Talk:Conservative vector field

message: this is a rotational vector in the figure and not an irrotational as it states !!  —Preceding unsigned comment added by 94.249.102.105 (talk) 11:03, 8 May 2009 (UTC)
replies: text: Indeed. I changed the caption. -- Crowsnest (talk) 12:26, 8 May 2009 (UTC)
thread_title:  error in the title of the vector field figure !??! 

message: If the vector length represents the magnitude, then the figure is wrong i THINK.  The vectors near the origin should continue to grow.
I suggest that if the vectors near the origin have been shortened for clarity, this should be stated somewhere.
It is a very nice figure nevertheless!  (prettier than IDL can do)
;; IDL code 
nx = 100  &  ny =100         ; dimensions
x=dblarr(nx,ny)
y=dblarr(nx,ny)
xv=dindgen(nx)/nx            ; x vector [0..99]/100 --> [0..1)
yv=dindgen(ny)/ny            ; y vector
for j=1,ny do x[*,j-1]=xv      ; replicate x vector -> x matrix
for i=1,nx do y[i-1,*]=yv       
; erroneous plot??
fx = -y/(x^2+y^2)
fy =  x/(x^2+y^2)
vel, fx, fy, title='Wiki',len=1   ; a curvy arrow vector randomly sampled field plot
Bdb112 (talk) 07:48, 15 January 2010 (UTC)
thread_title:   the vector field figure appears to be incorrect!??! 

message: I may be being a bit pedantic, but I'd like to change the first line from:
*In vector calculus a conservative vector field is a vector field which is the gradient of a function, known in this context as a scalar potential.
to
*In vector calculus a conservative vector field is a vector field which is the gradient of a function which is known as a scalar potential.
...or something similar, just to get across immediately what the article is referring to as a scalar potential.  — Preceding unsigned comment added by Larryisgood (talk • contribs) 17:37, 29 July 2011 (UTC)
thread_title:  Opening sentence 

message: Excellent page. Thanks to everyone who contributed. However, I found one error: 
The line:
  "(However, in any connected subregion of S, it is still true that it is conservative..."
I believe should say:
   "(However, in any SIMPLY connected subregion of S, it is still true that it is conservative..."
S itself is connected. A connected subregion encircling the Z axis wouldn't do at all. I hope this comment is helpful (and correct!).
Thanks.
  Catgod119  — Preceding unsigned comment added by Catgod119 (talk • contribs) 06:43, 16 January 2013 (UTC)
thread_title:  Correction - "connected" should be "simply connected" 

message: This statement seems to rely on the vector field being once partially differentiable in all its components with respect to all the spatial variables (in other words, the curl exists). The requirement that \nabla f = \mathbf{F} says nothing about whether the curl of F exists or not. The identity used, as well as Green's theorem, is naturally only valid when that curl exists. So the question is, must the curl of F necessarily exist for a conservative vector field? I don't think so. Surely there exist multivariate analogs of the Weierstrauss function or some other function, such that if F has components equal to those functions, then it has a once differentiable potential function but no curl (or divergence, for that matter).--Jasper Deng (talk) 05:33, 22 April 2014 (UTC)
replies: text: Now reading the discussion by  above about this article being poorly sourced (which it is), I can clearly see why my argument probably holds water.--Jasper Deng (talk) 05:55, 22 April 2014 (UTC)
replies: text: It is not clear to me that the partial derivatives of F must exist for curl F to exist.  For example, if f(x) = 0 if x1 < 0, and f(x) = x12 if x1 ≥ 0 (x1 being one of the components of the position vector x), then F = ∇f is continuous everywhere, and curl F = 0 everywhere, keeping in mind that it is defined as the limit of a path integral divided by the area bounded.  However, F is not differentiable where x1 = 0, which simply means that the expression for curl in terms of partial derivatives breaks, nevertheless keeping the answer to the question posed in this thread's heading as "yes", with or without sources.  —Quondum 06:23, 22 April 2014 (UTC)
replies: text: Yes, the requirement that \oint \mathbf{F}\ \cdot d\mathbf{r} = 0 seems to ensure that \nabla \times \mathbf{F} = \mathbf{0}, but the article on the curl does indeed state that the operator is from continuously differentiable vector fields to continuous vector fields. I'm rather sure that it is possible to construct a function whose gradient's curl is nowhere existent, and if not, it would interest me to know why. My textbook uses the partial derivative definition and "derives" the loop integral definition from Stokes' theorem, but all along assuming that the partial derivatives in the curl operator exist.--Jasper Deng (talk) 07:13, 22 April 2014 (UTC)
replies: text: In general, definitions in terms of integrals, like the definition of curl in terms of integrals around closed loops, are more forgiving of bad functions than definitions in terms of derivatives, like the definition of curl in terms of partial derivatives (which is in some sense the "real" definition, since it manifests it as the exterior derivative).  But sometimes one can rescue derivatives by replacing them with weak derivatives instead.  That may or may not be the case here (I haven't made any attempt to check) but I wouldn't find it too surprising.  Ozob (talk) 13:51, 22 April 2014 (UTC)
replies: text: I think that we should be careful of relying too closely on chains of definitions in WP – irrotational in terms of curl, curl in terms of integrals/partial derivatives.  The concept of irrotationality makes perfect sense in edge cases in that they are unambiguous (unlike some weak derivatives), and continue to satisfy (AFAICT) the statements made of them. The question then becomes one of whether they are sometimes (notably) defined to accommodate the non-differentiable edge cases. I would be unsurprised if various different approaches are used and accepted, some of them trying to be more general and hence more "forgiving", especially if this leads to questions such as posed in this thread having unconditional answers. —Quondum 22:26, 22 April 2014 (UTC)

text: I don't see any real disagreement (though I'm not classifying myself as a mathematician, and Ozob's comment about the exterior derivative bears thinking about). Using partial derivatives only is simpler, but does not deal with several useful cases. Electromagnetic field theory deals with distributions, for example a uniform surface charge density on a sphere. This leads to a non-differentiable EM field (when regarded as functions on a manifold rather than as distributions), but is a very accurate description of an everyday (classical) scenario. A suitably general treatment will include cases such as this. Choice of a set of definitions such that the statement "a conservative vector field is necessarily irrotation" does not hold would be an unnecessary artefact of an inadequate choice of definitions. It would be pointless to point out where this statement might fail when using certain definitions. The article on curl should be modified to remove the restriction to differentiable fields (I do not see it in the few books that I checked). The only sources that I've browsed (e.g. []) that look reasonably authoritative simply restrict the applicability of the definition of curl in terms of partial derivative to where the field is differentiable. Lack of applicability of a given definition of curl must not be confused with the nonexistence of the curl. —Quondum 16:15, 27 April 2014 (UTC)
thread_title:  Conservative vector fields are necessarily irrotational?