Physics Asked on May 13, 2021
I read a paper where there was written that "a Traceless Energy-momentum Tensor implies a massless field", so I did a bit of calculations but I seems not really true, is it true?
So now I’m considering 3 Lagrangian density:
$$mathcal{L}_1=-frac{1}{4}F^{rho sigma}F_{rho sigma}sqrt{-g}$$
$$Rightarrow T_1^{mu nu}=F^{mu sigma}F_{rhosigma}g^{rhonu}-frac{1}{4}g^{mu nu}F^{rho sigma}F_{rho sigma}$$
$$mathcal{L}_2=ibar{psi}left(gamma^{rho}partial_{rho}right)psisqrt{-g}$$
$$Rightarrow T_2^{mu nu}=ig^{mu nu}bar{psi}left(gamma^{rho}partial_{rho}right)psi-ibar{psi}left(gamma^{mu}partial^{nu}+gamma^{nu}partial^{mu}right)psi$$
$$mathcal{L}_3=frac{1}{2}partial_{rho}phi partial^{rho}phisqrt{-g}$$
$$Rightarrow T_3^{mu nu}=frac{1}{2}g^{mu nu}partial_{rho}phi partial^{rho}phi – partial^{mu}phipartial^{nu}phi$$
In 4 Dimension the Trace $T=g_{mu nu}T^{mu nu}$ is :
$$T_1=0$$
$$T_2=2ibar{psi}left(gamma^{rho}partial_{rho}right)psineq0$$
$$T_3=partial_{rho}phi partial^{rho}phineq0$$
$$mathcal{L}_1=-frac{1}{4}F^{rho sigma}F_{rho sigma}$$
$$Rightarrow tilde{T}_1^{mu nu}=F^{mu sigma}F_{rhosigma}g^{rhonu}+frac{1}{4}g^{mu nu}F^{rho sigma}F_{rho sigma}$$
But then in this case the energy-momentum tensor is no more traceless, in fact:
$$tilde{T}_1=2 F^{rho sigma}F_{rho sigma}$$
(For completeness this is the Simmetrized Belinfante Form, I could have not simmetrised it but I would have obtained $tilde{T}_1=frac{3}{2} F^{rho sigma}F_{rho sigma}neq 0$ anyway).
So what’s the meaning of the Trace $tilde{T}_1$ not being zero in the Classical Field Theory case?
Why $tilde{T}_1^{mu nu}neq T_1^{mu nu}$? Shouldn’t they be the same? Do they describe different physics?
So "a Traceless Energy-momentum Tensor implies a massless field" is it true only for the EM field in GR?
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