A Varying an action wrt a symmetric and traceless tensor

[binbagsss]

Consider a Lagrangian, #L#, which is a function of, as well as other fields #\psi_i#, a traceless and symmetric tensor denoted by #f^{uv}#, so that #L=L(f^{uv})#, the associated action is #\int L(f^{uv}, \psi_i)d^4x #.

To vary w.r.t #f^{uv}# , I write:

#f^{uv}=f_{symm}^{uv}-\frac{1}{d}\eta^{uv} tr(f)#, (1)

where #tr(f)=\eta_{uv}f^{uv}#, where #\eta_{uv}# is the metric associated to the space-time, and #f_{symm}^{uv}= (1/2) (f^{uv}+f^{vu})#.In (1), why is it to subtract the trace, I suspect adding the trace term is just as valid? (and perhaps there is a convention as to which consistent with the signature of the metric)? In the E-L equations, ofc, the variation w.r.t the deriviative of the field comes with the opposing sign, so it doesn't matter which is chose? thanks

 

[robphy]

Consider a Lagrangian, #L#, which is a function of, as well as other fields #\psi_i#, a traceless and symmetric tensor denoted by #f^{uv}#, so that #L=L(f^{uv})#, the associated action is #\int L(f^{uv}, \psi_i)d^4x #.

To vary w.r.t #f^{uv}# , I write:

#f^{uv}=f_{symm}^{uv}-\frac{1}{d}\eta^{uv} tr(f)#, (1)

where #tr(f)=\eta_{uv}f^{uv}#, where #\eta_{uv}# is the metric associated to the space-time, and #f_{symm}^{uv}= (1/2) (f^{uv}+f^{vu})#.In (1), why is it to subtract the trace, I suspect adding the trace term is just as valid? (and perhaps there is a convention as to which consistent with the signature of the metric)? In the E-L equations, ofc, the variation w.r.t the deriviative of the field comes with the opposing sign, so it doesn't matter which is chose?thanks

Use double-# for equations
https://www.physicsforums.com/help/latexhelp/

 

[haushofer]

Adjusted:

Consider a Lagrangian, L which is a function of, as well as other fields $\psi_i$, a traceless and symmetric tensor denoted by $f^{uv}$, so that $L=L(f^{uv})$, the associated action is $\int L(f^{uv}, \psi_i)d^4x$.

To vary w.r.t $f^{uv}$ , I write:

$f^{uv}=f_{symm}^{uv}-\frac{1}{d}\eta^{uv} tr(f)$, (1)

where $tr(f)=\eta_{uv}f^{uv}$, where $\eta_{uv}$ is the metric associated to the space-time, and $f_{symm}^{uv}= (1/2) (f^{uv}+f^{vu})$.In (1), why is it to subtract the trace, I suspect adding the trace term is just as valid? (and perhaps there is a convention as to which consistent with the signature of the metric)? In the E-L equations, ofc, the variation w.r.t the deriviative of the field comes with the opposing sign, so it doesn't matter which is chose?

--edit-- maybe I'm stupid, but why does the double # not work?

 

[haushofer]

Consider a Lagrangian, #L#, which is a function of, as well as other fields #\psi_i#, a traceless and symmetric tensor denoted by #f^{uv}#, so that #L=L(f^{uv})#, the associated action is #\int L(f^{uv}, \psi_i)d^4x #.

To vary w.r.t #f^{uv}# , I write:

#f^{uv}=f_{symm}^{uv}-\frac{1}{d}\eta^{uv} tr(f)#, (1)

where #tr(f)=\eta_{uv}f^{uv}#, where #\eta_{uv}# is the metric associated to the space-time, and #f_{symm}^{uv}= (1/2) (f^{uv}+f^{vu})#.In (1), why is it to subtract the trace, I suspect adding the trace term is just as valid? (and perhaps there is a convention as to which consistent with the signature of the metric)? In the E-L equations, ofc, the variation w.r.t the deriviative of the field comes with the opposing sign, so it doesn't matter which is chose?thanks

I'm not sure what you do here, but you subtract the trace because the trace transforms to itself. Usually, one can break up a tensor into a symmetric traceless part, an antisymmetric part and the trace. But since you don't give any reference and your tex is hard to read, I'm merely guessing.