Which mass stars dominate the volume of ionized natal star forming material?

[Piano man]

Homework Statement

In a particular star forming cloud the initial mass function (IMF) is given by
N(M_{*})=Cexp(-M_*^2)
where C is a normalisation constant. (The IMF describes the initial relative number of stars of different masses). Assuming that the number of hydrogen Lyman continuum photons created by stars of different mass, N_{Lyc}=10^{34}M_*^{32} which mass stars dominate the volume of ionized natal star forming material?

Homework Equations

I've found this equation while I was searching for a lead:

N=\int_{m_1}^{m_2} dN=\int_{m_1}^{m_2} \frac{dN}{dm}dm

but I'm not sure what values to use as integration limits.

Also, the problem sheet gives the following formula (including gamma and Riemann zeta functions) which is probably useful at some point, though I don't know where or how.

\int_0^\infty x^n\frac{1}{exp(x)-1}dx=\zeta(n+1)\Gamma(n+1)

The Attempt at a Solution

My first step was differentiating the expression I was given to get
\frac{dN}{dM}=-2M_*Cexp(-M_*^2)
which looks like something similar to the first equation, and also hints at the formula given in the problem sheet.
But I'm really not sure where to go from here, and how to incorporate the value given for N_{Lyc}.
Any help/pointers would be greatly appreciated.

 

[Piano man]

Looking at the expression I got for \frac{dN}{dM}, and assuming we're integrating from zero to infinity, the integral looks like
-2C \int_0^{\infty}M_{*}\frac{1}{exp(M_{*}^2)}dM_{*}
which looks quite like the formula given,
\int_0^\infty x^n\frac{1}{exp(x)-1}dx
except for the -1 on the bottom.

Any ideas as to how to move forward?

 

[gneill]

Hmm. Given a formula for the number of stars produced of a given mass M, and a formula for the number of (particular) photons produced by a star of a given mass M, what would be an expression for the total number of such photons produced by all stars of a given mass M?

 

[Piano man]

Well, I'd say that would be
(Total number of photons produced by all stars of given mass M) = (Number of photons produced by a star of mass M) x (Number of stars of mass M).
Or in symbols
N_{ph-tot}(M_*)=N_{Lyc}(M_*)*N(M_*)
And that would give
N_{ph-tot}(M_*)=10^{34}CM_*^{32}exp(-M_*^2)

Is that right?

If so, what's the next step, as it is still not in the form of the given integral?

 

[gneill]

Well, I'd say that would be
(Total number of photons produced by all stars of given mass M) = (Number of photons produced by a star of mass M) x (Number of stars of mass M).
Or in symbols
N_{ph-tot}(M_*)=N_{Lyc}(M_*)*N(M_*)
And that would give
N_{ph-tot}(M_*)=10^{34}CM_*^{32}exp(-M_*^2)

Is that right?

If so, what's the next step, as it is still not in the form of the given integral?

If that is an expression for the total number of photons as a function of M, I'd look for the M that gives that expression a maximum value... I wouldn't presuppose that the problem has to involve an integral that happens to appear on the problem sheet.

 

[Piano man]

Okay, so differentiating and setting to zero gives
\frac{dN}{dM}=10^{34}C\frac{d}{dM}\left[M^{32}exp(-M^2)\right]=0
giving
32M^{31}exp(-M^2)=2M^{33}exp(-M^2)
which simplifies to
M=4

So how do I interpret that? It only really makes sense if it's in units of Msun..

While that method seems to make sense (and thanks for your help) it's quite unusual for a formula to appear that isn't necessary for the problem.

Could there be an alternative method that requires the use of the given formula?

 

[gneill]

Okay, so differentiating and setting to zero gives
\frac{dN}{dM}=10^{34}C\frac{d}{dM}\left[M^{32}exp(-M^2)\right]=0
giving
32M^{31}exp(-M^2)=2M^{33}exp(-M^2)
which simplifies to
M=4

So how do I interpret that? It only really makes sense if it's in units of Msun..

Presumably the "normalisation constant", C, takes care of the mass units. I wouldn't be at all surprised if it were to be chosen to produce units of Msun

While that method seems to make sense (and thanks for your help) it's quite unusual for a formula to appear that isn't necessary for the problem.

Could there be an alternative method that requires the use of the given formula?

Nothing springs to mind.

 

[Piano man]

Thanks very much for your help :)