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Help understanding minkowski tensor and indices

  1. Mar 23, 2014 #1
    So I have just been introduced to indices, four vectors and tensors in SR and I'm having trouble knowing exactly what I am being asked in some questions.

    So the first question asks to write explicitly how a covariant two tensor transforms under a lorentz boost.

    Now I know that it transforms like [itex]A'_{\mu \nu}=(L^{-1})_{\mu}\ ^{\kappa}(L^{-1})_{\nu}\ ^{\lambda}A_{\kappa \lambda}[/itex]
    where L is the matrix with [itex]L_{00/11}=\gamma[/itex] and [itex]L_{12/21}=-\gamma \beta[/itex]

    Is this written explicitly or do I need to write it out in matrix form and if so what is the order of the matrix multiplication, how do I interpret from the indices what the order is?

    The next questions asks to show that the covariant form of the minkowski metric tensor is invariant under this lorentz boost. What I have so far is [itex]\eta'_{\mu \nu}=(L^{-1})_{\mu}\ ^{\kappa}(L^{-1})_{\nu}\ ^{\lambda}\eta_{\kappa \lambda}[/itex]

    [itex]\eta'_{\mu \nu}=(L^{-1})_{\mu}\ ^{\kappa}(L^{-1})_{\nu \kappa}[/itex]

    I'm not really sure how to proceed from there, I'm pretty sure there is some relationship between raising/lowering indices and the inverse of lorentz transformation I'm not aware of.

    Also I've seen the minkowski metric written both -1,1,1,1 and 1,-1,-1,-1 on various pdfs/notes online why is this?

    And finally is there a link to some proof that the L I described acts on contravariant components and not covariant, i.e somethat shows the basis vectors of minkowski space transform by L^-1?
     
    Last edited: Mar 23, 2014
  2. jcsd
  3. Mar 23, 2014 #2

    Bill_K

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    I can't provide an answer to a homework question directly, but I can help with this general question you raised:

    Index notation is more general than matrix notation, so some expressions can be written in matrix form and other cannot. Generally, writing an index expression as a matrix requires that summed indices be adjacent.

    (L-1)μκ (L-1)νλ Aκλ rewrite as (L-1)μκ Aκλ (L-1)νλ

    Now the κ's are adjacent, but the λ's are not. To get them adjacent, use a transpose: Vαβ = (Vβα)T. So

    (L-1)μκ Aκλ (L-1)νλ rewrite as (L-1)μκ Aκλ (L-1)Tλν

    and the matrix form of the expression becomes L-1 A (L-1)T.
     
  4. Mar 23, 2014 #3
    Can you give any indirect advice for the homework question or should I post it to the homework section of the site?
     
  5. Mar 23, 2014 #4

    pervect

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    I would suggest forgetting about matrix notion and trying to grasp directly with the tensor notation. You know about the Einstein convetion, that repeated indices are summed over, I assume?

    It may be painful, but if you have to , expand the shorthand tensor notation into the full sum, i.e.


    ##\left(x'\right)^a = \Lambda^{a}{}_{b} x^b ## in shorthand tensor notation is equivalent to the follow set of four equations written fully out:

    ##\left(x'\right)^0 = \Lambda^{0}{}_{0} x^0 + \Lambda^{0}{}_{1} x^1 + \Lambda^{0}{}_{2} x^2 + \Lambda^{0}{}_{3} x^3 ##

    ##\left(x'\right)^1 = \Lambda^{1}{}_{0} x^0 + \Lambda^{1}{}_{1} x^1 + \Lambda^{1}{}_{2} x^2 + \Lambda^{1}{}_{3} x^3 ##

    ##\left(x'\right)^2 = \Lambda^{2}{}_{0} x^0 + \Lambda^{2}{}_{1} x^1 + \Lambda^{2}{}_{2} x^2 + \Lambda^{2}{}_{3} x^3 ##

    ##\left(x'\right)^3 = \Lambda^{3}{}_{0} x^0 + \Lambda^{3}{}_{1} x^1 + \Lambda^{3}{}_{2} x^2 + \Lambda^{3}{}_{3} x^3 ##

    I've written the transformation matrix with indices as shown (northwest to southeast) because that's what my textbook (MTW) prescribes. I noticed yours was opposite, and it looks funny to me but I can't be positive that MTW"s conventions are universal.
     
  6. Mar 23, 2014 #5

    pervect

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    If it's an undergraduate course, it should be posted to the homework sections. Homework in graduate courses may be posted to the forums.
     
  7. Mar 26, 2014 #6

    vanhees71

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    You are right. It's very important not to write the indices on top of each other. By definition the contravariant components transform as you've written
    [tex]x'^a={\Lambda^a}_b x^b.[/tex]
    Now let's see, how the covariant components transform
    [tex]x'_a=\eta_{ab} x'^b=\eta_{ab} {\Lambda^{b}}_{c} x^c = \eta_{ab} {\Lambda^{b}}_{c} \eta^{cd} x_d.[/tex]
    Usually one defines
    [tex]\eta_{ab} {\Lambda^{b}}_{c} \eta^{cd}={\Lambda_a}^{d}.[/tex]
    Then you have
    [tex]x'_a={\Lambda_a}^b x_b.[/tex]
    Now let's check that this is indeed correct, i.e., that the covariant components indeed transform contragrediently to the contravariant ones as it must be.

    To that end we have to use the fact that we deal with Lorentz transformations, i.e., pseudo-orthogonal transformations wrt. the fundamental form [itex]\eta_{ab}[/itex]. That means that
    [tex]\eta_{ab} {\Lambda^a}_c {\Lambda^{b}}_d=\eta_{cd}.[/tex]
    Contracting with the inverse transformation, we get
    [tex]\eta_{ab} {\Lambda^a}_c = \eta_{cd} {(\Lambda^{-1})^{d}}_{b}.[/tex]
    Contracting with the fundamental form we find
    [tex]\eta_{ab} \eta^{cd} {\Lambda^a}_c={\Lambda_b}^{d}={(\Lambda^{-1})^d}_b.[/tex]
    QED.

    So it's
    [tex]x_a'={\Lambda_{a}}^b x_b={(\Lambda^{-1})^b}_a x_b.[/tex]

    This should also the OP's question (correcting the assumption in the very beginning!).
     
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