking, you are the one in the position to carry out these edits. All I can say is that wiki's are strange beasts, and articles such as this can develop a life of their own that can be disconnected from reality. For example, I'm the one who added most of the text on Laplace-Beltrami/Hodge/deRham, and this only out of frustration, after finding the article in much worse shape . I had a brief skirmish with another editor in the process; some of the things you object to are his fault :). I had no desire to write about the Euclidean space undergrad/engineering versions, so left that part mostly untouched.


 * Re comment 7: the wiki can be frustrating. I added what I thought was a nice, decent energy-functional/Laplace's equation derivation to geodesic, but 3-6 months later, a very strict Russian editor deleted it; apparently he thought it pandered to morons, or something like that; wasn't sufficiently obtuse for his tastes. :-(


 * Re 13) The article on Christoffel symbols gives expressions for the Laplacian in terms of Christoffel symbols. The "non-orthogonal" thing was a piece of crank science someone added; it has since been deleted, but apparently not the link. Now fixed. linas 06:48, 21 February 2006 (UTC)


 * Just to be clear: I doubt anyone will be making the suggested changes short-term; there's no committee here that "attends to things". It really is up to you. The way the process works is that if you wait 6 months or even a year or two, you may find the article vastly improved. Or possibly not.  People write only when they "feel like it", and rarely because "its the right thing to do". linas 07:15, 21 February 2006 (UTC)


 * Per Linas, have fun working on it. This is not a perfect article, but contains a great amount of information, and it much better and bigger than it was a year ago. So, make account, my dear, and help improve it. :) Oleg Alexandrov (talk) 16:34, 21 February 2006 (UTC)


 * In response to the above suggestions, I added a todo list to this entry. I also added a small section on the relation between the Laplacian and the energy functional. However, I still feel that this page need a lot of work. For example, there seems to be a "notational overflow" in the differential geometric section. Do we really need separate notation for $$\partial^i$$? Haseldon 12:09, 26 November 2006 (UTC)

pseudo Riemannian laplacian is hyperbolic
The article states that the Laplacian is hyperbolic on a pseudo-Riemannian manifold. But surely that's only true if n–s is odd? I mean Riemannian manifolds are pseudo-Riemannian manifolds with n=s, and their laplacians are elliptic. Maybe it would better say Lorentzian manifold. -lethe talk [ +] 18:53, 27 February 2006 (UTC)

Eigenfunctions / Spherical Coordinates
I suggest adding a new subsection regarding eigenfunctions and a small demonstration using a physically relevent function... let me explain.

I am told that the spherical harmonics are the eigenfunctions of the angular portion of the Laplacian operator in spherical coordinates (with eigenvalues l*(l+1)). I have found it difficult for people to explain exactly WHAT IS the angular portion of the Laplacian operator, i.e., does it include the factor of 1/r^2 or not? (Edit: $$\nabla^2 Y_{lm}(\theta,\phi) = -\frac{l(l+1)}{r^2}Y_{lm}(\theta,\phi)$$)

In physics, it is really very common to have functions of the form $$F(r,\theta,\phi) = f(r) Y_{lm}(\theta,\phi)$$ where Ylm(t,p) is a spherical harmonic. This is really quite nice because the "identity" listed in the article simplifies (due to orthogonality of the gradients AND because the spherical harmonic is an eigenfunction of the "angular part" of the laplacian.)

$$\nabla^2 \left(f(r)Y_{lm}(\theta,\phi)\right) = \left(\frac{d^2f(r)}{dr^2} + \frac{2}{r} \frac{df(r)}{dr} - \frac{l(l+1)}{r^2} f(r)\right)Y_{lm}(\theta,\phi)$$

66.41.8.178 00:07, 21 May 2006 (UTC)

66.41.8.178 16:31, 21 May 2006 (UTC)

OK. I made an addition, but I guess a subsection wasn't necessary. I think it makes for a nice note - something that is extremely useful both in electrostatics and in quantum mechanics - and it's something not touched on in the MathWorld Laplacian page. DrF 03:10, 22 May 2006 (UTC)

Interpretation of the Laplacian
I've heard about a possible interpretation of the laplacian as a measure of "how below the local average" the value is (i.e.: The laplacian of a minimum is positive, the laplacian of a maximum is negative). I don't have any references though. Does anyone knows it? Any other intuitive interpretations? -- Rend 05:02, 14 April 2007 (UTC)
 * The Laplacian at a minimum is non-negative actually (so it can be zero). This follows from the fact that for a function of one variable $$x$$ having a local minimum at $$a$$ one has


 * $$f''(a)=\lim_{h\to 0}\frac{f(x+a)-2f(a)+f(x-a)}{h^2}$$
 * $$=\lim_{h\to 0}\frac{(f(x+a)-f(a))+(f(x-a)-f(x))}{h^2}$$
 * $$\ge \lim_{h\to 0}\frac{0+0}{h^2}$$
 * $$=0$$


 * assuming that $$f$$ does have two derivatives at $$a.$$


 * Note that the converse is not true. Non-negative (or positive) laplacian does not imply local minimum. Oleg Alexandrov (talk) 05:38, 14 April 2007 (UTC)


 * Yeah, of course, but does the local average interpretation makes sense? I was thinking if the Laplacian isn't just a generalization of concavity, as $$f^{(2)}(x)$$ is usually taken. Rend 16:35, 20 May 2007 (UTC)

"The sign in front of the fourth term is negative, while it would have been positive in the Euclidean space"
The D'Alembert operator is often used to express the Klein-Gordon equation and the four-dimensional wave equation. The sign in front of the fourth term is negative, while it would have been positive in the Euclidean space. This seems wrong to me. The second fourth term is perfectly negative in Euclidean. What am I missing? Whaa? 09:12, 28 July 2007 (UTC)

Vote for new external link
Here is my site with laplace operator example problems. Someone please put this link in the external links section if you think it's helpful and relevant. Tbsmith 03:16, 29 December 2005 (UTC)

http://www.exampleproblems.com/wiki/index.php/PDE:Laplaces_Equation

Spherical laplacian and spherical harmonics
I think a good thing should be a clear (sub)section about spherical laplacian on the standard n-shpere and spherical harmonics defined once and for all as its eigenfunctions: here or in the article on spherical harmonics (that one is devoted to the case of the 2-sphere, however). Note that at the moment the Laplace-Beltrami on Riemannian manifolds has already been introduced here, together with some description in the case of the standard sphere. Thus this article has the sufficient level of generality for the natural definition of the spherical harmonics, with no need of talking about "angular portion of the Laplacian operator in spherical coordinates", which is not very clear as a main definition (what's "angular part"?), nor immediately useful (how to relate it to the spectral theory of compact symmetric operators? why are the spherical harmonics L^2 orthogonals? How to describe them in the general case of an n-dimensional sphere etc). The information a user would like to get is I guess

1. A clear definition of the Laplace-Beltrami operator in the special case of the standard n-dimensional sphere, also showing how it is related with the general definition in Riemann manifolds.

2. What are the eigenvalues of the spherical laplacian on the standard n-sphere. Why are the corresponding eigenfunctions exactly the (functions on the sphere that are restriction of the) harmonic homogeneous polynomials of degree k (in the n+1 cartesian coordinates). What is the multiplicity of the eigenvalues, lambda_k=-k(k+n-1) k=0,1,2.. 3. How the preceding facts fit into the general situation given by the spectral decomposition of a compact symmetric operator on an Hilbert space. How are spherical harmonics linked to the Poisson kernel.

PMajer (talk) 15:45, 17 November 2007 (UTC)

Laplace-de Rham - \nabla ^* \nabla
I just wanted to ask: One has the LC-connection on TM and one can get a connection $$\nabla$$ on $$\Lambda T^*M$$ from that via the Leibnitz-formula, can't one? One can then look at $$\nabla ^* \nabla$$ on forms, where $$\nabla ^*$$ is the formal adjoint of $$\nabla$$. But that's not the Laplace-de Rham operator, is it? Defining a Clifford algebra multiplikation on $$\Lambda T^*M$$ as $$X \omega := X\wedge \omega + X^b \llcorner \omega$$ gives one that the Dirac-operator $$D$$, fulfills $$\Delta = D^2 = \nabla ^* \nabla + R'$$, where $$R'$$ is some endomorphism on the fiber related to the curvature operator $$R$$. ... or am I confusing something here? -- JanCK (talk) 18:10, 6 December 2007 (UTC)


 * See Weitzenböck formula. You are correct that $$\nabla^*\nabla$$ is not the Laplace-de Rham operator.  This is (apparently) called the "Bochner Laplacian."  Silly rabbit (talk) 16:41, 22 December 2007 (UTC)

Spherical Laplacian
It seems to me that the expression for the spherical laplacian in the main article is incorrect. Namely, the $$\theta$$ and $$\phi$$ variables are switched in this expression. (In the sentence following the expression, it is explained that $$\theta$$ is the polar angle, which is the usual convention, so the expression itself cannot merely be a switching of convention.) I can cite: http://mathworld.wolfram.com/Laplacian.html and http://math.mit.edu/18.013A/HTML/chapter17/section03.html as supporting my computation. Since everything is swapped, IMHO there is a possibility that someone tampered with the equation deliberately. (i.e. one wrong variable might be a wrong transcription, but everything swapped is fishy.) In any case, someone should check this carefully and fix the main article! Ivan Blank PhD, Department of Mathematics Kansas State University. 24.248.230.218 (talk) 23:41, 21 January 2008 (UTC)
 * The formula looks ok to me. In the Mathworld link you give, &theta; is the azimuthal angle and &phi; the polar angle.  So that formula agrees with this one, since here &theta; is the polar angle and &phi; the azimuth.  Unless I'm losing it.  Silly rabbit (talk) 00:47, 22 January 2008 (UTC)
 * This is a general nuisance. Americans and Physicists both often get it Wrong, and put it the wrong way round compared to Sane People (i.e. Cambridge mathematicians). The terminology is confusing, and the lack of agreement can cause real confusion, but both sides are entrenched now, with each way a lot of literature to cite; there is unfortunately nothing to be done except disambiguate at the top of articles.— Kan8eDie (talk) 03:51, 24 February 2009 (UTC)
 * Or just say which convention is used, like in the current article. 72.95.229.48 (talk) 04:13, 24 February 2009 (UTC)

I recall vividly in undergrad that I had two textbooks which followed different conventions. I just got used to it -- instead of referring to theta and phi, you learn to mention which angle is the one from the pole, and which one is in the plane, or to interpret from context. No help for it. Ray (talk) 16:42, 24 February 2009 (UTC)

Section removed
I removed the following from the article. It seems to have no focus whatsoever, and is of only peripheral relevance to the article. The "identity" listed is an obvious one, and the spherical harmonic context is not terribly compelling. Spherical harmonics are already mentioned elsewhere in the article. Here is the offending section. Sławomir Biały (talk) 14:13, 2 September 2009 (UTC)


 * Identities
 * If f and g are functions, then the Laplacian of the product is given by


 * $$\Delta(fg)=(\Delta f)g+2((\nabla f)\cdot(\nabla g))+f(\Delta g).$$

Note the special case where f is a radial function $$f(r)$$ and g is a spherical harmonic, $$Y_{lm}(\theta,\phi)$$. One encounters this special case in numerous physical models. The gradient of $$f(r)$$ is a radial vector and the gradient of an angular function is tangent to the radial vector, therefore


 * $$2(\nabla f(r))\cdot(\nabla Y_{lm}(\theta,\phi))=0.$$

In addition, the spherical harmonics have the special property of being eigenfunctions of the angular part of the Laplacian in spherical coordinates.


 * $$\Delta Y_{\ell m}(\theta,\phi) = -\frac{\ell(\ell+1)}{r^2} Y_{\ell m}(\theta,\phi).$$

Therefore,


 * $$\Delta( f(r)Y_{\ell m}(\theta,\phi) ) = \left(\frac{d^2f(r)}{dr^2} + \frac{2}{r} \frac{df(r)}{dr} - \frac{\ell(\ell+1)}{r^2} f(r)\right)Y_{\ell m}(\theta,\phi).$$

Boundedness
Could someone add information about the boundedness of the Laplacian as a linear operator? As I understand it, it is unbounded when defined on the functions with bounded domain in R^n, but according to bounded operator it is bounded when the domain is R^n itself (and I presume other unbounded domains). Some mention of Sobolev spaces would be nice too. I will try to add what I know about these myself. —Preceding unsigned comment added by Slimeknight (talk • contribs) 21:31, 15 May 2010 (UTC)


 * This is wrong. As an operator on, e.g., L2(&Omega;) (of any domain &Omega; whatsoever&mdash;bounded, unbounded, all of Rn), the Laplacian is an unbounded densely-defined operator.  However, as an operator
 * $$\Delta : H^2(\Omega)\to L^2(\Omega) $$
 * it is bounded. But this statement has very little content: it is simply how the norm on the Sobolev space H2 is defined.   Sławomir Biały  (talk) 12:15, 16 May 2010 (UTC)


 * Ah, okay, I was confused about which norm was was being talked about, I see. That makes a lot more sense to me, as I originally made the comment after being confused as to why it wasn't always unbounded. Thanks very much. slimeknight (talk) 19:04, 16 May 2010 (UTC)

Azimuth and zenith
Since there is persistent confusion over the meaning of the words "zenith" and "azimuth", allow me to clarify to prevent further erroneous edits to the section on spherical coordinates. The angle &phi; measured with respect to the north pole is known as the zenith angle; the z axis itself is the zenith axis. The angle &theta; made between the positive x axis and the the orthogonal projection of the point into the xy plane is known as the azimuth angle, where the x axis itself is the azimuth axis. Now, the article at present says that
 * $$ \Delta f

= {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial f \over \partial r} \right) + {1 \over r^2 \sin \varphi} {\partial \over \partial \varphi} \left( \sin \varphi {\partial f \over \partial \varphi} \right) + {1 \over r^2 \sin^2 \varphi} {\partial^2 f \over \partial \theta^2}. $$ To see that this is correct for our conventions, consider the Laplacian of $$f(x,y,z)=z^2=r^2\cos^2\phi.$$ In Cartesian coordinates, the Laplacian is 2. Working this out in spherical coordinates also gives
 * $$ \begin{align}

\Delta f &= \Delta (r^2\cos^2\phi) = 6\cos^2\phi + (-4\cos^2\phi + 2\sin^2\phi) + 0\\ &=2 \end{align}$$ as expected. However, if instead we were to misunderstsand the meaning of "zenith" and "azimuth" and considered the function $$f=r^2\cos^2\theta,$$ then

\begin{align} \Delta f &= \Delta (r^2\cos^2\theta) = 6\cos^2\theta + 0 + (2\sin^2\theta-2\cos^2\theta)\\ &=2\cos^2\theta + 2 \end{align} $$ which is not equal to 2. Sławomir Biały (talk) 11:07, 27 July 2010 (UTC)
 * Comment: that is inconsistent with the usage in Spherical coordinate system, where &phi; is used for azimuth. ...ah, but reading on, I see that "However, some authors (including mathematicians) use φ for inclination (or elevation) and θ for azimuth, which 'provides a logical extension of the usual polar coordinates notation' Some authors may also list the azimuth before the inclination (or elevation), and/or use ρ instead of r for radial distance."
 * So, as so often, it's a case of "standards are wonderful, you can't have too many of them". Regards, JohnCD (talk) 16:31, 27 July 2010 (UTC)
 * Up until quite recently, we used the other convention. But I changed to this one because of this very issue.  People were still getting them mixed up even in the convention consistent with our spherical coordinate system article.  I think the current conventions are preferable, because we should use the same letter to designate the azimuth in the polar, cylindrical, and spherical coordinates.  I believe that this letter is almost always &theta; for the polar and cylindrical systems, and so we should use the convention for the spherical coordinate system in which it is also &theta;.  It would seem to me to be far too confusing to have &theta; an azimuth in one paragraph, and then a zenith in the very next part.  I thought changing to the "mathematical" convention would resolve the confusion, but apparently not.   Sławomir Biały  (talk) 17:50, 27 July 2010 (UTC)

varphi vs phi
Starting from the convention that theta is azmuthal and phi is latitude... I noticed you had two different versions of the equation, and one used \varphi for the latitude angle. I think these must have been intended to be the same angle (else there is no definition for the two), so I changed it. -Theanphibian (talk • contribs) 12:32, 14 May 2013 (UTC)

'Spheres centred at'...
Firstly it's not clear if it's talking about the surface or the solid (I know that it's surface technically but I don't know if the editor knew this).

Secondly, shouldn't really be in the intro, followed by zero follow up. Should probably be in the article somewhere, with at least some discussion of importance or derivation; I'm okay at vector calculus but I've never heard this interpretation before. — Preceding unsigned comment added by 86.131.51.193 (talk) 17:00, 12 July 2011 (UTC)
 * As usual in mathematics, "sphere" means the surface of the ball. This characterization was chosen for the lead paragraph since it seemed to be the simplest way to say what the Laplacian is for a general audience (no assumption of vector calculus).  Sławomir Biały  (talk) 17:53, 12 July 2011 (UTC)


 * Let's revisit this. The average of a function f over the sphere of radius r centered at the origin in $$\mathbb{R}^n$$ is
 * $$\overline{f}(r)=\frac{1}{\alpha_nr^{n-1}} \int_{S^{n-1}_r(0)} f(x)\,d\sigma(x).$$
 * Apply a change of variables to convert to spherical coordinates:
 * $$\overline{f}(r)=\frac{1}{\alpha_n} \int_{S^{n-1}(0)} f(r\xi)\,d\sigma(\xi).$$
 * Differentiating with respect to r gives
 * $$\overline{f}'(r) = \frac{1}{\alpha_n} \int_{S^{n-1}(0)} \nabla f(r\xi)\cdot\xi\,d\sigma(\xi) = \frac{1}{\alpha_nr^{n-1}} \int_{S^{n-1}_r(0)} \nabla f(x)\cdot\frac{x}{r}\,d\sigma(x).$$
 * Applying the divergence theorem gives
 * $$\overline{f}'(r) = \frac{1}{\alpha_nr^{n-1}}\int_{B^n_r(0)}\Delta f(x)\,dx.$$
 * So $$\overline{f}'(r)$$ is equal to $$r/n$$ times the average value of $$\Delta f$$ over the ball of radius r.  Sławomir Biały  (talk) 12:10, 13 July 2011 (UTC)


 * So... the introductory statement is incorrect? I can't see how the above is relevant. --81.152.176.83 (talk) 16:04, 13 July 2011 (UTC)
 * No, the precise statement is: $$\overline{f}(r)=f(0)+c_n r^2\Delta f(0) +o(r^2).$$  Sławomir Biały  (talk) 03:07, 14 July 2011 (UTC)


 * ...okay, for now, I'm removing the statement. The above in no way shows equality to 'the rate at which the average value of ƒ over spheres centered at p, deviates from ƒ(p) as the radius of the sphere grows', and as such is extremely unhelpful. — Preceding unsigned comment added by 86.131.49.172 (talk) 14:43, 19 July 2011 (UTC)
 * I've rephrased the statement. Sławomir Biały  (talk) 16:42, 19 July 2011 (UTC)
 * Rephrase it again. It is not even roughly equal 'up to a factor'; there is an r^2 term. The true statement is now so lacking in intuitive meaning and consequence that it seems to me totally bizarre having in the beginning as if it's either important or helpful. — Preceding unsigned comment added by 86.131.49.172 (talk) 15:57, 20 July 2011 (UTC)
 * Actually, the preamble was very enlightening to me as soon as I realized (from the harmonic oscillator page) that r was the radius of the test sphere.  I'll see if I can wordsmith that tidbit into the preamble) Dr. Crash (talk) 14:43, 8 May 2018 (UTC)

Invariance under orthogonal coordinate transformations
I think one important property worth mentioning is that if two coordinate charts are related as $$\overline{x} = U x$$, where $$U$$ is orthogonal (in general unitary), then $$\nabla^2 = \sum_i \frac{\partial^2}{\partial {x^i}^2} = \overline{\nabla}^2 = \sum_i \frac{\partial^2}{\partial {\overline{x}^i}^2}$$. This is conceptually important when transitioning from a differential equation interpretation to an operator field interpretation, for example in QM to QFT transition -- We need to have operators that are invariant under unitary coordinate transformations so that the differential equation governing the scalar/vector fields does not depend on the choice of the inertial frame. We thus have $$\nabla^2$$ in Schrödinger equation (or even the heat eqn), but a first order operator does not satisfy a similar property until we consider C&#x2113;1,3(C)-valued fields as in the Dirac equation (where it's satisfied with some gauge invariance). - Subh83 (talk &#124; contribs) 19:15, 1 April 2015 (UTC)


 * Agree, this should be added.  Sławomir Biały  (talk) 09:45, 25 April 2018 (UTC)


 * "The Laplacian is invariant under all Euclidean transformations: rotations and translations." I think the correct notion here is equivariance.Georg Muntingh (talk) 20:27, 9 November 2018 (UTC)


 * Invariance is correct for scalar fields.  Sławomir Biały  (talk) 02:18, 10 November 2018 (UTC)