Talk:Helmholtz decomposition/Archive 1

Archive 1

Something wrong?

Something's wrong here. As ${\mathcal {G}}$ is defined in the article on the Newtonian potential operator, ${\mathcal {G}}(\nabla \times \mathbf {F} )$ is a scalar field. How can you take the curl of it! --unsigned anon

I guess that in the formula

$\mathbf {F} =-\nabla \,{\mathcal {G}}(\nabla \cdot \mathbf {F} )+\nabla \times {\mathcal {G}}(\nabla \times \mathbf {F} )$

the quantity $\nabla \times \mathbf {F}$ is a vector, therefore, the quantity ${\mathcal {G}}(\nabla \times \mathbf {F} )$ is also a vector (applied componentwise to the components of $\nabla \times \mathbf {F}$). Oleg Alexandrov (talk) 04:32, 1 December 2007 (UTC)

recently added sentence in lead

I just removed the following sentence from the lead:

If $\mathbf {F}$ does not extend to infinity, but ends at a boundary, then its normal component at the boundary must be specified in addition to $\varphi$ and $\mathbf {A}$ in order for $\mathbf {F}$ to be unique.

This doesn't quite make sense: $\mathbf {F}$ is a given. It isn't "unique", it's specified --- it's an assumption. On the other hand, there isn't a unique scalar potential $\phi$ or vector potential $\mathbf {A}$. Modulo constants and potential fields, though, there is uniqueness. Or, in the case of a compactly supported $\mathbf {F}$, one can specify BCs for $\phi$ and $\mathbf {A}$ as an alternative. Is this what you're thinking? Lunch (talk) 20:37, 18 June 2008 (UTC)

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)

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)

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)

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)

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)