# Equivalence of Notation for Momentum Continuum

This wikipedia page for Magnetohydrodynamics lists the conservation of momentum in a continuum as:

$$\rho \left( \frac{\partial}{\partial t} + \vec v \cdot \nabla \right) \vec v = \vec J \times \vec B - \nabla p$$

I was rather intrigued with the notation used on the left hand side.

The form I'm more familiar with (and used in the Wikipedia derivation for the Navier-Stokes equation):

$$\rho \left( \frac{\partial \vec v}{\partial t} + \vec v \cdot \nabla \vec v\right) = \vec b$$

The difference in the right hand side is a result of what body forces are being considered ($\vec b = \vec J \times \vec B - \nabla p$).

Is this just bad practice/wrong and I'm getting worried over someone else's mistake/eagerness?

If not, how would I even go about interpreting the first equation as written?

Is it possible to translate the first equation to the second without using what I presume is an abuse of notation and distribute the $\vec v$ term? If this isn't an abuse of notation, why is this allowed?

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That it is an abuse of notation doesn't mean it's wrong. That's why it's called abuse and not misuse. – Javier Jun 21 '13 at 3:03
Fair enough if it is just an abuse of notation, but why abuse the notation here? I don't see an added interpretation value here, and I personally wouldn't consider factoring out a good simplification to take here. I also wanted to make sure there wasn't some theorem/operation which could translate 1 to 2, which it sounds like there isn't. – helloworld922 Jun 21 '13 at 3:30
This is "allowed" in a certain sense because $∂/∂t$ and $v\cdot$ are both linear operators, as is their sum. I think this falls under "operator theory" and "operator algebra", subjects I know little about. This "abuse of notation" is particularly useful in quantum mechanics. – Omnomnomnom Jun 21 '13 at 3:34
The idea to take away is that for operators $L,M$ (perhaps only under some constraints such as linearity, not sure here), it is possible to define addition over the operators so that $L+M$ satisfies $(L+M)(v)=L(v)+M(v)$. We could also take "operator multiplication" to be composition, so that $(L\cdot M)(v)$ = $L(M(v))$. This certainly works if $L$ and $M$ are some sort of differential operator. – Omnomnomnom Jun 21 '13 at 3:41