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I'm doing a past paper and this is the question I'm struggling with (it's a standalone question with no other information given outside of this screenshot):

enter image description here

To answer it, I used the Salpeter mass function $\psi$ to be (as I've been taught it):

$$\psi (m) = Cm^{-2.35} $$

where C is a normalising constant and m is mass.

Using this, I got 0.043, whereas the answer given is 0.05. To get to my answer, I ended up ignoring C because I'm not sure how to calculate it/what to do with it. I tried to use this to calculate it: $$ \int_{0}^{\infty} \psi (m) dm = 1$$

but I couldn't get to a solution.

Does anyone know how I would go about calculating C? Or am I going wrong and there's actually no need to calculate C for this question because it can be cancelled out in some way?

(This is a self-study exercise/past paper, not something I will be graded on. But I'm not sure if a full solution to any past paper question is allowed to be given, so even a hint as to how I could work it out would be so helpful.)

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The number of all stars with a mass between $m_1$ and $m_2$ will be$$\int_{m_1}^{m_2} \psi(m) \, dm.$$

The mass of all stars with a mass between $m_1$ and $m_2$ will be$$\int_{m_1}^{m_2} m \,\psi(m) \, dm.$$ Note the additional $m$ in the integral. This can be seen by noting that the mass of the stars in a range $[m, m+\Delta m]$ is approximately $m \psi(m) \Delta m$, summing over many "bins" of width $\Delta m$, and then taking the limit as the width of the bins goes to zero.

You are asked for the ratio of the number of stars in a particular range of masses to the total mass of the stars in that range. Can you see how to proceed from here?

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  • $\begingroup$ Ahh okay, I misunderstood what the question was asking for, thanks so much! $\endgroup$
    – user374355
    Commented Aug 21, 2023 at 12:44
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    $\begingroup$ Yeah, it's a little confusing at first. Particularly you can't actually calculate $C$ from the information given in the problem: any power-law distribution $x^{-\alpha}$ with $\alpha \geq 1$ can't be normalized without imposing a cutoff at some $x_\text{min} > 0$. $\endgroup$
    – Michael Seifert
    Commented Aug 21, 2023 at 13:38

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