Normalization of a four-velocity when constructing the metric from the EM tensor

Question

I've been following the MIT General Relativity Course. In lecture (14) the concept of linearized gravity is introduced. This is then demonstrated by showing how we can recieve the Newtonian limit from such a approach.

We assume we have the metric that can be locally written as

$$ g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$ where $\eta$ is the Minkowski metric and $h$ is a correction, we work only in the order $O(h^2)$. Using the correct gauge we recieve the equations $$ \Box \bar{h}_{\alpha\beta} = -16 \pi G \ T_{\alpha\beta}. $$ where $\bar{h}$ is the trace free part of $h$, i.e. $\bar{h}_{\alpha\beta}= h_{\alpha\beta}- \frac{1}{2}\eta_{\alpha\beta}h$.

Now the lecturer then says, if consider static non-relativistic perfect fluid solution, we have that $$T_{\mu \nu} = \rho u_\mu u_\nu, $$ since this is static the pertubation $h$ must be indepedent of the time parameter, we neglect the preassure part of $T$.

Now presumably if we consider the observer $u_\mu = \{1,0,0,0\}$ we get

$$ T_{00} = \rho $$

and all other terms zero, we then get the metric

$$ ds^2 = -(1+ 2 \Phi) dt^2 + (1- 2 \Phi) (dx^2 + dy^2 + dz^2), $$ with $\Phi$ being the Newtonian potential.

But I am really not convinced by this. How can we tell that a four-velocity is $u_\mu = \{1,0,0,0\}$? Such a four velocity is surely not normalized under the metric we receive later and thus should hold no physical meaning.

Answer 1

You are considering only weak fields, thus any corrections to $T_{00}=\rho$ due to normalization will be even weaker and will not be kept in the metric within linear approximation.

You can see this directly by adding normalization $u_\mu=(A,0,0,0)$, which will lead to the metric

$$ ds^2 = -(1+ 2 A^2\Phi) dt^2 + (1- 2 A^2\Phi) (dx^2 + dy^2 + dz^2), $$

From which you will get $$A^2=\frac{1}{4\Phi}\left[\sqrt{1+8\Phi}-1\right]=1-\Phi+o(\Phi^2)$$