Re: How to plot divergence of gradient as contour plot
- To: mathgroup at smc.vnet.net
- Subject: [mg124132] Re: How to plot divergence of gradient as contour plot
- From: Szymon Roziewski <szymon.roziewski at gmail.com>
- Date: Tue, 10 Jan 2012 06:00:20 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201201051100.GAA14785@smc.vnet.net> <201201060919.EAA26925@smc.vnet.net>
Hello once again,
Thank you very much for your ideas!
I haven't used that package before, so it's new to me.
It's weird that Div, Grad, Curl don't operate on x,y,z but on Xx,Yy,Zz.
It seems to be a programmer trick.
All best,
Szymon Roziewski
2012/1/6 Murray Eisenberg <murray at math.umass.edu>
> You need to use the VectorAnalysis package (carefully!), remembering that:
>
> (i) the default coordinate system is Cartesian[Xx, Yy, Zz]; and
> (ii) the result of using that package's Grad, Div, Curl is _not_
> really a "function" of the variables (more about this later).
>
> In your example:
>
>
> Div[Grad[Sin[Xx + Yy]]]
> -2 Sin[Xx + Yy]
>
>
> ContourPlot[
> Div[Grad[Sin[Xx + Yy]]] /. {Xx -> x, Yy -> y},
> {x, -5, 5}, {y, -5, 5}]
>
> (* or first define a function *)
>
> f[x_, y_] := Div[Grad[Sin[Xx + Yy]]] /. {Xx -> x, Yy -> y}
> ContourPlot[f[x, y], {x, -5, 5}, {y, -5, 5}]
>
> I find the whole paradigm of the VectorAnalysis package to be
> aggravatingly annoying. After all, from a mathematical perspective, if
> you're given a scalar-valued function with vector inputs, then its
> gradient is just a well-defined vector field -- and that has nothing
> whatsoever to do with so-called "coordinate systems". That is, one ought
> to be able to do something like this -- just like any standard modern
> text on multivariable calculus would show:
>
> phi[x_, y_] := Sin[x + y]
> Div[Grad[phi[x, y]]]
>
> In Mathematica, with the VectorAnalysis defaults, you'd get 0 as result
> there, since you didn't override the default variable names Xx, Yy, Zz.
>
> Then there's the whole nonsense of the gradient other coordinate
> systems. That, to my mind, is a very odd way to think about things.
> Mathematically, given a scalar field, it has an associated gradient
> vector field. Period. If you want to "express the gradient in terms of"
> another coordinate system, you're no longer finding that gradient;
> you're finding a composite of functions where you go first from that
> other coordinate system to cartesian coordinates, next take the
> gradient, and finally go back from cartesian coordinates in the result
> to the other coordinate system.
>
> (I trust others will disagree with the preceding argument!)
>
>
> On 1/5/12 6:00 AM, Szymon Roziewski wrote:
> > Hello there,
> >
> > I am concerning to plot something like that.
> >
> > ContourPlot[Divergence[Gradient[Sin[x + y]]], {x, -5, 5}, {y, -5, 5}]
> >
> > Plotting divergence of gradient scalar field.
> > How can I manage to do that?
>
> --
> Murray Eisenberg murray at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower phone 413 549-1020 (H)
> University of Massachusetts 413 545-2859 (W)
> 710 North Pleasant Street fax 413 545-1801
> Amherst, MA 01003-9305
>
>
--
Z wyrazami szacunku,
Szymon Roziewski
- References:
- How to plot divergence of gradient as contour plot
- From: Szymon Roziewski <szymon.roziewski@gmail.com>
- Re: How to plot divergence of gradient as contour plot
- From: Murray Eisenberg <murray@math.umass.edu>
- How to plot divergence of gradient as contour plot