<p>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.</p>

<p>I think Figure 2 <a href="https://www.physics.ncsu.edu/classes/astron/orbits.html" rel="nofollow noreferrer">here</a> is a good one to have in mind when thinking about this.</p>

<p>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.</p>

<p>V is something that is observable or known because it is just a Hubble flow.</p>

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

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

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

<p>That is the reason why you can only do the minimum mass from RV data.</p>

CC BY-SA 4.0
2018-12-29T18:21:15.490Z
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.
Carlos Vázquez Monzón
CC BY-SA 4.0
2018-12-29T21:49:59.220Z
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.
Kornpob Bhirombhakdi
CC BY-SA 4.0
2018-12-29T22:14:30.467Z
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...
Carlos Vázquez Monzón
CC BY-SA 4.0
2018-12-29T17:36:06.410Z
2018-12-29T17:44:10.450Z
2018-12-29T17:44:10.450Z
Kornpob Bhirombhakdi
<p>Let's assume we had a mathematical method that could separate the planet's mass from the inclination of the planet-star system.</p>

<p>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 <span class="math-container">$i = 0$</span>.</p>

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

<p>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.</p>

CC BY-SA 4.0
2018-12-31T11:46:18.657Z
2018-12-31T11:46:18.657Z
aventurin
<p>The formula of the radial velocity is:</p>

<p><span class="math-container">$$v_i= V - K(\sin(f_i+\omega)\ + e\sin\omega),$$</span> with
<span class="math-container">$$K = \frac{m_p}{m_s+m_p}\frac{na\sin i}{\sqrt{1-e^2}}.$$</span></p>

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

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

<p>But what if instead of estimate this set of parameters, we introduce another one? Just one with all the parameters in the equation: <span class="math-container">$V$</span>, <span class="math-container">$m_p$</span>, <span class="math-container">$i$</span>, <span class="math-container">$e$</span>, <span class="math-container">$T$</span>, <span class="math-container">$\omega$</span>, 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?</p>

<orbital-elements><radial-velocity>
Why is it impossible to estimate the inclination of an exoplanet with the Radial Velocity Method?
Carlos Vázquez Monzón
# Why is it impossible to estimate the inclination of an exoplanet with the Radial Velocity Method?

(Asked by: username_0 on 2018-12-29)

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?

**Tags:** orbital-elements, radial-velocity


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## 2 answers

### Answer by username_1 on 2018-12-29. Score: 3 | Accepted answer

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](https://www.physics.ncsu.edu/classes/astron/orbits.html) 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.

> **Comments:**
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: 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...  

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### Answer by username_2 on 2018-12-31. Score: 2

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.


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