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)