I'm currently in the process of familiarising myself with some basic concepts of general relativity and have stumbled upon a problem that is probanly quite simple. I'm referring to the book by Hobson, p.541 and 548, where the energy-momentum tensor for a simple matter field action $S$, $$T_{\mu\nu} = \frac{2}{\sqrt{-\det g}}\frac{\delta S}{\delta g^{\mu\nu}},$$ is calculated as an example:

$$ S = \int d^4x\sqrt{-\det g}(\frac{1}{2}g^{\mu\nu}(\nabla_\mu \Phi)(\nabla_\nu \Phi)-V(\Phi)) $$ $$ \delta(\det g) = \det g^{\mu\nu}\delta g_{\mu\nu} = -\det g g_{\mu\nu}\delta g^{\mu\nu} \Rightarrow \delta\sqrt{-\det g} = -\frac{1}{2}\sqrt{-\det g}g_{\mu\nu}\delta g^{\mu\nu} $$ \begin{eqnarray} \delta S &=& \int d^4x[\sqrt{-\det g}\frac{1}{2}\delta g^{\mu\nu}(\nabla_\mu \Phi)(\nabla_\nu \Phi) + \delta(\sqrt{-\det g})(\frac{1}{2}g^{\alpha\beta}(\nabla_\alpha \Phi)(\nabla_\beta \Phi)-V(\Phi))] \\ &=& \int d^4x\sqrt{-\det g}\frac{1}{2}[(\nabla_\mu \Phi)(\nabla_\nu \Phi) -g_{\mu\nu}(\frac{1}{2}g^{\alpha\beta}(\nabla_\alpha \Phi)(\nabla_\beta \Phi)-V(\Phi))]\delta g^{\mu\nu} \end{eqnarray}

and one has $T_{\mu\nu} = [...]_{\mu\nu}$

My problem now is: if I calculate $$T^{\mu\nu} = \frac{2}{\sqrt{-\det g}}\frac{\delta S}{\delta g_{\mu\nu}}$$ in the same way, using the first term for $\delta\sqrt{-\det g}$ instead of the second, I find the same term for $T$ as before (with the indices up, of course), but with a plus instead of a minus between the derivatives-term and the Lagrangian-term in $T$.

But this is wrong, isn't it? So what am I doing wrong?


1 Answer 1


It seems OP's question is rooted in the fact that infinitesimal variations of the inverse metric

$$\delta g^{\mu\nu}~=~-g^{\mu\lambda} ~\delta g_{\lambda\kappa} ~g^{\kappa\nu}$$

comes with a minus. Hence the stress-energy-momentum (SEM) tensor with upper and lower indices are defined with opposite signs. They are defined as

$$T^{\mu\nu}~=~\mp \frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}, \qquad T_{\mu\nu}~=~\pm \frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}}, $$ in the $(\pm,\mp,\mp,\mp)$ Minkowski signature convention, respectively, cf. this Phys.SE post.

  • $\begingroup$ If so, the upper indices stress energy tensor will not be obtaned from the lower indices one using metric tensor rasing indices. Is there any meaning to say it is a tenso? $\endgroup$
    – thone
    Aug 3, 2018 at 10:15
  • $\begingroup$ It is a tensor, and $T^{\mu\nu}=g^{\nu\lambda}T_{\lambda\kappa}g^{\kappa\nu}$ still holds. $\endgroup$
    – Qmechanic
    Aug 3, 2018 at 10:36
  • $\begingroup$ For the simplest action $$ -\frac{1}{2}\int d^4x \sqrt{|g|}g^{\alpha\beta}\partial_\alpha \phi \partial_\beta \phi ,$$ there seems no such relation as you write above. $\endgroup$
    – thone
    Aug 3, 2018 at 10:40
  • $\begingroup$ Using the definition, $$T_{\alpha\beta}=\partial_\alpha \phi \partial_\beta \phi +g_{\alpha\beta} L_M$$ and $$T^{\alpha\beta}=-\partial^\alpha \phi \partial^\beta \phi +g^{\alpha\beta} L_M$$ $\endgroup$
    – thone
    Aug 3, 2018 at 10:44
  • $\begingroup$ For the record, the sign in your scalar action, implies that you use the $(-,+,+,+)$ sign convention. Then $T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi -\frac{1}{2}g_{\mu\nu} \partial_{\lambda}\phi \partial^{\lambda}\phi$ and $T^{\mu\nu}=\partial^{\mu}\phi\partial^{\nu}\phi -\frac{1}{2}g^{\mu\nu} \partial_{\lambda}\phi \partial^{\lambda}\phi$. $\endgroup$
    – Qmechanic
    Aug 3, 2018 at 12:47

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