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Subscripts and superscripts in four vectors

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm having trouble with understanding four vectors in particle physics.
    I'm reading this wikipedia page,http://en.wikipedia.org/wiki/Einstein_notation, and its telling me that
    ## v^\mu= \begin{pmatrix} \mu_0 \\ \mu_1 \\ \mu_2 \\ \mu_3 \end{pmatrix} ##
    and
    ## v_\mu= \begin{pmatrix} \mu_0 & \mu_1 & \mu_2 & \mu_3 \end{pmatrix} ##
    But the raising indices equation gives
    ## v^\nu= \eta^{\mu\nu} v_\mu ##
    Which doesn't work if I do matrix multiplication with
    ## \eta^{\mu\nu}= \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} ##
    as it gives ## v^\nu## as a 1x4 not a 4x1 as I would expect from the initial definitions, which leads me to think my initial definitions of the superscripts and subscripts are wrong.

    Could anybody help me out?
    2. Relevant equations
    n/a

    3. The attempt at a solution
    n/a
     
  2. jcsd
  3. Feb 3, 2015 #2
    Where did you get these two formulas from? I do not see them at the quoted Wikipedia page.
     
  4. Feb 3, 2015 #3
    I assumed them from the vector representation section
     
  5. Feb 3, 2015 #4

    TSny

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    Most people write the raising operation as ## v^\nu= \eta^{\nu\mu} v_\mu ## (with the indices on ##\eta## switched from the way you wrote it.) It won't make any difference since the ##\eta## matrix is symmetric, but I thought you might want to know how it is usually written.

    I think of the equation as simply giving me the recipe for finding the value of ## v^\nu##. So, to me, the left hand side is not a 4-vector, but just the ##\nu##th component of the 4-vector. When I use the equation, I'm not thinking in terms of matrix multiplication. I just look at the right side and remember the Einstein summation convention for repeated indices. So,

    ## v^\nu= \eta^{\nu\mu} v_\mu = \eta^{\nu 0} v_0 + \eta^{\nu 1} v_1 + \eta^{\nu 2} v_2 + \eta^{\nu 3} v_3##. The right hand side is just a sum of certain terms and I don't need to visualize it as matrix multiplication.

    Of course you can get the same result by matrix multiplication. But then, you would need to write your covariant vector ##v_\mu## as a column vector rather than a row vector. Or, you could keep ##v_\mu## as a row vector that multiplies the ##\eta## matrix from the left.
     
  6. Feb 3, 2015 #5
    Ok I see how that works now I think. As there's no implicit sum its just the component of a vector. Am I right in thinking though that a subscript usually indicates a row vector, and a superscript a column?
     
  7. Feb 3, 2015 #6
    You wrote "I assumed them from the vector representation section". But there you must be careful. There are two different bases in this section. Unfortunately it is not clearly explained there what are these bases.
     
  8. Feb 3, 2015 #7
    "Am I right in thinking though that a subscript usually indicates a row vector, and a superscript a column?"

    That's correct.
     
  9. Feb 3, 2015 #8
    But the basis [itex]e_i[/itex] is a basis of column vectors, while the basis [itex]e^i[/itex] is a basis of row vectors! Can you understand the difference between components of a vector and the index numbering basis vectors? It is important.
     
  10. Feb 3, 2015 #9

    Fredrik

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    If we use the convention that the number on row ##\mu##, column ##\nu## of a matrix ##M## is denoted by ##M^\mu{}_\nu##, then the definition of matrix multiplication is ##(AB)^\mu{}_\nu=A^\mu{}_\rho B^\rho{}_\nu##. It can be convenient to use the notation ##M^\mu## for the ##\mu##th row of ##M## and ##M_\mu## for the ##\mu##th column of ##M##. But ##x^\mu## isn't the ##\mu##th column of a ##4\times 4##-matrix ##x##. It's just a number, the ##\mu##th component of the 4-tuple ##x##. ##x_\mu## is the ##\mu##th component of the 4-tuple ##\eta x##, where ##\eta## denotes the matrix with ##\eta_{\mu\nu}## on row ##\mu##, column ##\nu##. (The convention described above isn't used for ##\eta## or ##\eta^{-1}##).
     
    Last edited: Feb 3, 2015
  11. Feb 3, 2015 #10
    Yeah I think I understand that now, thanks a lot for your help, I see I need to be a lot more careful with what is just a component and what is a matrix. Hopefuly that will come with more practice of this notation.
     
  12. Feb 3, 2015 #11

    Fredrik

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    These earlier posts may be useful:

    https://www.physicsforums.com/threads/covariant-equation-intuition-confusing-me.792310/#post-4976147
    (Skip the first paragraph and start at "The modern...")

    https://www.physicsforums.com/threads/covariant-and-contravariant.789968/page-2#post-4963000

    https://www.physicsforums.com/threads/einstein-notation-notes.770129/#post-4847943

    Edit: If you're not interested in how various n-tuples transform under a change of ordered basis, then you won't find the first two very interesting. But the last one should be useful.
     
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