Talk:Helmholtz equation/Archive 1

Clarification needed
This article should define of link to operators such as grad or laplacian. Also, it should say where the equation acts. Presumably on R^3, but that is nowhere stated. There's not enough guidance here for an encyclopaedia article. Ewjw 08:05, 8 March 2006 (UTC)
 * Done, thanks. Oleg Alexandrov (talk) 06:24, 11 March 2006 (UTC)

Euclidean space and time independence

 * (First part moved from Oleg Alexandrov's talk page...)

I am not trying to be a smartass, but I am a bit confused by the phrase "defined on the three-dimensional space". Which three dimensional space are you talking about? Do you mean real, physical space in the sense that physicists, astronomers, and mechanical engineers use the term? Or do you mean Euclidean three-space, in the sense that mathematicians use the term? In other words, I am not sure you can talk about "the" three-dimensional space as much as you can talk about "a" three-dimensional space, unless you mean "real" physical space.

Also, I think it is important to make the distinction very clear between the domains over which a function is defined, as opposed to the domains over which the function varies. In many applications of the Helmholtz Equation, for example electromagnetic waves, the function A is defined in principle for all of space and all of time, but implicit in the derivation of the equation is the notion that the function A varies only with space, whereas it is independent of time (because we have separated A from the time-dependent function by assuing separability and a time-harmonic dependence). That is why, as I am sure you already know, the Helmholtz Equation is sometimes called the time-independent wave equation.

I wonder if there is a way to revise the intro paragraph to include these distinctions while maintaining good writing, grammar, and style. Any thoughts?

-- Metacomet 06:38, 12 March 2006 (UTC)


 * Good points. I never quite got over the "a" or "the" space. I guess if we are talking about Euclidean space, then it is "the" 3-dimensional Euclidean space. Or not? About the time dependence, I think it would confuse people if we start even talking about time, as it is not clear at the beginning what time has to do with anything; that shows up only later. But feel free to modify the article in any way you see fit, if I have comments I will write on its talk page. Oleg Alexandrov (talk) 16:28, 12 March 2006 (UTC)

On Euclidean space: I am not 100 percent sure. I looked at the first sentence of the article again, and as you can see, I changed the word "the" to "any" based mostly on intuition and connotation more than any definite logic. Take a look, see if it makes sense. If not, then let's try to figure something out. -- Metacomet 21:06, 12 March 2006 (UTC)


 * OK then, so how many 3-dimensional Euclidean spaces do we have? :) All I know is $$\mathbb R^3.$$ Oleg Alexandrov (talk) 21:52, 12 March 2006 (UTC)

The answer to your question depends: Are you a mathematician, a scientist, an engineer, none of the above, or all of the above? Think carefully before you answer... -- Metacomet 22:02, 12 March 2006 (UTC)


 * OK, I am a mathematician, an applied one... Oleg Alexandrov (talk) 22:36, 12 March 2006 (UTC)

Warning: I am about to make unfair generalizations and sterotypes about different groups of people and occupations. If you are easily offended, or if you do not have a sense of humor, please do not continue reading the present dialog...

In other words, the following is meant to be somewhat tongue-in-cheek. Please do not take it too seriously...

I am an engineer, which means among other things that I unfortunately have to learn the language and mental models of both scientists and mathematicians. Sometimes it seems like these two groups of people are speaking entirely different languages even when they are talking about the same thing, which is most of the time (and they don't even realize it).

to be continued...

-- Metacomet 00:24, 13 March 2006 (UTC)


 * Can you be more specific? There is only one 3D euclidean space I think, and both engineers and mathematicians live in the same one, no? Oleg Alexandrov (talk) 01:13, 13 March 2006 (UTC)

Again, it comes down to what we mean by the word "space". As an engineer, I would be inclined to agree with you, because by "space" I usually mean the physical space that encompasses the universe, in other words the "space" in Einstein's "space-time continuum." But a mathematician often means something different by the word space, as in the term "vector space". In that context, there are (I think) infinitely many three-dimensional Euclidean vector spaces, one of which we can designate as representing what scientists, engineers, and lay-people normally call "space". So from that perspective, I would prefer to deal with only the one 3D Euclidean vector space that we all use to represent the physical space we occupy. But pure mathematicians often do not like to worry about specific applications, and in particular, do not necessarily care what the variables actually represent other than abstract numbers.

Now a pure theoretical mathematician might argue that the Helmholtz equation, apart from any real-world application, is a differential equation involving a function defined over any arbitrary vector space $$\mathbb{R}^n$$ for any integer dimension n. Now as an engineer, I personally find this approach annoying, because I never use the Helmholtz equation for functions that depend on anything other than real, physical, three-dimensional space, but I realize that it can be generalized to work with vector spaces of dimension greater than 3. So, unfortunately, not only do we have to accept that there are multiple 3D Euclidean spaces, but there are also Euclidean spaces of higher dimension as well. Again, for my purposes, it is completely unnecessary and I would just as soon not know about it, but for others, it is very important.

-- Metacomet 01:27, 13 March 2006 (UTC)


 * Note that "Euclidean space" means not only a vector space, but also the distance function, topology, etc. Then, up to an isomorphism, there is precisely one 3D Euclidean space. Saying "any 3D Euclidean space" is confusing in my view. It is the same as saying that "on any real line", or on "any complex plane". Wonder what you think. Oleg Alexandrov (talk) 02:27, 13 March 2006 (UTC)

I don't really know. I think we are spending too much time discussing a single word "any" versus "the". Although it is an important distinction, it's only one word. If you want to change it back to "the", go ahead. I won't revert you on that point, unless I can think of a different way to approach the sentence that will give us a win-win situation.

Alternatively, we could simply eliminate the word entirely, and say it is "dfined on three dimensional Euclidean space" without specifying "any" or "the". Try it out, I think it might be okay.

-- Metacomet 03:31, 13 March 2006 (UTC)


 * OK, I will drop the word. But please note, a word can matter, as if it is a mistake, then it is a mistake. :) Oleg Alexandrov (talk) 03:56, 13 March 2006 (UTC)

No argument here. A single word can make a huge difference in meaning. But it just didn't "sound" right to me, and that is really an isue of style and usage. With the new revisions you have just made, I think it sounds fine, and it is also correct. That is what is known as a win-win. -- Metacomet 04:13, 13 March 2006 (UTC)

Harmonic solutions : Error in the article
The article states that
 * It is relatively easy to show that solutions to the Helmholtz equation will take the form:
 * $$A(\mathbf{r}) = C_1e^{i\mathbf{k}\cdot\mathbf{r}} + C_2e^{-i\mathbf{k}\cdot\mathbf{r}}.$$

While this is true in 1D, the solution can contain more harmonic components in 2D or 3D. For example, in 2D, the solution would be
 * $$A(\mathbf{r}) = \int d\theta \;C(\theta)e^{i\mathbf{k_\theta}\cdot\mathbf{r}},$$

where $$||\mathbf{k_\theta}|| = k$$ and $$\angle(\mathbf{e_x,k_\theta}) = \theta$$. I hope I am not saying a big nonsense here... --Mathieu Perrin (talk) 16:17, 15 October 2008 (UTC)

"Typically n = 1, 2, or 3, when the solution to this equation makes physical sense"
But isn't the case n = 4 the Wick-rotated Klein–Gordon equation? -- Army1987 – Deeds, not words. 21:39, 26 January 2009 (UTC)

Error in Hankel function for 2 dimensions?
The Green function solution for 2 dimensions is inconsistent with http://eqworld.ipmnet.ru/en/solutions/lpde/lpde303.pdf

There is a sign error and the Hankel function is type 2, not 1.

Should it be replaced by


 * $$G(x) = \frac{-i}{4}H^{(2)}_0(k|x|)$$

for n = 2, where $$H^{(2)}_0$$ is a Hankel function, and

? —Preceding unsigned comment added by 131.236.42.187 (talk) 08:42, 2 August 2010 (UTC)

82.168.114.204 (talk) 15:54, 7 September 2010 (UTC) This difference is merely a matter of choice, it is related to taking $$s = i\omega$$ or $$s = -i\omega$$ in the frequency domain representation for harmonic signals. In other words, in the first case the time domain expression for a function $$ f(t) $$ is found by taking $$ f(t) = \Re\{ f(\omega) \exp(i\omega t) \} $$ and in the second through $$ f(t) = \Re\{ f(\omega) \exp(-i\omega t) \} $$.


 * Whether to use H^{(2)} or H^{(1)} depends on the boundary condition imposed. So far the Green's functions in all dimensions have been written as outgoing waves which is H^{(1)}. H^{(2)} would correspond to an incoming wave. The prefactor is uniquely determined by the Delta-function in the differential equation for the Green's function. Some people use a positive sign there (or even a factor 4π) such that the sign of the Green's function changes. I have checked that everything is consistent and removed the CAUTION message. Fabian Hassler (talk) 07:37, 27 October 2010 (UTC)

Loss of generality
Why isn't there loss of generality in assuming the common constant value after separation of variables is -k^2? Unless k is complex, it assumes the constant is non-positive. —Preceding unsigned comment added by 130.215.63.211 (talk) 15:24, 30 June 2010 (UTC)

I agree, why is that? The article should include a short explanation for this. 80.232.11.13 (talk) 12:49, 3 January 2012 (UTC)

Opening Notation
Is it just me or is the opening equation



(\nabla^2 + k^2) A = 0 $$

confusing?

I thought the Helmholtz equation was $$\nabla^2 A + k^2 A = 0$$, where nabla squared is that Laplacian operator and k is a constant. If the present notation is going to be used, it should have an up-front explanation so that us plebeians can understand. IMO, confusing people with the first equation is not a step in the right direction. —Preceding unsigned comment added by ChrisChiasson (talk • contribs) 02:34, 4 October 2007 (UTC)

(This issue has been fixed. Ovikholt (talk) 13:06, 5 January 2012 (UTC))

Solving the Helmholtz equation using separation of variables
The transition from the section Solving the Helmholtz equation using separation of variables to the subsection Vibrating membrane is not smooth, i.e. the logical flow could be better: After having read the first text, one expects to find a solution by the separation of variables method immediately following, however the following subsection starts describing the particular case of a vibrating membrane and its history... Ovikholt (talk) 13:10, 5 January 2012 (UTC)