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Re: cross-product in cylindrical problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg51264] Re: [mg51258] cross-product in cylindrical problem
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 11 Oct 2004 01:25:23 -0400 (EDT)
*References*: <200410100952.FAA25275@smc.vnet.net> <3B366122-1AC0-11D9-BEB1-000A95B4967A@mimuw.edu.pl>
*Sender*: owner-wri-mathgroup at wolfram.com
I forgot to include the line:
SetCoordinates[Cylindrical[Ï?, Ï?, z]];
Without that JacobianMatrix[] should be replaced by
JacobianMatrix[Cylindrical[Ï?, Ï?, z]].
Andrzej
On 10 Oct 2004, at 22:28, Andrzej Kozlowski wrote:
> On 10 Oct 2004, at 18:52, news wrote:
>>
>> I'm really puzzled by this behavior of Mathematica, I have two
>> vectors in
>> cylindrical coordinates and would like to take their cross-product in
>> cylindrical, but it seems to give me incorrect answer, see below:
>>
>> define parametric path {r,phi,z}
>>
>> In[110]:=
>> f[\[Rho]_, \[Phi]_] = {\[Rho], \[Phi], 0}
>> Out[110]=
>> {\[Rho], \[Phi], 0}
>>
>> take derivates of path w.r.t. r then w.r.t phi, get {1,0,0}, and
>> {0,1,0}
>>
>> In[113]:=
>> v1 = D[f[\[Rho], \[Phi]], \[Rho]]
>> v2 = D[f[\[Rho], \[Phi]], \[Phi]]
>> Out[113]=
>> {1, 0, 0}
>> Out[114]=
>> {0, 1, 0}
>>
>> then cross them in cylindrical coords, and should get {0,0,1}, but
>> instead
>> get wrong answer below
>>
>> In[117]:=
>> n = CrossProduct[v1, v2, Cylindrical[\[Rho], \[Phi], z]] //
>> FullSimplify
>> Out[117]=
>> {0, 0, 0}
>>
>> As you can see, when I cross {1,0,0} with {0,1,0} in cylindrical
>> coords, I
>> get {0,0,0}, when I should be getting {0,0,1}.
>>
>> Can anyone help?
>>
>>
> The problem is with the meaning of "two vectors in cylindrical
> coordinates". Even if your default coordinate system is cylindrical,
> Mathematica still represents tangent vectors using Cartesian
> coordinates. In other words, the meaning of a vector {a,b,c} in
> cylindrical coordinates is the vector given by:
>
> << Calculus`VectorAnalysis`
>
> CoordinatesToCartesian[{a,b,c},Cylindrical]
>
> Thus, the vectors {1,0,0} and {0,1,0} are turned to:
>
>
> CoordinatesToCartesian[{1, 0, 0}, Cylindrical]
>
> {1, 0, 0}
>
> and
> CoordinatesToCartesian[{0,1,0},Cylindrical]
>
> {0,0,0}
>
> So you get the zero vector (since {0,1,0} in cylindrical coordinates
> represents just the origin) and hence the result you get.
> I am not hundred percent sure what you meant. But I suppose thatt by
> {1,0,0} and {0,1,0} in cylindrical coordinates you meant the basic
> vector fields in the tangent bundle of R^3 (in other words vectors
> changing from point to point). Presumably (I am guessing) the vector
> (field) {1,0,0} is
>
> JacobianMatrix[] . {1, 0, 0}
>
> {Cos[Ï?], Sin[Ï?], 0}
>
> and the vector (field) {0,1,0} is
>
> JacobianMatrix[] . {0, 1, 0}
>
> {(-Ï?)*Sin[Ï?], Ï?*Cos[Ï?], 0}
> in Cartesian coordinates.
>
> In that case
>
> Simplify[Cross[JacobianMatrix[] . {1, 0, 0}, JacobianMatrix[] . {0, 1,
> 0}]]
>
> {0, 0, Ï?}
>
> which is the vector field sometimes denoted by {0,0,1} in cylindrical
> coordinates.
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.akikoz.net/~andrzej/
> http://www.mimuw.edu.pl/~akoz/
>
>
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