<p>You need to divide 60,000 by the integral of this function from zero to infinity. That will then give you the <em>fraction</em> of stars with a velocity  (I think it is probably speed, not velocity?) between 0 and 1000.</p>
<p>The reason you need to do this is that the distribution function as you have given is given as a proportionality and does not have the correct normalisation constant at the front.</p>

CC BY-SA 4.0
2022-08-03T09:13:04.983Z
Thank you very much.
Elena Greg
CC BY-SA 4.0
2022-08-03T08:26:20.467Z
2022-08-03T08:26:20.467Z
ProfRob
<p>The fraction <span class="math-container">$f$</span> of stars with velocities <span class="math-container">$v$</span> between <span class="math-container">$v$</span> and <span class="math-container">$v + \Delta v$</span> is given by
<span class="math-container">\begin{equation}
f(v) \propto 4\pi v^2 e^{-\frac{m \phi(\vec{x}) + \frac{mv^2}{2}}{k_\mathrm{B}T}}
\end{equation}</span></p>
<p>What does it mean when we integrate this function, please? I integrated from 0 to 1000 and obtained about 60 000. Is it the total number of stars?</p>

<star-cluster>
integration of velocity distribution f(v)
Elena Greg
# integration of velocity distribution f(v)

(Asked by: username_0 on 2022-08-03)

The fraction $f$ of stars with velocities $v$ between $v$ and $v + \Delta v$ is given by
\begin{equation}
f(v) \propto 4\pi v^2 e^{-\frac{m \phi(\vec{x}) + \frac{mv^2}{2}}{k\_\mathrm{B}T}}
\end{equation}

What does it mean when we integrate this function, please? I integrated from 0 to 1000 and obtained about 60 000. Is it the total number of stars?

**Tags:** star-cluster

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## 1 answers

### Answer by username_1 on 2022-08-03. Score: 3 | Accepted answer

You need to divide 60,000 by the integral of this function from zero to infinity. That will then give you the *fraction* of stars with a velocity (I think it is probably speed, not velocity?) between 0 and 1000.

The reason you need to do this is that the distribution function as you have given is given as a proportionality and does not have the correct normalisation constant at the front.

> **Comments:**
> **username_0**: Thank you very much.  

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