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Message boards : General discussion : What Caused the Big Bang / Big Bounce

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Message 117948 - Posted: 15 May 2018 | 9:20:55 UTC
Last modified: 15 May 2018 | 10:13:09 UTC

As far as I know, the mechanism of the Big Bang / Big Bounce is open to speculation, with all sorts of unproven physical theories being bandied about.

However, there is a simple and reasonable explanation with known physics for how a Big Crunch causes a Big Bang. Simply put, coalescing black holes suddenly emit Hawking radiation at a higher rate than the combined rate of the pre-collision black holes. This will be shown below.

Black holes grow by 2 methods: 1) by gradual accretion of ordinary infalling matter (whose incremental mass is very small in relation to that of the black hole), and 2) by coalescing black holes as observed by the LIGO experiment.

In relation to method 1, it is conceivable that a black hole grows so large that it reaches equilibrium with its environment, where the rate of infalling mass equals the rate of mass ejection through Hawking radiation. This doesn't lead to an explosion, merely a modulating rate of Hawking radiation as rates of infalling matter vary.

However, method 2 leads to quite another story.

Coalescing black holes SUDDENLY emit mass through Hawking radiation at a rate that exceeds the total rate of mass ejection of the individual black holes. Here is the proof, and it's mathematical of course.

I assume (so can someone verify this?) that the absolute rate of Hawking radiation (H) from a black hole is directly proportional to the surface area (A) of the black hole at the event horizon. The event horizon has Schwarzschild radius (R). So H = K (A) = K 4 Pi (R)^2 with some constant K. If K is not constant, this theory is in trouble, but it it still far simpler to assume a constant value of K than to invent completely new physics as is the current fashion in cosmology.

Now consider the case of coalescing black holes, which has been shown to happen with the LIGO experiment.

Given 2 equal mass black holes, each of mass M,

the Schwarzschild radius of each black hole is (R1) = 2 GM / c^2

and surface area at the event horizon of each black hole is the sphere (A1) = 4 Pi (R1)^2.

So the 2 black holes have combined surface area equal to 2 (A1) and the combined rate of Hawking radiation (Hc) is thus (Hc) = 2 K (A1).

When the 2 equal mass black holes coalesce, the mass of the resulting black hole is (2M) and its Schwarzschild radius is (R2) = 2 G(2M) / c^2 = 2 (R1).

The surface area of the coalesced black hole is (A2) = 4 Pi (R2)^2 = 4 Pi (2 (R1))^2 = 16 Pi (R1)^2 and its rate of Hawking radiation is thus (H2) = K (A2).

How much is (H2)? Well, the ratio of the rates of Hawking radiation "after" divided by "before" coealescing is (H2) / (Hc) = (K A2) / (2 K A1) = (A2) / (2 A1) = 16 Pi (R1)^2 / (2 * 4 Pi (R1)^2) = 2.

Exactly double! The coalesced black hole emits Hawking radiation at double the combined rate of the two original (same size) black holes together. Note that this relation holds for any size black hole.

For "small" black holes of dozens of solar masses, we have seen a lot of energy released as gravitational waves. For black holes at equilibrium with Hawking radiation the masses are vastly larger, and the energy emitted is both gravitational waves and via Hawking radiation. Mathematically we have shown that the rate of Hawking radiation doubles when 2 equal sized black holes merge, without consideration for the gravitational energy release.

At equlibrium a black hole ejects as much matter as it accumulates. Two such masses coalescing will eject the mass of both black holes in the time it takes them to coalesce. We know this is pretty short, 0.2 second in the first LIGO observation.

That's a real BIG BANG, caused by the merger of 2 equal size black holes, and it doesn't necessarily require the entire mass of the "known universe", just 2 sufficiently large black holes at Hawking radiation equlibrium. I'll let someone else figure out how much mass is required for a black hole to be at Hawking radiation equilibrium, and the value of K.

EDIT: because the margin is too small for the rest of the proof.

EDIT EDIT: Oh yes, I have to say that I assume the rate of Hawking radiation production has no upper limit, and that it increases faster than the rate of mass increase, so that eventually a black hole grows to Hawking radiation equilibrium and no larger. This is where you may prove me wrong. But I doubt it, since I have shown that doubling the mass of a black hole doubles its radius and the Hawking radiation assumption is proportional to the surface area.

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Message 117953 - Posted: 15 May 2018 | 10:58:56 UTC
Last modified: 15 May 2018 | 11:14:26 UTC

This Big Bang theory has an interesting implication.

Black holes are localized, so it suggests that Big Bangs are localized phenomena.

Who has seriously suggested that the cosmic microwave background (CMB) radiation is not a residue from "our" Big Bang? We don't know how far away the CMB is. We can't measure it's distance. We only see events in front of it that are 14 billion years old, so we assume that the CMB also came from that event.

That's a stretch. It could be the radiation from many other far more distant bangs that is reaching us now from a lot longer than 14 billion years ago.

The standard line we hear in cosmology is the CMB is from "our" Big Bang, and theories like "inflation" have been proposed to explain its uniformity. That's pretty self-centred thinking, and we have a long and respected history of that - like positing that the stars are part of our sky and turning around the Earth.

All I know is, nothing has ever worked better than thinking "it's not just us, there's more of the same".

EDIT: I just remembered the main support for the CMB distance/dating is red shifting the 21 cm emission line from hydrogen. I retract this particular post. However, the Big Bang theory stands.

EDIT EDIT: Uh, wrong again. It's a thermal argument. I should just stick to defending the original post.

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Message 117954 - Posted: 15 May 2018 | 11:05:42 UTC - in response to Message 117952.


Here you also have the single point, next for also the singularity itself, if not also the line, area for that of a square, and next also the volume for that of the cube, except for the notion of time as well.

That's not the point. Hawking radiation happens away from the singularity, at the event horizon. Matter materializes from the quantum vacuum energy and is separated from antimatter by the gravity of the black hole. Ergo, the larger the event horizon, the more matter can materialize and be expelled. Maybe there's an estimate on the upper limit of the rate of spontaneous materialization?

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Message 117974 - Posted: 15 May 2018 | 20:35:51 UTC - in response to Message 117948.

At equlibrium a black hole ejects as much matter as it accumulates. Two such masses coalescing will eject the mass of both black holes in the time it takes them to coalesce. We know this is pretty short, 0.2 second in the first LIGO observation.


Interesting theory.
But I don't undestand why the new single black hole would have to eject the mass (Hawking radiation) so quickly.
The coalescence is a very quick process and converts an incredible amount of mass into gravitational-wave energy. This is a sort of gravitational fusion.
But the merged black hole may be relatively stable. It has slightly less than twice the mass of a black hole at equilibrium. It will decay to equilibrium but this process may be long.

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Message 117985 - Posted: 16 May 2018 | 4:41:42 UTC - in response to Message 117974.

At equlibrium a black hole ejects as much matter as it accumulates. Two such masses coalescing will eject the mass of both black holes in the time it takes them to coalesce. We know this is pretty short, 0.2 second in the first LIGO observation.


Interesting theory.
But I don't undestand why the new single black hole would have to eject the mass (Hawking radiation) so quickly.
The coalescence is a very quick process and converts an incredible amount of mass into gravitational-wave energy. This is a sort of gravitational fusion.
But the merged black hole may be relatively stable. It has slightly less than twice the mass of a black hole at equilibrium. It will decay to equilibrium but this process may be long.

You have a point of contention. The only hypothesis that is supported by the assumption is that the rate of mass ejection is proportional to the surface area. Since the surface area is doubled (approximately), so is the rate of conversion of matter from the vaccuum energy.

The equilibrium state says nothing about the rate of mass ejection except that it balances the rate of mass injection. If K is very very small, then R at equilibrium is huge. Doubling the surface area creates a big surplus in the rate mass ejection over equilibrium. Releasing that much mass in a short time is an explosion. However, if K is fairly large, then R at equilibrium will not be so big, and the rate of excess mass ejection will be smaller, appearing to be a somewhat slower decay as you suggest.

The largest black hole detected so far is in NGC 1277 and it is estimated to contain 17 billion solar masses. We can surmise that this black hole is not larger than equilibrium size, or else it would be a very bright source of radiation. The size of this black hole is about 11 times the "size of the orbit of Neptune". It is also suggested to have been hanging around since the big bang, so it has had a long time to decay to it's present size if it is now at equlibrium. https://www.space.com/18668-biggest-black-hole-discovery.html.

But we are supposing that Hawking radiation is the mechanism of the Big Bang, so the total mass of the coalesced black hole would be in the range of 1000 billion billion stars by current estimates of stars in the universe. https://helios.gsfc.nasa.gov/qa_star.html What would be the Schwarzschild radius for half this much mass, and then what is the implication for magnitude of K?

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Message 118001 - Posted: 16 May 2018 | 12:17:02 UTC - in response to Message 117985.

But we are supposing that Hawking radiation is the mechanism of the Big Bang, so the total mass of the coalesced black hole would be in the range of 1000 billion billion stars by current estimates of stars in the universe. https://helios.gsfc.nasa.gov/qa_star.html What would be the Schwarzschild radius for half this much mass, and then what is the implication for magnitude of K?

Note that if the expansion of our universe is the Hawking radiation of a huge black hole, we may not have to suppose that the "Big Bang" was 'fast'. The matter comes from the surface of a black hole, an area where the time is highly distorted and the laws of physics are unknown!

But how to explain the acceleration of the expansion of our universe with it? https://map.gsfc.nasa.gov/universe/bb_concepts_exp.html

And the expansion is a 3D phenomenon (we see the "Big Bang" in any direction)... the "wave" of a black hole is a surface???

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Message 118003 - Posted: 16 May 2018 | 13:43:42 UTC - in response to Message 118001.

Note that if the expansion of our universe is the Hawking radiation of a huge black hole, we may not have to suppose that the "Big Bang" was 'fast'.


You are correct. The speed of expansion of the "Big Bang" is a theory which is based on the need for inflation, which is another invention to explain the uniformity of the CMB. But the surface of a black hole is uniform. Minute variations in the CMB could be the result of ripples (ringing gravitational waves) emanating from the merger of the 2 super-size black holes.


But how to explain the acceleration of the expansion of our universe with it? https://map.gsfc.nasa.gov/universe/bb_concepts_exp.html


Maybe the primordial black hole is still sending off matter? Then we are not in a Big Bang universe but in one of continuous creation.


And the expansion is a 3D phenomenon (we see the "Big Bang" in any direction)... the "wave" of a black hole is a surface???


Yes, and we are inside the bubble of expansion.

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Message 118007 - Posted: 16 May 2018 | 23:03:22 UTC - in response to Message 118003.


And the expansion is a 3D phenomenon (we see the "Big Bang" in any direction)... the "wave" of a black hole is a surface???


Yes, and we are inside the bubble of expansion.

Inside or on the surface?

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Message 118008 - Posted: 17 May 2018 | 1:02:33 UTC
Last modified: 17 May 2018 | 1:07:03 UTC

This is baffling. These authors (Premalatha, Balamurugan, Kannimuthu) say that radiation from a black hole is inversely proportional to its mass, which is the exact opposite of my postulate.

The thermal spectrum temperature is proportional to the surface gravity of the black hole, which for a Schwarzschild black hole, is reciprocally proportional to the mass. Recent Developments in Intelligent Nature-Inspired Computing, edited by Patnaik, Srikanta (google books)

Is that "surface" something other than the event horizon? We all know that the event horizon is the sphere around the singularity where the escape velocity matches the speed of light.

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Message 118009 - Posted: 17 May 2018 | 3:17:09 UTC - in response to Message 118008.

This is baffling. These authors (Premalatha, Balamurugan, Kannimuthu) say that radiation from a black hole is inversely proportional to its mass, which is the exact opposite of my postulate.

The thermal spectrum temperature is proportional to the surface gravity of the black hole, which for a Schwarzschild black hole, is reciprocally proportional to the mass. Recent Developments in Intelligent Nature-Inspired Computing, edited by Patnaik, Srikanta (google books)

Is that "surface" something other than the event horizon? We all know that the event horizon is the sphere around the singularity where the escape velocity matches the speed of light.

So Hawking radiation is quantum mechanical tunnelling from the singularity to the event horizon, making it less likely as the radius grows? That's a different process than virtual particles born from vacuum energy fluctuations, which happens everywhere in space; and being separated by the gravity gradient at the event horizon rather than returning to the vacuum, which is the basis for my assumption. Then I have been using the wrong name for this latter type of radiation.

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Message 118066 - Posted: 19 May 2018 | 9:25:05 UTC

This is funny. I started by questioning whether new physics is needed to explain the Big Bang, and ended up requiring new physics (non-Hawking black hole emission proportional to surface area at the event horizon) to achieve a hypothesized equilibrium state.

Since the observable universe contains about 10^54 kg and the largest suspected black hole so far contains around 10^40 kg (within a factor of 10^4 either way) there is still a lot of room for the possible existence of this emission mechanism, with a constant factor of proportionality so tiny that we haven't seen it yet.

It's pretty difficult to hide a black hole as big as needed to see it emitting radiation, so the proposed physics is wrong, or there is no such black hole, or we haven't associated one with a known source of radiation. On the other hand, at equilibrium we're not going to see one, will we?

Message boards : General discussion : What Caused the Big Bang / Big Bounce

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