Contracting Tensors: Why G^αβ?

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Why would [itex]g^{\alpha \beta} \partial_{\beta} T_{\beta \rho}[/itex] become [itex]\partial^{\alpha} T_{\beta \rho}[/itex] and not [itex]\partial^{\alpha} T_{\rho}^{\alpha}[/itex] or could it be either?
 
pleasehelpmeno said:
Why would [itex]g^{\alpha \beta} \partial_{\beta} T_{\beta \rho}[/itex] become [itex]\partial^{\alpha} T_{\beta \rho}[/itex] and not [itex]\partial^{\alpha} T_{\rho}^{\alpha}[/itex] or could it be either?

The first formula is ambiguous because a bound index β occurs an odd number of times.
 
There's something wrong with that expression. β isn't supposed to appear three times. So do you mean ##g^{\alpha \gamma} \partial_{\gamma} T_{\beta \rho}## or ##g^{\alpha \gamma} \partial_{\beta} T_{\gamma \rho}## or something different from both of these?

In situations where you're thinking about moving a ##g^{\alpha\beta}## to the right of a ##\partial_\gamma##, you must ask yourself if the components of the metric are constant in the coordinate system you're using. If they're not, you would have to use the product rule for derivatives.
 
I was trying to contract [itex]R^{\sigma}_{\mu \nu \rho}[/itex] to [itex]R_{\mu \nu}[/itex],

and i thought the best way to do it would be [itex]\eta^{\rho \alpha} \eta_{\alpha \sigma}R^{\sigma}_{\mu \nu \rho}[/itex] but perhaps that is wrong
 
oops yeah sorry, it is from the book by hobson