# Stellar mass function

1. Oct 3, 2011

### RHK

1. The problem statement, all variables and given/known data
Stars in a globular cluster are distributed as a mass function of: $\phi(M)=K M^{-2}$, such that $dN=\phi(M) dM$ is the stars number in the infinitesimal mass interval. Masses are between a lower limit $M_{inf}=0.3 M_{sun}$ and an upper limit $M_{sup}$, unknown.
The constant K is equal to $200 M_{Sun}$.
Let's assume also that the relation between the bolometric luminosity and the stars masses is $L(M)=L_{Sun}(\frac{M}{M_{Sun}})^{3.5}$. Requests:

(i) to show the necessity of the existence for the upper limit in the mass distribution, to avoid that the cluster mass diverge toward infinite.

(ii) to calculate the value of such limit so that the total mass of the cluster is $600 M_{Sun}$;

(iii) to calculate the cluster bolometric luminosity in solar units, and the corresponding absolute magnitude.

EDIT: the absolute bolometric magnitude for the Sun is given in the exercise text: $m_{sun}=4.75$

2. Relevant equations
$M_{TOT}=\int_{M_{inf}}^{M_ {sup}} M \phi(M) dM$

3. The attempt at a solution
$M_{TOT}=\int_{M_{inf}}^{M_ {sup}} K M^{-1} dM = K log M$

Is it the right way?
Can i have a hand please?

Last edited: Oct 3, 2011
2. Oct 3, 2011

### Jimmy Snyder

Your solution is a good start, but it has a problem. On the one hand, it looks like an indefinite integral because the limits of integration don't appear on the right hand side. On the other hand, if it were an indefinite integral, it would need a constant of integration to go along with it. The way to go here is definite integral because of the limits of integration in the middle term of your solution. Once you do that, I think the answer to part (i) will pop out at you.

3. Oct 3, 2011

### RHK

Yes, ok. I wrote a definite solution just to have some feedback...
I will do that and then i will post it to have other help for the other parts.
Thanks

4. Oct 3, 2011

### RHK

Well, the first two points are quite straightforward:
$M_{TOT}= K log \frac{M_s}{M_i}$,
so the existence of the upper limit is imposed from the otherwise divergent result.

Imposing $M_{TOT}= 600 M_{Sun}$, it's easy to find Msup=$300 M_{Sun}$, that is not so realistic.

But i can not proceed for the third point...

5. Oct 3, 2011

### Jimmy Snyder

You are right, it is not realistic. 3 or 4 such stars would be the whole cluster and that is no cluster. You are assuming that the log is base 10. However, the formula you are using for the integral uses a different base.

6. Oct 3, 2011

### RHK

That is right!
It was suspect, indeed.

Any suggest for the third point too?

7. Oct 3, 2011

### Jimmy Snyder

What work have you done so far on part (iii)?

8. Oct 3, 2011

### RHK

None: i can't handle it...

9. Oct 3, 2011

### Jimmy Snyder

It's a rule here. we're not allowed to help you until you take the first step. Surely you have something you can show us.

10. Oct 3, 2011

### RHK

Maybe: $L_{bol}=\int L(M) dM = \int L_{Sun} (\frac{M}{M_{Sun}})^{3.5} dM$

so the bolometric luminosity in solar unit is

$\frac{L_{bol}}{L_{Sun}}=\frac{1}{M_{Sun}^{3.5}} \frac{M^{4.5}}{4.5}$

where M is MTOT.
Is this plausible?

11. Oct 3, 2011

### Jimmy Snyder

I see a couple of problems with this solution. For one thing, you have not taken into account the number of stars having a given luminosity. This shouldn't be hard to do though since you know the luminosity in terms of mass and you sort of know how many stars are of a given mass. Another problem is that you have once again taken an indefinite integral (without the constant of integration) when what is wanted is a definite integral.

12. Oct 3, 2011

### RHK

If i understand your suggestion, the right integral is:

$L_{bol}=\int_{M_{inf}}^{M_{sup}} L(M) \phi(M) dM$

This make more plausible the solution, indeed.
But i wait further confirms to proceed with calculations.

13. Oct 3, 2011

### Jimmy Snyder

This is correct. It is instructive to compare this formula to the one given as a relevant equation in the OP.

14. Oct 3, 2011

### Jimmy Snyder

I am taking my son to basketball practice now. Perhaps someone else can pick up from here. It seems there isn't much left to do.

15. Oct 3, 2011

### RHK

Thanks a lot!

From the right integral it's possible to find the bolometric luminosity in solar unit.
Then the absolute magnitude is:

$m_{bol} = -2.5 \frac{L_{bol}}{L_{sun}} + m_{sun}$ where $m_{sun}$ is the absolute bolometric magnitude for the Sun, and it's given in the exercise text $m_{sun}=4.75$

Last edited: Oct 3, 2011
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