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The scalar product of fields in curved space-time is defined by (Birrel, Davies)

$$\left(\phi_{1}, \phi_{2}\right)\equiv-\mathrm{i} \int_{\Sigma} \phi_{1}(x) \overset{\leftrightarrow} {\partial_\mu}\phi_{2}^{*}(x)\left[-g_{\Sigma}(x)\right]^{\frac{1}{2}} \mathrm{~d} \Sigma^{\mu}$$

where $\mathrm{d} \Sigma^{\mu}=n^{\mu} \mathrm{d} \Sigma$, with $n^\mu$ a future-directed unit vector orthogonal to the spacelike hypersurface $\Sigma$ and $\mathrm{d}\Sigma$ is the volume element in $\Sigma$. If $\Sigma$ is taken to be a Cachy surface in the globally hyperbolic spacetime, then the scalar product is independent of $\Sigma$.

I don't know how to apply this formula to a particular example. For instance, consider a Schwarzschild spacetime. How will this formula look like when I choose $\Sigma$ to be the past light-like infinity (of course, one can choose from many coordinate systems)?

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  • $\begingroup$ Light-like infinity is not a spacelike surface. But anyway, you do the same as with any surface integral: you parametrize with some coordinates. $\endgroup$
    – Javier
    Dec 26, 2020 at 20:26

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First of all that is not a scalar product but it is a symplectic form.

Yes you can use the lightlike infinity provided

(a) no information escapes through the timelike infinity,

(b) you have rewritten that integral in the language of differential forms ,

(c) the field is massless (otherwise it vanishes too fast before reaching the past light infinity).

I used lots of times that mathematical technology in the past. See for instance this couple of papers https://arxiv.org/abs/gr-qc/0512049 and https://arxiv.org/abs/gr-qc/0610143 I wrote in the past (around Eq.(47) in the former there is an explicit discussion about the expression of the symplectic form in terms of differential forms) and the rigorous construction of the Unruh state in the Kruskal-Schwarzshild manifold I obtained in collaboration with two colleagues https://arxiv.org/abs/0907.1034. Also this brief monography https://doi.org/10.1007/978-3-319-64343-4 may be useful. In these papers you can find the explicit version, in terms of forms, of that symplectic form in a manner that can be used for light like 3-surfaces.

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  • $\begingroup$ Doesn't a null surface have the problem that the metric determinant is zero? $\endgroup$
    – Javier
    Dec 27, 2020 at 13:56
  • $\begingroup$ No, using the volume form that is not a problem: the form remains regular even there. You can find the formula in the first paper I quoted. Now I am using my phone and it is difficult to write complicated formulae. $\endgroup$ Dec 27, 2020 at 13:59
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    $\begingroup$ Sorry, the formula is here instead arxiv.org/abs/gr-qc/0512049, see the discussion around formula (47).. $\endgroup$ Dec 27, 2020 at 14:05
  • $\begingroup$ This might be a very basic question, but if you are on null infinity and set $\Omega = 0$, shouldn't we then have that $d\Omega = 0$, and the metric has a zero eigenvalue? $\endgroup$
    – Javier
    Dec 27, 2020 at 14:30
  • $\begingroup$ No, $d\Omega\neq 0$, because $\Omega$ vanishes exactly at the null infinity which is a 3d submanifold, whereas it does not vanish in a neighborhood of that submanifold and the differential sees all directions. $\endgroup$ Dec 27, 2020 at 15:04

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