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Re: Re: RE: Re: Dot Product in Cylindrical Coordinates
*To*: mathgroup at smc.vnet.net
*Subject*: [mg69365] Re: [mg69351] Re: [mg69301] RE: [mg69276] Re: Dot Product in Cylindrical Coordinates
*From*: Pratik Desai <pratikd at wolfram.com>
*Date*: Sat, 9 Sep 2006 03:26:42 -0400 (EDT)
*References*: <NDBBJGNHKLMPLILOIPPOKEFIFCAA.djmp@earthlink.net>
David Park wrote:
> Pratik,
>
> Your example does not work for me.
>
> Needs["Calculus`VectorAnalysis`"]
> SetCoordinates[Cylindrical[r, theta, z]];
>
> g = {gr[r, theta, z], gtheta[r, theta, z], gz[r, theta, z]};
> f = {fr[r, theta, z], ftheta[r, theta, z], fz[r, theta, z]};
>
> gr[r_, theta_, z_] := r
> gtheta[r_, theta_, z_] := theta
> gz[r_, theta_, z_] := z
> fr[r_, theta_, z_] := r^2
> ftheta[r_, theta_, z_] := theta
> fz[r_, theta_, z_] := Cos[z]
>
> If we use CrossProduct and DotProduct along with a standard identity using
> Div and Curl, which should always be true, then it does not work.
>
> Div[CrossProduct[g, f]] ==
> DotProduct[Curl[g], f] - DotProduct[Curl[f], g] // Simplify
> (r^2 + 3*r^4*z^2 - 5*r^3*z*Cos[z] +
> Sqrt[r^2*((-r)*z + Cos[z])^2] + r^2*Cos[2*z])/
> (r*Sqrt[r^2*((-r)*z + Cos[z])^2]) ==
> (theta*(-z + Cos[z]))/r
>
> However if we use the standard Dot and Cross functions the identity is
> verified.
>
> Div[Cross[g, f]] == Dot[Curl[g], f] - Dot[Curl[f], g] // Simplify
> True
>
Hello David
I think this was a very poorly conceived example. Sorry about that.
Regards,
Pratik
> The point here is that the package functions DotProduct and CrossProduct are
> rather special and have NOTHING to do with standard vector calculus. Users
> should not stumble into using them where they do not apply.
>
> In standard vector calculus, in a coordinate system, an orthonormal frame is
> erected at each point in the space. The axes of the frame point along the
> coordinate directions. The components of vectors are specified in terms of
> this orthonormal frame. Since the frame is orthonormal, at any point we can
> simple use the standard Dot and Cross product for combining two vectors at
> that point. Curl and Div demand vector components in terms of an orthonormal
> frame.
>
> If businessmen can jump on computerized inventory control, lasers and
> barcodes and DVDs and other technology all within a decade or so of their
> invention, why or why can't engineering and physics schools dump the
> misbegotten vector calculus of over a century ago?
> David Park
> djmp at earthlink.net
> http://home.earthlink.net/~djmp/
>
>
> From: Pratik Desai [mailto:pratikd at wolfram.com]
To: mathgroup at smc.vnet.net
>
> I think it is quite imperative to note here that
>
> "There are often conflicting definitions of a particular coordinate
> system in the literature. When you use a coordinate system with this
> package, you should look at the definition given below to make sure it
> is what you want." --Mathematica Documentation
>
> So for cylindrical coordinate system one must define the system as:
>
> g = {g?[r, theta, z], g?[r, theta, z], gz[r, theta, z]}
> f = {f?[r, theta, z], f?[r, theta, z], fz[r, theta, z]}
> g?[r_, theta_, z_] = r
> g?[r_, theta_, z_] = theta
> gz[r_, theta_, z_] = z
> f?[r_, theta_, z_] = r^2
> f?[r_, theta_, z_] = theta
> fz[r_, theta_, z_] = Cos[z]
>
> Then everything works fine:
>
> In[20]:=
> Div[CrossProduct[g,f]]===DotProduct[Curl[g],f]-DotProduct[Curl[f],g]
>
> Out[20]=
> True
>
>
> Hope this helps
>
> Pratik
>
>
>
>
>
--
Pratik Desai
Wolfram Research, Inc
Technical Support
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