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I am currently finding myself doing lots of applied mathematics, e.g. fluid dynamics, and of course this involves a lot of vector calculus amongst other things. This had me thinking about proper notation. Prompted by Chappers answer to this question (I could not find additional posts on this website discussing this particular irregularity), I decided to probe you guys a bit further. Disregarding common conventions in the applied fields, I would like to hear your perspectives. Knowing that people have different preferences when it comes to notation, I was wondering if you have any (as there often are) good arguments for using one notation over the other:

Let $f$ be a scalar field and $\mathbf{F}$ a vector field.

Do you write $\text{grad} f$ or $\nabla f$? $\text{div } \mathbf{F}$ or $\nabla \cdot \mathbf{F}$? $\text{curl } \mathbf{F}$ or $\nabla \times \mathbf{F}$? Etc. And why?

How about when generalizing to tensor fields?

I'm looking forward to hearing your responses.

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closed as primarily opinion-based by naslundx, Hans Lundmark, user370967, Ethan Bolker, user223391 Apr 27 '18 at 12:54

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I see the textual notations more as describing a concept while the notations with nabla ($\nabla$) are formulas for how to compute them. $\endgroup$ – md2perpe Apr 25 '18 at 6:19
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I, for myself, prefer the spelled-out variants, at least when doing analysis: $$\mathop{\mathrm{grad}} f, \quad \mathop{\mathrm{div}} \mathbf F, \quad \mathop{\mathrm{rot}} \mathbf F $$ The variant $\nabla f$ for the gradient is ok, but I like consistency, and using the same "style" underlines the connection between the three operations (more below). Sometimes, however, especially when doing physics, I need to revert to the "physicist $\nabla$ notation" because some formulae look a lot more familiar that way, e.g. $$\nabla \cdot \mathbf E = \frac{\rho}{\varepsilon_0} $$ For the laplacian, instead, I very much favor the notation $\Delta f$, since I find it easy to confuse $\nabla^2 f$ with $\nabla f$ or the bilaplacian $\Delta^2 f$ at first glance. Plus the big triangle is something specific to the laplacian, which makes it recognizable to me – it screams "heat equation". Same for the dalembertian $\square f$ (which I write in the mathematician's way, without the $^2$). $$\Delta u = \beta \frac{\partial u}{\partial t}, \qquad \square u = \frac 1 {c^2} \frac{\partial^2u}{\partial t^2} - \Delta u = 0 $$ It all changes when I need to do tensors on manifolds. All the first-order derivative operators are instances of the exterior derivative on forms (of different degree), so it is best to keep it simple: $$F = F_{\mu\nu} \mathop{}\mathrm{d}x^{\mu}\wedge \mathop{}\mathrm{d} x^\nu = B + E \wedge \mathop{}\mathrm d x^0, \qquad \mathrm d F = 0 $$ If you want to be acting on vector fields or scalar fields on three-dimensional varieties specifically, though, you may "redefine" the operations from the beginning with the Hodge star, the external derivative, and the musical isomorphisms: $$\mathop{\mathrm{grad}} f = (\mathrm d f)^\sharp, \quad \mathop{\mathrm{div}} X = \star \mathrm d{\star}(X^\flat), \quad \mathop{\mathrm{rot}} X= \left(\star \mathrm d (X^\flat) \right)^\sharp$$ Heck, you may even redefine the laplacian as $\mathop{\mathrm{grad}}(\mathop{\mathrm{div}})$: $$\Delta f = \star \mathrm d {\star} \mathrm d f$$ Also, in the context of differential geometry I prefer to denote the standard real-analytic partial derivatives as $\partial_k$, leaving the $\partial / \partial x^k$ to be abstract vectors on a manifold with respect to a chart $x$.


Bonus. Here is a diagrammatic depiction of the action of the first order differential operators on $p$-forms over a $3$-dimensional manifold $M$: enter image description here Here $\Omega^p$ denotes the space of $p$-forms $\Omega^p(M)$, while $\Gamma = \Gamma(TM)$ is the space of smooth sections of the tangent bundle, i.e. vector fields. $0$ is just the space $\Omega^4= \Omega^5 = \dots$, which is a trivial vector space.

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    $\begingroup$ Thanks for your detailed input! Definitely something to think about. Love the musical isomorphisms haha. And great bonus! Thanks for the effort! $\endgroup$ – JackHummingbirder Apr 26 '18 at 4:40
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In contrast to giobrach, I prefer the nabla because it clearly indicates that $\nabla f$, $\nabla \cdot \mathbf{F}$, $\nabla \times \mathbf{F}$ and $\nabla^2 f$ are vector, scalar, vector and scalar respectively. I never use $\Delta$ for Laplacian.

Also, from a dimensional analysis point of view each occurrence of $\nabla$ carries with it dimensions of reciprocal length. This information is hidden if spelt operators are used. When Cartesian coordinates are specified (and no curls are being taken) I find myself writing $\partial/\partial\mathbf{x}$ in place of $\nabla$, which makes the dimensionality even clearer, and e.g. with scaling $\hat{\mathbf{x}} = \mathbf{x}/L$ we have

$$ \frac{\partial}{\partial\mathbf{x}} = \frac{1}{L} \frac{\partial}{\partial\hat{\mathbf{x}}}, \qquad \frac{\partial}{\partial\hat{\mathbf{x}}} = L \frac{\partial}{\partial\mathbf{x}}. $$

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