(The quotation below is given in pure ASCII, but at the end of this
post you will also find a rendition with some Unicode being used.)
(The quotation below is given in pure ASCII, but at the end of this
post you will also find a rendition with some Unicode being used.)
I have read the following derivation in a chapter on SR.
(0) We define:
X := p_"mu" p^"mu",

(1) from this, by Eq. 2.36 we get:
= p_"mu" "eta"^"mu""nu" p_"mu",

(2) from this, using matrix notation we get:
 ( 1 0 0 0 ) ( p_0 )
= ( p_0 p_1 p_2 p_3 ) ( 0 1 0 0 ) ( p_1 )
 ( 0 0 1 0 ) ( p_2 )
 ( 0 0 0 1 ) ( p_3 ),

(3) from this, we get:
= p_0 p_0  p_1 p_1  p_2 p_2  p_3 p_3,

(4) using p_1 p_1  p_2 p_2  p_3 p_3 =: p^"3vector" * p^"3vector":
= p_0 p_0  p^"3vector" * p^"3vector".
. Now, I used to believe that a vector with an upper index is
a contravariant vector written as a column and a vector with
a lower index is covariant and written as a row. I'm not sure
about this. Maybe I dreamed it or just made it up. But it would
be a nice convention, wouldn't it?
Anyway, I have a question about the transition from (1) to (2):
In (1), the initial and the final "p" both have a /lower/ index "mu".
In (2), the initial p is written as a row vector, while the final p
now is written as a column vector.
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?
Here's the same thing with a bit of Unicode mixed in:
(0) We define:
X ≔ p_μ p^μ

(1) from this, by Eq. 2.36 we get:
= p_μ η^μν p_ν

(2) from this, using matrix notation we get:
 ( 1 0 0 0 ) ( p₀ )
= ( p₀ p₁ p₂ p₃ ) ( 0 1 0 0 ) ( p₁ )
 ( 0 0 1 0 ) ( p₂ )
 ( 0 0 0 1 ) ( p₃ )

(3) from this, we get:
= p₀ p₀  p₁ p₁  p₂ p₂  p₃ p₃

(4) using p₁ p₁  p₂ p₂  p₃ p₃ ≕ p⃗ * p⃗:
= p₀ p₀  p⃗ * p⃗
. TIA!
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?
ram@zedat.fuberlin.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 clearsightedly 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.
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