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BH to Sun, back of the napkin calculation of solar mass

  1. Oct 29, 2014 #1
    Yesterday I had a bizarre idea. I supposed that a black hole (which I assumed to be a sufficiently dense sphere) became our sun. I know this is completely wrong, but please humor me and see what I've done.

    Suppose we have a sphere sufficiently dense, so that the Schwarzchild radius, [itex]r_s[/itex], is greater than the radius of the sphere, [itex]r[/itex]. Since the escape velocity is greater than c, let's assume this object will continue to 'trap' masses which enter it and grow until [itex]r=r_s[/itex]. Then when [itex]r \geq r_s[/itex], lets assume photons and such will no longer be trapped, and the sphere will appear as a fireball, somewhat like the sun. So now lets ask the question, what is the mass of the sphere at this point?

    Lets say this sphere is as dense as a nucleus ([itex]\rho \approx 2.3x10^{17}[/itex] kg/m3), or the core of a neturon star, roughly ([itex]\rho \approx 8x10^{17}[/itex] kg/m3). Let's further assume that [itex]\rho[/itex] is constant.

    The Schwarzchild radius is given by [itex]r_s=\frac{2Gm}{c^2}[/itex]. However we can say [itex]m=\rho V[/itex], and since we have assumed a sphere, [itex]V=4/3\pi r^3[/itex]. Since we are interested in the case [itex]r=r_s[/itex], this yields

    [itex]r=\sqrt{\frac{3c^2}{8\pi \rho G}}[/itex], and then the mass of this sphere is [itex]m=4/3\rho \pi r^3[/itex]. (An interesting note here is that as [itex]\rho[/itex] increases, r decreases.)

    Running the numbers for [itex]\rho = 2.3x10^{17}[/itex] kg/m3 yields [itex]m =1.78x10^{31}[/itex] kg (roughly 8.9x the solar mass),

    Using [itex]\rho = 8x10^{17}[/itex] kg/m3 yields [itex]m =9.55x10^{30}[/itex] kg (roughly 4.8x the solar mass).


    I was shocked yesterday to find that these numbers were in the ball park of the solar mass. What gives? How could such a rough calculation (not to mention based on a completely wrong foundation) give a relatively accurate answer? What are your thoughts?

    Could a star form in such a way?

    Thanks!
     
  2. jcsd
  3. Oct 30, 2014 #2

    Vanadium 50

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    I don't think you have shown anything other than that a 1 solar mass neutron star is not enormously bigger than a 1 solar mass black hole.
     
  4. Oct 30, 2014 #3

    Ken G

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    Yet there is something important in this fact. The density of a neutron star, and the density of a nucleus, is essentially the density of matter when all its particles are degenerate (not just the electrons, as in white dwarfs). The mass of a black hole is set by the requirement that the object be able to become relativistic before it becomes degenerate. Neutron stars are perched between these. It is a remarkable fact that objects with the largest masses that can reach that density, and not be black holes, are stellar masses. This may very well be the reason that stars have this mass, which is remarkable because the stage of a star's life where its mass gets set is long before the object goes either degenerate or relativistic, but there is the genesis of a kind of competition between these effects even that early on in the star's life.
     
    Last edited: Oct 30, 2014
  5. Oct 30, 2014 #4

    Bandersnatch

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    Take the Schwartzschild radius equation. It shows that doubling the mass increases the rs by the same factor. At the same time, doubling the mass of a uniform sphere of material increases its radius by ##\sqrt[3]{2}##.

    So, even if you start with the assumption that physical state of the material beyond the event horizon remains unchanged, a black hole accreting more mass will have its Schwartzschild radius grow much faster than the radius of the hypothetical uniform sphere of matter contained within.

    Your calculations merely show that for a given density, there is a radius of a sphere made of that material when it becomes a black hole. The less dense the material, the higher the radius. It's no surprise that for a neutron star-like material you get the kind of ballpark mass required for black hole formation.
    You can use this to find out how dense e.g., an object of the size of the Milky way would have to be for that to happen. It's also one of the reasons Star Wars-style dogfights make so little sense, as you can calculate just how soon would an air-like medium-filled stretch of space end up as a black hole.

    On the other hand, there were some recent papers on possible evolution of black holes resulting in essentially very similar outcome to what you postulated. Search the forum or arxiv for "planck stars".
     
    Last edited by a moderator: Oct 31, 2014
  6. Oct 30, 2014 #5
    Have I made a mistake? If [itex]r_s[/itex] grew faster than [itex]r[/itex] for constant density, then [itex]r \geq r_s[/itex] could not occur when starting with [itex]r < r_s[/itex], right?

    by the way, thanks for the mention of planck stars. I will be looking into that.
     
  7. Oct 30, 2014 #6

    Bandersnatch

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    That's correct. If you start with all mass within its ##r_s## then it'll never grow out of it due to accretion. However, if you start with a sphere of mass whose radius is greater than ##r_s##, and allow it to accrete, then it'll eventually turn into a black hole.
    As far as I can see the only mistake you've made is with the interpretation of the equations.
     
  8. Oct 30, 2014 #7
    Ok, so I checked to see what [itex]r[/itex] would have to be for [itex]\frac{dr_s}{dm} \leq \frac{dr}{dm}[/itex] with constant [itex]\rho[/itex]. This yielded

    [itex]r \leq (\frac{8\pi G \rho}{c^2})^{-1/2}[/itex]

    and by substituting in the expression for [itex]r[/itex] where [itex]r = r_s[/itex] (from above) in order to solve for a value of [itex]\rho[/itex] which satisfies [itex]r = r_s[/itex] and [itex]\frac{dr_s}{dm} \leq \frac{dr}{dm}[/itex] , I found it is only true for when [itex]3 \leq 1[/itex].

    Therefore I admit defeat.

    Thanks for your comments!
     
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