Assuming Prof. Poplawski's "universe in a black hole" ( and at Arxiv here ) concept is true--that our universe is the result of a nonsingular Big Bounce from a supermassive black hole (SMBH) under Einstein-Cartan gravitation--then it seems possible to calculate the volume of this ur-object using the following method:

  1. Assume all black holes have the same mass density.
  2. Assume there is some median black holes size.
  3. Assume every galaxy in our universe contains a Super Massive Black Hole (SMBH) at its center.
  4. This means there are 2 trillion such objects (the same number as the number of galaxies).
  5. Two trillion times the size (not mass) of the median SMBH gives us the volume of the ur-SMBH which can then expanded to include everything in our universe.
  6. Packing that volume into a single object gives us (hopefull) the size of the Ur-SMBH.

What size is that? enter image description here

  • 3
    $\begingroup$ There's numerous issues with this process and the assumptions involved, not least of which that it is untrue that you can take the two volumes of separate black holes, combine them and get the new volume by adding the volumes of the individual black holes. That's just not how black holes work. $\endgroup$
    – zephyr
    Mar 19 '18 at 14:06
  • $\begingroup$ @zephyr heck, you can't even add the volumes of H2O and C22H12O11 when you combine them. :-) $\endgroup$ Mar 19 '18 at 14:18
  • 2
    $\begingroup$ Tom, your question should include links to published (refereed) articles by your named source. Further, you need to provide some justification for claiming there was any measureable parameter (spatial or temporal) prior to the BigBang. $\endgroup$ Mar 19 '18 at 14:20
  • 3
    $\begingroup$ I'm voting to close this question as off-topic because this appears to be in the realm of unscientific theory. $\endgroup$ Mar 19 '18 at 14:21
  • 1
    $\begingroup$ See N. Poplawski publications $\endgroup$
    – Mike G
    Mar 19 '18 at 16:31

Some problems :

  • The energy total of the universe is predominantly made up of Dark Matter and Dark Energy, with only 5% for Baryonic matter ("normal" matter). But as the proposed idea seems to completely change the model for the "dark" matter and energy values, it's hard to relate to that model. We either need Dark Matter or some radical changes in physics to explain e.g. stellar motions in galaxies and I don't think the "ur-SMBH" is going to do that. So we still need all that mass and energy.

  • The SMBs at the center of galaxies are a small proportion of the normal matter in a galaxy. For the Milky Way, our SMB mass is about $4\times 10^6 M_0$ whereas the Milky Way mass is about $10^{12}M_0$ which makes the SMB almost insignificant in calculating the mass total for your black hole.

  • Note you can't ignore energy. GR tells us that energy contributes to gravitational fields, so it's not just about mass. Every photon is significant, every neutrino, and that (again) means we need to include dark matter and dark energy.

So your estimate of the ur-SMBH mass (and volume) would be omitting the most significant contributions to mass and energy in the Universe.

Limiting the remainder to the idea of how you calculate the size of combined black holes.

The simplest black hole model is a Schwarzchild black hole : non rotating and alone in the universe.

The radius of such a black hole is :

$$R = \frac {2GM}{c^2}$$

So the volume and mass are not proportional. Instead we get :

$$V = \frac {4\pi}{3} \left(\frac {2GM}{c^2}\right)^3$$

So when we add two black hole together (and ignoring mass loss due to gravitational waves which I gather from this answer on Physics SE is between about 5% and 11%), we find that :

$$M_{total} = M_1 + M_2 $$

$$V_{total} = \frac {4\pi}{3} \left(\frac {2G(M_1+M_2)}{c^2}\right)^3$$

$$V_{total} = \frac {4\pi}{3} \left( R_1+R_2 \right)^3$$

$$V_{total} = \frac {4\pi}{3} \left( R_1^3+R_2^3+3R_1^2R_2+3R_1R_2^2 \right)$$

$$V_{total} = V_1 + V_2 + 4\pi \left( R_1^2R_2+R_1R_2^2 \right)$$

So :

$$V_{total} = V_1 + V_2 + 3V_1^{\frac 2 3}V_2^{\frac 1 3} + 3V_1^{\frac 1 3}V_2^{\frac 2 3}$$

So if you were to merge two identical black holes of equal volume $V_0$ together you'd get a black hole of volume $8V_0$.

And if you combined $N$ black holes of equal size together you'd get :

$$V_{total} = N^3V_0$$

This of course is a very naive calculation, but it does show why you can't just add the volumes together as you are suggesting.

  • $\begingroup$ If the mass of the universe is $\sim 1.5 \times 10^{53} \mathrm{kg}$, I get a Schwarzschild radius of about 23 billion light years, or half the radius of the observable universe. $\endgroup$
    – Mike G
    Mar 19 '18 at 22:52
  • $\begingroup$ Thank-you for going at it. I do NOT want to combine anything. I merely want to know what the additive volume of all SMBHs would be. Since, as you point out, there is a lot more mass than just SMBHs, O.K. add them in. The only point of this inquiry is not to debate the physics of black holes, but to try to estimate what the size of the Poplawski SMBH would be before it expanded/exploded to create the current universe. $\endgroup$
    – Tom Holzel
    Mar 20 '18 at 1:12
  • $\begingroup$ Stephen G. Thanks again. So, are you suggesting that the Ur-SMBH out of which this universe was created (in Papolawski's theory) is not a tiny fraction of the size of our current universe, but a major fraction of it? $\endgroup$
    – Tom Holzel
    Mar 20 '18 at 1:15
  • 1
    $\begingroup$ @TomHolzel If we use the Milky way as a model, the central black hole is about 400 million solar masses and the galaxy about 700 billion, so that would put a rough estimate of (very ballpark) 1/1000th the total mass that StephenG came up with if you just want the mass of the combined supermassive black holes. I don't see what insight that estimate provides, personally. $\endgroup$
    – userLTK
    Mar 20 '18 at 8:04
  • 1
    $\begingroup$ Bit the bullet. and with some help, my calculations show the Paplowski Ur-SMBH to have been about 1/10th the size of our galaxy. i.e., with a diameter of 10 LYs. $\endgroup$
    – Tom Holzel
    Apr 7 '18 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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