# The largest black hole [duplicate]

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?

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

• Possible duplicates: physics.stackexchange.com/q/23118/2451 and links therein. – Qmechanic Apr 10 at 4:27
• The only issue with your approach, is that you said it ''hits'' zero, hopefully my answer below demonstrates it only approaches zero. – Gareth Meredith Apr 11 at 19:35
• 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. – old lowboy Apr 13 at 1:39

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.

• I'm deleting this since it doesn't appear to answer the question. – David Z Apr 10 at 8:14
• 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.) – rob Apr 11 at 19:29