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Commutator-like notation, index notation |
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| Nov29-09, 03:44 PM | #1 |
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Commutator-like notation, index notation
1. The problem statement, all variables and given/known data
There are some equations in the notes on field theory I am reading with notation I have never come across before. Someone told me it was a way of ensuring that the expression was anti-symmetric. I can't find it used the same anywhere else but no explanation is provided in the notes which makes me think maybe it is just an ordinary commutator but I am not sure. 2. Relevant equations The equations are... i. [tex]$\Phi_{\mu\nu} = 2x_{[\mu}\partial_{\nu]}\phi[/tex] ii. [tex]$\Lambda^{\rho}_{\mu\nu} = 2x_{[\mu}\delta^{\rho}_{\nu]}L[/tex] iii. [tex]$\partial_{[\mu}F_{\nu\rho]} = 0[/tex] F is the electromagnetic tensor if it helps. 3. The attempt at a solution does the first mean... [tex] \left(2x_\mu\partial_\nu\phi - 2x_\nu\partial_\mu\phi\right) [/tex] but then what do the others mean? For the second I was thinking of ignoring the superscript index - the rho - and doing the same as with the first but then what about the third one or am I not even close? Thanks |
| Nov29-09, 04:13 PM | #2 |
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not really.
you have to divide by the factorial of the number of terms in square brackets and then you write out all the possible permutations of the terms in square brackets (where even permutations are positive and odd permutations are negative), so [latex]2x_{[\mu} \partial_{\nu]} \phi=2 \frac{1}{2!} \left(x_{\mu} \partial_{\nu} - x_{\nu} \partial_{\mu} \right) \phi = \left(x_{\mu} \partial_{\nu} - x_{\nu} \partial_{\mu} \right) \phi=[/latex] not this can get much more complicated. for example in your third case you'll have 6 permutations, 3 odd and 3 even. recall a permutation is odd if it can be written as an odd number of transpositions and even if it can be written as an even number of transpositions. |
| Nov29-09, 04:20 PM | #3 |
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awesome! thanks
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| commutator, field theory, index, index notation, notation |
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