Can sound distort the fabric of space?

[John Clement Husain]

Is it possible for Low Frequency to distort the fabric of space? If so, how?

 

[PeterDonis]

Low Frequency

Low Frequency what?

 

[John Clement Husain]

Low Frequency what?

of Sound

 

[PeterDonis]

of Sound

Sound requires a medium, such as air. The medium will have nonzero stress-energy, so it will curve spacetime, yes. The sound itself will contain some energy, but that will be included in the stress-energy tensor of the medium if we properly evaluate that tensor with the sound present.

 

[PeterDonis]

Moderator's note: I have edited the title of this thread to be more descriptive of the specific question.

 

[John Clement Husain]

Are there any equations for it? I haven't seen any study done online yet...

 

[PeterDonis]

Are there any equations for it?

The stress-energy tensor for a perfect fluid is well known; see, for example, here:

https://en.wikipedia.org/wiki/Perfect_fluid

The presence of sound waves just means the pressure $p$ and density $\rho$ of the fluid are functions of position and time.

 

[John Clement Husain]

OH THANK YOU VERY MUCH

 

[Paul Colby]

Actually, I've just completed a maxima calc for the transverse traceless gravitational radiation from any (well in principle) general time harmonic stress tensor. Basically, only shear waves in a material emit gravitational waves. When I feed my calc a diagonal pressure wave, I get 0 gravitational radiation. So my answer is no. Compressional sound waves do not radiate. Shear waves will however. [edit] This is in the linearized theory of course.

 

[PeterDonis]

Compressional sound waves do not radiate.

That's not what he asked. He asked if sound waves curve spacetime. Emitting gravitational radiation is not the only way for something to curve spacetime.

 

[Paul Colby]

That's not what he asked. He asked if sound waves curve spacetime. Emitting gravitational radiation is not the only way for something to curve spacetime.

However, in the time harmonic case for weak fields I bet by far the most significant contribution are the near field effects which go like $1/(kr)^{5}$. Only the $1/kr$ terms are radiation proper. So for a sound wave I think this is the answer to his question. If not I'd like to know. [edit] even these near fields are 0 for scalar pressure waves.

 

[PeterDonis]

by far the most significant contribution are the near field effects

No, by far the most significant contribution is the energy density of the medium, including the energy density contained in the sound waves. That is going to be many, many, many orders of magnitude larger than any gravitational radiation.

The next most significant contribution by far is going to be the pressure of the medium, including the pressure caused by the sound waves. That, while many orders of magnitude smaller than the energy density, is still going to be many, many orders of magnitude larger than any gravitational radiation.

 

[Paul Colby]

Yes, $mc^2$ will always win for the static component. Not my point.

 

[PeterDonis]

$mc^2$ will always win for the static component. Not my point.

Not mine either. My point is that the energy density will cause many, many, many orders of magnitude more spacetime curvature than the gravitational waves you are talking about (with pressure in second place, still by many, many orders of magnitude). And the OP was asking about spacetime curvature.

Also, the energy density and pressure in question are not static. They are time-varying, because there are sound waves present.

 

[hilbert2]

You'd need to have a quite high-amplitude pressure wave to cause relativistic effects...

 

[John Clement Husain]

Say, if space is like water,
then can we safely say that if we shoot a low frequency sound in a single point it will be able to create a spatial whirlpool?

 

[Ibix]

Say, if space is like water,

It's spacetime that curves in general relativity, not space, and it's not like water.

What Peter was saying is that a pressure wave in water or air (or whatever medium) means changes in the energy and stress distribution in the matter, and that means changes in spacetime curvature. He wasn't saying that spacetime itself supports sound waves.

You can get waves in spacetime. These are called gravitational waves, but they don't have much in common with sound waves.

 

[John Clement Husain]

Hmm, I do understand that space is basically the thing that encompasses everything (Like why are you shaped like that? Why is a door shaped rectangular? etc.) but what I meant was about the matter in space. Is it PLAUSIBLE, if not possible, to create such phenomenon in my previous statement?

 

[Ibix]

Spacetime is the relevant thing. It isn't generally possible to separate it into space and time in a unique way.

I'm not sure what you are trying to describe, but I would just point out that gravitational effects due to the rotation of the entire Earth are only detectable after months of integration time (look up Gravity Probe B). So I very much doubt that a mass of air in motion at soundwave speeds would produce any gravitational effect detectably different from a stationary mass.

 

[John Clement Husain]

Ah, I see... thank you!

 

[PeterDonis]

I very much doubt that a mass of air in motion at soundwave speeds would produce any gravitational effect detectably different from a stationary mass

This is true, yes. My responses to Paul Colby were only pointing out that even this every small effect (too small to detect with our current technology) is still many, many orders of magnitude larger than the effect of gravitational radiation due to the sound waves in the air.

 

[Paul Colby]

This is true, yes. My responses to Paul Colby were only pointing out that even this every small effect (too small to detect with our current technology) is still many, many orders of magnitude larger than the effect of gravitational radiation due to the sound waves in the air.

My confusion is that I get identically zero gravitational radiation from a pressure wave in an ideal fluid when the TT gauge is used. This makes sense to me since weak GW are shear waves and the stress energy for an ideal fluid has no shear. Identically zero is always many orders smaller than any other non-zero effect one cares to name. If this is known to be false a reference would be helpful.

 

[PeterDonis]

I get identically zero gravitational radiation from a pressure wave in an ideal fluid when the TT gauge is used.

For an ideal fluid, yes, AFAIK this is correct. However, you did not just mention pressure waves in an ideal fluid. You also mentioned shear waves, which will be present in actual air since air is not quite an ideal fluid. You also mentioned near field effects going like $1 / (kr)^5$. My general point was that the spacetime curvature due to the stress-energy tensor of the fluid (energy density and pressure--not pressure changes but just pressure itself) is many, many orders of magnitude larger than any of these effects.

 

[Paul Colby]

For an ideal fluid, yes, AFAIK this is correct.

Thanks.

 

[Andy SV]

I wonder if the shock front of a nuclear blast would have a significantly greater curve than the max amplitude of sound I think it would but how would distance effect it near the center has the most energy but it would collect more density as it propagated

Not a serous question but the conversation had me thinking it

 

[Ibix]

If memory serves, nuclear weapons convert 1-2% of their mass to energy. That means that, even at detonation, the energy density associated with the explosion is around two orders of magnitude smaller than the energy density associated with the bomb a moment earlier. And that number falls extremely rapidly as the explosion expands.

Given that the mass of fissile material in a nuclear weapon isn't hugely more than the masses used in the Cavendish experiment, I'm going to go ahead and say that standard Newtonian gravity effects would only be detectable at ground zero (and not until after the shockwave passed you, so probably not actually at all), and relativistic corrections to that are indetectable.

 

[JLowe]

If memory serves, nuclear weapons convert 1-2% of their mass to energy. That means that, even at detonation, the energy density associated with the explosion is around two orders of magnitude smaller than the energy density associated with the bomb a moment earlier. And that number falls extremely rapidly as the explosion expands.

Given that the mass of fissile material in a nuclear weapon isn't hugely more than the masses used in the Cavendish experiment, I'm going to go ahead and say that standard Newtonian gravity effects would only be detectable at ground zero (and not until after the shockwave passed you, so probably not actually at all), and relativistic corrections to that are indetectable.

How dense does the fuel become the moment before explosion? I've only read things like "extremely hot and dense", and temperatures of millions of degrees.

 

[Drakkith]

How dense does the fuel become the moment before explosion? I've only read things like "extremely hot and dense", and temperatures of millions of degrees.

I found the following quote (underlining mine) from a website that talks about nuclear weapon design: http://nuclearweaponarchive.org/Nwfaq/Nfaq4-1.html (Section 4.1.2)

In the table below I give some illustrative values of c, total cross section, total mean free path lengths for the principal fissionable materials (at 1 MeV), and the alphas at maximum uncompressed densities. Compression to above normal density (achievable factors range up to 3 or so in weapons) reduce the MFPs, alphas (and the physical dimensions of the system) proportionately.

If I understand the article correctly, the core is compressed to a density of upwards of 3x normal density. But I admit I only briefly scanned the article, so I could be utterly and completely wrong here.

 

[PeterDonis]

a website that talks about nuclear weapon design

This is for a fission weapon--more precisely, an implosion type fission weapon. But those kinds of weapons don't convert 1% or so of their fuel mass to energy; it's less than 1/10 of that. To get about 1% conversion rate, you need a fusion weapon. Those are discussed in a different part of the same website you linked to:

http://nuclearweaponarchive.org/Nwfaq/Nfaq4-4.html#Nfaq4.4.3

As far as I can tell, this page doesn't give explicit numbers for the compression ratio, but it seems to imply that the fusion fuel is compressed to degenerate matter densities (which would make sense), which are many orders of magnitude higher than the densities of ordinary solids, which is what the fusion fuel is before detonation. So it looks like the compression ratios for fusion weapons are much higher than those for fission weapons.

 

[PeterDonis]

it seems to imply that the fusion fuel is compressed to degenerate matter densities

Hm, actually this appears to be wrong. Section 4.4.3.4.3 (gotta love those numbering schemes ) estimates a compression ratio of 197 for the "Mike" device (one of the early tests done by the US in the 1950s) and 878 for the W-80 (one of the standard weapons in the modern US arsenal). Still much larger than the ratio of 3 estimated for fission weapons, but nowhere near degenerate matter densities.

 

[Andy SV]

Well yes but that is for what starts the explosion.
What does the edge of the expanding void do to space time as it sweeps up
The air and converts it to a dense shell of plasma. It probably contains a great deal of mass in a comparatively thin shell before it starts to implode.

 

[PeterDonis]

What does the edge of the expanding void do to space time

Not much in a practical sense. Even at 1000 times the density of ordinary solids, which is larger than the largest compression ratio I can get from the web site, the gravitational effects of the matter in the weapon would only be detectable by sensitive devices like torsion balances. Any analysis of the weapon's effects could assume flat spacetime in the weapon's vicinity as an extremely good approximation.