- #1
binbagsss
- 1,254
- 11
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
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