3
$\begingroup$

The formula of the radial velocity is:

$$v_i= V - K(\sin(f_i+\omega)\ + e\sin\omega),$$ with $$K = \frac{m_p}{m_s+m_p}\frac{na\sin i}{\sqrt{1-e^2}}.$$

Being $V$the systematic velocity of the system, $m_p$ the mass of the planet, $m_s$ the mass of the star, $n = \frac{2\pi}{P}$ the mean motion and $P$ is orbital period of planet, $a$ the length of the semi-major axis of the planet, $i$ the inclination of the orbital plane with the ecliptic, $e$ the eccentricity of the planet, $f_i$ the true anomaly at time $t_i$ and $\omega$ the longitude of periastron.

It's well known the usual procedure. You have data about the instances $t_i$, the radial velocities $v_i$ at time $t_i$ and then you also have the measurements uncertainties $\sigma_i$. With that data, you can estimate 6 parameters: $V$, $K$, $e$, $\omega$, $T$ (the periastron passage time) and $P$. After you estimate these parameters, you can calculate (if you know the mass of the star and you assume that $m_p<<m_s$), $m_p\sin i$, which is known as the minimum mass. So, as far as this way shows, you got no way to estimate the mass of the planet and the inclination separately.

But what if instead of estimate this set of parameters, we introduce another one? Just one with all the parameters in the equation: $V$, $m_p$, $i$, $e$, $T$, $\omega$, etc.? Couldn't that be possible? But nonetheless, it's always assumed that it's impossible to know the inclination with the RV data. Is there some reason why?

$\endgroup$
3
$\begingroup$

Essentially the problem comes down to geometry: radial velocity provides a 1-dimensional view of a 3-dimensional system. See the diagram below:

RV diagram

The observer sees only the component of the velocity vectors along the line from the observer to the star. That means that they cannot distinguish between the two blue velocity vectors: they will see only the purple (radial) component. This means a larger velocity vector that is tilted further away from the line-of-sight can masquerade as a smaller vector in the line of sight.

No matter how you define your variables, you will always run into this issue because it is inherent to the geometry of the situation.

This does break down in the case of planet-planet interactions, which can break the degeneracy here (a good example of this is Gliese 876).

$\endgroup$
  • $\begingroup$ Yes, you are partially true, but there is a problem with that: that only happens in the radial velocity method. Even when the line of the sight is what it is, with the transit method is possible to estimate the inclination. Besides, the geometry is not a problem once you have the model and the parameters. You have n parameters to fit m data. Why can't you do that? Treat i as a parameter like the others? Because if what you are saying was a problem in itself, you would also have that problem with the transit method, I mean, you're watching that eclipse because you are in a certain perspective. $\endgroup$ – Carlos Vázquez Monzón Dec 29 '18 at 18:14
  • 1
    $\begingroup$ You asked "why is it impossible to estimate the inclination of a planet with the radial velocity method?", so I don't know why you are bringing transits into this - transits are a completely different method of detection to RV. When you have radial velocities, the measurements are only measuring the system in 1D and are completely insensitive to transverse velocities. It doesn't matter how many data points you have. $\endgroup$ – antispinwards Dec 29 '18 at 18:28
2
$\begingroup$

I am not sure what you meant by "introduce another one." The system of equations is derived simply from Newtonian mechanics, and it is already completed in itself. So, my guess answer is, no that is not possible.

I think Figure 2 here is a good one to have in mind when thinking about this.

The nature of RV data is that you have a bunch of data points specifying (vi,ti). From there, you can get P straightforwardly. There was a post I saw recently asking how to get P from RV data, you might want to check it out. Or, simply google.

V is something that is observable or known because it is just a Hubble flow.

Let me express the equation like this: $v_i = V - K * f(t_i,P(\cdot))$ where $P(\cdot)$ is the period which is a function of some parameters that is irrelevant here. Since we know $(t_i,P)$, $f(\cdot)$ is known, regardless of $K$ because it is just a multiplicative constant. However, we know $(v_i,V,f(\cdot))$, we also know $K$.

Since we want $i$, we actually have to break the degeneracy in $K = K(m_p,m_s,e,P,a,i)$. We can get $e$ from $f(\cdot) = f(t_i,e,\omega) = g(t_i,\omega) + h(e,\omega)$ by choosing $\omega$ that fits the equation because we can treat $h(\cdot)$ as a constant. Once we get $\omega$, we resolve $h(\cdot)$ and get $e$.

So far from $K(m_p,m_s,e,P,a,i)$, we know $e,P$. I am not sure how to estimate $a$ but it does not matter because the leftover degeneracy between $m_p, i$ at the nominator of $K$ cannot be broken unless you can estimate one of the two (given that $m_s$ is known as you mentioned).

That is the reason why you can only do the minimum mass from RV data.

$\endgroup$
  • $\begingroup$ The thing is: I'm not trying to change the system of equations, just the set of parameters. What I'm saying is that: forgetting about K. In the denominator, by the way, what is done is Mp + Ms = Ms, so that you could indeed break i. So, a is estimated once you have mp, ms and P, and in the usually case is like that too. $\endgroup$ – Carlos Vázquez Monzón Dec 29 '18 at 18:21
  • $\begingroup$ You cannot break $i$. That is your problem to begin with. Also, you don't know $m_p$, so from what you said then you cannot get $a$. I think I am not sure now what you are talking about. $\endgroup$ – Kornpob Bhirombhakdi Dec 29 '18 at 21:49
  • $\begingroup$ Well, of course you don't know $m_p$, first you got to estimate them. You estimate $m_p$, $P$ and $i$. In the case of $a$ and $n$ you just derive it. Also, I don't see how you can't break $i$ if you can estimate the rest of the parameters... $\endgroup$ – Carlos Vázquez Monzón Dec 29 '18 at 22:14
1
$\begingroup$

Let's assume we had a mathematical method that could separate the planet's mass from the inclination of the planet-star system.

Then we could derive the mass of the planet and the inclination independently from each other. In particular we would be able to determine the mass of the planet in the case of inclination $i = 0$.

But this cannot be true for the Radial Velocity Method since for inclination $i = 0$ ($sin~i = 0$) the planet's motion in the line of sight is $0$ and hence the radial velocity measurements give the same result ($0$) for all planet masses.

This shows that we cannot separate the planet's mass from the inclination with the Radial Velocity Method. As a corollary it follows that it is impossible to determine the inclination of an exoplanet with the Radial Velocity Method.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.