Re: Vector notation?

On 2024-08-01 11:13:59 +0000, Stefan Ram said:

ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

When, in (1), both "p" are written exactly the same way, by what
reason then is the first "p" in (2) written as a /row/ vector and
the second "p" a /column/ vector?

   In the meantime, I found the answer to my question reading a text
  by Viktor T. Toth.
   Many Textbooks say,
               ( -1  0  0  0 )
eta_{mu nu} = (  0  1  0  0 )
              (  0  0  1  0 )
              (  0  0  0  1 ),
   but when you multiply this by a column (contravariant) vector,
  you get another column (contravariant) vector instead of
  a row, while the "v_mu" in
 eta_{mu nu} v^nu = v_mu
   seems to indicate that you will get a row (covariant) vector!
   As Viktor T. Toth observed in 2005, a square matrix (i.e., a row
  of columns) only really makes sense for eta^mu_nu (which is just
  the identity matrix). He then clear-sightedly explains that a
  matrix with /two/ covariant indices needs to be written not as
  a /row of columns/ but as a /row of rows/:
 eta_{mu nu} = [( -1 0 0 0 )( 0 1 0 0 )( 0 0 1 0 )( 0 0 0 1 )]
   . Now, if one multiplies /this/ with a column (contravariant)
  vector, one gets a row (covariant) vector (tweaking the rules for
  matrix multiplication a bit by using scalar multiplication for the
  product of the row ( -1 0 0 0 ) with the first row of the column
  vector [which first row is a single value] and so on)!

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. Using abstract
indices ix clearer. Just note that if an index is used twice in lower
position the inverse "eta" must be used. For SR the upper index position
is not really necessary.
--
Mikko