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I have just read a short line phrase (published on Instagram) which states this:

"If you could produce a sound louder than $1100$ dB, you would create a black hole, and ultimately destroy the galaxy".

Can you tell me if this phrase is true, and why? What would mean $1100$ dB of sound, what would be the real effect?

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    $\begingroup$ I have no idea what the (unknown) article you say meant, but please read this question about the loudest sound possible and related links. Anything about 191 dB is not considered a sound as such. $\endgroup$ – StephenG Aug 11 '17 at 12:23
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    $\begingroup$ A possible answer: since sounds have energy density a loud enough sound would imply enough mass-energy to implode. Decibel is power rather than energy density, but given a volume you get a density from sound energy passing through. Exactly what density is needed for implosion is a bit uncertain, but since 1100 db is about 10^100 W, which is above the Planck power it seems reasonable. $\endgroup$ – Anders Sandberg Aug 11 '17 at 12:46
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The definition for acoustic decibels is

$$L = 20\log_{10}\frac{P}{P_0}$$

where the reference pressure is $P_0=20\, \mu \mathrm{Pa}$ in air. Thus $L=1100\,\textrm{dB}$ would give

$$P = 2\times 10^{50}\,\mathrm{Pa}.$$

There is no physics up to here, just definitions. The gist of the claim, I guess, is to naively apply acoustic, even though that pressure is too high to make any sense. The energy density of a wave would be

$$w = \frac{P^2}{\rho c_s^2}$$

where $\rho$ is the mass density and a $c_s$ the sound speed. For air, $\rho \approx 1\, \mathrm{kg}/\mathrm{m}^3$ and $c_s\approx 300\, \mathrm{m}/\mathrm{s}$, so

$$w \approx 10^{98}\,\mathrm{J}/\mathrm{m}^3.$$

What to do with that number? Not sure. A black hole forms when 3-4 solar masses collapse. The corresponding total energy, naively using $E=mc^2$, is $E_\bullet \approx 10^{48}\,\mathrm{J}$. Clearly, as @AndersSandberg found out too, this acoustic wave energy is way higher than this threshold. So collapse, yes, but the specific number 1100 dB had me believe that this would be a threshold.

Another idea, would be to consider how small a volume would get us to the threshold of black hole collaspse: if the above energy density $w$ is contained in a volume $V=E_\bullet/w=10^{-50}\,\mathrm{m}$, we are there. That would be a cube of dimension $\approx 10^{-17}\,\mathrm{m}$, which is 1/100th of a proton radius. This makes no particular sense.

We can run it the other way around by taking a volume of $V=1\, \mathrm{m}^3$, and require $w = E_\bullet/V\approx 10^{48}\,\mathrm{J}$, which using the acoustic formula for $w$ gives $P\approx10^{26}\,\mathrm{Pa}$, and therefore a level of $\approx 600\,\mathrm{dB}$. So from that perspective, the claim should say 600 dB instead of 1100 dB. Note this is not the same thing as what @AndersSandberg calculated.

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  • $\begingroup$ Note that if you have 10^98 J you have 10^50 solar masses per cubic meter. That sounds very collapsible. $\endgroup$ – Anders Sandberg Aug 11 '17 at 17:10
  • $\begingroup$ Yes, sure. I interpreted the claim reported by the OP as a threshold though. But that does not work. I should have been clearer. I was working on my answer while you posted yours, so I did not notice it, by the way. $\endgroup$ – user154997 Aug 11 '17 at 19:24
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The phrase is not true: it looks like the sound cannot form a black hole.

A sound of intensity $P$ Watts per square meter has a sound power level $L=10\log_{10}(P/10^{-12})$ decibel. If we turn the equation around, $P=10^{(L/10)-12}$ Watt. So a 1100 dB sound has intensity $10^{98}$ Watt per square meter.

The Planck intensity, where the energy level is enough to cause gravitational effects, is $1.4\cdot 10^{122}$ Watts per square meter.

So we are about 24 orders of magnitude below the point where the sound will start affecting spacetime. Making black holes this way doesn't seem to work. We need 1340 dB!

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  • $\begingroup$ Note that sound intensity is frequently reported in dB SPL, which is the sound pressure referred to a reference level of $20\,\mu\mathrm{Pa}$. $\endgroup$ – Massimo Ortolano Aug 11 '17 at 13:15
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You can't get sound in air louder than around $190dB$. The reason is the rarefied or minimum part of the wave becomes a vacuum. A sound wave louder must be in a pressurized vessel. People do actually work on these things and I read some years ago about a $600dB$ sound in such a thing. The other way to get something louder is to have a shock wave. As seen in the calculations above you need enormous pressure to generate a black hole.

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  • $\begingroup$ You can't get a sound wave louder than 190dB. however, you can create a shock with a peak pressure almost as high as you like. Whether you feel it is valid to measure its intensity in dB as though it was a sound wave may be another question. $\endgroup$ – Cort Ammon Aug 12 '17 at 1:29

protected by Qmechanic Aug 11 '17 at 20:55

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