Newtonian approximation of the metric tensor I was reading Dirac's General Theory of Relativity. In chapter 16, the Newtonian approximation, we start with

Let us consider a static gravitational field and refer it to a static coordinate system. The $g_{\mu\nu}$ are then constant in time, $\frac{\partial g_{\mu\nu}}{\partial x^0}=0$. Further, we must have
\begin{equation}
g_{m0}=0, \text{    }(m=1,2,3)
\end{equation}
This leads to
\begin{equation}
g^{m0}=0, \text{    }g^{00}=(g_{00})^{-1}
\end{equation}

How does having a static gravitational field lead to $g_{m0}=0$?
 A: A stationary spacetime $M$ is one which admits a complete timelike Killing vector field $\boldsymbol \xi = \xi^a \frac{\partial}{\partial x^a}$.  The integral curves of $\boldsymbol \xi$ define a flow which preserves the metric.  That is, if you define a map $\phi_t:M\rightarrow M$ which takes some point $p\in M$ and "flows" along the vector field $\boldsymbol \xi$ for a parameter distance $t$, then the metric doesn't change.  This is captured by the fact that the Lie derivative of the metric along $\boldsymbol \xi$ vanishes:
$$\mathcal L_{\boldsymbol \xi} \mathbf g = 0 \iff \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = 0$$
The implication of this can be seen by noting that if such a vector field exists, then in any neighborhood where $\boldsymbol \xi \neq 0$, we can choose a coordinate system in which we use the flow parameter $t$ as the temporal coordinate $x^0$. If we do so, then $\boldsymbol \xi = \frac{\partial}{\partial t}$ and the Lie derivative reduces to $(\mathcal L_{\boldsymbol \xi} \mathbf g)_{\mu\nu} =\partial_t g_{\mu\nu} = 0$, i.e. a spacetime is stationary if there exists a coordinate system in which none of the metric components depend on time.

A static spacetime is one which is stationary, but has the additional requirement that there exist a spacelike hypersurface $\Sigma$ which is everywhere orthogonal to the orbits of $\phi_t$.  In a neighborhood of such a surface, every point $p$ lies on an integral curve of $\mathbf \xi$ which intersects $\Sigma$ at a unique point $p_0$, and can therefore be labeled by coordinates $(t,x^1,x^2,x^3)$ where $t$ is the Killing flow parameter (which we set equal to $0$ at $p_0$) and $\{x^a\},a=1,2,3$ are arbitrarily chosen coordinates on $\Sigma$.
The requirement that $\Sigma$ be orthogonal to the orbits of $\phi_t$ means that for any vector $\mathbf v=v^a \frac{\partial}{\partial x^a},a=1,2,3$  (that is, any vector tangent to $\Sigma$), we have that $\mathbf g(\boldsymbol \xi, \mathbf v) = v^a g_{ta} = 0$ for any choice of $v^a, a=1,2,3$.  This immediately implies that $g_{ta}=0$, and therefore that
$$\mathbf g = -\alpha(x^1,x^2,x^3) dt^2 + \beta_{ab}(x^1,x^2,x^3) dx^a dx^b\qquad a,b=1,2,3$$

From a physical standpoint, one should interpret a stationary spacetime as one which possesses time-translation symmetry and a static spacetime as one which possesses both time-translation and time-reversal symmetries (in our judiciously-chosen coordinate system); note that $t\rightarrow -t$ leaves a stationary metric unchanged iff there are no $dt \ dx^a$ cross terms.
Roughly, stationary-but-not-static spacetimes are those for which, given any spacelike hypersurface $\Sigma$, the timelike Killing flow $\phi_t$ pushes points along $\Sigma$ as well as forward in time.  In other words, the infinitesimal time-translation map $\phi_{\delta t}$ induces a change in the spatial coordinates as well as the time coordinate.  This is manifested as frame-dragging, and is characteristic of spacetimes around rotating bodies - see e.g. the Kerr spacetime, which is stationary (no $t$-dependence in the metric components) but not static.
A: I think the key phrase is "and refer it to a static coordinate system."
It isn't so much that a static spacetime implies the metric takes a particular form, but that you can choose a coordinate system such that $g_{m0}=0$.
A good exercise would be to start with with a metric where every component was time independent, and $g_{m0}$ was nonzero, and explicitly construct a coordinate transformation which set the $g_{m0}$ to zero. Essentially this should amount to a kind of space-time dependent boost.
