Talk:Divergence

Divergence is more general than described
I changed the maths rating from maths rating|frequentlyviewed=yes| field = analysis| importance = high| class = C

to maths rating|frequentlyviewed=yes| field = basics| importance = high| class = C

This article should be about divergence in a more general setting (as opposite of convergence than about it in an analysis alone,

Hope someone will improve it WillemienH (talk) 11:13, 5 February 2015 (UTC)

Non-general definition
The definition of divergence, it seems to me, should be defined for any amount of variables. Am I wrong in this, or can a four variable function have a divergence as well? Fresheneesz 20:48, 10 February 2006 (UTC)
 * The divergence is of course defined in any dimension. But it makes most sense in 3D only, as ther it behaves nicely with the curl, and has a physical interpretation. I will now add a mention of the general case. I would be opposed to rewriting this article from the n-dimenional perspective, as a lot of the physics would be lost, I think. Oleg Alexandrov (talk) 02:59, 11 February 2006 (UTC)

Initial example
The example of water in a bathtub can be confusing. Water does not 'vanish', and the divergence of water flow is in fact very close to zero. I've changed this to air expanding to some level which is a better example. PCM —This unsigned comment is by Paul Matthews (talk • contribs).
 * On further reflection, I think you are right. Oleg Alexandrov (talk) 02:09, 15 March 2006 (UTC)

actually i want to ask that if ▼.B is equal to B.▼.... if it is i just cant understand its physical interpretation..as ▼.B means a operator is operating over B but what then B.▼ implies?? 210.212.8.61 17:18, 6 February 2007 (UTC)
 * You're wrong, ▼.B != B.▼ — Preceding unsigned comment added by Thedoctar (talk • contribs) 08:22, 7 January 2013 (UTC)

Also, there is no example of how to calculate an easy case, like on the 'gradient' page. Jobonki (talk) 17:08, 10 January 2019 (UTC)

definition
I think definition should be as general as possible. therefore,

$$\nabla\cdot\mathbf{F}=\lim_{V\rightarrow 0} \frac{\oint_{s}\mathbf{F}d\mathbf{s}}{V}$$

Eventually, in 3D orthogonal system, this expression becomes the known one.

Nevo taaseh 11:32, 8 April 2007 (UTC)
 * I think it is preferrable that definitions start simple, not general. Things can be generalized later I believe, once the basic concept is clear. No? :) Oleg Alexandrov (talk) 15:06, 8 April 2007 (UTC)


 * I definitely agree with the concept of simpler things first. But, I would argue that an arbitrary V is actually simpler then that of a sphere.  Barring that a rectangular parallel piped of Volume dxdydz would be simpler.  This volume has the advantage that it is much easier to demonstrate the divergence theorem with then spherical coordinates. TStein (talk) 13:29, 10 April 2009 (UTC)

Abuse of notation
Is &nabla;·F actually an abuse of notation? &nabla; is a vector and can operate on scalar or vector quantities. If it operates on a vector, then one needs to know if it performs a scalar product or a cross product, see curl (&nabla;xF).

I agree that it is not an abuse of notation. ALittleSlow 02:02, 30 September 2007 (UTC)

I think it's totally an abuse of notation. The dot product is just an inner product, so the elements on the left and right of the dot must be in the same vector space. Only when you choose an orthonormal basis for R^n is the dot product defined by the familiar formula given here. &nabla; is clearly not an element of R^n, which is why this expression is an abuse of notation. 68.73.194.10 (talk) 18:14, 6 May 2012 (UTC)


 * I get the logic that the representation of an operator should ideally be something like div(f), but is it not still perfectly true to write that div(f) = &nabla;.f, and also div(f) = &nabla;.f ?
 * Can someone expand on the comment above that "&nabla; is clearly not an element of R^n"? Specifically, on the linked WP article on del, it is currently written:  Though this page chiefly treats del in three dimensions, this definition can be generalized to the n-dimensional Euclidean space Rn.
 * —DIV (137.111.13.4 (talk) 23:43, 1 April 2014 (UTC))

Visualisation
This article could do with a nice visualization as in the gradient article —Preceding unsigned comment added by 137.205.125.78 (talk) 09:47, 26 October 2007 (UTC) sad —Preceding unsigned comment added by 207.148.171.2 (talk) 22:15, 7 April 2008 (UTC)

WikiProject class rating
This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 09:48, 10 November 2007 (UTC)

More definition concerns
From the definition section: "Let x, y, z be a system of Cartesian coordinates.." AFAIK the divergence operator can be defined in other coordinates, right?Mathchem271828 (talk) 06:59, 1 June 2008 (UTC)

scalar field
Should it say somewhere that the result is a scalar field not just a scalar? --Kupirijo (talk) 23:27, 4 October 2008 (UTC)

The terms 'source' and 'circulation'
Is there a better standard for the terms 'source' and circulation. For instance if
 * divF = D, and
 * curlF = C,

then what do we call D and C?

If we call D the 'source' then what do we call C? The term 'circulation' seems to be used already for the line integral around a given path.

I am biased toward using 'outflow source' for D and 'circulation source' for C as that fits well with how I teach the Electricity and Magnetism course I teach. (In this pedagogical approach the divergence is defined by the divergence theorem (fundamental theorem of the divergence) as the outflow density while the curl is defined as a circulation density by the fundamental theorem of the curl. Even I must admit, though, that this approach is both wordy and to some extent incomplete.  For it is generalizing the term source as anything that 'creates' or perhaps 'induces' the field F yet there can be a field F even when D and C are both zero.  (This field is as far as I can tell with my math background limited to physics math is due to some sort of 'strain' type 'zero-trace symmetric tensor source' which can be determined everywhere by the boundary conditions.)

I would like to be able to contribute more of a physics view to the article, but I don't want to ruin it for the mathematicians. Any help with this would be appreciated. TStein (talk) 13:55, 10 April 2009 (UTC)


 * I may be missing something, but isn't the standard term for D the divergence and C the curl? Why introduce new terms? I think the terms 'source' and 'circulation' should actually refer to the parts of F with zero curl and divergence respectively (actually the zero divergence part is usually called solenoidal). This is made precise through the Helmholtz, or Hodge, or Hodge-Morrey, or some other decomposition. One way to see that D and C both zero does not imply that F is zero is to note that for any harmonic function (harmonic means zero Laplacian) h, the gradient of h has zero curl and zero divergence. It is also possible to make a similar construction based on 2-tensors. This is all related to the deRahm cohomology of the underlying domain. 97.126.82.176 (talk) 17:51, 9 May 2009 (UTC)

Visual aids
I think this article could use some more visual aids. There should, in particular, be a diagram with the graph of a vector field and a circle showing vectors entering and exiting. I can make some Sage pictures if people want. — Preceding unsigned comment added by Wham Bam Rock II (talk • contribs) 21:45, 6 January 2013 (UTC)

Questions about the cartesian coordinates expression for 2nd order tensors
I might be mistaken, but I've seen different cartesian expressions for the divergence of second order tensors from several academic and educational sources. Using Einstein's notation, the one in the article would be
 * $$\left( \overrightarrow{\operatorname{div}}\,\mathbb{A} \right)_{i}

= \frac{\partial A_{ji}}{\partial x_j}$$ It is coherent with the one used by Poinsot and Veynante (trusted authors in combustion sciences) in their combustion textbook. In other sources, such as Sébastien Candel's fluid mechanics textbook (renowned and trusted physicist, known for his contribution to the fields of combustion and acoustics)
 * $$\left( \overrightarrow{\operatorname{div}}\,\mathbb{A} \right)_{i}

= \frac{\partial A_{ij}}{\partial x_j}$$ Both expressions are equivalent for symmetric tensors (which is true for the viscous stress tensor) but not for nonsymmetric ones, of course. I'm guessing the first expression, more coherent with the nabla notation, is the correct one. Would it be relevant to warn readers about this frequent and confusing "mistake"? (Which is, in fact, not exactly one, since it is correct in the cases considered.) M4urice (talk) 10:53, 17 May 2013 (UTC)

The example for a 2nd order tensor in the main article was based on the second convention but the link to the reference was broken. I have found a more recent version of the reference where I could verify that they also suggest the first convention which in my opinion is clearly more common. So I have fixed the example for the divergence of a 2nd order tensor to use the first convention and updated the broken reference. --Logari81 (talk) 07:34, 11 November 2015 (UTC)

History
Why does this, and related articles, have no mention of the history. Who developed the concept originally? Who devised the notation? —DIV (137.111.13.4 (talk) 23:45, 1 April 2014 (UTC))

More Visualization
One of the problems that I have with the Wikipedia on math topics is they quickly ascend to "Equation Land" leaving virtually everyone else behind on the topics they are trying to describe.

I'm not a math guy but I think that rather than the uber-complex Calculus equations we see something like this may be more relevant to the general masses: http://www2.sjs.org/raulston/mvc.10/topic.6.lab.1.htm

I agree that documenting the precise definitions are important but if we leave the general population behind more high-school and undergrads will abandon the field because they just don't get it.

For example check out "Field 12"... I see the beginning of the understanding of wing tip vortices (yes I understand that this article is about electromagnetics).

— Preceding unsigned comment added by Lesds (talk • contribs) 06:40, 26 February 2015 (UTC)

Divergence is invariant under linear transformations
Divergence is invariant under any invertible linear transformation of the vector field and its domain, not just orthogonal transformations. Let $$M$$ be an invertible linear transformation applied to vector field $$f$$ and its domain. Define $$F=Mf.$$ The precise relationship between the transformed vector field $$F$$ and the original vector field $$f$$ is $$(F\circ M)x = (M\circ f)(x)$$ (since the domain of $$f$$ is also transformed by $$M$$). So the chain rule shows that the relationship between the Jacobian matrices of $$F$$ and $$f$$ is


 * $$DF = M(Df)M^{-1}.$$

So the trace of $$DF$$ must equal the trace of $$Df.$$ Jrheller1 (talk) 19:23, 10 August 2016 (UTC)

External links modified
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This article's title is potentially confusing
This article is about divergence in vector calculus, but its title is simply Divergence, which has various other common meanings (such as divergence (statistics) and divergence (infinite series)). Should we rename this article to Divergence (vector calculus) so that the links to this article can be disambiguated more easily? Jarble (talk) 00:09, 16 July 2017 (UTC)

I made images (available in wikimedia commons)
I made these svg images (based on a I found on wikicommons by Biolynk):

I'm almost certain that the 3rd example is correct (the vectors have no x-component, and the y-component is a function of only x, hence doesn't change in the y-direction). If someone could confirm that (I haven't passed Calc III yet), that would make me feel a whole lot more confident adding it to an article. Bfoshizzle1 (talk) 00:16, 24 February 2019 (UTC) Hello, nice work on the image and it also seems to me to be correct. I would, however, say that it could bring to confusion, because the sink and source images both stem from a single point while the null divergence image shows a "complete" field with null divergence... maybe that could be clarified in the legend? Louis (talk) 04:56, 11 July 2019 (UTC)

Divergence of mixed tensor
I'm not sure I understand the definition of the divergence of a mixed tensor: $$( \operatorname{div} T ) ( Y_1, \ldots , Y_{q-1}) = {\operatorname{trace}} \Big( X \mapsto \sharp ( \nabla T ) ( X , \cdot , Y_1 , \ldots , Y_{q-1} ) \Big)$$

Which index does the sharp apply to? What does the dot stand for? Also, why should the divergence of a (p,q)-tensor be a (p, q-1)-tensor? Shouldn't it be a (p-1, q)-tensor, e.g., the divergence of a vector field is a function?


 * Maybe its the musical isomorphism? 67.198.37.16 (talk) 14:37, 1 November 2020 (UTC)

Divergence versus del dot
The source cited quotes that "some authors" choose to define "del dot" as the divergence. This distinction is nonsensical; "del dot" is a mnemonic, an abuse of notation devoid of any real meaning, and it's common to denote divergence by "del dot" even for higher-order quantities. This is not sufficient sourcing for this situation in my opinion.--Jasper Deng (talk) 21:24, 19 February 2020 (UTC)

The "Del dot" has symbolic sense and is widely used (and defined) - especially with areas where we use non-symetric tensor: Kamil Kielczewski (talk) 20:02, 20 February 2020 (UTC)
 * in current article, definition NOT describes how to calculate $$\nabla\cdot T$$ or $$\mathrm{div} T$$ where T is 2nd order tensor (it is focused only on vector field for which $$\mathrm{div}\vec v = \nabla\cdot \vec v$$).
 * source (full book) author email provide following general/tensor definitions of "div" and "del dot" (page 119) (my changes here]) we can define this divergence operation on tensor T in two ways :
 * $$\mathrm{div}T=\mathrm{grad}T:I

= \frac{\partial T}{\partial x_i}e_i = \frac{\partial (T_{jk}e_j\otimes e_k)}{\partial x_i}e_i = \frac{\partial T_{ij}}{\partial x_j}e_i $$
 * $$\nabla\cdot T

= e_i\frac{\partial }{\partial x_i}\cdot(T_{jk}e_j\otimes e_k) = \frac{\partial T_{ji}}{\partial x_j}e_i $$
 * this two are not equal but we have $$\nabla\cdot T=\mathrm{div}(T^T)$$


 * source (MIT) author use "del dot" in explicte form of NS (page 2 top) $$\nabla\cdot\tau=\frac{\partial \tau_{ji}}{\partial x_j}e_i$$


 * source auhors among others from Department of Mathematics and Statistics (page 66, formula 1.660, and p. 68 table 1.4 with explicit formula): $$\nabla\cdot\tau =\sum_{i=1}^3\sum_{j=1}^3\frac{\partial \tau_{ij}}{\partial x_i}e_j$$


 * source pdf here Non-newtonian fluids; (page 326, eq. 1 ) stress tensor in NS equation is expressed as $$\frac{\partial\tau_{ji}}{\partial x_j}$$


 * source non-Newtonian Flows - equation E4 and E5 where $$\nabla\cdot\sigma$$ is given explicite in cylindrical coordinates (it is easy to se that indexes change like in 'del dot' (not like in 'div')


 * source numerical non-Newtonian fluids 2D case page 56 eq. 4: $$\nabla\cdot\sigma$$ is expressed in explicite form in eqatons 20 and 21 as $$[\frac{\partial\tau_{xx}}{\partial x} + \frac{\partial\tau_{yx}}{\partial y}, \frac{\partial\tau_{xy}}{\partial x} + \frac{\partial\tau_{yy}}{\partial y}]^T$$ which is 2D case of $$\frac{\partial\tau_{ij}}{\partial x_i}$$ (Einstein sum-notation)


 * source (Cambridge) Cauchy momentum equation - page 133 eq. (6.3-6) and eq. (6.3-7) 3D explicit formula for $$\frac{\partial\sigma_{ji}}{\partial x_j}e_i= \lim_{\Delta V\rightarrow 0}\frac{\Delta F}{\Delta V}=\nabla\cdot\sigma$$


 * source Cauchy momentum equation for asymmetric stress tensor - page 4 (13 pdf) author use $$\rho\frac{D u_i}{D t}=\rho G_i + \sigma_{ki,k}$$, where $$\sigma_{ki,k}=\frac{\partial\sigma_{ki}}{\partial x_k}=\nabla\cdot\sigma$$ (according to author explanation below equation)


 * source PhD Thesis page 17 (pdf 37) for stress tensor explicit form of $$\frac{\partial\tau_{ij}}{\partial x_i}$$ is given.


 * source Nematic liquid crystals where in hydrodynamic flow equation stress tensor is used as follows (page 2 top) $$\partial_j(-p\delta_{ji} + \sigma^d_{ji} + \sigma^f_{ji} + \sigma^'_{ji} )$$ which is form of $$\nabla\cdot\sigma=\frac{\partial\sigma_{ji}}{\partial x_j}e_i$$



As you can see in all that sources authors use/define the $$\nabla\cdot T$$ version of divergence operator (defined in second 'bullet' in above list) instead of $$\mathrm{div} T$$. '''In general, the authors who searched deeply found these formulas. Others (most) stopped at the symmetrical tensor case where''' $$\mathrm{div} T = \nabla\cdot T$$ (which is appropriate in most cases) (but i.e. gives wrong form for NS equation for non-symetric stress tensor). Navier-Stokes equations assumes that fluid is Newtonian and this assumption impose that T is simetric - so they are using div and "del dot" 'randomly' because they are equivalent for symetric tensors (but if we deal with non-newtonian fluid with non-symetric stress tensor $$\sigma$$ or $$\tau$$ there is difference - and the $$\nabla\cdot T$$ is right form). I'm not the first who notice that distinction - here is another wiki article: Tensor_derivative_(continuum_mechanics). People who deal with simetric tensors (like in Navier-Stokes equetions for Newtonian fluid) don't explicite explain which of this two operators they use (because for symmetric tensor it doesn't matter). Only small amount of authors who need to deal with some cases of not-Newtonian fluids (where it matters) distinct this two divergence operators. Kamil Kielczewski (talk) 22:30, 20 February 2020 (UTC)


 * The del-dot notation is used exclusively in pretty much all undergraduate physics textbooks. The $$\mathrm{div}$$ notation more-or-less never used. In higher math, e.g. on tensors, see Hodge star and pushforward (differential). 67.198.37.16 (talk) 14:43, 1 November 2020 (UTC)