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

AI Thread Summary
The discussion explores the hypothetical scenario of a black hole transforming into a sun-like object by examining the relationship between density, Schwarzschild radius, and mass. Calculations suggest that a sphere with densities comparable to neutron stars yields masses in the range of solar mass, raising questions about the formation of stars. Participants highlight that the mass of a black hole is influenced by its ability to become relativistic before degeneracy occurs, emphasizing the significance of density in stellar evolution. The conversation also touches on the implications of black hole accretion and the dynamics of mass and radius growth. Overall, the discussion reveals intriguing insights into the interplay between black holes and stellar formation.
elegysix
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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 Schwarzschild radius, r_s, is greater than the radius of the sphere, r. Since the escape velocity is greater than c, let's assume this object will continue to 'trap' masses which enter it and grow until r=r_s. Then when r \geq r_s, let's assume photons and such will no longer be trapped, and the sphere will appear as a fireball, somewhat like the sun. So now let's ask the question, what is the mass of the sphere at this point?

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

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

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

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

Using \rho = 8x10^{17} kg/m3 yields m =9.55x10^{30} 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!
 
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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.
 
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.
 
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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".
 
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Bandersnatch said:
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
Have I made a mistake? If r_s grew faster than r for constant density, then r \geq r_s could not occur when starting with r < r_s, right?

by the way, thanks for the mention of Planck stars. I will be looking into that.
 
elegysix said:
Have I made a mistake? If r_s grew faster than r for constant density, then r \geq r_s could not occur when starting with r < r_s, right?
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.
 
Ok, so I checked to see what r would have to be for \frac{dr_s}{dm} \leq \frac{dr}{dm} with constant \rho. This yielded

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

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

Therefore I admit defeat.

Thanks for your comments!
 
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