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

In summary, the conversation discussed the idea of a black hole becoming our sun and the mass and density calculations involved. It was determined that for a given density, there is a radius of a sphere made of that material when it becomes a black hole. However, for a black hole accreting more mass, its Schwarzschild radius would grow faster than the radius of the hypothetical uniform sphere of matter contained within, making it impossible for r = r_s. The conversation also touched on the concept of "Planck stars" and the mistake made in the interpretation of the equations.
  • #1
elegysix
406
15
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, [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], 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 ([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 Schwarzschild 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!
 
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  • #2
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.
 
  • #3
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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  • #4
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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  • #5
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 [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.
 
  • #6
elegysix said:
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?
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.
 
  • #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!
 

1. How do you calculate the solar mass using the back of the napkin method?

The back of the napkin method is a quick and simple way to estimate the solar mass using a few basic equations. First, you need to know the distance between the black hole and the sun, which can be obtained from astronomical observations. Then, you can use the equation M = r^3/T^2, where M is the solar mass, r is the distance between the black hole and the sun, and T is the orbital period of the sun around the black hole. This equation is derived from Kepler's third law of planetary motion.

2. What is the accuracy of this calculation method?

The back of the napkin method is a rough estimate and should not be used for precise calculations. It is best used for quick estimations and to gain a general understanding of the solar mass. The accuracy of this method depends on the accuracy of the input values, such as the distance and orbital period.

3. Can this method be used for any black hole and sun system?

Yes, this method can be used for any black hole and sun system, as long as the distance and orbital period can be accurately determined. However, it is important to note that this method assumes a circular orbit, so it may not be accurate for systems with highly elliptical orbits.

4. How does the back of the napkin method compare to other methods of calculating solar mass?

The back of the napkin method is a simplified version of more complex methods used by scientists. While it may not be as accurate as these methods, it provides a quick and easy way to estimate the solar mass. It is also a good way to double-check the results of more complex calculations.

5. Are there any limitations to using the back of the napkin method for calculating solar mass?

As mentioned before, the back of the napkin method assumes a circular orbit and may not be accurate for systems with highly elliptical orbits. It also does not take into account any other factors that may affect the orbital motion, such as the presence of other massive objects. Additionally, the input values used in the calculation must be accurate in order to obtain a reasonable estimate of the solar mass.

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