What is the Initial Mass Function problem and its solution?

[grahammtb]

Hi, I've got a solution to this problem but I don't know if it's correct. My lecturer hasn't given us any examples but I had a go and the answer seems fine. Here it is:

The IMF of a cluster of stars is: dN$$\propto$$m^-2.5dm
There are 5 stars in the cluster with mass greater than 10 solar masses.
What is the number of stars with mass greater than 2 solar masses?

I put in 5 for dN, and solved the integral from 10 to infinity, to get the proportionality constant: 237.
Now using the constant, I solved the integral from 2 to 10 to find N in this range of masses. Then I added the 5 stars which are more massive than 10 solar masses.
Final answer: 56 stars.

Seems plausible to me, but I have no way right now of checking the answer.
Thanks a lot for any help!

 

[Redbelly98]

Welcome to PF

I agree with your math, though I am not familiar with mass distributions of stars and will take your word for it that part is valid.

FYI, you could also do the integral from 2 to ∞, and that will include the 5 stars with m>10.

 

[grahammtb]

thank you for your opinion. I've been working away at a few different problems and I seem to have them sorted out now
I found some examples on my physics department website which helped.

I did agree with you about solving the integral from 2 to $$\infty$$ but I tried it and it doesn't give the right answer. I think this is since the question only says there are 5 stars above 10 solar masses, and so these could be very massive or only a little more than 10 solar masses. So I think the method is to integrate from 2 to 10 then simply add on the remaining 5 to give the correct answer

thanks again

 

[Redbelly98]

That's weird, I definitely get 56 using 2 to ∞ for the integral.

http://www.google.com/search?hl=en&q=237*(2^(-1.5))%2F1.5&btnG=Search

 

[grahammtb]

That's weird, I definitely get 56 using 2 to ∞ for the integral.

http://www.google.com/search?hl=en&q=237*(2^(-1.5))%2F1.5&btnG=Search

Oh, I probably made a silly mistake when I did the integral. At least it confirms my answer of 56