       Re: cross-product in cylindrical problem

• To: mathgroup at smc.vnet.net
• Subject: [mg51646] Re: cross-product in cylindrical problem
• From: "news" <symbio at s.dn.com>
• Date: Wed, 27 Oct 2004 23:43:06 -0400 (EDT)
• References: <200410100952.FAA25275@smc.vnet.net> <3B366122-1AC0-11D9-BEB1-000A95B4967A@mimuw.edu.pl> <ckd5nu\$50d\$1@smc.vnet.net>
• Reply-to: "news" <symbio at s.dn.com>
• Sender: owner-wri-mathgroup at wolfram.com

```hello Andrzej,
the symobls are suppposed to be in: SetCoordinates[Cylindrical[Ï?, Ï., z]];

thanks very much!

"Andrzej Kozlowski" <akoz at mimuw.edu.pl> wrote in message
news:ckd5nu\$50d\$1 at smc.vnet.net...
>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:=
>>> f[\[Rho]_, \[Phi]_] = {\[Rho], \[Phi], 0}
>>> Out=
>>> {\[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:=
>>> v1 = D[f[\[Rho], \[Phi]], \[Rho]]
>>> v2 = D[f[\[Rho], \[Phi]], \[Phi]]
>>> Out=
>>> {1, 0, 0}
>>> Out=
>>> {0, 1, 0}
>>>
>>> then cross them in cylindrical coords, and should get {0,0,1}, but
>>>
>>> In:=
>>> n = CrossProduct[v1, v2, Cylindrical[\[Rho], \[Phi], z]] //
>>> FullSimplify
>>> Out=
>>> {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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