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Homework Statement
[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]
Homework Equations
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 ?