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Homework Help: Interval and a general tensor

  1. Apr 7, 2013 #1
    I'm trying to understand what kind of relation the metric can have with a general tensor B.

    [tex]d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}=d{{s}^{2}}[/tex]
    [tex]\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=1[/tex]
    [tex]\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=\frac{1}{D}g_{a}^{a}[/tex]
    [tex]\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}g_{a}^{d}}{d{{s}^{2}}}=\frac{1}{D}g_{d}^{a}g_{a}^{d}[/tex]
    [tex]\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}g_{a}^{d}}{d{{s}^{2}}}A_{m}^{n}=
    \frac{1}{D}g_{d}^{a}g_{a}^{d}A_{m}^{n}[/tex]
    Define: [tex]B_{am}^{dn}=g_{a}^{d}A_{m}^{n}[/tex]
    substitute in
    [tex]\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}}{d{{s}^{2}}}B_{am}^{dn}=\frac{1}{D}g_{d}^{a}B_{am}^{dn}[/tex]
    [tex]\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}}{d{{s}^{2}}}B_{am}^{dn}=\frac{1}{D}B_{am}^{an}[/tex]

    It all looks Ok to me. Does all of the following look reasonable, or is there a problem somewhere?
     
  2. jcsd
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