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Much spectrum analysis on many chemical compounds are mapped and categorized at Earth atmosphere (1000 millibar) or thereabouts.

But many celestial objects are of greater ranges of surface pressure.

And many chemical compounds reach supercriticality at higher pressure.

One of those chemical compound is water molecule. Water molecule exhibit a different ro-vibrational in the diffraction spectral analysis at deeper pressure.

Are we compensating and broadening our their search parameters for super-criticality of such chemical compounds when doing this passive spectral analysis of and for their search of exoplanets?

Was reading this paper on “Mid-IR spectroscopy of supercritical water: from diluted gas to dense fluid”.

https://bpb-us-w2.wpmucdn.com/voices.uchicago.edu/dist/2/2305/files/2019/11/Hestand-et-al.-2019-Mid-IR-spectroscopy-of-supercritical-water-From-d.pdf

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  • $\begingroup$ I believe that the shown differences discussed in the linked paper are rather minute compared to the available accuracy of exoplanet spectra. $\endgroup$ Commented Apr 17, 2023 at 10:59
  • $\begingroup$ “shown differences” or “more diffused” (fuzzier)? $\endgroup$ Commented Apr 17, 2023 at 11:11
  • $\begingroup$ More details. webbtelescope.org/contents/articles/… $\endgroup$ Commented May 3, 2023 at 19:54
  • $\begingroup$ More on biomarkers on exoplanets. quantamagazine.org/… $\endgroup$ Commented Mar 25 at 14:52

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You wouldn't see the supercritical water: Typical solar-like molecular gas mixtures become optically thick at around 0.1-1 bars, so that's where the emission is going to come from, i.e. where most lines are going to be formed.

When you see water at this pressure you don't know how much sits below, i.e. you can't really determine the total amount of water in the atmosphere, or the total atmospheric mass.

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  • $\begingroup$ right, forgot that we cannot see rock composition either of exoplants, just the part of gaseous form surrounding such exoplanets. So, hyperpressurized chemical component should not be viewable (yet) by spectral analysis. Accepting this as an answer. $\endgroup$ Commented Apr 18, 2023 at 17:27

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