Mikko <
mikko.levanto@iki.fi> wrote or quoted:
Matrices do not match very well with the needs of physics. Many physical
quantities require more general hypermatrices. But then one must be
very careful that the multiplicatons are done correctly.
In the meantime, I have written about this for the case of a ( 0, 2 )
tensor, i.e., a bilinear form (such as "eta"). It turns out that for
this case, a simple single rule for matrix multiplication suffices.
To give it the right context, my following text starts with a small
introduction into the linear algebra of vectors and forms and
arrives at the actual matrix multiplication only near the end:
(If one is not into bilinear algebra, one may stop reading now!)
It's about the fact that the matrix representation of a ( 0, 2 )-
tensor should actually be a row of rows, not a row of columns,
as you often see in certain texts. A row of columns, on the
other hand, would be suitable for a ( 1, 1 )-tensor. I got this
from a text by Viktor T. Toth. All errors here are my own though.
But since I want to start with the basics, this matrix
representation will only be dealt with towards the end of
this text, where impatient readers could of course jump to.
In this text, I limit myself to real vector spaces R, R^1,
R^2, etc. For a vector space R^n, let the set of indices
be I := { i | 0 <= i < n }.
Forms
The structure-preserving mappings f into the field R are precisely
the linear mappings of a vector to R.
I call such a linear mapping f of a vector to R a /form/ or a
/covector/.
Let f_i be n forms. If the tuple ( f_i( v ))ieI of a vector v
is equal to the tuple ( f_i( w ))ieI of a vector w if and only
if v=w, I call the tuple ( f_i )ieI a /basis/ of the vector space.
The numbers v^i := f_i( v ) are the /(contravariant) coordinates/
of the vector v in the basis ( f_i )ieM.
I call the vector e_i, for which f_j( e_i ) is 1 for i=j and 0
for i<>j, the i-th /basis vector/ of the basis ( f_i( v ))ieI.
If f is a form, then the tuple ( f( e_i ))ieI are the /(covariant)
coordinates/ of the form f.
Matrices
We write the covariant coordinates f_i of a form f as a "horizontal"
1xn-matrix M( B, f ):
( f_0, f_1, ..., f_(n-1) ).
The contravariant coordinates v^i of a vector v we write in a
basis B as a "vertical" nx1-matrix M( B, v ):
( v^0 )
( v^1 )
( . . . )
( v^( n-1 )).
The application f( v ) of a form to a vector then results from
the matrix multiplication M( B, f )X M( B, v ).
Rule For The Matrix Multiplication X
.-----------------------------------------------------------------.
| The /multiplication X/ of a 1xn-matrix with an nx1-matrix is |
| a sum with n summands, where the summand i is the product of |
| the column i of the first matrix with the row i of the second |
| matrix. |
'-----------------------------------------------------------------'
( 0, 2 )-Tensors
We also call the forms (covectors) "( 0, 1 )-tensors" to express
that they make a scalar out of 0 covectors and one vector linearly.
Accordingly, a /( 0, 2 )-tensor/ is a bilinear mapping (bilinear
form) that makes a scalar out of 0 covectors and /two/ vectors.
Matrix representation of ( 0, 2 )-tensors
According to Viktor T. Toth, for us, the matrix representation
of a ( 0, 2 )-tensor f is a horizontal 1xn-matrix M( B, f ),
whose individual components are horizontal 1xn-matrices of
scalars. The scalar at position j of component i of M( B, f )
is f( e^i, e^j ), where the superscripts here do not indicate
components of e but a basis vector.
(PS: Here I am not sure about the correct order "f( e^i, e^j )"
or "f( e^j, e^i )", but this is a technical detail.)
Let's now look at the case n=3 and see how we calculate the
application of such a tensor f to two vectors v and w with
the matrix representations!
( v^0 ) ( w^0 )
( (f_00,f_01,f_02)(f_10,f_11,f_12)(f_20,f_21,f_22) ) X ( v^1 ) X ( w^1 )
( v^2 ) ( w^2 )
We start with the first product:
( v^0 )
( (f_00,f_01,f_02) (f_10,f_11,f_12) (f_20,f_21,f_22) ) X ( v^1 )
( v^2 ).
According to our rule for the matrix multiplication X, this is the
sum
v^0*(f_00,f_01,f_02)+v^1*(f_10,f_11,f_12)+v^2*(f_20,f_21,f_22)=
(v^0*f_00,v^0*f_01,v^0*f_02)+
(v^1*f_10,v^1*f_11,v^1*f_12)+
(v^2*f_20,v^2*f_21,v^2*f_22)=
(v^0*f_00+v^1*f_10+v^2*f_20,
v^0*f_01+v^1*f_11+v^2*f_21,
v^0*f_02+v^1*f_12+v^2*f_22).
This is again a "horizontal" 1xn-matrix (written vertically
here because it does not fit on one line), which can be
multiplied by the vertical nx1-matrix for w according to
our rules for matrix multiplication X:
(v^0*f_00+v^1*f_10+v^2*f_20,
v^0*f_01+v^1*f_11+v^2*f_21, ( w^0 )
v^0*f_02+v^1*f_12+v^2*f_22) X ( w^1 )
( w^2 ).
According to our rule for the matrix multiplication X, this
results in the number
w^0*(v^0*f_00+v^1*f_10+v^2*f_20)+
w^1*(v^0*f_01+v^1*f_11+v^2*f_21)+
w^2*(v^0*f_02+v^1*f_12+v^2*f_22).
So, the multiplication of the given matrix representation
of a ( 0, 2 )-tensor with the matrix representations of
two vectors correctly results in a /number/ using the single
uniform rule for the matrix multiplication X.
In the literature (especially on special relativity),
the "Minkowski metric", which is a (0,2)-tensor, is written as
a row of /columns/. The application to two vectors would then be:
( f_00, f_01, f_02 ) ( v^0 ) ( w^0 )
( f_10, f_11, f_12 ) ( v^1 ) ( w^1 )
( f_20, f_21, f_22 ) ( v^2 ) ( w^2 ) =
( f_00 * v^0 + f_01 * v1 + f_02 * v^2 ) ( w^0 )
( f_10 * v^0 + f_11 * v1 + f_12 * v^2 ) ( w^1 )
( f_20 * v^0 + f_21 * v1 + f_22 * v^2 ) ( w^2 )
Now the product of /two column vectors/ appears, which is
not defined as a matrix multiplication! (Matrix multiplication
is not the same as the dot product of two vectors.)
Vector notation?
Re: Vector notation?
