There is quite simple formula that will give you the tidal-locking half-time
$$T = C, frac{a^6 R mu}{ M_s M_p^2}$$
- $a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters
- $R$ - planet radius in meters
- $M_s$ - planet mass in kg
- $M_p$ - parent planet/star mass in kg
- $mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.
- $C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. (See the discussion.)
Solving the problem of interaction between the parent star and between the planets will be difficult. Generally, if tidal-locking half-time between the planets is much shorter than the tidal-locking half-time between the star and the planet, the planets will tidal-lock to each other rather than to the star. So we ask, whether
$$ frac{A^6}{ M_s^3} ll frac{a^6}{ M_s M_p^2};,$$
where $A$ is the semi-major axis of the orbit of the double-planet. For typical planet in the habitable zone of the red dwarf star, $a$ = 9,000,000,000 m (0.06 AU), while the distance of the double-planet could be for example $A$ = 30,000,000 m. $M_s = 6times 10^{24};mathrm{kg}$ and $M_p = 2.9times 10^{29};mathrm{kg}$. We get
$$2cdot 10^{-41}ll 6.3cdot 10^{-36};,$$
which is well-satisfied and the planets will tidal lock to each other rather than to the parent star. Were the distance between the planets $A$ = 300,000,000 m, the inequality would yield
$$2cdot 10^{-35}ll 6.3cdot 10^{-36};,$$
which is not satisfied and the tidal forces of the star would probably disrupt the double-planet. (The planets would not even be in the Hill sphere, which approximately gives the stable orbits.)
So the answer is: a very close double planets can tidal-lock to each other rather than to the parent red-dwarf star, but they would have to be much closer than the Earth to the Moon.
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