<p><span class="math-container">$\sigma^2$</span> is the variance and the (general) definition of variance (for any set of data) is <span class="math-container">$$\sigma^2=\frac{\sum x^2}{N} - \bar x^2$$</span> That is the mean of the squares minus the square of the mean.</p>
<p>But if you have subtracted the systemic velocity from the data, then <span class="math-container">$\bar x$</span> is zero, so the variance in the radial velocity is the average of the squares of the residual velocities after the mean has been subtracted.</p>
<p>The fact that
<span class="math-container">$v^{2}_{3D} = 3v^{2}_{1D}$</span> is just an application of Pythagoras' theorem.  If a star (or group of stars) has residual velocity <span class="math-container">$v$</span> towards Earth, then assuming it has the same velocity in the two tangential directions, then by Pythagoras, it's 3d velocity relative to the system is <span class="math-container">$$v_{3D}=\sqrt{v^2+v^2+v^2}.$$</span></p>

CC BY-SA 4.0
2023-06-08T05:42:02.487Z
Thank you for the clear explanation!
Nownuri
CC BY-SA 4.0
2023-06-07T22:13:52.943Z
2023-06-07T22:13:52.943Z
James K
<p>This <a href="http://spiff.rit.edu/classes/phys440/lectures/glob_clus/glob_clus.html" rel="nofollow noreferrer">article</a> explains how to calculate the mass of a globular cluster. The first equation is
<span class="math-container">\begin{equation}
v^{2}_{1D} = \sigma^{2}_{r}
\end{equation}</span></p>
<p>Can someone please elaborate why this is the case? I get that you can measure the radial velocities of many stars and calculate the average their velocity-squared. So, I assume <span class="math-container">$v_{1D}^{2}$</span> means</p>
<p><span class="math-container">\begin{equation}
v_{1D}^{2} = \frac{1}{N}\sum_{i=1}^{N} v_{r,i}^{2} 
\end{equation}</span>
where <span class="math-container">$v_{r,i}$</span> is the magnitude of the radial velocity for star <span class="math-container">$i$</span>, <span class="math-container">$N$</span> is the number of stars (let's say <span class="math-container">$N = 10^{6}$</span>).</p>
<p>Also, if we define <span class="math-container">$\hat{x}, \hat{y}$</span> directions to be perpendicular to the radial directions (<span class="math-container">$\hat{r}$</span>), then
<span class="math-container">\begin{equation}
v_{1D}^{2} = \frac{1}{N}\sum_{i=1}^{N} v_{r,i}^{2}  = \frac{1}{N}\sum_{i=1}^{N} v_{x,i}^{2} = \frac{1}{N}\sum_{i=1}^{N} v_{y,i}^{2}.
\end{equation}</span></p>
<p><strong>Question 1. Then what is the physical meaning of this <span class="math-container">$\sigma_{r}$</span> and why is this equal to <span class="math-container">$v_{1D}^{2}$</span>?</strong></p>
<p>The second equation also puzzles me.
<span class="math-container">\begin{equation}
v^{2}_{3D} = 3v^{2}_{1D}
\end{equation}</span></p>
<p><strong>Question 2. What would be the definition of <span class="math-container">$v^{2}_{3D}$</span> and why does the equation hold?</strong></p>

<radial-velocity><globular-clusters>
Average velocity of stars in globular cluster
Nownuri
# Average velocity of stars in globular cluster

(Asked by: username_0 on 2023-06-07)

This [article](http://spiff.rit.edu/classes/phys440/lectures/glob_clus/glob_clus.html) explains how to calculate the mass of a globular cluster. The first equation is
\begin{equation}
v^{2}\_{1D} = \sigma^{2}\_{r}
\end{equation}

Can someone please elaborate why this is the case? I get that you can measure the radial velocities of many stars and calculate the average their velocity-squared. So, I assume $v\_{1D}^{2}$ means

\begin{equation}
v\_{1D}^{2} = \frac{1}{N}\sum\_{i=1}^{N} v\_{r,i}^{2}
\end{equation}
where $v\_{r,i}$ is the magnitude of the radial velocity for star $i$, $N$ is the number of stars (let's say $N = 10^{6}$).

Also, if we define $\hat{x}, \hat{y}$ directions to be perpendicular to the radial directions ($\hat{r}$), then
\begin{equation}
v\_{1D}^{2} = \frac{1}{N}\sum\_{i=1}^{N} v\_{r,i}^{2} = \frac{1}{N}\sum\_{i=1}^{N} v\_{x,i}^{2} = \frac{1}{N}\sum\_{i=1}^{N} v\_{y,i}^{2}.
\end{equation}

**Question 1. Then what is the physical meaning of this $\sigma\_{r}$ and why is this equal to $v\_{1D}^{2}$?**

The second equation also puzzles me.
\begin{equation}
v^{2}\_{3D} = 3v^{2}\_{1D}
\end{equation}

**Question 2. What would be the definition of $v^{2}\_{3D}$ and why does the equation hold?**

**Tags:** radial-velocity, globular-clusters

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## 1 answers

### Answer by username_1 on 2023-06-07. Score: 3 | Accepted answer

$\sigma^2$ is the variance and the (general) definition of variance (for any set of data) is $$\sigma^2=\frac{\sum x^2}{N} - \bar x^2$$ That is the mean of the squares minus the square of the mean.

But if you have subtracted the systemic velocity from the data, then $\bar x$ is zero, so the variance in the radial velocity is the average of the squares of the residual velocities after the mean has been subtracted.

The fact that
$v^{2}\_{3D} = 3v^{2}\_{1D}$ is just an application of Pythagoras' theorem. If a star (or group of stars) has residual velocity $v$ towards Earth, then assuming it has the same velocity in the two tangential directions, then by Pythagoras, it's 3d velocity relative to the system is $$v\_{3D}=\sqrt{v^2+v^2+v^2}.$$

> **Comments:**
> **username_0**: Thank you for the clear explanation!  

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