<p>The only reference plane for a binary is that perpendicular to the line of sight. The angle between that and the orbital plane is the inclination. This is the same angle as that between the normal vectors of those planes: your <span class="math-container">${\bf N}$</span> and the line of sight.</p>
<p>All that matters then is the angle between <span class="math-container">${\bf N}$</span> and the line of sight. (i.e. The inclination angle is <span class="math-container">$\cos^{-1}({\bf N}\cdot {\bf u})$</span>, where <span class="math-container">${\bf N}$</span> and <span class="math-container">${\bf u}$</span> are unit vectors perpendicular to the orbital plane and along the line of sight respectively).</p>
<p>That is the definition of inclination. In general, if you change your line of sight then you change the inclination and the measured radial velocity.</p>

CC BY-SA 4.0
2024-11-21T23:25:04.133Z
Thank you. If I understand correctly, rotating in the original diagram in the linked question about the z-axis would change the angle with respect to the line of sight and thus the inclination. But isn't that a rotation by $\Omega$, which is a separate orbital element? How do I reconcile the fact that rotating by a different orbital element affects the inclination?
theta
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2024-11-21T23:27:16.743Z
There is no z-axis in the linked question. @theta
ProfRob
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2024-11-21T23:43:50.003Z
by z-axis, I mean an axis vertically upward from CM @ProfRob
theta
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2024-11-22T00:12:15.060Z
So you mean change ${\bf N}$? If you do that, the inclination changes, which is defined as per my answer @theta If you rotate the binary around an arbitrary z axis  you will in general change $i$ and $\Omega$.
ProfRob
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2024-11-22T01:07:31.627Z
Okay, that makes sense to me. But aren’t inclinations usually defined with respect to a reference plane rather than a single line? E.g. the picture here (https://en.m.wikipedia.org/wiki/Orbital_inclination). Changing $\Omega$ here shouldn’t change the inclination, or should it? I believe you, I’m just trying to reconcile all the different pictures I’ve seen for inclinations. @ProfRob
theta
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2024-11-21T23:14:24.910Z
2024-11-22T07:11:33.883Z
2024-11-22T07:11:33.883Z
ProfRob
<p><a href="https://i.sstatic.net/oTfZBehA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oTfZBehA.png" alt="enter image description here" /></a>Let's say that I have some planetary orbit and some line of sight, such as that illustrated here: <a href="https://astronomy.stackexchange.com/questions/57475/relative-inclination-with-respect-to-line-of-sight">Relative inclination with respect to line of sight</a>. If I rotate the orbit azimuthally (about an axis that extends vertically upwards from CM), does the radial velocity signal change?</p>
<p>As a follow-up question, I want to think about this in the 3D sense (see my diagram). Let's say I have some planetary orbit in space, with some weird physics causing the vector <span class="math-container">$\vec{N}$</span> to be oriented anisotropically over the sphere in some fixed coordinate system. If I observe the system from isotropic lines of sight, do I observe a relative inclination distribution that is isotropic or anisotropic?</p>
<p>I think the answer depends on whether you consider the inclination the difference between polar angles, or the total 3D angle between your line of sight and the vector <span class="math-container">$\vec{N}$</span>. If the former, you see something anisotropic. If the latter, isotropic (since 3D angle is preserved). But I'm not sure which is correct for the case of measuring inclinations of planetary orbits with respect to lines of sight, or if this is the correct way to think about this problem. See the diagram attached with two different lines of sight that are at the same z-value, but have different x-y components. <span class="math-container">$\vec{N}$</span> is the in the y-z plane. Do these two lines of sight measure different inclinations? They have the same polar angle with respect to <span class="math-container">$\vec{N}$</span> but different 3D angles. Is the third Euler angle needed to successfully answer the question?</p>

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2024-11-21T00:17:34.530Z
what defines the azimuth for you in the linked diagramme? The observer's relative to their LOS? The planet's relative to its orbital plane? Also preferentially include the diagramme directly here as it seems central to your question.
planetmaker
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2024-11-21T00:26:31.073Z
Thank you, I have clarified this and added a diagram.
theta
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2024-11-21T22:35:23.677Z
@ProfRob First, I didn't mean that your statement was incorrect. I meant it was "correct, as the statement suggests" but I have fixed the grammar. I realize now that you are talking about rotations around the line of sight. I am talking about rotations around the azimuth (a line vertically upward from CM). Does the radial velocity measurement then change? Second, I am working with simulations with a distribution of $\vec{N}$ over time, so I am curious what an observer sees if I isotropically distribute those simulations. This would be equivalent to observing one system isotropically.
theta
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2024-11-21T22:37:00.683Z
This is my specific question: Pretend just the single vector $\vec{N}$ is oriented exactly as in my diagram. And for my hypothetical, pretend there are two telescopes on two different planets observing the vector from the lines of sight that I have drawn. Do they see different radial velocity signals? I.e. is the inclination different because they have different 3D angles with respect to $\vec{N}$ or is it the same because $\vec{N}$ has the same inclination with respect to the x-y plane? If I rotate $\vec{N}$ about the z-axis, does the radial velocity that either planet would measure change?
theta
CC BY-SA 4.0
2024-11-21T23:14:14.740Z
My apologies then, I misunderstood. My question still stands.
theta
<orbit><orbital-elements><inclination>
Relative inclinations between line of sight and orbit in 3D space
theta
# Relative inclinations between line of sight and orbit in 3D space

(Asked by: username_0 on 2024-11-21)

[![enter image description here](https://i.sstatic.net/oTfZBehA.png)](https://i.sstatic.net/oTfZBehA.png)Let's say that I have some planetary orbit and some line of sight, such as that illustrated here: [Relative inclination with respect to line of sight](https://astronomy.stackexchange.com/questions/57475/relative-inclination-with-respect-to-line-of-sight). If I rotate the orbit azimuthally (about an axis that extends vertically upwards from CM), does the radial velocity signal change?

As a follow-up question, I want to think about this in the 3D sense (see my diagram). Let's say I have some planetary orbit in space, with some weird physics causing the vector $\vec{N}$ to be oriented anisotropically over the sphere in some fixed coordinate system. If I observe the system from isotropic lines of sight, do I observe a relative inclination distribution that is isotropic or anisotropic?

I think the answer depends on whether you consider the inclination the difference between polar angles, or the total 3D angle between your line of sight and the vector $\vec{N}$. If the former, you see something anisotropic. If the latter, isotropic (since 3D angle is preserved). But I'm not sure which is correct for the case of measuring inclinations of planetary orbits with respect to lines of sight, or if this is the correct way to think about this problem. See the diagram attached with two different lines of sight that are at the same z-value, but have different x-y components. $\vec{N}$ is the in the y-z plane. Do these two lines of sight measure different inclinations? They have the same polar angle with respect to $\vec{N}$ but different 3D angles. Is the third Euler angle needed to successfully answer the question?

**Tags:** orbit, orbital-elements, inclination

> **Comments:**
> **username_1**: what defines the azimuth for you in the linked diagramme? The observer's relative to their LOS? The planet's relative to its orbital plane? Also preferentially include the diagramme directly here as it seems central to your question.  
> **username_0**: Thank you, I have clarified this and added a diagram.  
> **username_0**: @username_2 First, I didn't mean that your statement was incorrect. I meant it was "correct, as the statement suggests" but I have fixed the grammar. I realize now that you are talking about rotations around the line of sight. I am talking about rotations around the azimuth (a line vertically upward from CM). Does the radial velocity measurement then change? Second, I am working with simulations with a distribution of $\vec{N}$ over time, so I am curious what an observer sees if I isotropically distribute those simulations. This would be equivalent to observing one system isotropically.  
> **username_0**: This is my specific question: Pretend just the single vector $\vec{N}$ is oriented exactly as in my diagram. And for my hypothetical, pretend there are two telescopes on two different planets observing the vector from the lines of sight that I have drawn. Do they see different radial velocity signals? I.e. is the inclination different because they have different 3D angles with respect to $\vec{N}$ or is it the same because $\vec{N}$ has the same inclination with respect to the x-y plane? If I rotate $\vec{N}$ about the z-axis, does the radial velocity that either planet would measure change?  
> **username_0**: My apologies then, I misunderstood. My question still stands.  

---

## 1 answers

### Answer by username_2 on 2024-11-21. Score: 2 | Accepted answer

The only reference plane for a binary is that perpendicular to the line of sight. The angle between that and the orbital plane is the inclination. This is the same angle as that between the normal vectors of those planes: your ${\bf N}$ and the line of sight.

All that matters then is the angle between ${\bf N}$ and the line of sight. (i.e. The inclination angle is $\cos^{-1}({\bf N}\cdot {\bf u})$, where ${\bf N}$ and ${\bf u}$ are unit vectors perpendicular to the orbital plane and along the line of sight respectively).

That is the definition of inclination. In general, if you change your line of sight then you change the inclination and the measured radial velocity.

> **Comments:**
> **username_0**: Thank you. If I understand correctly, rotating in the original diagram in the linked question about the z-axis would change the angle with respect to the line of sight and thus the inclination. But isn't that a rotation by $\Omega$, which is a separate orbital element? How do I reconcile the fact that rotating by a different orbital element affects the inclination?  
> **username_2**: There is no z-axis in the linked question. @username_0  
> **username_0**: by z-axis, I mean an axis vertically upward from CM @username_2  
> **username_2**: So you mean change ${\bf N}$? If you do that, the inclination changes, which is defined as per my answer @username_0 If you rotate the binary around an arbitrary z axis  you will in general change $i$ and $\Omega$.  
> **username_0**: Okay, that makes sense to me. But aren’t inclinations usually defined with respect to a reference plane rather than a single line? E.g. the picture here (https://en.m.wikipedia.org/wiki/Orbital_inclination). Changing $\Omega$ here shouldn’t change the inclination, or should it? I believe you, I’m just trying to reconcile all the different pictures I’ve seen for inclinations. @username_2  

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