Del operator - order of operations

Reading through an old scientific paper, I've come across some potentially ambiguous notation using the del operator. In this example, $f$, $g$, and $x$ are scalars, and $\textbf{V}$ is a vector. The equation is:

$$x = \nabla \cdot f \textbf{V} g.$$

Should this be interpreted to indicate:

$$x = \nabla \cdot \left(f g\textbf{V} \right),$$

or does it possibly indicate:

$$x =g \nabla \cdot \left(f \textbf{V}\right)?$$

Is the answer indicated by a strict order of operations argument, or is the first form some sort of commonly known shorthand notation? I'm not sure why the two scalars would be separated if they were both intended to be part of the dot product.

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My guess would be that the two scalars have originated from separate expressions and it's written that way to indicate that. Still, I don't often see scalars written on the right, but in commutative context it's not much of a difference... –  tomasz Nov 26 '12 at 22:46

See the discussion on the Del operator in Wikipedia: it helps to disambiguate notation involving the $\nabla$ operator. See in particular the section entitled "Precautions". Then you might want to scroll to the top of the webpage to read the entire entry.

See also the Product Rules when using the $\nabla$ operator.

The use of $\nabla$ and how it operates depends to some degree on the context in which it is used. The "three possible meanings—gradient, divergence, and curl—can be formally viewed as the product of scalars, dot product, and cross product, respectively, of the del "operator" with the field. These formal products do not necessarily commute with other operators or products."

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The Wikipedia page is useful, and I consulted it before asking this question. However nothing in the "precautions" seems to apply here. –  JCOidl Nov 26 '12 at 22:26
@amWhy: He's not asking about how to manipulate the expression, but how to parse it in the first place. –  Henning Makholm Nov 26 '12 at 22:37
Hopefully @HenningMakholm cleared up the confusion. My example doesn't match any of the product rule examples because it has no parenthesis. The lack of parenthesis is exactly the problem. –  JCOidl Nov 26 '12 at 22:38
I certainly didn't mean to imply that you haven't done your research, or that Wikipedia will answer your question. It simply suggests how to make sense of the expression. No doubt, the expression is ambiguous. –  amWhy Nov 26 '12 at 22:38

Are these things fields on a manifold? I so, the paper may be following the common convention in differential geometry that "vector field" on a manifold means a linear operator that takes scalar fields to scalar fields and satisfies certain conditions. Morally you can think of it as

$$\mathbf V g \equiv \mathbf V \cdot (\nabla g)$$

which means that $\mathbf V g$ is a scalar field and $f \mathbf V g = f(\mathbf Vg) = (f\mathbf V)g$ is a different scalar field (made by pointwise multiplication of $f$ with $\mathbf V g$.

On the other hand, if that is the meaning, then the dot after the explicit $\nabla$ in your expression is a bit mysterious, and the result of the entire expression ought to be a (co)vector rather than a scalar. So you'll have to decide for yourself from the context whether this suggestion makes sense at all.

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The dot following $\nabla$ isn't necessarily mysterious. What is mysterious is the lack of parentheses. –  amWhy Nov 26 '12 at 22:41
Thanks for the suggestion. This is actually from a fluid mechanics text, so I don't think this applies here. I hope your answer is useful to someone else down the road though! –  JCOidl Nov 26 '12 at 22:42