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This article 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?

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$\sigma^2$ 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 $$v_{3D}=\sqrt{v^2+v^2+v^2}.$$

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  • $\begingroup$ Thank you for the clear explanation! $\endgroup$
    – Nownuri
    Jun 8, 2023 at 5:42

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