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Mass-luminosity relation problem

  1. Oct 16, 2009 #1
    I have to solve a question about figuring out the luminosity of a globular cluster using the given mass-luminosity relation L~M4 (where 4 is the power of M of course). But the problem here is that this relation does not apply to massive clusters but only to a certain mass limit. Would L~M would be the right relation in this case. So with respect to the suns luminosity (say l) and mass(m) we would have (we already know everything except L)

    L/l = M/m4 (as for sun l ~ m4)
    is this the right way to go? (I used this relation but still couldn't get the right answer)
  2. jcsd
  3. Oct 16, 2009 #2


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    Not sure I understand your notation, but usually in ASCII we write ^ (shift-6) for power.
    Anyway, if the relation is L~M^4, then L/l=(M/m)^4.
  4. Oct 26, 2009 #3
    i used the equation but still not getting the answer.
  5. Oct 26, 2009 #4


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    Any more details? (it's pretty hard to help you based in the limited info you've given). What answer are you expecting? What answer do you get (and how did you get it)?
  6. Oct 26, 2009 #5
    well, I' ll try to do my best.
    I have to find the luminosity of a globular cluster using the mass luminosity relationship (L ~ M^4) . The globular is supposed to be composed of main sequence stars (10^4 - 10^6 in number). The mass of this cluster is 10^6 times sun's mass.
    The equation should look something like this

    L/l = (M/m)^4

    l and m being sun's luminosity and mass respectively.

    the answer given is 2 X 10^8 times sun's luminosity (l)
  7. Oct 26, 2009 #6


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    In principle, to answer this question you need to assume a mass function for the stars (i.e. what fraction of the mass in the cluster exists in stars between a mass of M and M + dM for all M), along with the luminosity function ( the luminosity of an individual star of a given mass). You need to integrate over these functions to get the total luminosity of the cluster. This also therefore requires the assuming of cutoffs in the upper and lower mass for stars (in order to get the integration limits).

    What is the background to your question? If you are taking a course on this, do you recall encountering the mass function ( a popular mass function for clusters is the Saltpeter mass function)?
  8. Oct 26, 2009 #7
    this question is of four parts (this one is the second part), the first two parts are follows

    a. consider a newly formed globular cluster, with a total mass 10^6 times mass of the sun and an initial mass function dN/dm = a.M^-2.35 in the mass range 0.1 - 20 times sun's mass, where m = M/sun's mass.
    find the constant a. (answer - 1.9 X 10^5)

    b. find the total luminosity of the cluster, assuming that all its stars are on the main sequence, and a mass-luminosity relation L ~M ^4. What fraction of the luminosity contributed by the stars more massive than 5 X sun's mass. (answer - 2 X10^8times sun's luminosity and 0.98)

    could you explain this in a bit detail.. this is my first time.
  9. Oct 26, 2009 #8


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    There you go! The first question has given you the mass function and mass limits that you need. So, you need to first combine the mass function and the luminosity function together such that you know the contribution to the total luminosity (dL) for the mass range between M and M + dM, then integrate that over the mass limits given (0.1 - 20 times the sun's mass).

    The answer the second part, change the lower mass limit to 5 times solar and then compare that luminosity to the total.
  10. Oct 26, 2009 #9
    thank you very much .. i'll try to do that
  11. Oct 26, 2009 #10
    I am trying to figure it out but its all over my head right now. Could you please explain it in more detail as the first question is also a problem for me.
  12. Oct 26, 2009 #11


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    How good is your calculus? If you are not comfortable with integration then this question will be a little tricky for you, there's no way around that (except to brush up your calculus of course!). To start you off, look at the first question. The total mass of the cluster is given by the integration of the number of stars in a given dM of mass, multiplied by the mass of the stars in that range, e.g.

    [tex] M = \int_{M_{min}}^{M_{max}} \frac{dN}{dM} M dM [/tex]

    If you do this integration, substitute for the things you know (the mass function, the total mass, the mass limits) and then re-arrange you will find the constant a. I just did this quickly and actually found [tex] a = 1.6 \times 10^5 [/tex] but I did it quickly and might have made a mistake somewhere.
  13. Oct 26, 2009 #12
    so if i plug in the values

    M = 10^6 sun's mass

    dN/dM = a.m^-2.35

    M(max) = 20
    M(min) = 0.1

    what about the other M in the integral is it the total mass of the cluster ?
  14. Oct 26, 2009 #13


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    Sorry, yes I was a little sloppy in my notation. Here is a clearer version, with substitutions already made

    [tex] 10^6 m = a\int_{0.1m}^{20m}M^{-2.35}MdM [/tex]

    where m=solar mass.

    Note that the first step is to combine the extra factor of M with the one from the luminosity function, and then integrate the result. The extra M factor comes about because of the definition of the mass function which tells you the number of stars in a given M -> M + dM mass range. If you want to find the contribution to the total mass this range makes, you need to multiply by M. I hope this is making sense, sorry if it is unclear.
  15. Oct 26, 2009 #14
    wouldn't i get a negative result by integrating m^-1.35dm
    which would be m^-0.35/-0.35
    I hope i'm not wrong.
  16. Oct 26, 2009 #15


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    Remember this is a definite integral (you've written the answer done above as an indefninte integral), apply the integration limits and you will see that the result is not negative.
  17. Oct 26, 2009 #16
    Now I get it ... so after getting solving the first question how does that fit into the second problem

    L/l = (M/m)^4


    L = l(M/m)^4

    or for a small luminosity within the mass limit

    dL = l(dM/m)^4

    integrating the rhs (dM) within the limits of M(min) to M(max)
    still can't figure where the extra M fits in as you said

    is this correct ??

  18. Oct 26, 2009 #17


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    Remember that the luminosity function just tells you how much light stars of a given mass give off. To answer the question you also need to consider how many stars there are as a function of stellar mass, hence you need to include the mass function as well.

    In the first question you have a function, [tex]\frac{dN}{dM}M[/tex] which tells you how much contribution to the total mass comes from the range M -> M + dM. Therefore if you integrate over the whole mass range you get the total mass.

    In the second question, what you need is instead the total luminosity. So you need a function that tells you the luminosity produced by all the stars in the mass range M -> M + dM, which you again integrate over the mass range.

    Think about where the factor of M came from in question 1. You had a function ([tex]\frac{dN}{dM}[/tex]) that told you what proportion of the stars sat in a mass range around M. To find the total mass they contribute, you simply multiplied that proportion by M. In this case, you still have that same proportion function, but now you need to instead multply not by the mass, but by the luminosity that stars of that mass have, which is the luminosity function you have. You think integrate that function.

    Putting it together you get

    [tex] L_{total} = \int_{M_{min}}^{M_{max}} \frac{dN}{dM}l(\frac{M}{m})^4 [/tex]
  19. Oct 26, 2009 #18
    Thank you very much for your help. I think I troubled you enough today.. the problem is with garbage of a book that I have. It doesn't explain anything.
  20. Oct 26, 2009 #19


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    No worries. I strongly suspect that the book is assuming the reader is comfortable with calculus and hence doesn't explain these details step by step.
  21. Oct 26, 2009 #20
    Well its called 'Astrophysics In A Nutshell' . The calculus isn't the main problem for me the physics concepts should be properly explained from the point of view of the exercises at the back of each chapter. It is completely unrelated in many cases and I have to google most of the time or just stumble onto the answer by trying again and again.
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