Scalar geometric product ambiguity involving derivatives

Question

I'm working on deriving the Euler equations for a perfect fluid from the conservation of the energy-momentum tensor and I've found some difficulties with the notation.

Given two vector fields, $v(x)$, $a(x)$ and the vector derivative $\nabla$. If you form an expression like

\begin{equation} (v \cdot \nabla) a = (v^\mu \nabla_\mu ) a, \end{equation}

it would mean that $(v \cdot \nabla)$ is a scalar differential operator (the directional derivative) acting on $a$.

However, if you construct the expression

\begin{equation} ( \nabla \cdot v) a = (\nabla_\mu v^\mu ) a, \end{equation}

it would mean that you are multiplying the divergence of $v$ with $a$.

Considering that in Geometric Algebra the scalar product between vectors is symmetric, should not both expressions be equivalent?

Answer 1

The symmetry of the inner product is not particular to geometric algebra, it is something from standard vector calculus. In both contexts, the two mean very different things as $\nabla$ operates on $v$ in one case and $a$ in the other; in fact, you may see some authors prefer to write the directional derivative as, $$ \left(v\cdot\nabla\right)a=v\cdot\left(\nabla a\right)\tag{1} $$ to indicate that $\nabla$ acts on $a$ and is dotted with $v$ (these are indeed equivalent, though I prefer using the first form).

I suspect that your confusion arises when trying to write these as, $$ \left(v\cdot\nabla\right)a=\langle v\nabla\rangle_0a=\langle\nabla v\rangle_0a=\left(\nabla\cdot v\right)a\tag{2} $$ and assuming they are equal due to the symmetry of the grade-0 product. However, $\nabla$ in this case is an operator, rather than an ordinary vector, so there are particular rules one needs to assert when considering expressions utilizing vectors:¹

1. In the absence of brackets, $\nabla$ operates on the object to its immediate right.
2. When $\nabla$ is followed by brackets, it operates on all terms in the brackets.
3. When $\nabla$ acts on an object to which it is not adjacent, use overdots to indicate scope.

Thus, you would write (2) using Rule #3 as: $$ \left\{\left(v\cdot\nabla\right)a=\langle v\dot\nabla\rangle_0\dot a=\langle\dot\nabla v\rangle_0\dot a\right\}\neq\left\{\langle\dot\nabla\dot v\rangle_0a=\left(\nabla\cdot v\right)a\right\} $$ which makes it more clear that the two expressions are not equivalent.

^{1. I have seen some authors use notation like $\langle v\nabla^aa\rangle_1$ to indicate that $\nabla$ acts on $a$. The rules here are also given by Hestenes; I believe that both Bromborsky and Doran & Lasenby also follow these rules.}