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Integral manipulation

  1. Apr 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume the Milky Way contains about 3 × 10^11 stars that were formed all at once,
    with an initial mass function:

    [tex]\frac{dN}{dM}\propto M^{-2.35}[/tex]

    in the range 0.1–100 solar masses.

    How many stars in the Galaxy are less massive than the Sun? How much mass
    do these stars make up? What fraction of the total stellar mass is it?

    2. Relevant equations

    [tex]N_{total}= \int \frac{dN}{dM} dm[/tex], with an upper limit of 100 solar masses, and a lower limit of .1 solar masses.

    3. The attempt at a solution

    Since dN/dM is proportional to M^-2.35...

    Set the integral up as...

    [tex]3x10^{11}=\int CM^{-2.35} dm[/tex], with the same boundaries as before, 100, and .1.

    Solving for C, i got 1.81x10^10.

    Changing the boundaries to 1 solar mass and .1 to find how many stars have a mass lower than the sun i got..

    N= 1.81x10^10[-.741+16.6] = 2.86x10^11 stars

    Now where im having trouble is finding the mass fraction. The total mass would just be the sum of the mass of all the stars, but im having trouble figuring out how to manipulate the integral to give me the total mass. Any help is appreciated.
    Last edited: Apr 22, 2009
  2. jcsd
  3. Apr 23, 2009 #2
    Dear nissanztt90 , I wish I could do some help. Hints are given below in details, yet, you have to get the final results on your own. And please translate the standard tex codes yourself.

    The problem says that, the stars haver a continuous mass distribution ranking from $0.1m$ to $100m$, where the constant $m$ is the mass of sun. The distribution function of probability density is:
    q = k · M^{-2.35}
    where the constant $k$ is to be determined. Hence the probability for a star to carry a mass from $M$ to $M+dM$ is
    P(M)= \int k · M^{-2.35} dM = \frac{k}{-1.35} M^{-1.35} (Eq. 1)
    For $M \in [0.1m, 100m]$, the probability for a star carrying mass from $0.1m$ to $100m$ is 1. Hence
    [ \frac{k}{-1.35} M^{-1.35} ] _{0.1m} ^{100m} =1
    k · (100m)^{-1.35} - (0.1m)^{-1.35} = -1.35
    k=\frac{1.35}{ (0.1m)^{-1.35}-(100m)^{-1.35} } (Eq. 2)
    Now, we can get down to your three problems.----

    1. What fraction of the total stellar mass is it? "

    The fraction of the stars less massive than the Sun, is the probability of carrying mass from $0.1m$ to $m$. Just set the upper and lower limit of the integral in Eq.1 to be $m$ and $0.1m$, respectively.

    2. How many stars in the Galaxy are less massive than the Sun?

    As you have got the probability in subproblem 1, just multiply the total number of stars ($N=3 · 10^11$) with the probability.

    3. How much mass do these stars make up?

    As you have got the number of stars in subproblem 2, just multiply the number of stars with mass $M$ and their coreesponding mass $M$, i.e. calculate the integral (with $k$ in Eq.2)
    \int _{0.1m} ^{m} N · k · M^{-2.35} · M dM

    1. The first step towards the solution is constructing the probability density function for continuous distribution. I would like to draw your attention to Maxwell Speed Distribution of Gas Molecular for an inspiring comparison.

    2. It's interesting that we take the opposite order to solve the three subproblems.
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