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Deriving the number of stars as a function of flux and bolometric magnitude.

  1. Feb 1, 2013 #1
    QUESTION
    A star of bolometric luminosity L at a distance r will exhibit a bolometric energy flux F given by

    F = L/4πr2

    in the absence of obscurity.

    A. Assume that all stars have the same bolometric luminosity L and that stars are uniformly distributed in space with a number density n. Derive the number of stars N of bolometric energy flux greater than F as a function of F, i.e. derive N(F).

    B. Express the result of part A in terms of bolometric magnitude by deriving the number of stars N' of bolometric magnitude less than m as a function of m, i.e. derive N'(m).

    C. Now assume that stars are distributed in luminosity according to a luminosity function n(L), where n(L)dL is the number density of stars of luminosity between L and L+dL. How does this change the results of part A? Namely, derive N(F) in this case.

    ATTEMPT
    PART A:
    Ftotal = NL/4πr2.
    n = N/V
    ∴ Ftotal = nVL/4πr2

    nV = (F4πr2)/L = N(F)

    I'm really guessing here. I know it shouldn't be a difficult question, but I'm not even sure how to fulfill the requirement which states that the bolometric energy flux should actually be greater than F...
    Any hints are much appreciated, thank you. :(
     
  2. jcsd
  3. Feb 1, 2013 #2

    cepheid

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    If all stars have the same luminosity, then their measured flux depends only on their distance. So the requirement that flux be greater than F gives us a constraint on the distance to the star:

    L/(4πr2) > F

    L/(4πF) > r2

    So, the distance r must be LESS than a certain value. This makes sense. Only stars within a certain radial distance will have fluxes equal to or brighter than "F".

    Given that the stars are uniformly distributed through space, how many of them lie within a distance of +√[L/(4πF)] from us?
     
  4. Feb 2, 2013 #3
    That certainly makes sense.

    I feel as though I'm still overcomplicating things, however.
    That would mean the smaller flux is L/4π(√([L/(4πF)]))2??...multiplied by nV (= N) would give you the total smaller flux...F' = N4πF2/L, and then solve for N?
    That's way too convoluted to be correct, haha.
     
  5. Feb 2, 2013 #4

    cepheid

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    I don't know what you mean by "the smaller flux." I also don't understand why you're trying to calculate some "total" flux F_total of all the stars with flux > F. This is not part of the problem.

    We showed that if a star of luminosity L is to have a measured flux greater than F on Earth, its distance must be ##r \leq \sqrt{L / (4\pi F)}##

    The problem is asking: HOW MANY stars are there that satisfy this condition? In other words, how many stars are there within this distance of Earth?
     
  6. Feb 2, 2013 #5
    Er, I meant greater flux. I calculated the flux with r < sqrt[L/(4piF)] for N stars and just solved for N because that would give you the number of stars...? I honestly am clueless, I'm sorry. :\
     
  7. Feb 2, 2013 #6

    cepheid

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    Stars are evenly distributed in space (n per unit volume). Okay, so how many of them are within a given radius? Well, what is the geometry of a region of space defined by all points within a radius r from Earth? How do you figure out the volume of this region? Based on that, how many stars are in it?

    I didn't really want to say the above...gave away too much, I think.
     
  8. Feb 2, 2013 #7
    So...since r is greater than or equal to √[L/(4πF]...the volume of the region of space defined by all points w/in that radius would simply be the volume of a sphere, (4/3)πr3.

    N=nV=n*(4/3)πr3...and then just plug in r, which does contain F so N would be a function of F. Correct?
     
  9. Feb 2, 2013 #8

    cepheid

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    Yes.
     
  10. Feb 2, 2013 #9
    haha I feel silly. Thanks so much for being patient.
     
  11. Feb 2, 2013 #10
    PART B:

    In order to express my result in terms of the bolometric magnitude, wouldn't I first have to know what it is, which would require comparison with another number of stars of different luminosity, as per

    m2-m1=-2.5log(L2/L1)?

    Or I can use m_apparent = m_bol - 5(1-log(r))...but that would require knowing m_apparent???
     
  12. Feb 2, 2013 #11

    cepheid

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    You are correct that magnitude needs a reference flux. For instance, the difference in magnitude between two stars is

    m1 - m2 = -2.5log(F1/F2)

    So the usual way of defining the magnitude of *a* particular star is to choose a reference flux Fref that defines what magnitude 0 is:

    m = -2.5log(F/Fref)

    You can see that if F = Fref, then you have log(F/Fref) = log(1) = 0.

    Of course, using properties of logs, this equation can be written as

    m = -2.5[log(F) - log(Fref)] = -2.5log(F) + some arbitrary constant.

    The reference flux is chosen arbitrarily. For visual magnitudes, I think Fref used to be the measured flux of the star Vega in the visual band, meaning that Vega had mV = 0 by definition. Now the reference flux has changed. In any case, it's arbitrary. For absolute magnitude (which measures luminosity) it's the same thing. Wikipedia says that the reference luminosity Lref for absolute bolometric magnitude is just the luminosity of the sun Lsun i.e.:

    Mbol = -2.5log(L/Lsun) = -2.5log(L) + some constant

    The key point is WHO CARES what "some arbitrary constant" is? In the scenario in your problem, all stars have the same luminosity, L, which means that all stars have the same absolute magnitude. So I would be inclined to just define Lref = L so that every star has absolute magnitude

    Mbol = -2.5log(L/L) = 0

    Then you can just find the apparent bolometric magnitude mbol from the absolute bolometric magnitude using the standard relation between apparent and absolute magnitudes (the "distance modulus").

    If you choose a different reference point for your magnitude scale (e.g. say you decide to set Mbol = 10), it doesn't matter, because it will just add a constant offset to all of your apparent magnitudes (i.e. shift them all by the same amount).
     
  13. Feb 2, 2013 #12
    Okay, got it. Thanks again!!
     
  14. Feb 2, 2013 #13
    any advice on part c?
     
  15. Feb 5, 2013 #14
    Anything on part C?
     
  16. Feb 5, 2013 #15

    cepheid

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    Wow, there sure is a lot of demand for a solution to this problem. For this part, I think that instead of going out to a fixed maximum distance r, you'll have to go out to r = ∞. The reason is, now that there is a whole distribution of luminosities, theoretically, at any distance r, there could be a star with luminosity L bright enough to produce flux ≥ F.

    This will almost certainly take the form of an integral. After all

    dN = ndV

    Since there is spherical symmetry, you should be able to express this in terms of dr. I.e. the number density should depend only on radial distance, not on the direction you look. However, the number density n is going to be a function of r, n(r). The reason is that, for each distance r, there is going to be a certain number of stars with luminosity > L, where L is the value required to produce flux = F at that distance. So, at each value of r, to get n(r), you're going to have to integrate the luminosity function n(L) over the range of values of L that produce flux ≥ F at that distance.

    Make sense? Let me know how it goes!
     
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