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Bracketed subscripts

  1. Sep 20, 2005 #1
    I. What does

    A_{[\alpha} B_{\beta]}

    mean? How do you expand this?

    II. How do you put TeX in board posts?
     
  2. jcsd
  3. Sep 20, 2005 #2

    robphy

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    [ tex ] A_{[\alpha} B_{\beta]} [ /tex ] (remove the spaces in the tags)
    [tex] A_{[\alpha} B_{\beta]} [/tex] (click the on the equation to see)

    [tex] A_{[\alpha} B_{\beta]} =\frac{1}{2!}\left( A_{\alpha} B_{\beta} - A_{\beta} B_{\alpha} \right)[/tex] the "antisymmetric part of [tex] A_{\alpha} B_{\beta} [/tex]"
     
  4. Sep 21, 2005 #3

    dextercioby

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    It's not necessary to use the numerical factor (we call it "weght"). See for example the em. tensor in vacuum. U can use the "no number convention" (i use it)

    [tex] F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:\partial_{[\mu}A_{\nu]} [/tex]

    or the one Rob exemplified, when an ugly 2 comes up

    [tex] F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:2 \partial_{[\mu}A_{\nu]} [/tex]

    Daniel.
     
  5. Sep 21, 2005 #4

    dextercioby

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    And for symmetrizing, we use round brackets. For example, the linearized graviton field has the irreducible Lagrangian infinitesimal gauge transformations

    [tex] \delta_{\epsilon}h_{\left(\mu\nu\right)}(x) =\partial_{(\mu}\epsilon_{\nu)}(x) [/tex]

    or with the "2", if you use an analogue convention Rob used.

    Daniel.
     
  6. Sep 21, 2005 #5

    robphy

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    While the combinatorial factor I used may be conventional, I believe it is the preferred convention. See, for example, http://mathworld.wolfram.com/AntisymmetricTensor.html

    Note that the "symmetric part" of a matrix is [tex]\frac{1}{2}(A+A^T)[/tex] and the "antisymmetric part" of a matrix is [tex]\frac{1}{2}(A-A^T)[/tex]. (Similarly, the "real part of a complex number z" is [tex]\frac{1}{2}(z+\bar z)[/tex] and "imaginary part of a complex number z" is [tex]\frac{1}{2i}(z-\bar z)[/tex].)

    So, one can write the matrix equation
    [tex] A= A_{SYM} + A_{ANTISYM} [/tex]
    and an analogous tensorial equation
    [tex] \begin{align*}
    A_{ab} &=
    \frac{1}{2}( A_{ab} + A_{ba} )+
    \frac{1}{2}( A_{ab} - A_{ba} )
    \\
    &=
    A_{(ab)} + A_{[ab]} \\
    \end{align*}
    [/tex]

    Note, if A is antisymmetric, then we can write
    [tex] A_{ab} = A_{[ab]} [/tex].



    I think there is an inconsistency in your use of the brackets in the "no number convention" [tex] F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:\partial_{[\mu}A_{\nu]} [/tex]

    If "bracket" means "sum the alternating permutations without dividing by the combinatorial factor", then you must write for an antisymmetric F:
    [tex]F_{\mu\nu}=\frac{1}{2}F_{\left[\mu\nu\right]}=\frac{1}{2}(F_{\mu\nu}-F_{\nu\mu})[/tex] or [tex]2F_{\mu\nu}=F_{\left[\mu\nu\right]}[/tex]
     
  7. Sep 23, 2005 #6

    dextercioby

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    I didn't in any place claim that

    [tex] F_{[\mu\nu]}=F_{\mu\nu}-F_{\nu\mu} [/tex]

    So no inconsistency. Whatsoever.

    Daniel.
     
  8. Sep 23, 2005 #7

    robphy

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    So, maybe I am misunderstanding what your "no number convention" is.
    What does [tex]\partial_{[\mu}A_{\nu]}[/tex] mean in your convention?
    [tex]\partial_{[\mu}A_{\nu]}\stackrel{?}{=}\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}[/tex] or
    [tex]\partial_{[\mu}A_{\nu]}\stackrel{?}{=}\frac{1}{2!}\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)[/tex]?
     
  9. Oct 3, 2005 #8

    dextercioby

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    Obviously the first.

    Daniel.
     
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