There have been other similar questions, but none of the answers address this point.

  • Inflation ends after the first 10^-32 seconds. Our observable universe was the size of a grain of sand that contained a mass-energy ~10^80 GeV, which is much denser than the critical density that would presumably make it collapse into a black hole.
  • since the mass-energy in the universe was evenly distributed at the end of inflation, there was no over-density in one region that resulted in black hole. This can be explained by Newton's shell theorem.
  • separately, radiation exerted an outward pressure that prevented over-dense regions from forming. Newtonian dynamics goes a long way towards explaining how the outward pressure of radiation balanced the collapsing force of gravity.

However, it seems to me that if the observable universe was all there was (just a 10^80 GeV grain of sand surrounded by vaccuum), then the universe would have collapsed to a black hole. The universe must have been much larger than our observable grain of sand, and mass-energy must have been smoothly distributed outside of our observable grain of sand.

  • 1
    $\begingroup$ Re your last paragraph. If there is just a 'grain of sand' surrounded by vacuum, it still isn't the case that the grain of sand is all there is. For the grain of sand to be literally all there is, there would be nothing apart from the grain of sand. A vacuum that surrounds is something and not nothing. A Schwarzschild black hole is surrounded by vacuum (The Schwarzschild black hole solution is called vacuum solution for that reason) but the Universe (spacetime) that it describes includes the vacuum exterior to the hole. $\endgroup$ Apr 5, 2018 at 1:37
  • $\begingroup$ To me isn't clear if outside there was even vacuum. Isn't the sand grain just our to become observable universe? But perhaps surrounded by other grains? I got the sand and grapefruit sizes are referring to what is now our observable universe and part beyond, not all there is out there. Am I wrong? $\endgroup$
    – Alchimista
    Apr 5, 2018 at 8:43
  • $\begingroup$ Ok, for sake of discussion, outside of the grain of sand, there was infinite space filled with cold, very low density matter. $\endgroup$ Apr 6, 2018 at 19:40
  • $\begingroup$ I guess what I'm struggling with is that FLRW assumes homogeneity and isotropy everywhere. A homogeneous, isotropic grain of sand surrounded by nearly empty space no longer satisfies the FLRW constraints. $\endgroup$ Apr 6, 2018 at 19:47

1 Answer 1


There is a reason why the current model is the Big Bang model. In an explosion, the kinetic energy of the particles exploding is larger than the gravitational force pulling them back in.

When going to the four dimensions of general relativity, singularities can also be "implosive", as in black holes, or "explosive" in generation of universes, depending on the model.

There are different types of singularities in General Relativity and black holes and the Big Bang do not belong to the same type.

This class of singularities is large enough to contain isotropic singularities, warped-product singularities, including the Friedmann-Lemaitre-Robertson-Walker singularities, etc. Also a Big-Bang singularity of this type automatically satisfies Penrose's Weyl curvature hypothesis.

It ain't simple.

  • $\begingroup$ Help me understand the FLRW equations. I was hoping to find an equation that says "if temperature is below this threshold, then big bang should collapse instead of explode". Instead, there is the fluid equation $\dot{\rho} = -3\frac{\dot{a}}{a}(\rho + p)$ The fluid equation and acceleration equations seem to say that the 'velocity' of the scale factor has inertia in and of itself. Does this mean that the big bang would have expanded even if the temperature (after inflation) was 0? $\endgroup$ Apr 6, 2018 at 19:22
  • $\begingroup$ I am assuming that if temperature is 0, then $p$ is 0, and the fluid equation becomes $\dot{\rho} = -3\frac{\dot{a}}{a}(\rho)$ Is it incorrect for me to think of 'kinetic energy' and 'temperature' as the same thing here? $\endgroup$ Apr 6, 2018 at 19:29
  • $\begingroup$ Sorry, my background does not go further than the answer above. I just wanted to point out that there are several types of singularities in General Relativity depending on the various metrics , not one for both implosion and explosion. (true for classical mechanics too, due to boundary conditions, but it gets more complicated when the space itself is involved) $\endgroup$
    – anna v
    Apr 7, 2018 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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