3
$\begingroup$

What is the name of this indentity and which mathematician did derive this?

$$\frac{1}{2}\mathbf{\nabla (u \cdot u) = u \times (\nabla \times u ) + (u \cdot \nabla)u}$$

$\endgroup$
5
$\begingroup$

Vector analysis identities typically do not have special names and appear as exercises in textbooks, they can be derived in too routine a manner to take notice who was first to do it. The first of such textbooks, Vector Analysis, a Text-book for the Use of Students of Mathematics and Physics has a Wikipedia article all to itself and was written by the co-founder (with Heaviside) of vector analysis Gibbs together with Wilson in 1901. The book went through seven editions afterwards, but it was based on Gibbs's 1881-1884 lectures Elements of Vector Analysis that were printed as a short pamphlet.

The last edition of the text is available on Gutenberg, where an even more general identity appears as formula (46)′ on p.161:

$$\mathbf{\nabla (u \cdot v)_u = (u \cdot \nabla)v+ u \times (\nabla \times v )}$$

Earlier on p. 159 the authors explain: "The notation $\bf\nabla (u \cdot v)_u$ will be used to denote that in applying the operator $\bf\nabla$ to the product $\bf(u\cdot v)$ the quantity $\bf u$ is to be regarded as constant. That is, the operation $\bf\nabla$ is carried out only partially upon the product $\bf(u\cdot v)$". Apparently, Gibbs and Wilson felt very strongly about this notation since they promote it in a footnote on the same page:"This idea and notation of a partial $\bf\nabla$ so to speak may be avoided by means of the formula (41). But a certain amount of compactness and simplicity is lost thereby. The idea of $\bf\nabla (u \cdot v)_u$ is surely no more complicated than $\bf(u \cdot \nabla)v$ or $\bf v\times(\nabla \times u)$".

By inspection from the coordinate formulas on the next page, $\bf\nabla (u \cdot v)_u$ is equal to $\frac{1}{2}\mathbf{\nabla (u \cdot u)}$ when $\bf v=u$. There is a lot of applying this identity on the subsequent pages, but I am afraid no name for it other than "(46)′ ".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.