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Re: Re: coordinate transformation

  • To: mathgroup at
  • Subject: [mg13308] Re: [mg13169] Re: [mg13117] coordinate transformation
  • From: Sean Ross <seanross at>
  • Date: Fri, 17 Jul 1998 03:19:09 -0400
  • References: <>
  • Sender: owner-wri-mathgroup at

David Withoff wrote:
> > S.-Svante Wellershoff wrote:
> > >
> > > Can someone explain me, how to use mathematica for transformations
> > > between
> > > different coordinate systems?
> > >
> > > Like (x,y,z) in carthesian equals (r sin t cos p, r sin t sin p, r cos
> > > t) in spherical system.
> > >
> > > Thanks, Svante
> > >
> > > p.s.: hope its no faq!
> > >
> > > ---------------------------------------------------------------------
> > >   S.-Svante Wellershoff      svante.wellershoff at
> > >                    
> > > ---------------------------------------------------------------------
> > >   Institut fuer Experimentalphysik
> > >   Freie Universitaet Berlin         phone +49-(0)30-838-6234 (-6057)
> > >   Arnimallee 14                     fax   +49-(0)30-838-6059
> > >   14195 Berlin - Germany
> > > ---------------------------------------------------------------------

> The remark about transformations that are "valid only for vectors of
> infinitesimal spatial extent centered at the origin" is confusing, but
> may refer to the fact that, when working with a vector field, the field
> vectors are typically described using a locally cartesian coordinate
> system derived from the coordinate system that is used for the host space.
> You can get into all sorts of trouble if you get the host space (or the
> coordinate system that is used to describe it) mixed up with the spaces
> (or their coordinate systems) used for the field vectors.  If you aren't
> working with vector fields, then none of that is relevant, of course, and
> it isn't relevant to the Calculus`VectorAnalysis` package in any case,
> since the functions in that package only deal with one space at a time,
> and are not intended for transformations between, say, one locally
> cartesian space and another.
> Dave Withoff
> Wolfram Research

I seem to have the talent for confusing people at Wolfram with the nasty
little subtleties of physical mathematics.  Here is another go round. 

This whole issue becomes painfully clear with the following exercise:
Take two cartesian vectors and their cross product.  Transform the
vectors into cylindrical or spherical coordinates using the transforms
like those in the mathematica package and take their cross product in
cylindrical/spherical coordinates and transform the result back to
cartesian and compare the two.  My hat off to anyone who gets the
correct answer the first two or three times trying.

But to explain my original comment about certain transformations only
being valid for vectors of infinitesimal spatial extent or those
centered on the origin:

Take the standard tensor definition of cross product:

            g[i,j] epsilon[i,j,k] B[j] A[k]

Where g is the coordinate metric, epsilon is the Levi-Cevita tensor and
A and B are vectors.  The coordinate metric is the identity matrix in
cartesian coordinates, but is equal to  {{1,0,0},{0,1/r,0},{0,0,1}} for
contravariant vectors in cylindrical coordinates, for example.

So, suppose I have a vector A={x1,y1,z1} and B={x2,y2,z2} and I
transform them into cylindrical coordinates and get A={r1,theta1,z1}
and B={r2,theta2,z2} and now I want to take the cross product in
cylindrical coordinates and compare it to the cross product in
cartesian coordinates.  Which value of r (r1 or r2?) do I use in the
coordinate metric????  The answer is that I must use the value of r at
the point at which the vectors exist.  If the vectors are length
vectors, then there are many r values along which they exist, so I
actually end up needing to do an integral just to correctly perform a
cross product!

Now suppose a different example.  Let
A={Ex[{x,y,z}],Ey[{x,y,z},Ez[{x,y,z}]]} and

so that A and B are no longer length vectors, but are field vectors!  In
this case I have chosen electric field and electric displacement.  The
spatial extent of a field vector at a point is infinitesimal.  Now the
standard tensor definition of a cross product is valid as are all those
wonderful transformations in the Mathematica package.  Notice that the
two vectors A and B exist at the same point in space, which is
necessary  for all those vector operations to work and be valid.

My warning was intended to remind people that you can't just take two
vectors A and B and take their cross product, transform them, take
laplacians and divergences etc. until you have firmly established
exactly what kind of vectors they are.  I am sure that the
mathematicians have different words to describe them than the ones I
use, but physically, it matters if the vector represents a length
vector, a field vector or a vector operator.  Perhaps there are also
classes of abstract vectors that represent different types with which I
am unfamiliar.  They all transform and operate differently.  The vector
transformations in the mathematica package are valid for all field
vectors and for all length vectors centered on the origin (radial
vectors), but nothing else.

The other thing to watch out for is that the unit basis vectors are
position dependent in all non-cartesian coordinate systems, so if you
transform the components of a vector, you don't have the entire
picture. Vectors in non-cartesian coordinate systems require 6 pieces
of information (3 components and 3 basis vectors or location of vector)
to fully describe them, while cartesian vectors only require the 3

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