Interpretation of an anticorrelation between 𝐻0 and log10(𝜔𝐵𝐷)

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SUMMARY

The discussion centers on the anticorrelation between H0 and log10BD), derived from the equation ΩBD = (ωBD/6)(F0/H0)2 - (F0/H0). The correct formulation for H0 is established as H0 = (-F0 + √(F02 + (2ΩBDωBDF02/3))) / (2ΩBD), which confirms the anticorrelation when ωBD increases. The derivation of H0 was initially incorrect, leading to confusion regarding the expected correlation.

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I get below the following contours of a MCMC run with the main cosmological parameters for Brans-Dickce's theory without introducing a cosmological constant (##\Lambda=0##) and considering only baryonic matter component.
triplot_TCL_HDF5_REFERENCE.png


Could you justify the anticorrelation that I get between ##H_0## and ##\omega_{BD}## (actually ##\log10(\omega_{BD}##) ?

If we take the relation :

##\Omega_{B D}=\frac{\omega_{B D}}{6}\left(\frac{F_0}{H_0}\right)^2-\frac{F_0}{H_0} ##, then I can express ##H_0## as a function of ##\omega_{BD}## :

##H_0=\frac{-F_0+\sqrt{F_0^2+\frac{2 \Omega_{B D \omega_{B D} F_0^2}}{3}}}{2 \Omega_{B D}} .##

From this relation, we are expected to have a correlation instead of an anti-correlation since if ##\omega_{BD}## increases, then, ##H_0## will increase.

If someone could help me to justify my result (if it is true), this would be great.
 
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Solution : the derivation of ##H_0## is wrong and the valide one is :

##\begin{aligned}
& \Omega=\frac{\omega}{6}\left(\frac{F}{H}\right)^2-\frac{F}{H} \\
& \frac{E}{H}=A \\
& \Omega=\frac{\omega}{6} A^2-A \\
& \frac{\omega}{6} A^2-A-\Omega=0 \\
& A_{1 / 2}=\frac{1 \pm \sqrt{1+\frac{4 \omega \Omega}{6}}}{2}, A>0 \\
& \frac{F}{H}=\frac{1+\sqrt{1+\frac{4 \omega \Omega}{6}}}{2}=\frac{1}{2}+\sqrt{4+\frac{\omega \Omega}{6}} \\
& H=\frac{F}{\frac{1}{2}+\sqrt{4+\frac{\omega \Omega}{6}}}
\end{aligned}
##

So there is an anticorrelation between ##H_0## and ##\omega_{BD}##
 
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