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Conservation of mass - Stellar equation

  1. May 28, 2013 #1
    Conservation of mass -- Stellar equation


    I was reading over a article on hydrostatic equilibrium of stars. I came across a chapter stating Conservation of Mass, where there is a sphere: r distance from the sphere, density as a function of radius ρ(r). Let m be the mass interior to r then the conservation of mass shows:

    dm=4 pi r^2 ρdr.

    Written in differential form:

    dm/dr=4 pi r^2ρ

    Now, I might be asking a silly question: I understand 4 pi r^2 is the surface area of a sphere. The law of conservation of mass states that mass of the system remains constant. Then how did the above equation arrives?

    Can somebody please explain me, the equation how it arrives and its' connection to the law?

  2. jcsd
  3. Jun 6, 2013 #2


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    Hi shounakhbatta,

    I was hoping somebody more knowledgeable would answer your question, but since nobody had, then you'll have to do with me.

    As far as I understand, the "conservation of mass equation" is called so, because it follows from the assumption that mass of a star remains constant with time.
    It seems to be useful in analysing the state of a star, as mass outflows are typically low.
    So you use polar coordinates, linking mass to radius, so that the mass contained in a thin shell is given by the area of the shell at radius r, times density, times the infinitesimally small thickness of the shell. Hence: [itex]dm=4\pi r^2ρ(r)dr[/itex]
    Integrating from 0 to R would give you the total mass of the star.
  4. Jun 6, 2013 #3

    "...it follows from the assumption that mass of a star remains constant with time".

    Huh? Last I heard, stars constantly lose mass via e = mc^2. How have I gone wrong here?
  5. Jun 7, 2013 #4


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    Fusion reactions convert negligible amounts of mass into radiation(less than 0.7% over the whole lifetime on the main sequence). The more significant losses are due to solar wind, which in some cases can be too strong to neglect. But then you can't use mass conservation equation.

    Sure, even for the typical main sequence stars, the equation is just an approximation, just as e.g., the ideal gas equation is, but it's a useful one. You obviously wouldn't be using it if you were studying mass loss, or high accuracy were needed, but as with ideal gas, you're better off using it whenever you can, as without constancy of mass analysing the stellar structure is not so simple.
  6. Jun 7, 2013 #5


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    Currently, the sun has a mass of about 2*10^27 tonnes. Of this mass, about 73.5% is hydrogen, which amounts to about 1.47*10^27 tonnes.

    At the core, the current fusion rate (mass to energy) is about 4.26*10^6 tonnes/s. In a year's time, about 1.34*10^14 tonnes of mass are converted to energy within the sun's core, at its current fusion rate. One can see that although the quantities of mass involved are huge, the relative percentage of mass converted to energy within the sun is very small.
  7. Jun 7, 2013 #6
    Thanks for the responses! I had not realized that the mass loss was so slight.
  8. Jun 7, 2013 #7


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    With E = mc^2, the c^2 is a huge number. Even so, the amount of energy produced by the sun at its core is estimated to be about 275 W/m^3. This is about equal to the energy output of the metabolism of a reptile. When you have a large volume like the core of the sun has, then the watts add up and you get something that shines like the sun.

    See: http://en.wikipedia.org/wiki/Sun
  9. Jun 7, 2013 #8
    SteamKing -- That is about the strangest relationship that I've ever heard of. We have a pet lizard, which thanks to you has acquired some new found respect. Maybe I'll start calling him "Sunny."
  10. Jun 10, 2013 #9
    Hello Bandersnatch,

    Sorry for the late reply. The conservation of mass is now clear to me. Actually the catch was relating mass to radius. From my poor concept of physics, mass is constant. The surface area of sphere 4pir^2, how it gets connected to the mass.......I was not getting to it. As you have written mass contained=area of the shell at r radius * density.....

    In that case, I would like to ask you one thing: any spherical object's mass =area of the shell at r * density?
  11. Jun 10, 2013 #10


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    In the most general terms, any object's mass is the collection of all infinitesimal masses dm, which is the same as infinitesimal volumes dV that make up the object, times density.
    When thinking of counting the mass of an object like that, we just need to come up with such a clever division of the object into dV's so as to be able to easily add them up(i.e., integrate).

    So the difficult part is to find the volume of the object as a function of just a single variable. The mass is just ρ*V.

    If it were a cylinder(radius r, height h), then you could for example divide it into very thin discs, so that the dV equals the area of the disc times the infinitesimally small bit of height of the cylinder. It'd look like this [itex]dV=\pi r^2 dh[/itex] and you integrate from 0 to h to get the total volume.
    You could also divide it into cylindrical shells, like very thin pipes, so that [itex]dV=2\pi r h dr[/itex]* and you add all the shells by integrating from 0 to r.

    With spheres, the most intuitive division is into shells of thickness dr, but if you were very determined, you should be able to divide it along, say, x axis, into many thin discs of varying radius. You'd have to come up with a way to connect the radius' of the discs with x, which would probably be a sine function.
    It's so much simpler with shells, though.

    *if it's not clear to you, here's how you get this equation: http://www.stewartcalculus.com/data...texts/upfiles/3c3-Volums-CylinShells_Stu .pdf
    (the first derivation at the top of th page)
    Last edited: Jun 10, 2013
  12. Jun 10, 2013 #11
    Hello Bandersnatch,

    Thank you very much for this amazing reply. You described things so easily, yet technically that it just cleared everything. I need more people like you.

    One thing: the surface area of a cylinder is 2pir^2+2pi.r.h. You have written dV=2pirdh. Here, are you differentiating the infinitesimals height into dh? Where the formula r^2 goes?
  13. Jun 10, 2013 #12


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    Well spotted, that's a silly mistake on my part. The surface area of a circle is of course [itex]\pi r^2[/itex], so the volume of a disc dV should be: [itex]dV=\pi r^2dh[/itex]

    Sorry for confusing you like that.
  14. Jun 10, 2013 #13
    Ok. That's not an issue. But the surface area of the cylinder is 2.pi.r^2. How does the 2 vanish?
  15. Jun 10, 2013 #14


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    A cylinder has two ends, so to get the total area you add the area of both ends.
    But here, you're not interested in the total area of a cylinder, but in the volume. To get the total volume of a cylinder, you normally multiply the height times the area of just one end, not both. Same here. You're just adding together the volumes of infinitely many tiny cylinders stacked on top of each other, or, in the pipe-shell approach, infinitely many thin pipes of increasing radius sitting one over the next.
  16. Jun 10, 2013 #15
    Great. Tot.volume=area (pi.r^2) x d.height.

    Now, once conceptual question. Whenever, we are calculating any area, volume, density by differentiating i.e. calculating the infinitesimals we do the same thing in maths right? The traditional formulas get transferred through differentiation. Say, if we want to calculate the area of a triangle = base.height/2, differentiating we would get dbase.dheight/2. Something like that?
  17. Jun 10, 2013 #16


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    No, not really. I mean, you could divide the triangle into infinite amount of small triangles like that and then integrate over two variables, but it's kind of missing the point.

    I don't want to go into any more detail here, as we have strayed far enough from the topic of astrophysics. You'll be better off asking questions about maths in the maths section. Or simply grab a book or some other resource about integral calculus - in a way, all it's concerned with is calculating areas and volumes under line graphs and surfaces.
  18. Jun 10, 2013 #17
    Ok. Thanks a lot Bandersnatch for clearing all the doubts.
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