Sujet : Re: Vector notation?
De : ram (at) *nospam* zedat.fu-berlin.de (Stefan Ram)
Groupes : sci.physics.relativityDate : 01. Aug 2024, 12:13:59
Autres entêtes
Organisation : Stefan Ram
Message-ID : <vector-20240801121332@ram.dialup.fu-berlin.de>
References : 1
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)!