Integrating a scalar function on a manifold So I have the following action in Minkowski spacetime $(M, \eta)$:
$
S[\phi] = \int \eta^{\alpha \beta}(\partial_{\alpha} \phi)(\partial_{\beta} \phi)\sqrt{-\eta}d^2x
$
Now, I have the following two charts $(x^{\alpha})$ and $(\xi^{\alpha})$ related as :
$
(x^{\alpha}(\xi^0,\xi^1)) = (t,x) = (\frac{1}{a}e^{a\xi^1}sinh(a\xi^0), \frac{1}{a}e^{a\xi^1}cosh(a\xi^0)). 
$
and also useful
$
ds^2=dt^2-dx^2=e^{2a\xi^1}((d\xi^0)^2-(d\xi^0)^2)
$
The two chart overlap only in the region of the manifold that would correspond to $x>|t|$ and $\xi^0,\xi^1 \in R$. If I recast the action for $(x^{\alpha})$ I would get
$
S[\phi] = \int_{x>|t|} ((\partial_{t} \phi)^2-(\partial_{x} \phi))d^2x
$
and for the other chart Im getting 
$
S[\phi] = \int_{R^2} ((\partial_{\xi^0} \phi)^2-(\partial_{\xi^1} \phi))d^2\xi.
$
I would argue I did something wrong as the two integrals should yield the same answer as we are integrating $\mathcal{L}$, which doesnt care about diffeos. On the otherhand Im following Mukhanov "Quantum effects in gravity" (yes this is the unruh radiation) and he gets the same two integrals but doesnt specify the integration limits. 
What am  I missing?
Thanks!
 A: I ended up finding the answer, I will leave the comment I wrote on my MSc. thesis for someone else to learn. If some experienced mathematician finds this answer incorrect let me know.
"Here is where the standard abuse of notation shines with malice. Expressions above naively suggest the actions are everything but equal: same integral but different integration bounds. Nothing further away from reality, we have been careless,  remember that the field $\phi: \mathcal{M} \rightarrow \mathbb{R}$, but $(\xi^{\alpha}), (x^{\alpha}) \notin \mathcal{M}$. Hence $\phi$ in expressions above really should be
\begin{equation}
    \left( \phi \circ \psi^{-1} \right) _{\psi(p)} \quad  \mathrm{and} \quad  \left( \phi \circ \theta^{-1} \right) _{\theta(p)}
\end{equation}
respectively, where $p \in U$, and $\psi$ and $\theta$ are the inertial and accelerated charts and $\psi^{-1}$ and $\theta^{-1}$ their inverses, i.e. $\psi : p \in U \mapsto (x^{\alpha})$. Thus it is the field itself that differs from the two integrals. We conclude $S_I = S_A$ if they integrate over the same region in $\mathcal{M}$ (which they do)."
