Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

With crude calculations following densities can be approximated:

Given that radius of proton is $1.75×10^{−15} m$ and it's mass is $1.67 × 10^{-27}kg$, this gives density of proton to be $\dfrac {1.67 × 10^{-27}kg} {\frac{4}{3}(1.75×10^{−15} m)^3}=\dfrac {3}{4} \dfrac{1.67 × 10^{-27}}{5.36×10^{−45}} \dfrac{kg}{m^3}=2.34 \times 10^{15} \dfrac{kg}{m^3}$

Density of the heaviest naturally occurring solid $\mathrm U^{238}$: 1 cubic meter of Uranium is about $2 \times 10^4 kg$, and since there are 238 protons+neutrons in Uranium, this gives the density $8 \times 10^{-3} \times 10^4 \dfrac{kg}{m^3}=8\times10 \dfrac{kg}{m^3}$

This means the protons are dispersed by a factor of $2.92\times10^{13}$, in other words there are $2.92\times10^{13}$ empty space units to each unit of space filled with proton.

Density of water $=\frac{1}{20}\times10 \dfrac{kg}{m^3}=5\times10^{-1}\dfrac{kg}{m^3}$

The density in the core of neutron star is $8×10^{17} kg/m^3$, and the density of a black hole is supposedly infinite according to Wikipedia, but then again before transitioning to a black hole there must have been some finite mass distributed over a non zero volume of space therefore having a finite density.

So far on logarithmic scale (base 10) followings look obvious for $\operatorname {log}_{10}(density \dfrac {kg}{m^3}) = \operatorname {log}_{10} \dfrac{kg}{m^3} + \operatorname {log}_{10}density$

Let $ d =\operatorname {log}_{10}density,\quad \operatorname {log}_{10} \dfrac{kg}{m^3}= \text{what is one to make of this}?$

Then following crude observations can be made:

$-1\leq\mathcal O(d)\leq1$: Order of density of life as we know it.

$\mathcal O(d)\approx 15$: Order of density of matter packed space

$\mathcal O(d)\approx 17$:Order of density at the core of a Neutron star

$\mathcal O(d)\geq x$:Order of density of black holes, (collapse of space to contain matter?) what is $x$?

My question is: does these calculations make sense and is there anything more to the order of densities than these naïve observations?

share|improve this question
2  
Sorry but what is your question? Of course one can crudely define any ratio of mass per volume as a density, and yes, Neutron stars are very dense but what is your point? –  Alexander Nov 25 '13 at 13:51
    
my point is at what density, the fabric of space breaks down and at beyond that point only black holes with infinite density can exist? in other words what is the upper limit for density before space break down? What is the value of $x$? –  Arjang Nov 25 '13 at 23:41
    
Related: physics.stackexchange.com/q/5888/2451 and links therein. –  Qmechanic Nov 26 '13 at 8:40
    
@Qmechanic : not really that question is about the density of black hole, this question is about density threshold where the space collapses and black hole forms. –  Arjang Nov 26 '13 at 8:51
add comment

1 Answer 1

up vote 4 down vote accepted

I think you're asking if there is some special cutoff density after which spacetime "collapses" and forms a black hole. If this is your question then the answer is no, there is no specific cutoff.

Density unites are $\frac{\mathrm{mass}}{\mathrm{volume}}$ but the size of black holes is dependent on the mass and the size is not proportional to the volume but to the radius. That is, the Schwarzchild radius $r_s$ of a black hole of mass $m$ is $$r_s = \frac{Gm}{c^2}$$

What this means is that as a black hole becomes more massive, the critical density $d_c$ needed to form a black hole is $$d_c = \frac{3m}{4\pi r_s^3}$$

That is, as the black hole mass increases the density needed to make it form is reduced. In principle you can make the density arbitrary small which means there is no fundamentally critical density.

share|improve this answer
1  
For this post to be correct, I'm assuming the Hoop Conjecture –  Brandon Enright Nov 26 '13 at 2:15
    
Thank you Brandon, yes, that is the question I updated the title. –  Arjang Nov 26 '13 at 2:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.