Square Brackets in Partial Derivatives: Meaning & Examples

[countable]

What do the square brackets represent in the expressions below:

\partial_{[\mu}\partial_{\nu}A_{\rho ]}

\partial_{[\mu}F_{\nu\rho ]}

I'm guessing they aren't commutators?

thanks.

 

[dextercioby]

Index total antisymmetrization. This means that the <objects> in the LHS of

T_{\mu\nu\rho} = \partial_{[\mu}\partial_{\nu}A_{\rho ]}

and

R_{\mu\nu\rho} = \partial_{[\mu}F_{\nu\rho ]}

are totally antisymmetric in the 3 indices they have.

http://en.wikipedia.org/wiki/Antisymmetric_tensor

 

[countable]

Index total antisymmetrization. This means that the <objects> in the LHS of

T_{\mu\nu\rho} = \partial_{[\mu}\partial_{\nu}A_{\rho ]}

and

R_{\mu\nu\rho} = \partial_{[\mu}F_{\nu\rho ]}

are totally antisymmetric in the 3 indices they have.

http://en.wikipedia.org/wiki/Antisymmetric_tensor

Thanks for the reply. So they are both antisymmetric tensors - by this do we simply mean that F_{\mu\nu}=-F_{\nu\mu}?

Also would I be right in saying that the square brackets mean cyclic permutations thus:

T_{\mu\nu\rho} = \partial_{[\mu}\partial_{\nu}A_{\rho ]} = \partial_\mu \partial_\nu \partial_\rho + \partial_\nu \partial_\rho \partial_\mu + \partial_\rho \partial_\mu \partial_\nu

 

[dextercioby]

This last part is not true, because the desired antisymmetrization is not achieved: if you shuffle indices in the RHS, you can't obtain that, e.g.

T_{\mu\nu\rho} = - T_{\nu\mu\rho}.

In other words, you normally have to have terms with - (minus) in the sum in the RHS.