General relativity: gauge fixing

In his lectures professor Hamber said that the metric tensor is not unique, just like the 4 vector potential is not unique for a unique field in electrodynamics. Since the metric tensor is symmetric, only ten components of the metric tensor are unique.

However, as the covariant divergence of the Einstein's tensor is zero, 4 more constraints are imposed and hence the number of independent components of metric tensor now has come down to 6. Finally he says that only two are unique.

How did he arrive at the final result of 2 unique components of metric tensor. Can you please explain tis me ? Also, what is the physical difference between Ricci tensor and Reimann tensor ?

As in the case of Electromagnetism, in General Relativity you can perform two gauge transformations (diffeomorphisms). The first one being $x'^\mu(x)=\Lambda^\mu(x)$ you have four gauge functions which fix the gauge, for example the harmonic one $$\Gamma^\mu=\partial_\alpha(\sqrt{g}g^{\alpha\mu})=0.$$ Now it can be shown that exist a residual gauge transformation such that the latter gauge fixing still hold, for example, one for which $g^{\mu\nu}\partial_\mu\partial_\nu\Lambda^\alpha=0$. And so $$10-4-4=2$$ that is the number of degrees of freedom of the metric tensor.
• Think about classic EM where one perform $A_\mu\mapsto A_\mu+\partial_\mu\Lambda$ and if you fix the gauge (Lorentz for example) you still have again a gauge transformation freedom. – yngabl Sep 9 '14 at 11:23