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  • Matrix Multiplication in SR

    From Stefan Ram@21:1/5 to All on Tue Jul 30 23:44:26 2024
    .
    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",

    [[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]]

    |
    |(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. We thus can
    write (0) in two-dimensional notation:

    ( p^0 )
    = ( p_0 p_1 p_2 p_3 ) ( p^1 )
    ( p^2 )
    ( p^3 )

    So, 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?

    Let's write p_"mu" "eta"^("mu""nu") p_"mu" with two row vectors,
    as it should be written:

    ( 1 0 0 0 )
    = ( p_0 p_1 p_2 p_3 ) ( 0 -1 0 0 ) ( p_0 p_1 p_2 p_3 )
    ( 0 0 -1 0 )
    ( 0 0 0 -1 )

    . AFAIK, the laws for matrix multiplication just do not define
    a product of a 4x4 matrix with a 1x4 matrix, because for every
    row of the left matrix, there has to be a whole column of the
    right matrix of the same size. Does this show there's something
    off with that step of the calculation?

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  • From Stefan Ram@21:1/5 to Stefan Ram on Thu Aug 1 07:01:38 2024
    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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