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Proof - epsilon permutation and metric tensor relation

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\mbox{Prove that}\,g^{ij} \epsilon_{ipt}\epsilon_{jrs}\,=\, g_{pr}g_{ts}\,-\,g_{ps}g_{tr}[/tex]
    Notation :
    [tex]e_{ijk}\,=\,e^{ijk}\,=\,\left\{\begin{array}{cc}1,&\mbox{ if ijk is even permutation of integers 123...n }\\-1, & \mbox{if ijk is odd permutation of integers 123...n}\\0&\mbox{in all other cases} \end{array}\right[/tex]

    [tex] \epsilon_{ijk}\,=\,\sqrt{g}e_{ijk} [/tex]
    [tex] \epsilon^{ijk}\,=\,\frac{1}{\sqrt{g}}e^{ijk} [/tex]

    [tex] \mbox{where}\,g\, = | g_{ij}| \mbox{ value of determinant formed by metric components of space }[/tex]


    2. Relevant equations



    3. The attempt at a solution

    [tex] \epsilon_{ipt}\epsilon_{jrs}\,=\,ge_{ipt}e_{jrs} [/tex]

    [tex]g^{ij}\epsilon_{ipt}\epsilon_{jrs}\,=\,g^{ij}ge_{ipt}e_{jrs}\,=\,g^{ij}\,g\left| \begin{array}{ccc}\delta_{ij}&\delta_{ir}&\delta_{is}\\ \delta_{pj}&\delta_{pr}&\delta_{ps}\\ \delta_{tj}&\delta_{tr}&\delta_{ts}\end{array}\right|[/tex]

    [tex] = g^{ij}\,\left| \begin{array}{ccc}g_{ij}&g_{ir}&g_{is}\\ g_{pj}&g_{pr}&g_{ps}\\ g_{tj}&g_{tr}&g_{ts}\end{array}\right|[/tex]


    [tex] = g^{ij}g_{ij} ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, g^{ij}g_{ir} ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, g^{ij}g_{is} ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )[/tex]

    [tex] = ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, \delta_r^j ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, \delta_s^j ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )[/tex]

    [tex] = g_{pr}g_{ts} \, - \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts} \, + \, g_{ps}g_{tr} \, + \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts} [/tex]

    [tex]g^{ij}\epsilon_{ipt}\epsilon_{jrs}\, = \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts}[/tex]

    [tex]g^{ij}\epsilon_{ipt}\epsilon_{jrs}\, = \, - ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} )[/tex]

    Why am I getting unexpected -ve sign ?
     
  2. jcsd
  3. Jun 25, 2010 #2
    Got the correction ! The modified solution is as follows -

    From above solution continuing up to step -

    [tex] g^{ij}\epsilon_{ipt}\epsilon_{jrs}\,=\, g^{ij}g_{ij} ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, g^{ij}g_{ir} ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, g^{ij}g_{is} ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )[/tex]

    then -

    [tex]g^{ij}\epsilon_{ipt}\epsilon_{jrs}\,=\delta_j^j( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, \delta_r^j ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, \delta_s^j ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} ) [/tex]

    [tex]= 3 ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, \delta_r^j ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, \delta_s^j ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )[/tex]

    [tex]= 3( g_{pr}g_{ts} ) \, - \,3( g_{ps}g_{tr} ) \, - \, g_{pr}g_{ts} \, + \, g_{ps}g_{tr} \, + \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts}[/tex]

    [tex]g^{ij}\epsilon_{ipt}\epsilon_{jrs}\, = \, g_{pr}g_{ts} \, - \, g_{ps}g_{tr} [/tex]

    which is same as required !
     
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