4
$\begingroup$

In the stress-energy tensor (SET) for free scalar and vector fields, any references to the connection $\Gamma^\lambda_{\mu\nu}$ in the kinetic terms appear to either be absent ($\nabla_\mu \phi = \partial_\mu \phi$) or cancel each other out ($F_{\mu\nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu = \partial_\mu A_\nu - \partial_\nu A_\mu$). This makes sense to me in that one shouldn't need to know the derivatives of the metric/vielbein in order to compute the source of Einstein's field equations (EFE) at a point. Starting with the symmetrized Lagrangian for a Dirac field,

$$\mathcal L = \frac{i}{2}\bar\psi \gamma^\mu \partial_\mu \psi - \frac i2 \bar\psi \overset{\leftarrow}\partial_\mu \gamma^\mu \psi - m \bar\psi\psi,$$

and promoting partial derivatives to covariant derivatives, I get the following form for the SET:

$$ T^{\mu\nu}_{\text{Dirac}} = \frac{i}{2} \bar\psi ( \gamma^\mu \overset\rightarrow\nabla^\nu + \gamma^\nu \overset\rightarrow\nabla^\mu ) \psi - \frac{i}{2} \bar\psi ( \overset\leftarrow\nabla^\nu \gamma^\mu + \overset\leftarrow\nabla^\mu \gamma^\nu ) \psi - g^{\mu\nu} \mathcal L_{\text{Dirac}} $$

In this expression for the SET of a free Dirac field, I can't seem to get the spin connection terms ($\nabla_\mu \psi - \partial_\mu \psi$) in the covariant derivatives to cancel out, so this tensor depends explicitly on the derivatives of the vielbein $e^a_\mu$. Do those spin connection terms in fact cancel each other out and reduce covariant derivatives to ordinary partial derivatives?

If not, is this not a problem when plugging into Einstein’s field equations? Since Dirac fields also obey the Klein-Gordon equation, can we write down a Klein-Gordon-like Lagrangian (throwing away information about spin) and use that to compute a connection-independent SET?

$\endgroup$
2
  • 1
    $\begingroup$ Even if curvature is "absent" in the stress tensor for a Maxwell field and a scalar, I would say it is still there implicitly in the system of equations. The equations of motion for a scalar and the Maxwell field do know about curvature. Thus the on-shell values for the Maxwell field and the scalar will reflect the presence of curvature in the way they source Einstein's equations. $\endgroup$ Commented Jul 18, 2016 at 3:26
  • $\begingroup$ Thanks! I agree (I think) – curvature does influence the evolution of the stress tensor and equations of motion implicitly. My question is limited to computing the stress tensor and curvature on a time slice in terms of (a) the local field values and their partial derivatives only vs. (b) needing to include covariant derivatives to account for curvature. $\endgroup$
    – rossng
    Commented Jul 18, 2016 at 4:00

1 Answer 1

4
$\begingroup$

As is wellknown, the EFE is a PDE for the metric $g_{\mu\nu}$ / vielbein $e^a{}_{\mu}$, which plays the role of the dynamical fields of GR.

OP is apparently pondering the following question.

Is the EFE's source term (i.e. the matter SEM tensor $T^{\mu\nu}$) independent of the GR fields (i.e. the metric $g_{\mu\nu}$ / vielbein $e^a{}_{\mu}$) and derivatives thereof?

Answer: No, this needs not be the case.

Examples:

  • OP is considering matter composed of Dirac fermions. The vielbein generalization of the Hilbert SEM tensor does depend on the spin connection, cf. my Phys.SE answer here.

  • Already the Maxwell SEM tensor in curved space depends on the metric $g_{\mu\nu}$.

  • A similar situation takes place in scalar QED. Here the EFEs are replaced with Maxwell's equations. The dynamical fields are now $A_{\mu}$. One may show that the source term (the electric $4$-current $j^{\mu}$) in this case depends on $A_{\mu}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.