Talk:Helmholtz decomposition

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Suggestions

Suggestions:

1)To make the introduction a bit gentler, I suggest moving the equation,

\mathbf {F} =-\nabla \varphi +\nabla \times \mathbf {A}

to the beginning of the presentation.

I agree Berland 13:18, 18 January 2007 (UTC)[reply]

2)The Helmholtz decomposition takes a particularly simple form if one first takes the Fourier transform of $\mathbf {F}$ . Perhaps discuss this? It is widely used in electromagnetism.

Suggested wording change:

Change "where \mathcal{G} represents the Newtonian potential" to "where \mathcal{G} represents the Newtonian potential *operator*".

Mirko vukovic 17:24, 1 June 2007 (UTC)[reply]

BE PRECISE: how fast does a vector field have to decay for the Helmholtz decomposition to be valid? 83.30.183.19 16:22, 20 August 2007 (UTC)[reply]

The more precise statement can in fact be seen from the Hodge decomposition (which this article refers to as a generalisation). Fast decay seems to be introduced here in order to eliminate the null-space of the Laplacian. This article lacks rigour as it does not state that it only considers vector fields on

R^{3}

(in the first part at least). The section on the weak form repairs this. Bas Michielsen (talk) 23:51, 8 May 2008 (UTC)[reply]

3) Shouldn't this really say:

If ${\mathcal {G}}(\nabla \cdot \mathbf {F} )=0$ (or even $\nabla {\mathcal {G}}(\nabla \cdot \mathbf {F} )=0$ ) we say F is solenoidal...?

as well as:

If ${\mathcal {G}}(\nabla \times \mathbf {F} )=0$ (or even $\nabla \times {\mathcal {G}}(\nabla \times \mathbf {F} )=0$ ) then F is said to be curl-free...?

As it stands now, the article is assuming that if the arguments to the Newtonian potential are zero, the potential itself, or the gradients and curl thereof, are also zero. If this is true in all cases, the reason why should be discussed. If not, then this would look to be an error in the article which holds true in general, only for a d=1 Newtonian potential, where ${\mathcal {G}}(x)=c_{1}\left|x\right|$ .

Jay R. Yablon

Is this decomposition unique? It is worthwhile to mention this.

Further suggestion: Wasn't there a proof of this theorem here before? Anyway, I would like to suggest that one would be added. There is one in http://farside.ph.utexas.edu/teaching/em/lectures/node37.html NAkyoLyBabeguTe (talk) 14:39, 10 September 2008 (UTC)[reply]

Uniqueness and generalizations

The subject of uniqueness still needs to be addressed; I'll see if I can find a good source for that. (It's probably pretty trivial, but....) Also, Griffiths (the source I put in a little while back) has this footnote: "any vector function can be expressed as the gradient of a scalar plus the curl of a vector. But only if it goes to zero sufficiently rapidly at infinity are the scalar and vector potentials given by [the Newtonian potential operator]." Does anyone have a source for that? Annoyingly, Griffiths gives no source in his footnote. -- Spireguy (talk) 03:07, 3 April 2009 (UTC)[reply]

Uniqueness follows from the L² orthogonality of the decomposition, which indeed is trivial (on sufficiently smooth vector fields). Any source on the Hodge decomposition theorem would cover this aspect of things, and probably it is in any decent mathematical source that specifically addresses the Helmholtz decomposition, although I admit that I'm not that familiar with such sources. The Newtonian potential operator isn't well-behaved for functions that are not compactly supported (or at least rapidly decreasing). In many cases of practical interest, the Newtonian potential is not well-defined. I think the standard approach for more general classes of functions (e.g., L^p) is to use the Riesz transforms instead. Sławomir Biały (talk) 15:53, 5 August 2010 (UTC)[reply]

Thanks for your reply. Since R³ is not compact, one can't just quote the standard references on the Hodge decomposition, which assume that the manifold is compact. So what I'm asking for is a source that does rigorously and precisely address the case of R³. I wasn't able to find that in what I have. I'm not familiar with the Riesz transform approach, and I'm not sure if that addresses this particular problem. -- Spireguy (talk) 17:05, 6 August 2010 (UTC)[reply]

Compactness is not really necessary for the uniqueness. This is just the L² orthogonality of the three spaces in the Hodge decomposition, and that much holds regardless of compactness. As for a book that treats the case of Rⁿ specifically, I have Iwaniec and Martin, Nonlinear analysis and geometric function theory, although this seems to be the only book that I have that singles out the Rⁿ case for special consideration. Sławomir Biały (talk) 19:56, 6 August 2010 (UTC)[reply]

I found the following reference which should give general precise conditions on the Helmholtz decomposition. Unfortunately it is in German:

von Wahl, W. (1990), "Abschätzung für das Neumann-Problem und die Helmhotz-Zerlegung von L^p", Nachr. Akad. Wiss. Göttingen, Math.-Phys., II (2).

--Sławomir Biały (talk) 20:04, 6 August 2010 (UTC)[reply]

Hello, I just stumbled across the uniqueness question myself. The German version of the Helhomholtz page lists a good reference where this question is adressed: G. P. Galdi, An introduction to the mathematical theory of the Navier-Stokes equations. Vol. I, Springer Tracts in Natural Philosophy, vol. 38, Springer-Verlag, New York, 1994, ISBN 0-387-94172-X

Theorem III.1.2 states that each vector field in L^q(U), 1<q<\infty can uniquely be decomposed into two vector fields in L^q(U) where the one is a gradient and the other one is divergence free, provided that either the boundary of U is sufficiently smooth or U is the whole space or a half-space. If q=2 Theorem III.1.1 states that the decomposition holds for every domain in R^n.

-- Joachim

Disappearance rate?

The first section says it must disappear faster than 1/r, the second says it must disappear faster than (1/r)^2
Which one? 71.209.145.7 (talk) 05:32, 19 February 2010 (UTC)[reply]

Rapidly decaying vector field

What is a rapidly decaying vector field ? —Preceding unsigned comment added by 193.157.197.135 (talk) 12:01, 28 February 2011 (UTC)[reply]

Clarified final section slightly

Excellent article! Just split some lines in the final section so that we're not equating scalar fields with vector fields. Even if both are zero, they are the zeros of different spaces... — Preceding unsigned comment added by Sukarsono (talk • contribs) 16:58, 4 April 2011 (UTC)[reply]

It's undoubtedly more rigorous to tell two zeroes apart, but you may regard writing the same as an abuse of notation. Could you imagine a situation where the difference between two zeroes will actually matter, say, affect the result of calculations or validity of a theorem? --Netheril96 (talk) 01:11, 5 April 2011 (UTC)[reply]

Helmholtz decomposition is wrong

The Fundamental theorem of vector calculus, (Helmholtz decomposition) states that any sufficiently smooth, rapidly decaying vector field in three dimensions ${\mathbf {F} }$ can be constructed with the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field (scalar potential $\varphi$ and a vector potential ${\mathbf {A} }$ )

${\mathbf {F} }=-\operatorname {grad} \psi +\operatorname {rot} {\mathbf {A} }\Rightarrow {\mathbf {F} }=\operatorname {grad} \varphi +\operatorname {rot} {\mathbf {A} }$ (1)

However, the gradient of scalar function does not form the vector field. As well known from textbook [1, p. 15] « … under co-ordinate change the gradient of function transforms differently from a vector »: hence the theory requiring (1) must be false. The next unpleasant things we can see for such well-known classical rules. In mathematics and physics the rot (or curl) is an operation which takes the vector field ${\mathbf {A} }$ and produces another vector field $\operatorname {rot} {\mathbf {A} }$ . However it is well-known that $\operatorname {rot} {\mathbf {A} }$ is an Antisymmetric Tensor . Therefore under co-ordinate change the tensor $\operatorname {rot} {\mathbf {A} }$ transforms differently from a true vector. For elimination of these contradictions the Fundamental theorem of vector calculus can be written as follows:

${\vec {F}}=\operatorname {grad} \varphi +\operatorname {rot} \operatorname {rot} {\vec {A}}$ . (2)

This formula completely corresponds to transformed Navier–Stokes equations(NSE) for incompressible fluids ( $\operatorname {div} {\dot {\vec {u}}}=0$ )

$\rho {\vec {F}}-\operatorname {grad} p+\mu \nabla ^{2}{\dot {\vec {u}}}=\rho {\ddot {\vec {u}}}\Rightarrow \rho ({\vec {F}}-{\ddot {\vec {u}}})=\operatorname {grad} p+\operatorname {rotrot} \mu {\dot {\vec {u}}}$ . (3)

Here, ${\vec {F}}={\vec {F}}_{1}+{\vec {F}}_{2}+...$ -¬¬ vectors sum of a given, externally applied forces (e.g. gravity ${\vec {F}}_{1}$ , magnetic ${\vec {F}}_{2}$ and other), $p$ - pressure (scalar function), ${\dot {\vec {u}}}$ - velocity vector, ${\ddot {\vec {u}}}=d{\dot {\vec {u}}}/dt$ - acceleration vector, $\rho$ - density (const), $\mu$ - viscosity (const), $\nabla ^{2}$ - Laplace operator.

Equations (3) and (2) are consistent. Hence there is no reason to say that the theory requiring (2) must be false. As we can see from NSE the sum - $\operatorname {grad} p+\mu \nabla ^{2}{\dot {\vec {u}}}=-(\operatorname {grad} p+\operatorname {rotrot} \mu {\dot {\vec {u}}})$ forms the vector field.

Note that we will receive the formula (2) also after similar transformation of the Navier–Stokes for a compressible fluid and after transformation of the Lame equations for an elastic media.

From this brief note follows that Helmholtz decomposition is wrong and demands major revision.

Therefore let's try to formulate the text for editing of this article.

1.B. A.; Fomenko, A. T.; Novikov, Sergeĭ Petrovich (1992). Modern Geometry--methods and Applications: The geometry of surfaces, transformation groups, and fields] (2nd ed.) . Springer. (p. 15).ISBN 0387976639.

Alexandr--188.163.102.90 (talk) 21:30, 24 January 2012 (UTC)[reply]

I think it would be good if you make an account, it is easier to talk that way. Another place to request opinions at is Wikipedia talk:WikiProject Mathematics. Oleg Alexandrov (talk) 16:22, 26 January 2012 (UTC)[reply]

Thanks for a quick reply! However it is not clear what account you mean? Alexandr--188.163.30.98 (talk) 18:31, 26 January 2012 (UTC)[reply]

In 3D Euclidean space, the gradient of a scalar field is a vector field. (Covectors and vectors can be identified.) And if F is a vector field than A is a pseudovector field. This does not disprove the theorem.

A more fundamental point: The Helmholtz decomposition is 100 years old and is well understood by thousands of mathematicians, covered in hundreds of textbooks, and taught each year to thousands of students. It is unlikely to be "wrong". It is much more likely that you have made a mistake in your own thinking or understanding, or that wikipedia has a typo. Either way, you should not be inventing your own corrected version (see WP:NOR). You should be looking up in a book what the correct version is and how to understand it. :-) --Steve (talk) 15:28, 27 January 2012 (UTC)[reply]

Dear Steve! Thanks for your comments!

“:In 3D Euclidean space, the gradient of a scalar field is a vector field.”

It is only assumption! As well known from textbook Modern Geometry «… under co-ordinate change the gradient of function transforms differently from a vector».

“The Helmholtz decomposition is 100 years old and is well understood by thousands of mathematicians, covered in hundreds of textbooks, and taught each year to thousands of students”.

These classical representations in majority textbooks have not been strictly proved. It follows from comparison of (3) and (2). Why you do not make comments on the formula (3)? Look attentively: the formula (2) does not contradict classical representations! The formula (2) follows as a special case of the formula (1) if a gradient and a rotor to consider as vectors! It is necessary to make replacement А=rot Ặ. A surprising case, You agree?

“Either way, you should not be inventing your own corrected version (see WP:NOR)”.

This version follows from comparison of two articles(in various articles in Wikipedia). Therefore I hope, that you support this editing of article.

For improvement of my claim I can give additional arguments, if You agree. Alexandr--94.27.85.94 (talk) 12:13, 29 January 2012 (UTC)[reply]

Dear Steve! Thanks for quick editing of http://en.wikipedia.org/wiki/Helmholtz_decomposition#External_links ! However I hope for your variant of article editing taking into account this discussion http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics/Archive/2012/Mar#Helmholtz_decomposition_is_wrong --Alexandr (talk) 11:33, 10 May 2012 (UTC)[reply]

No. I hope you read WP:NOR. If you have already read it, I hope you read it again! --Steve (talk) 12:28, 10 May 2012 (UTC)[reply]

I read WP:NOR. But, these OR are comparison of two Wikipedia articles (http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations and http://en.wikipedia.org/wiki/Helmholtz_decomposition ). These articles conflict with each other. In Wikipedia all contradictions must be deleted. If it not maybe that such contradictions can not be hidden. It is necessary to write about them. Otherwise it is Wikipedia disrupting --Alexandr (talk) 18:34, 22 May 2012 (UTC)[reply]

Just to butt in here: the problem here seems to be terminology. First of all: the gradient of a scalar field is a 'covector field'. But this is a synonym in some contexts for a 1-form. Second: the curl of a vector is a 'pseudovector'. But both of these don't really have any effect on the statement of the article. The article seems to be using vector field to mean a function from a vectorspace into itself. In that case, there is no argument; the Helmholtz decomposition IS a vector field in this sense by definition. Furthermore, this implies that the vector potential is a pseudovector. As for the conservative component being a "covector", this is adding non usefull meaning into a (for the purposes of the article) unrelated term.--217.84.63.84 (talk) 12:34, 27 July 2012 (UTC)[reply]

The problem is that you aren't supposed to synthesize information from different sources to edit Wikipedia. If you would like to edit it to say that there are problems with the Helmholtz decomposition, then you need to find a reliable source that directly concludes as much. Anything else is WP:SYN. PiT (The Physicist) (talk) 05:03, 18 October 2012 (UTC)[reply]

Dear Alexandr, this is Taka. Thank you for letting me know this talk page. I want to bring out this problem from another view points. This thorem is Ok for electromagnetics, but as a theorem for hydrodynamics I have found some problems on this theorem. I think this theorem has been established on many mistakes.

mistake 1.

We can calculate the distribution of ( $\operatorname {div} \mathbf {F}$ )　and ( $\operatorname {rot} \mathbf {F}$ ) from any flow $\mathbf {F}$ . But it does not mean that the flow $\mathbf {F}$ . can be separated into two kinds of flows.

If ( $\operatorname {div} \mathbf {F}$ )　and ( $\operatorname {rot} \mathbf {F}$ ) were calculated from two of separated flows, you could calculate a scalar potential $\varphi$ and a vector potential ${\mathbf {A} }$ .

Here, I want to define $\mathbf {G}$ =- $\operatorname {grad} \psi +\operatorname {rot} \mathbf {A}$ .

Helmholtz Decomposition theorem says that $\mathbf {G}$ = $\mathbf {F}$

But, it is not needed to separate $\mathbf {F}$ into two flows to get those distribution of ( $\operatorname {div} \mathbf {F}$ )　and ( $\operatorname {rot} \mathbf {F}$ ) .

To be able to get uniquely these distributions is not an evidence for that Helmholtz Decomposition is right.

If $\mathbf {F}$ has the component which play roles both of ( $\operatorname {div} \mathbf {F}$ )　and ( $\operatorname {rot} \mathbf {F}$ ) ,

Generally, $\mathbf {F}$ ≠ $\mathbf {G}$ .

There are $\operatorname {grad} \psi$ and $\operatorname {rot} \mathbf {A}$ in $\mathbf {G}$ , not in $\mathbf {F}$

This comment would be easy to understand, if you would see some illustration posted in my blog.

Many hydrodynamicists have misunderstood ${\mathbf {G} }$ with ${\mathbf {F} }$

mistake 2

Why can I say that there are not any scalar potential nor any vector potential in any flow which has both curls and divergences?

There are two of basically mathematical theorem ^[1] on any flow(vector function).

One of them says that if a flow has culr component, there is no velocity potential $\psi$ , and if there are ^[2] $\psi$ in a flow, the flow has not curl.

That is, not to have curl component is the necessary and sufficient condition for existing of velocity potential $\psi$ .

And other says that if a flow has divergent component in the flow, there is no ^[3] $\mathbf {A}$ , and if there are vector potentials $\mathbf {A}$ in the flow, the flow has no divergence. That is, not to have divergence in the flow is the necessary and sufficient condition for exiting of vector potential $\mathbf {A}$ .

So, I can definitively say that there is no $\psi$ and　 $\mathbf {A}$ in the flow which has curl and divergence.

But, almost of authorities of Meteorology and Hydrodynamics think that there are both a scalar potential and a vector potential in any flow.

I can show you some good example.

I would like to liken all kinds of flows to container boxes.

Then, there are just two kinds of shape of container, one of them is spherical shape, and other is cubic. A cubic shape container means a curl-free flow, and a spherical shape container means a flow including curl.

Then, I check all of containers(flow), and if it’s a cubic container(curl-free), I put cotton( $\psi$ ) into it. And if it’s a spherical container(including curl), I confirm that there is no cotton( $\psi$ ) in it.

And then, all container is painted with only two colors, red or black. Red one means a divergence-free flow, and black one means a flow including divergence.

Then, I check all of containers(flow), and if it’s a red container(divergence-free), I put star ornaments( ${\mathbf {A} }$ ) into it. And if it’s a black container(including divergence), I confirm that there is no star ornament( ${\mathbf {A} }$ ) in it.

Aren’t you sure that there is no cotton nor star ornament in any spherical black container.

I can’t believe that those clever persons think that there are cotton and star ornament in a spherical black container.

Why do they think so? I think they confuse ${\mathbf {G} }$ with ${\mathbf {F} }$ in < mistake 1 >

The spherical black container is ${\mathbf {F} }$ , not ${\mathbf {G} }$ . There is not $\psi$ nor　 ${\mathbf {A} }$ in ${\mathbf {F} }$ .

mistake 3

If you want to prove that any vector ${\mathbf {F} }$ can be devided into a irrotational vector ${\mathbf {F} }$ e and a solenoidal vector ${\mathbf {F} }$ r, you need to find some identity which can be described by like ${\mathbf {F} }$ ＝ ${\mathbf {F} }$ e + ${\mathbf {F} }$ r.

As far as I know, the only equation is [vector triple product identity]” .

$\mathbf {U} \times$ （ $\mathbf {V} \times \mathbf {W}$ ）=( $\mathbf {U} \cdot \mathbf {W}$ ) $\mathbf {V}$ –( $\mathbf {U} \cdot \mathbf {V}$ ) $\mathbf {W}$

If you replace $\mathbf {U}$ and $\mathbf {V}$ with $\nabla$ , you may get

$\nabla \times$ （ $\nabla \times \mathbf {W}$ ）= $\nabla$ ( $\nabla \cdot \mathbf {W}$ ) – $\nabla$ ² $\mathbf {W}$

So,you can get

${\mathbf {F} }$ = - $\nabla$ ² $\mathbf {W}$ =- $\nabla (\nabla \cdot \mathbf {W}$ )+ $\nabla \times \nabla \times \mathbf {W}$ = - $\nabla \psi +\nabla \times \mathbf {A}$

In above equation, the first term of the right hand shows a curl-free flow, and the second one shows a divergence-free flow.

So, you might say that Helmholtz decomposition is perfectly proved.

But, the third mistake is that they take - $\nabla$ ² $\mathbf {W}$ as any flow.

Any flow $\mathbf {F}$ certainly exist. But $\mathbf {W}$ is not guaranteed to exist. You need to make sure that there exists $\mathbf {W}$ for any flow $\mathbf {F}$ .

And if $\mathbf {W}$ exists, at the same time, - $\nabla \psi$ 　and $\nabla \times \mathbf {A}$ are decided uniquely with $\mathbf {W}$ , and they need to appear together.

I can show you that there are many flows which do not include $\mathbf {W}$ .

You have got following expression from the vector triple product identity.

${\mathbf {F} }$ = - $\nabla \psi +\nabla \times \mathbf {A}$

It is given on terms and conditions as required by $\psi$ = $\nabla \cdot \mathbf {W}$ , $\mathbf {A}$ = $\nabla \times \mathbf {W}$

That is, $\psi$ and $\mathbf {A}$ are deriven from the common functin $\mathbf {W}$ .

If $\psi$ is decided from $\mathbf {W}$ , then $\mathbf {A}$ should be decided uniquely at the same instance.

Assuming that Helmholtz decomposition theorem is correct. you can consider two flows as following.

${\mathbf {F} }$ ₁ = - $\nabla$ ² $\mathbf {W}$ ₁ = - $\nabla \psi$ ₁ + $\nabla \times \mathbf {A}$ ₁

${\mathbf {F} }$ ₂ = - $\nabla$ ² $\mathbf {W}$ ₂ = - $\nabla \psi$ ₂ + $\nabla \times \mathbf {A}$ ₂

$\psi$ ₁ and $\mathbf {A}$ ₁ are functions which are derived from $\mathbf {W}$ ₁ and, $\psi$ ₂ and $\mathbf {A}$ ₂ are derived from $\mathbf {W}$ ₂ .

Here, because an arbitrary flow (vector function) must be possible, you can consider the flow ${\mathbf {F} }$ ₃ which includes divergent component of (- $\nabla \psi$ ₁) and rotational component of ( $\nabla \times \mathbf {A}$ ₂).

${\mathbf {F} }$ ₃ = - $\nabla$ ² $\mathbf {W}$ ₃ = - $\nabla \psi$ ₁ + $\nabla \times \mathbf {A}$ ₂

Here, $\psi$ ₁ = $\nabla \cdot \mathbf {W}$ ₁, ${\mathbf {A} }$ ₂= $\nabla \times \mathbf {W}$ ₂,

I must say that again ${\mathbf {F} }$ ₃ should have ( $\psi$ ₁) as divergent component, and have ( $\nabla \times \mathbf {A}$ ₂) as rorational component.

But according to Helmholtz decomposition, ${\mathbf {F} }$ ₃ can be decomposed into two flows only by the vector triple product identity.

Then, ${\mathbf {F} }$ ₃ = - $\nabla$ ² $\mathbf {W}$ ₃ = - $\nabla \psi$ ₃ + $\nabla \times \mathbf {A}$ ₃

Therfor, ${\mathbf {F} }$ ₃ has (- $\nabla \psi$ ₃) as divergent component, and has ( $\nabla \times \mathbf {A}$ ₃) as rorational component.

But, because ${\mathbf {F} }$ ₃≠ ${\mathbf {F} }$ ₁, and ${\mathbf {F} }$ ₃≠ ${\mathbf {F} }$ ₂,

$\mathbf {W}$ ₃≠ $\mathbf {W}$ ₁ $\mathbf {W}$ ₃≠ $\mathbf {W}$ ₂

So, ${\mathbf {F} }$ ₃ = - $\nabla$ ² $\mathbf {W}$ ₃≠- $\nabla (\nabla \cdot \mathbf {W}$ ₁ ) + $\nabla \times$ ( $\nabla \times \mathbf {W}$ ₂ )

Here, you must say that ${\mathbf {F} }$ ₃ which has combined with divergent component of ( $\nabla \psi$ ₁) and rotational component of ( $\nabla \times \mathbf {A}$ ₂) can not decomposed into divergent component of ( $\nabla \psi$ ₁) and rotational component of ( $\nabla \times \mathbf {A}$ ₂).

Or we may have to say that there is no $\mathbf {W}$ functions in ${\mathbf {F} }$ ₃.

I can show you many flows like this kind of flow.

We have to say that "The vector triple product identity" shows the special cace in which a vector function can decomposed into two special vector functions, and Helmholtz Decomposition theorem that says any flow can be decomposed into an irrotational flow and a non-divergent flow is wrong.

More strictly the problem Helmholtz Decomposition Contradictions is covered here http://analysis3.com/Helmholtz-decomposition-contradictions-download-w117376.html --Alexandr (talk) 14:25, 1 December 2014 (UTC)[reply]

i agree with Aleksandr and Taka. Vector analysis was derived from the earlier quaternion calculus in a way that cut degrees of freedom. Vectors are pure quaternions, versors are unit quaternions. In full quaternion calculus the interaction between scalar and vector part is richer than in vector analysis (gradient (scalar to co-vector), divergence (vector to scalar) and curl (vector to pseudo-vector)). From this cutting of degrees of freedom emerge the inconsistencies that surround vector calculus and also the Helmholtz theorem. — Preceding unsigned comment added by 2A02:2450:102C:38A:B169:D7CF:64FF:177D (talk) 12:10, 5 October 2019 (UTC)[reply]

References

Also called Helmholtz–Weyl decomposition

This decomposition also seems to be called the Helmholtz–Weyl decomposition or even just the Weyl decomposition, and is a special case of the de Rham-Hodge-Kodaira decomposition on manifolds.

Probably the alternate names & generalizations should be mentioned in the lede. Googling these terms uncovers many, many references, so it is only a matter of selecting the appropriate ones.

— Steven G. Johnson (talk) 18:36, 16 July 2013 (UTC)[reply]

Intro needs work

It's not clear that the Helmholtz decomposition works by splitting into an operation on a scalar field and an operation on a vector field. It is worded as "This implies" rather than stated, or explained why. — Preceding unsigned comment added by 45.49.18.32 (talk) 21:55, 31 May 2015 (UTC)[reply]

Better now? Sławomir Biały (talk) 22:55, 31 May 2015 (UTC)[reply]

What do you think of changing the title by "Helmholtz Decomposition Theorem"?

I think that "Helmholtz Decomposition Theorem" would be more appropriated for the title because in many places it is referred to the "Helmholtz Theorem" or the "Helmholtz Theorem of electromagnetism" etc.

Best regards. Physicist137 (talk) 21:27, 29 April 2020 (UTC)[reply]

Misleading formulae

The formula for $\Phi$ and $A$ in the first section involve surface integrals. But these integrals are obviously null because F, being continuous and null outside V, should be null at the boundary of V. This is misleading. And if these surface integrals are removed, V can be replaced by $\mathbb {R} ^{3}$ , which is better Thus, after proving the result with the help of a finite domain V, the proof can be extended to all functions that decay at least as fast as 1/r, by just multiplying F by a function that is equal to 1 on the ball of radius R, to 0 outside the ball of radius R+1, and sufficiently smooth in between. That's just Lebesgue convergence formula. .maimonid (talk) 21:57, 7 May 2022 (UTC)[reply]

[1] [1]

[2] [potentials]

[3] vector potential

[1]

[2]

[3]