# How are the apparently unbound degrees of freedom in Einstein field equations filled?

Something that has always bothered me in general relativity is the annoying fact that there seems to be too few information in the Einstein field equations themselves.

In order to solve a system we need to calculate both the stress energy tensor $T^{\mu \nu}$ and the metric $g^{\mu \nu}$ for all points in spacetime. These tensors have 10 independent components each, so 20 in total. Now the available laws of physics provide us with two bunches of equations.

Einstein field equations $$R^{\mu \nu} - \frac{1}{2} Rg^{\mu \nu} = \frac{8 \pi G}{c^4} T^{\mu \nu}$$

Conservation laws $$T^{\mu \nu}_{; \mu} = 0$$

These amount to 14 equations falling short 6. In concrete calculations in textbooks this is always countered by strong assumptions on the metric or stress energy tensor, but these seem ad hoc solutions.

What am I not seeing here? Why is this not a fundamental problem?

• Does it bother you that the $6$ dofs in $F_{\mu\nu}$ and $4$ in $j^\mu$ don't all die to $\partial_\mu F^{\mu\nu}=j^\nu$ and $\partial_\mu j^\mu=0$? Do you see how that's analogous? – J.G. May 6 '18 at 21:29
• I see the analogy, but yes it kind of bothers me... – Daan May 7 '18 at 7:39

The energy-momentum tensor has to be generated by some matter content of the theory. Suppose the energy momentum tensor is generated by a (free, massless) scalar field $\phi$ satisfying
$$g^{\mu\nu} \nabla_\mu \nabla_\nu \phi = 0.$$
$$T_{\mu\nu} = \nabla_\mu \nabla_\nu \phi$$.
• Note the $6$ dofs in the second-order derivative are exactly what we need. – J.G. May 7 '18 at 8:22