What does $g_{tt}=0$ in the metric tensor mean? For example, in the Einstein-Rosen bridge metric, $g_{44}$ vanishes but it is pointed out that it is not a singularity.
 A: As long as the metric is non-singular (i.e., its determinant is nonzero), all it means for a diagonal component to be zero is that the corresponding coordinate vector is a null vector.  The closest that you can get to extracting any physical meaning from that is this: there exists at least one null vector in your spacetime.  That's a pretty trivial result, though.  This is because the metric relates physical things like distances to unphysical things like the units you use to measure those distances, or the particular basis you've chosen for a vector space — which means that no particular component of the metric has any physical significance.
[Note: I'll use the OP's apparent convention of the $t$ component being element 4 (starting from 1), as opposed to the usual element 0 (starting from 0).]
Even in Minkowski space with coordinates $(x, y, z, t)$ and metric
\begin{equation}
  \begin{pmatrix}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & -1
  \end{pmatrix},
\end{equation}
you can transform the coordinates by defining $u = (t-z)/\sqrt{2}$ and $v = (t+z)/\sqrt{2}$.  Vectors in the $u$ and $v$ direction are just along the usual null cone, and are null vectors.  But you can now transform the metric to the $(x, y, u, v)$ coordinate system, and you get
\begin{equation}
  \begin{pmatrix}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 0 & -1 \\
    0 & 0 & -1 & 0
  \end{pmatrix}.
\end{equation}
The determinant here is nonzero.  In fact, the determinant is -1, just like the usual Minkowski metric — though any negative number could be physically identical.  And since the $g_{uu}$ and $g_{vv}$ components are 0, we explicitly see that vectors in the $u$ and $v$ directions are null.
But we can still reconstruct the original $t$ and $z$ vectors as $t = (v+u) / \sqrt{2}$ and $z = (v-u) / \sqrt{2}$, and contracting those with the metric still gives us $-1$ and $+1$, respectively.  So it's really the same physical content, just in a different coordinate system.
