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maximum orbital frequency formula derivation #108

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toribonidie opened this issue Apr 25, 2023 · 4 comments
Open

maximum orbital frequency formula derivation #108

toribonidie opened this issue Apr 25, 2023 · 4 comments

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@toribonidie
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Hello! I am writing to you to ask for some clarification in your documentation on how you derive your expression for maximum orbital frequency. I am using this code to search for giant planets around red clump stars, and the minimum period I calculate is very different than what TLS algorithm is using to search for transits.

In your paper, you say that the most short- period (high-frequency) circumstellar orbit is given by

$$f_{max} = \frac{\sqrt{GM_{s}/(3R_{s}})^{3}}{2\pi}$$

where the $3R_{s}$ term is calculated from the Roche limit, assuming a pessimistic density of $\rho_{p} = 1 g/cm^{3}$. I do not see a derivation for this formula in your paper, but I assume you're using Kepler's 3rd law

$$f_{max} = 1/P = \frac{\sqrt{GM_{s}/(a})^{3}}{2\pi}$$

And you are using the roche limit to calculate the semi major axis. For a rigid transiting body, the roche limit is given by,

$$a = R_{s} (2 \frac{\rho_{s}}{\rho_{p}})^{1/3}.$$

Or, for a fluid transiting body, the roche limit is given by,

$$a \approx 2.44R_{s} (\frac{\rho_{s}}{\rho_{p}})^{1/3}.$$

It says that you are assuming $\rho_{p} = 1 g/cm^{3}$, but I do not see any assumption about $\rho_{s}$. Now, if use this assumption of planet density and I plug in my value for a typical density of a RC star (0.002 $g/cm^{3}$), I get

$$a^{3} = 0.004 R_{s}^{3}$$

for a rigid body and

$$a^{3} = 0.03 R_{s}^{3}$$

These are both extremely different from the value than the $a^{3} = (3R_{s})^3$ from your paper. Are you inputing a typical density for a solar mass star (~1.4 $g/cm^{3}$ for the sun)? Because that would produce an $a^{3}$ term that is roughly $(3R_{s})^3$.

I don't see anything in the paper that makes this assumption, so I could be misinterpreting something, but this formula is restricting the period range that TLS searches and I believe shorter periods are physically possible.

Please let me know if this makes sense. Thank you so much!

@hippke
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hippke commented Apr 26, 2023

Hi @toribonidie, thank you for your question. This seems less like a code question, assuming the code replicates the paper correctly. Instead, I believe your question is better answered with physical reasoning. I'm a little tight on time at the moment, but I'll try to ping my co-author René for an answer.

@toribonidie
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Hi @hippke ! Thank you for getting back to me. Please take your time to respond. I am able to override the default period grid that TLS provides while performing my searches, but I just wanted to make sure I'm interpreting this correctly so that I am not searching for periods that are physically impossible.

So, no rush! But I look forward to hearing back from you or René for clarification.

@reneheller
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Hello @toribonidie! Your calculations look alright to me. Our formulae have been adopted from Ofir (2014), see his Sect. 3.1 therein for a few more details. In brief, yes, fmax is derived from Kepler's 3rd Law as you illustrate. The factor of 3 Rs comes from the division of the stellar density by the planetary density and then taking the third root. For a fluid-like Sun-like star and very low-density planet, we have 2.44 × (ρs/ρp)1/3 ~ 2.44 × (1.4 g cm-3 / 1 g cm-3)1/3 ~ 2.44 × 1.12 ~ 2.73 ~ 3.

I checked our paper and you're right, we should have explicitly mentioned that we assumed a solar-type star. In fact, for low-mass stars the density is higher than for solar-type stars and the Roche limit can move out to ~ 10 Rs. Michael (@hippke), maybe this is something that we should note in the TLS documentation for users to be aware that it might make sense to adjust the Roche limit and therefore fmax?

Tori, for your particular case of an extremely low-density star you end up with a Roche radius inside the star, did you notice? a3 between 0.004 Rs3 and 0.03 Rs3 is equivalent to a between 0.16 Rs and 0.3 Rs. I would expect any planet at that "distance" to the star to lose angular momentum and fall into the stellar core within < 1 million years. What do you think?

@toribonidie
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Hi René!

Thank you so much for getting back to me and clarifying. I think adding that to the documentation might be useful in the future since I know of other people using this algorithm to look for planets around hot subdwarfs and m dwarfs, which would both have very different Roche radii.

And yes, for my case, since the Roche radius is inside the star, I am just setting the minimum semi major axis to be $a = R_{star}$, so my period grid now spans from 6-32 days.

Thank you again!

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