# What are the properties of metric tensor? [duplicate]

It's frequently said that graviton has spin-2, so its wave function should have $$5$$ independent components. The metric tensor has $$n^2=16$$ components, but it obeys the following property:

$$$$g_{\mu\nu}=g_{\nu\mu}$$$$

So we have only $$\frac{n(n+1)}{2}=10$$ components

What are other properties and constraints in metric tensor that reduce number of independent components to $$5$$?

• "It's frequently said that graviton has spin-2, so its wave function should have 5 independent components." [citation needed], see also this question about why photons don't have 3 spin states even though they are "spin-1" Aug 11, 2022 at 15:54
• Possible duplicates: physics.stackexchange.com/q/68824/2451 and links therein. Aug 11, 2022 at 17:07

The graviton only has two degrees of freedom, which are essentially its two helicity states (like the photon).

Your first step is correct; 6 components are not independent due to the metric being symmetric. However, on top of that you have the fact that the metric is coordinate independent i.e. when you perform a general coordinate transformation:

$$$$g_{\mu \nu} = g_{\alpha \beta} \frac{\partial x^{\alpha} }{\partial x^{\mu}} \frac{\partial x^{\beta}}{\partial x^{\nu}}$$$$

you might end up with some off-diagonal elements in one coordinate basis which might not have existed before. These extra non-zero components are therefore unphysical/not true degrees of freedom, and you get one such contribution for each spacetime dimension. In total, there are four further not independent components.

Finally, notice that any metric satisfies the Bianchi identities:

$$$$R_{\mu \nu \rho \sigma \, ; \lambda} + R_{\mu \nu \lambda \rho \, ; \sigma} + R_{\mu \nu \sigma \lambda \, ; \rho} = 0$$$$

where the semicolon indicates the covariant derivative:

$$$$R_{\mu \nu \rho \sigma \, ; \lambda} = \nabla _{\lambda}R_{\mu \nu \rho \sigma}$$$$

In the contracted form, one may rewrite them as:

$$$$R^{\mu \nu}{}_{;\mu} - \frac{1}{2} g^{\mu\nu} \, R _{;\mu} = 0$$$$

where we implicitly used that the covariant derivative of the metric tensor trivially vanishes. Notice how this is none other than the covariant derivative of the lhs of the Einstein field equations in the absence of a cosmological constant:

$$$$R^{\mu \nu} - \frac{1}{2} g^{\mu \nu} \, R = 8\pi G \, T^{\mu \nu}$$$$

So the contracted Bianchi identities are equivalent with the conservation of energy and momentum:

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

The result is that some of the Einstein equations act as constraints. These are further four implicit constraints on the metric tensor, meaning four less independent components. Summing it all up, you end up with just two degrees of freedom.