How total mass of universe is calculated? [closed]

Question

I think that Total mass of universe can be calculated using below formula.

Total mass of universe = (Age of Universe) × (Planck mass / Planck time)

= (4.35×10^17 ) × (2.18×10 ^−8 / 5.39×10^−44 ) Kg

= 1.75×10^53 Kg Which matches the current predictions to great extents.

I don't have fully developed theory as yet. and it will create more question than answers. But still let me start with this.Heisenberg’s Uncertainty Principle is the reason of expansion from initial ZERO; Now regarding time, we can say that: past and future does not exist and present can exist only for minimum amount of time: “the quanta of time” and it value can be only equal to Planck’s time ( 5.391247(60)×10^−44 s ). This is very small time hence uncertainty principle comes into play. In this small time duration the vacuum energy is generated as: Δ E = h / 2πΔ T –

Or Δ E = 6.62607015×10^−34 / (2 * 2π *5.391247(60)×10^−44 ) So energy generated in smallest quanta of time Δ E = 9 × 10^8 J Now estimated age of universe; that is around 13.82 billion years, is equal to 4.3 × 10^17 seconds If this age of Universe is divided in quanta of time so total intervals so far: 4.3 × 10^17 / 5.391247(60)×10^−44 s = 8 × 10^60 and total energy produced in Universe in these intervals is 9 × 10^8 J * 8 × 10^60 = 7.2 × 10^69 J

Would like to see your feedback about this.

Answer 1

This is a cute observation, but is essentially just dimensional analysis. It follows from doing a rough order of magnitude estimate where you only keep track of the Hubble constant plus fundamental constants.

To estimate the mass of the observable Universe at an order of magnitude level, we multiply the density by the volume of the observable universe. The mass density is $\rho \sim \frac{H_0^2}{8\pi G}$, and the volume is $\sim R_H^3$, where the Hubble radius is $R_H\sim c/H_0$. Then total mass is the density times the volume, which is approximately $M \sim \rho D^3 \sim \frac{c^3}{8 \pi G H_0}$. Since the age of the Universe $T\sim H_0^{-1}$, where $H$ is the Hubble constant today, this can be rewritten as $M \sim\frac{c^3}{8 \pi G} T$.

Now, dimensionally, $c^3/8\pi G$ has to have units of mass / time. But, the only dimensional quantities involved are $c$ and $G$, so it has to be expressible as Planck mass / Planck time. And indeed, it is, since, $$ \frac{c^3}{8\pi G} = \sqrt{\frac{\hbar c}{8 \pi G}} \sqrt{\frac{c^5}{\hbar 8 \pi G}} = \frac{t_{\rm Pl}}{M_{\rm Pl}} $$ (where I'm using the "reduced Planck mass").

On the one hand, it is gratifying that a simple estimate gives a roughly correct answer.

However, the appearance of Planck units here is kind of fake -- note that $\hbar$ only appears because we multiply by $1=\hbar/\hbar$. It ultimately is a statement that any combination of $G$ and $c$ could be rewritten in terms of Planck units if we wanted to -- you can always do a change of variables from the set of constants $\{G, c, \hbar\}$ to the equivalent set of constants $\{M_{\rm Pl}, \ell_{\rm Pl}, t_{\rm Pl}$}. This is not deep, just simple algebra.

Additionally, the form of the answer isn't surprising in the sense that we have made a very crude estimate of the mass, where the only parameter besides fundamental constants we have allowed for is the Hubble constant. The fundamental constants have no choice but to arrange themselves into an appropriate collection of Planck units that make the equation dimensionally consistent.

If we were to perform a more sophisticated calculation, there would be many more parameters that would appear in the final answer, such as the density parameters giving the relative amounts of dark energy, dark matter, baryonic matter, and radiation. There would be more complicated numerical factors coming from, eg, doing a more realistic calculation of the volume, accounting for redshift effects.

This is what you should generically expect. If you make a naive estimate with only one parameter, you get something very simple looking, but as you add in more realistic complications, the answer becomes more complicated. Interesting and deep relationships would go against this pattern, by having potential complications cancel out of the final answer, leaving an answer that is somehow simpler than it naively should have been. An example of this kind of surprising simplicity happening is the spacetime for a black hole, which naively could depend on exactly what kind of matter was used to build the black hole and how the black hole formed, but in the end only depends on the black hole's mass, charge, and angular momentum. However, it does not happen with your example.

Answer 2

If the universe is flat (which our universe seems to be within the margins of error) or hyperbolic the total mass is infinite, since finite density times infinite volume equals infinite mass.

If the universe is closed, the volume is $\rm V=2 \pi^2 r_k^3$ (the surface area of a 4D sphere) with the curvature radius $\rm r_k=c/(H \sqrt{-\Omega_k})$ and the curvature parameter $\rm \Omega_k=1-\Omega_r-\Omega_m-\Omega_{\Lambda}$, where $\rm \Omega_k$ is negative if the universe is closed.

So multiply that volume with the density of your model, for example the matter density (including bright and dark matter) in our current universe is around $\rm 3E-27 \ kg/m^3$.

If you're just confusing the whole flat universe with the observable universe take the euclidean volume inside the particle horizon (which is currently around $\rm 3 \times$ larger than the Hubble radius) and multiply that with your density.