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, Thank you for your reply, but I can't read your fonts, can you explain what 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[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/ >> >> > >
- Follow-Ups:
- Re: Re: cross-product in cylindrical problem
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: Re: cross-product in cylindrical problem
- References:
- cross-product in cylindrical problem
- From: "news" <anonym@bamboo.com>
- cross-product in cylindrical problem