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This question already has an answer here:

The Schwarzschild radius involves an expression in terms of Newton's constant $G$, the mass $M$ inside a radius $r$, and the speed of light squared $c^2$. Current estimates of the universe's matter density are about six protons per cubic meter. But, the $M$ inside a sphere goes up as $r^3$, while the "time curvature" coefficient is $1 - \frac{2GM}{c^2\, r}$. So this coefficient is bound to hit zero for $r$ large enough. The $M$ outruns the denominator as a function of $r$.

I calculated that this coefficient hits zero when $r$ equals 13.54 billion light years. Question: Is this any evidence for our universe being one very large black hole?

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marked as duplicate by Emilio Pisanty, David Z Apr 10 at 8:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/23118/2451 and links therein. $\endgroup$ – Qmechanic Apr 10 at 4:27
  • $\begingroup$ The only issue with your approach, is that you said it ''hits'' zero, hopefully my answer below demonstrates it only approaches zero. $\endgroup$ – Gareth Meredith Apr 11 at 19:35
  • $\begingroup$ My post mentions the coefficient of the time differential in the space-time metric hitting zero on the horizon; it does not refer to temperature. Why temperature in your comment? Also, I believe that black holes we can observe are a complete vacuum inside--no protons, no nothing, just a gravitational singularity at the center. So, next question: are there any black holes that are not empty of matter? I can think of only one---our observable universe. $\endgroup$ – old lowboy Apr 13 at 1:39
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Aruns weak equivalence is an argument which goes like this: To make the temperature of a black hole go down, you need to add matter to the system. Using the following approximation we have

$m \rightarrow \infty$

Then the temperature goes to zero

$T \rightarrow 0$

And for a black hole with infinite mass, the curvature tends to zero as well!

$K \rightarrow 0$

As I have stated before though, you cannot really have a system like a vacuum reach absolute zero, when the vacuum is not perfectly Newtonian. To add to his extended weak equivalence, assume the following ~

The radius of a black hole is found directly proportional to its mass $R \approx M$. The density of a black hole is given by its mass divided by its volume $\rho = \frac{M}{V}$ and since the volume is proportional to the radius of the black hole to the power of three $V \approx R^3$ then the density of a black hole is inversely proportional to its mass radius by the second power $\rho \approx M^2$.

What does all this mean? It means that if a black hole has a large enough mass then it does not appear to be very dense, which is more or less the description of our own vacuum: it has a lot of matter, around $3 \times 10^{80}$ atoms in spacetime alone - this is certainly not an infinite amount of matter, but it is arguably a lot yet, our universe does not appear very dense at all.

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  • $\begingroup$ I'm deleting this since it doesn't appear to answer the question. $\endgroup$ – David Z Apr 10 at 8:14
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    $\begingroup$ After some discussion among the mods, it's a little clearer how this relates to the question, and we've un-deleted it. (Gareth, feel free to flag these comments as "no longer needed" once you've seen them.) $\endgroup$ – rob Apr 11 at 19:29

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