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MathGroup Archive 1997

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Re: Wrong behavior of CrossProduct

  • To: mathgroup at smc.vnet.net
  • Subject: [mg7976] Re: [mg7958] Wrong behavior of CrossProduct
  • From: Sergio Rojas <sergio at scisun.sci.ccny.cuny.edu>
  • Date: Wed, 30 Jul 1997 02:37:46 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

 Hello:

 How Mathematica implement the Cross Product of two vectors?

  As far as I know the basic definition of the Cross Product between
two vectors in ANY ORTHOGONAL coordinated system is as follows:

(a1,a2,a3)X(b1,b2,b3) = (a2*b3 - a3*b2,
                         a3*b1 - a1*b3,
                         a1*b2 - a2*b1)

  In physics this is usually illustrated by taking any three
  UNIT vectors (u[1],u[2],u[3]) with the orthogonal property:

   u[i].u[j] = Delta[i,j]  where Delta[i_, j_] := If[i==j, 1, 0]

   u[1]xu[2] =  u[3] ; u[2]xu[3] =  u[1] ; u[3]xu[1] =  u[2]
   u[2]xu[1] = -u[3] ; u[3]xu[2] = -u[1] ; u[1]xu[3] = -u[3]
   u[1]xu[1] =   0   ; u[2]xu[2] =   0   ; u[3]xu[3] =   0

  Then, the above result follows by expanding:

(a[1]*u[1] + a[2]*u[2] + a[3]*u[3])x(b[1]*u[1] + b[2]*u[2] + b[3]*u[3])

 and using the orthogonal property of the unit vectors.

  In my example,

a = ( 0, 0,1) ; b = (a1,a2,0)

axb = ( a2, -a1, 0 ) .

  This should be true in ANY ORTHOGONAL coordinated system. However,
the meaning of each symbol need to be adjusted accordingly.

  Please, let me know what I am missing if the above is not true.

Rojas

E-mail: sergio at scisun.sci.ccny.cuny.edu


On Fri, 25 Jul 1997, David Withoff wrote:

> > (* Hello fellows:
> > 
> >    After playing a little bit with the Mathematica construction for the cross
> >    product of two vectors, implemented by the function CrossProduct of the
> >    package VectorAnalysis, I strongly believe that CrossProduct do not
> >    work properly on Mathematica ... *)
> > 
> > In[1]:= Needs["Calculus`VectorAnalysis`"];
> > In[2]:= SetCoordinates[Cylindrical[r,phi,z]];
> > In[3]:= V = {a1,a2,0};
> > In[4]:= U = {0, 0, 1};
> > In[5]:= CrossProduct[U,V]
> > 
> >                 2        2     2        2
> > Out[5]= {Sqrt[a1  Cos[a2]  + a1  Sin[a2] ], 
> >  
> > >    ArcTan[-(a1 Sin[a2]), a1 Cos[a2]], 0}
> > 
> > In[6]:= PowerExpand[Simplify[%]]
> > Out[6]= {a1, ArcTan[-(a1 Sin[a2]), a1 Cos[a2]], 0}
> > 
> > In[7]:= ?ArcTan  
> > ArcTan[z] gives the inverse tangent of z. ArcTan[x, y] gives the inverse
> >    tangent of y/x where x and y are real, taking into account which quadrant
> >    the point (x, y) is in.
> > 
> > 
> > 	(* Using Mathematica definition for ArcTan[x, y], Out[6] can be 
> >            rewritten as {a1,-ArcTan[Cot[a2]],0}. This answer is obviously
> >            wrong as far as the Cross Product of V and U concern *)
> 
> I think that the result from CrossProduct[U,V] is correct.
> ArcTan[-(a1 Sin[a2]), a1 Cos[a2]] is not equivalent to -ArcTan[Cot[a2]].
> One way to see that is to insert numerical values for a1 and a2 and
> observe that the results are not the same.
> 
> In[1]:= ArcTan[-(a1 Sin[a2]), a1 Cos[a2]] /. {a1 -> 1, a2 -> 1.3}
> 
> Out[1]= 2.8708
> 
> In[2]:= -ArcTan[Cot[a2]] /. {a1 -> 1, a2 -> 1.3}
> 
> Out[2]= -0.270796
> 
> In this example:
> 
> > In[1]:= Needs["Calculus`VectorAnalysis`"];
> > In[2]:= SetCoordinates[Spherical[r,theta,phi]];
> > In[3]:= V = {a1,a2,0};
> > In[4]:= U = {0, 0, 1};
> > In[5]:=  CrossProduct[U,V]
> > Out[5]= {0, 0, 0}
> > 	         (* Again, wrong result. Same results were obtained on *)
> > In[1]:= $Version
> > Out[1]= SPARC 2.2 (December 15, 1993)
> > 
> > Rojas
> > 
> > E-mail: sergio at scisun.sci.ccny.cuny.edu
> 
> U is a zero vector (r is 0), so it seems that the result {0, 0, 0} for
> CrossProduct[U,V] is correct.
> 
> Dave Withoff
> Wolfram Research
> 


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