ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:
[[Mod. note -- I think that last subscript "mu" should be a "nu".
That is, equations (0) and (1) should read (switching to LaTeX notation)
$X := p_\mu p^\mu
= p_\mu \eta^{\mu\nu} p_\nu$
-- jt]]
Thanks for that observation!
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)!
Exercise Work out the representation of eta^{mu nu} in the same
spirit.
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