Density / Schwarschild radius

Questions
1

If the only information available is the density of an object, is it possible to calculate its schwarzschiid radius?

2

No, it’s proportional to the mass.

3

if the mass of an object is known its schwarzschild radius can be calculated. The volume of a sphere with that radius can then be determined.
Then that mass per that volume provides a density. Therefore, every schwarzschild radius corresponds to a specific density. Is it just a simple algebraic equation to go from density to Schwarzschild radius?

4

The Schwarzschild radius of a mass is not the radius of the object. It is the radius below which if a non-rotating spherically symmetric distribution with that mass were contained within it it would constitute a black hole. Frank’s answer is correct.

Now let’s get back to our originally scheduled broadcast and purpose for this forum: Wolfram technologies. :wink:

5

I am obviously not communicating my question well. Let’s try this: for a substance with a given density what volume of that substance contained within a non-rotating symmetric sphere would be required to create a black hole?

6

Ok, I think that I can give you a more sensible answer.

Here is the expression for the Schawrtzschild radius in terms of the Mass:

r=\frac{2 G M}{c^2}

Now substitute for the mass in terms of the density (M=\frac{4 \pi r^3}{3}) and get:

r=\frac{8 \pi G \rho r^3}{3 c^2}

Now solve for r,

Solve[r == (8 G \[Pi] r^3 \[Rho])/(3 c^2), r]

and get

{{r -> 0}, 
{r -> -((c Sqrt[3/(2 \[Pi])])/( 2 Sqrt[G] Sqrt[\[Rho]]))},
 {r -> (c Sqrt[3/(2 \[Pi])])/( 2 Sqrt[G] Sqrt[\[Rho]])}}

and the last solution is clearly the physical one you are looking for:

\frac{\sqrt{\frac{3}{2 \pi }} c}{2 \sqrt{G} \sqrt{\rho }}

Is this the correct interpretation of your question?
I hope this helps…

7

Exercise for the interested person: put this in Mathematica form. This is, after all a Wolfram Technologies forum…

8

See my revised answer above…
By the way, if you forget the expression for the Schwartzschild radius you can get it from

FormulaData[{"BlackHoleEventHorizonRadius", "Standard"}]
9

In my post above there needs to be a rho in the expression for the mass in terms of the radius. The community technology seems to screw up the TeX code if I try to edit it, dropping various parts in the edit window and I do not want to have to repost the entire thing. The final answer is still right.

10

Apropos of my computation above here is the value one gets if one uses the density of water:

UnitConvert[
  (Quantity["SpeedOfLight"] Sqrt[3/(2 \[Pi])])/(2 Sqrt[
      Quantity["GravitationalConstant"]] Sqrt[
      Quantity[1, "Grams"/"Centimeters"^3]]), "Kilometers"]/
 UnitConvert[PlanetData["Earth", "Radius"], "Kilometers"]

which gives

6.297 10^4

times the radius of the earth…

11

Planck mass

mp = Sqrt[\[HBar] c/G];

 Planck length

lp = Sqrt[\[HBar] G/c^3];

 Schwarzschild radius of Planck mass is 2 Planck lengths

rs = 2*G*mp/c^2;

rs/lp // PowerExpand

2
12

It’s worth noting in passing that my simple calculation above is rather unrealistic. That the density is uniform throughout the sphere (assuming that the radius is less than the Schawrtzschild radius so the object is not a black hole and so that we can then discuss its internals) is not physically reasonable. Assuming that the sphere is made of the same “stuff” throughout, one needs an equation of state to determine what its steady state composition is–particularly what the density is as a function of radius, and this will not be constant.

13

Thanks for the help. David, you last comment is definitely accurate but for what I am working on it shouldn’t make a difference. Once I have worked with the info you provided I may have a question that needs to be put into Mathematica format. Your comments and time are very much appreciated.

14

Frank that thank you was for you as well.