if \epsilon_{ijk} P_{kj}=0 how do we show P_{ij}=P_{ji}?
Apr 26, 2009 #1 latentcorpse 1,444 0 if [itex]\epsilon_{ijk} P_{kj}=0[/itex] how do we show [itex]P_{ij}=P_{ji}[/itex]?
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Apr 26, 2009 #3 latentcorpse 1,444 0 do you mean just substitute in numbers and see what happens because that works - i was just wondering if i could do it without explicitly using numbers (i.e. all of it in index notation).
do you mean just substitute in numbers and see what happens because that works - i was just wondering if i could do it without explicitly using numbers (i.e. all of it in index notation).
Apr 26, 2009 #4 dx Homework Helper Gold Member 2,011 18 For i = j, clearly A_{ij} = A_{ji}, so you only need to check the cases in which i ≠ j. For example, setting i = 1 in ε_{ijk}A_{kj} = 0, you get ε_{123}A_{32} + ε_{132}A_{23} = A_{32} - A_{23} = 0 ⇒ A_{32} = A_{23}. You just need to do the same with i = 2 and i = 3.
For i = j, clearly A_{ij} = A_{ji}, so you only need to check the cases in which i ≠ j. For example, setting i = 1 in ε_{ijk}A_{kj} = 0, you get ε_{123}A_{32} + ε_{132}A_{23} = A_{32} - A_{23} = 0 ⇒ A_{32} = A_{23}. You just need to do the same with i = 2 and i = 3.