# Difference between $\mathop{\text{div}} \vec x$ and $\vec \nabla \cdot \vec x$?

I often see the following notations for the divergence in theoretical Physics: $$\mathop{\text{div}} \vec x \qquad \left\langle \vec \nabla, \vec x \right\rangle \qquad \vec \nabla \cdot \vec x$$

Similarly, $\mathop{\text{curl}} \vec x$ ($\mathop{\text{rot}} \vec x$ in German) and $\vec \nabla \times \vec x$.

Is there a difference to those notations or is it just personal preference? Are there good reasons to use either of the notations? I see that $\vec \nabla \times$ is international, but “curl” and “rot” are different in each language.

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It's personal preference - all three notations mean the same operator.

I tend to use "$\nabla\cdot x$" when I'm working in coordinates (usually in flat $\mathbb{R}^n$) and "$\operatorname{div}x$" when I'm working on a general manifold or without coordinates. The inner product notation is helpful for remembering what the coordinate formula for the divergence, but if I want to work without coordinates, it just gets in the way.

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The notation $\vec \nabla$ means $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z})$.

Then you can see why we say $\textrm{div}( \vec x) = \vec \nabla . \vec x$. Same for curl.

This notation is very usefull to remember that curl (curl) = grad(div) - $\Delta$.

Writting $$\nabla \wedge \nabla \wedge x = \nabla ( \nabla. x)- \nabla^2 x$$ is the same formula as for the usual exterior product $\wedge$ : $$a \wedge (b \wedge c) = b (a.c) - (a.b)c.$$

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I think the reason people like $\text{div}$ is because $\nabla$ isn't really a vector, and $\nabla\cdot$ isn't really a inner product either. That being said, I think $\nabla$ is far more readable and concise, and that most people understand that it's just a convenient shorthand notation.

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Is $\nabla$ really a covector then? – Martin Ueding Jan 27 '13 at 13:26
No not really. If $\nabla$ is a functional, what is $\nabla\cdot$? Again - this is just notation candy - nothing more. – nbubis Jan 27 '13 at 13:40
@queueoverflow On page 44 of the following german text (essentially the last page), there are given some examples why $\nabla$ is best thought of as just a notational convenience. However, the gradient ($\operatorname{grad} f=\nabla f$) is a covector. – Thomas Klimpel Jan 27 '13 at 14:31