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Re: Wrong behavior of CrossProduct
*To*: mathgroup at smc.vnet.net
*Subject*: [mg7996] Re: [mg7958] Wrong behavior of CrossProduct
*From*: seanross at worldnet.att.net
*Date*: Wed, 30 Jul 1997 23:57:49 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
Sergio Rojas 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]:= $Version
> Out[1]= DEC OSF/1 Alpha 2.2 (September 9, 1994)
>
> In[2]:= Needs["Calculus`VectorAnalysis`"];
> In[3]:= V = {a1,a2,0};
> In[4]:= U = {0, 0, 1};
>
> In[5]:= CrossProduct[U,V]
> Out[5]= {-a2, a1, 0}
> (* This result is correct *)
> In[6]:= CoordinateSystem
> Out[6]= Cartesian
> (************ Quit and start again ************)
>
> 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 *)
>
> In[7]:= CoordinateSystem
> Out[7]= Cylindrical
>
> (************ Quit and start again ************)
>
> 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
What has happened is that some very important vector issues have been
swept under the rug by your physics professors. I ran into the same
problem you just found when I took my graduate-level classical
mechanics. Here is an exercise for you: Take two arbitrary spherical
vectors{r,theta,phi}, take their cross product as you have been taught
in spherical coordinates. Then convert both vectors to cartesian, take
their cross product in cartesian coordinates and convert the result back
to spherical. You won't get the same result if you do it casually. In
any event, you will find out a lot about vectors that nobody usually
mentions. The first is that there is a huge difference between
displacement vectors and field vectors. I think the technical terms are
covariant and contravariant vectors. Anyway, a displacement vector
represents the difference between two vectors and does not have a single
location in space that it occupies. A field vector has a value at a
single point in space. The formulas that you have been given for cross
products are only valid for field vectors, not displacement vectors.
When you specify an {r,theta,phi} vector, by definition, you mean a
displacement vector(otherwise you would have had to specify
{Vr,Vtheta,Vphi} at point {r,theta,phi}). Looking at the problem you
gave mathematica in spherical coordinates you specified V=(a1,a2,0},
which is a displacement vector beginning at the origin and going to the
point V. You then wanted to cross it with the vector U={0,0,1}, which
is a displacement vector beginning at the origin and ending at the
origin, so you took a cross product between two vectors, one of which
had a zero magnitude. The answer given by mathematica was correct for
DISPLACEMENT VECTORS. This makes perfect mathematical sense, but is
ludicrous from a physical standpoint since all cross-products that
appear in physical equations are for field vectors, not displacements.
What you intended probably was that U and V were not displacement
vectors at all, but were field vectors at some unspecified point in
space. The rules for those kind of cross products are the same in all
coordinate systems and the cartesian cross product would give you the
correct answer for the spherical field vectors provided you gave it the
correct metric, which does not seem to be built in to mathematica.
This is another case where somebody at Wolfram made a choice which was
not the one that most of the books teach or what the average person
doing classical mechanics expects, but is arguably "correct". Similar
cases are found in the definition of both symbolic and discrete Fourier
transforms which have an infinite number of mathematically correct
definitions, but only one which optics and electronics and nature
believes in. Wolfram, of course, went with a common mathematical
definition, rather than the physical one.
Since symbolic tensor determinants are built-in, it shouldn't be hard to
write your own field vector cross product function. The latest version
of the CRC handbook, even has the metric for all 11 separable coordinate
systems in it. Good Luck. Sean Ross.
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