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From: ram@zedat.fu-berlin.de (Stefan Ram)
Newsgroups: sci.physics.relativity
Subject: Vector notation?
Date: 28 Jul 2024 09:27:30 GMT
Organization: Stefan Ram
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  (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^"3-vector" * p^"3-vector":
|= p_0 p_0 - p^"3-vector" * p^"3-vector".

  . 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!