Del operator - order of operations

Hey! Is it true that when you dot the del-operator on another vector, the differentiation has priority over the dot-product? That's why you get all those weird formulas for the divergence in circular and cylindrical coordinates (which are very different to the Cartesian ones)?

So in the case of a 2 dimensional vector in cartesian coordinates it actually goes like this?

$$\nabla \cdot \vec{V} = i \frac{\partial \vec{V}}{\partial x} + j \frac{\partial \vec{V}}{\partial y}$$

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D H
Staff Emeritus
Remember when you were taught that it's best to look at ##\frac {dy}{dx}## not as a fraction but as some operator ##\frac d{dx}## applied to the function ##y(x)## ?

The same applies here. It's best to look at ##\nabla\cdot## as an operator. Or maybe as a convenient abuse of notation. (The same applies to ##\nabla\times##) Don't look at these as ##\nabla\cdot\vec v## as involving a pair of operators that are somehow magically combined with some weird order of operations rule.

thanks for the reply :), but I don't fully understand your point.

Are you saying I should be rigorous when applying (ie taking the dot-product) the del-operator on a vector? Ie, I should think like I did in the latex-equation in the OP, instead of just saying

## \nabla \cdot \vec{V} = \frac{\partial V_x}{\partial x} + \frac{\partial V_y}{\partial y} ##?

And what exactly do you mean by
It's best to look at ##\nabla\cdot## as an operator
?

CompuChip
Homework Helper
Consider an operator, let's call it "D" which acts on a vector, and is defined by
$$\operatorname{D}( \vec V ) = \frac{\partial \vec V}{\partial x} + \frac{\partial \vec V}{\partial y}.$$

No wait, let me change my mind and instead of "D" let me call it "div". Also, I will leave out the brackets as this is pretty common for operators; so I have defined
$$\operatorname{div} \vec V = \frac{\partial \vec V}{\partial x} + \frac{\partial \vec V}{\partial y}.$$

This is what DH is trying to tell you: view it as an operator, defined in this way; how you call it is irrelevant. Actually, we can change our mind again, and instead of "D" or "div" we can expand our alphabet and call it ##\operatorname{\Large\spadesuit \tilde\ast}## and write
$$\operatorname{\Large\spadesuit \tilde\ast} \vec V = \frac{\partial \vec V}{\partial x} + \frac{\partial \vec V}{\partial y}$$
or we could call it ##\nabla \cdot## and write
$$\nabla \cdot \vec V = \frac{\partial \vec V}{\partial x} + \frac{\partial \vec V}{\partial y}.$$
Hmm, actually that last one uses a symbol we already know. Conveniently it has some things in common with what we normally use it for, so let's stick to this one.

(Actually, "div" is also used quite regularly; probably by authors who try not to confuse people like yourself :-) )

Similarly, you could define an operator "grad" by
##\operatorname{grad} \vec V = \hat i \frac{\partial V}{\partial x} + \hat j \frac{\partial V}{\partial y} + \hat k \frac{\partial V}{\partial z}##. If you are too lazy to write "grad" and prefer shortening it to a scribble looking like ##\nabla## be my guest (in fact, most of us mathematicians do that). But again, view it as shorthand notation and not as some vector ##\nabla = \hat i \frac{\partial}{\partial x} + \hat j \frac{\partial}{\partial y} + \hat z \frac{\partial}{\partial z}## forming some strange product with ##\vec V = \hat i V_x + \hat j V_y + \hat k V_z##.

jtbell
Mentor
Here's how I think of ##\nabla \cdot \vec V## (in Cartesian coordinates), for what it's worth:

$$\nabla \cdot \vec V = \left( \hat i \frac {\partial}{\partial x} + \hat j \frac {\partial}{\partial y} + \hat k \frac {\partial}{\partial z} \right) \cdot \left( \hat i V_x + \hat j V_y + \hat k V_z \right)\\ = \frac {\partial V_x}{\partial x} + \frac {\partial V_y}{\partial y} + \frac {\partial V_z}{\partial z}$$

[Added] To see this (if necessary), distribute the dot product completely and note that ##\hat i \cdot \hat i = 1##, ##\hat i \cdot \hat j = 0##, etc.

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vanhees71
Gold Member
This only works, because in Cartesian coordinates the basis vectors are constant in space.

It's much better to remember the coordinate-free definitions of the operators. For the divergence it reads
$$\vec{\nabla} \cdot \vec{V}=\lim_{\Delta V \rightarrow 0} \int_{\partial \Delta V} \mathrm{d}^2 \vec{F} \cdot \vec{V},$$
where the limit is taken in the sense of that the volume $\Delta V$ is shrunk to the point at which you want to get the divergence.

To get the expression for specific coordinates, you take $\Delta V$ to be spanned by the coordinate lines in the neighborhood of the point in question.

arildno
Homework Helper
Gold Member
Dearly Missed
It is quite possible to establish a calculative formalism (which I learnt was called "dyadic formalism"), in which you may derive the correct relationships for arbitrary coordinate systems, as long as you let "differentiation" have precedence to other operations.
Let us for example, see how the Laplacian operator for 2-D polar coordinates can be derived:
We have:
$$\nabla=\vec{i}_{r} \frac{\partial}{\partial{r}}+\vec{i}_{\theta} \frac{\partial}{r\partial\theta}$$
The Laplacian operator can now be written as:
$$\nabla^{2}=\nabla\cdot\nabla=\vec{i}_{r}\cdot\frac{\partial}{\partial{r}}(\vec{i}_{r}\frac{\partial}{\partial{r}} + \vec{i}_{\theta}\frac{\partial}{r\partial\theta}) + \vec{i}_{\theta}\cdot\frac{\partial}{r\partial\theta}(\vec{i}_{r}\frac{\partial}{\partial{r}} + \vec{i}_{\theta}\frac{\partial}{r\partial\theta})$$
Now, computing all differentiations, including most importantly, those of unit vectors, we gain the correct result for the Laplacian operatior:
$$\nabla^{2}=\frac{\partial^{2}}{\partial{r}^{2}}+\frac{1}{r} \frac{\partial}{\partial{r}}+\frac{\partial^{2}}{r^{2}\partial\theta^{2}}$$

Suitably formulated, this can be used for other coordinate systems as wll, for example 3-D spherical coordinates.

When calculating the divergence or cross product of a vector, proceed in a perfectly similar way, but DO remember to differentiate those pesky unit vectors prior to the product operations; otherwise, you'll get wrong results.

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Obviously, this calculation method doesn't constitute any "proof", or strict derivation on its own, but is a clever formalism that gives the correct results that, for example, the method vanhees71 describes, yields in a more rigorous manner.

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