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

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


1 Answer 1


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)′ ".


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