Gravitational wave or "pure gauge"? Given the following linearized metric tensor ($c=1$)
$$h_{\mu\nu}=\begin{bmatrix}
    0 & f(t-z) & 0 & 0 \\
    f(t-z) & 0 & 0 & -f(t-z) \\
    0 & 0 & 0 & 0 \\
    0 & -f(t-z) & 0 & 0 \\
\end{bmatrix},$$
how can one know whether it describes a gravitational wave or a "pure gauge"?
 A: Consider a linearised perturbation of Minkowski spacetime
$$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$
where $|h_{\mu\nu}|\ll1$. 
Consider a diffeomorphism $\phi_{t}$ generated by a vector field $X$:
$$(\phi_{-t})_{*}(g)=g+\mathcal{L}_{X}g+\ldots=\eta+h+\mathcal{L}_{X}\eta+\ldots$$
where the ellipsis denotes higher order quantitites. From this you deduce that $h$ and $h+\mathcal{L}_{X}\eta$ describe physically equivalent spacetimes. In other words, linearised GR has the gauge symmetry $$h_{\mu\nu}\rightarrow h_{\mu\nu}+\partial_{\mu}X_{\nu}+\partial_{\nu}X_{\mu}.$$
The question you are asking is whether $h_{\mu\nu}$ can be set to zero by a gauge transformation, i.e. if there exists an $X$ such that
\begin{align}
h_{01}+\partial_{0}X_{1}+\partial_{1}X_{0} &=0\\
h_{13}+\partial_{1}X_{3}+\partial_{3}X_{1} &=0\\
\partial_{2}X_{3}+\partial_{3}X_{2} &=0\\
\partial_{0}X_{3}+\partial_{3}X_{0} &=0\\
{\rm etc...}
\end{align}
If such an $X$ exists, then $h$ is pure gauge.
In fact, you can show that 
$$X=-F(t-z)\frac{\partial}{\partial x}$$
where
$$ F'(Z)=f(Z)$$
is a solution to the above.
