Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determinat formula in Einstein notation

  1. Jul 10, 2012 #1
    Hi all,

    I've been looking around at formulae for determinants (using them for tensor densities) and I just want to clarify that the expression below is correct (i.e. formulae are correct):
    [tex] |M| = \sum^n_{a_1,a_2, \ldots ,a_n = 1} \epsilon_{a_1a_2 \ldots a_n} M_{1a_1}M_{2a_2} \ldots M_{na_n} = \sum^n_{a_1,a_2, \ldots ,a_n = 1} \epsilon_{a_1a_2 \ldots a_n} M_{a_11}M_{a_22} \ldots M_{a_nn}[/tex]
    The reason I ask is that the second formulae lends itself to Einstein notation:
    [tex] |M| = \epsilon_{a_1a_2 \ldots a_n} M^{a_1}{}_1M^{a_2}{}_2 \ldots M^{a_n}{}_n[/tex]

    As an aside question, is this correct in the sense that there are unmatched indices on each side of the equation? I have found the following formula which seems to correct this:
    [tex] \epsilon_{b_1b_2 \ldots b_n}|M| = \epsilon_{a_1a_2 \ldots a_n} M^{a_1}{}_{b_1}M^{a_2}{}_{b_2} \ldots M^{a_n}{}_{b_n}[/tex]
    I think they are both correct...?

    Cheers
     
  2. jcsd
  3. Jul 10, 2012 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Try

    [tex]|M| = \frac{1}{n!} \varepsilon^{b_1 \ldots b_n} \varepsilon_{a_1 \ldots a_n} M^{a_1}{}_{b_1} \ldots M^{a_n}{}_{b_n}[/tex]
     
  4. Jul 10, 2012 #3
    OK, that's good. Putting [itex]\epsilon_{b_1 \ldots b_n}[/itex] on both sides gives my final formula.

    Also, according to http://en.wikipedia.org/wiki/Levi-Civita_symbol#Determinants does this not imply that my second formula is equivalent to yours?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determinat formula in Einstein notation
  1. Is that a Formula? (Replies: 7)

Loading...