I Second Tonga shockwave - why a drop in the pressure?

[Borek]

TL;DR

Weather station recording pressure at my home registered the shockwave from Tonga eruption twice, as it was traveling around the globe. That's expected, but what is the physics behind the fact second shockwave registered as a drop in the pressure, not as a spike?

I already posted this image in GD as a curiosity, but there is one thing that I don't get. This is pressure recorded by my weather station near Warsaw in Poland after the Tonga eruption (the description says '24 h' but it is wrong, I forgot to modify the script generating the plot). Around 20:05 there is a first pressure spike, then, several hours later, around 2:51 there is a second change in the pressure. In both cases timing and the distance agree with the expected propagation of the shockwave with the speed of sound, so there is no doubt these are both related and there is doubt first event is the shockwave propagating from the epicenter and the second event is the shockwave traveling around the globe back to Pacific. Identical (or at least reasonably similar) patterns were observed by other people in Poland at around the same time (give or take several minutes related to the distance).

But why is the second event a drop in the pressure? I expected it to be a spike as well, not a drop. In GD @Jonathan Scott suggested it can be effect of the wave going in a growing circle, but somehow I am not convinced.

(in case you wonder: I haven't seen the third shockwave, but at the time when it was expected there was a violent atmospheric front getting through here, so there was plenty of noise in the pressure measurements)

 

[Jonathan Scott]

Similar patterns were observed in several places in the UK as well at similar times.

For now, I stand by my idea that the shockwave would displace a huge ring of air "inwards" the first time and "outwards" the second (on the way back) and hence overall temporarily compress it one way and expand it the other. I admit it sounds as if it would be too small an effect to generate such a visible result, but I think when you consider the huge speeds and volumes involved, it does seem possible.

Let's assume the circle is moving at 1100km/h, the observation point is about 16000km from the start and the circumference of the Earth is about 40000km.

The distance left to the furthest point from the start is about 4000km, one fifth of the way, so the radius of the circle is not very much less. For the circumference to change by (say) about 1 part in 1000, resulting in a similar change of pressure, the radius must therefore change by about 4km. This seems plausible to me as the sort of order of magnitude by which the ring of air could be displaced by the eruption shock wave.

Edit: The wave would travel this distance in less than 15 seconds.

 

[Frabjous]

Interesting observation.
Maybe @Astronuc knows.

 

[boneh3ad]

If the "negative" spike is indeed related to the eruption, then it is not a shock wave. A so-called expansion shock violates the second law of thermodynamics except in some very specific and uncommon scenarios when gas behavior departs from ideal. The atmosphere is very nearly an ideal gas. Further, a shock wave propagates faster than the local speed of sound in the medium into which it is propagating.

Off the top of my head, I don't know what that negative spike is from, but it is not a shock.

 

[DaveC426913]

 

[Baluncore]

I would expect the gas explosion to have been a superheated groundwater steam explosion. After the initial impulse due to boiling on release, there is a depression (following a few seconds later) due to the outward momentum of the steam. Then there follows a regional condensation period over an hour or so, when the steam returns to liquid and falls as rain. That condensation phase can be seen in the pressure record following the high pressure event.

 

[Jonathan Scott]

The initial shock wave from the eruption will have degenerated into a pressure wave traveling at the speed of sound (like, as you might guess, sound, so one might expect some sound from the eruption to accompany the wave). Such a pressure wave on a small scale would normally have no overall effect on the pressure on the time scale measured by weather instruments, as the changes would be too fast and would average out. So the small overall pressure change shown for a few minutes after the wave passes is a different effect.

Any explanation for this change needs to explain why the same pressure wave causes a temporary positive change when first going past, then a temporary negative change when the same wave passes in the other direction. I think my explanation covers this, although if someone's expertise already covers this area I'd like to hear from them.

I assume that since the speed of sound near sea level is higher, the lower part of the pressure wave would be largely deflected up to a level where the speed of sound is approximately constant with height, so the effects at ground level would be tiny compared with those at height.

 

[Orthoceras]

The vulcano has its antinode somewhere in the Sahara. I wonder whether the shape of the pressure wave is distorted at the antinode. Ideally, the pressure wave at the antinode should be identical to the original pressure at the eruption. However, the original pressure might have been so extreme that the propagating pressure wave is severely distorted when approaching the antinode, perhaps affecting the positive peak more than the negative peak.

The pressure wave of the Krakatau that traveled around the world in 1883 seems to show the same effect: at the second visit of the wave, the negative peak is more conspicuous.

 

[Jonathan Scott]

The speed of sound would vary with temperature around the globe, so I presume the pressure wave would reach the opposite point at slightly different times from different directions, causing the effect to be "out of focus" at that point but probably still quite powerful.

 

[Orthoceras]

But why is the second event a drop in the pressure? I expected it to be a spike as well, not a drop. In GD @Jonathan Scott suggested it can be effect of the wave going in a growing circle, but somehow I am not convinced.

For now, I stand by my idea that the shockwave would displace a huge ring of air "inwards" the first time and "outwards" the second (on the way back) and hence overall temporarily compress it one way and expand it the other.

I don't fully understand your explanation for the second event being a drop. I would think that p, the amplitude of the pressure wave, is a function of θ, the angular distance to the vulcano, along the great circle. If energy is conserved, and pressure is energy density, p should be inversely proportional to sin θ. The angle θ is identical for the first and second visit of the pressure wave to a city. Why would inwards/outwards (equivalent to sinθ decreasing/increasing) matter?

 

[Jonathan Scott]

The pressure wave itself is the same both ways apart from some loss of energy, and if you could record sufficiently fast you would see a similar pressure profile as it passes. However, the wave knocks a lot of air temporarily in the forwards direction as it passes, which then has to get back to equilibrium afterwards. If you imagine that occurring on a flat disk, it is clear that displacing a volume of air inwards will create temporary high pressure, and displacing it outwards will create temporary low pressure.

 

[Orthoceras]

What would that imply for a pressure recording in Australia? The first visit a low pressure, and the second visit a high pressure?

 

[Jonathan Scott]

What would that imply for a pressure recording in Australia? The first visit a low pressure, and the second visit a high pressure?

Yes, that's what I would expect (as a smoothed effect on the scale of minutes). Does anyone know of any such recordings?

 

[Frabjous]

The pressure wave itself is the same both ways apart from some loss of energy, and if you could record sufficiently fast you would see a similar pressure profile as it passes. However, the wave knocks a lot of air temporarily in the forwards direction as it passes, which then has to get back to equilibrium afterwards. If you imagine that occurring on a flat disk, it is clear that displacing a volume of air inwards will create temporary high pressure, and displacing it outwards will create temporary low pressure.

I do not follow what you are saying. A blast wave ideally looks like this

In a convergent geometry the peak increases. In a divergent geometry (which I believe is the case we are concerned with for both waves) the peak decreases.

 

[Jonathan Scott]

At a sufficient distance from the origin (which may admittedly rule out NZ and Australia), I would expect the pressure wave itself to be barely detectable on a scale of minutes because of being too rapid to be recorded by weather instruments, and the average deviation due to the pressure wave itself would be zero if it was moving with a straight wavefront. The effect I'm expecting to see is not due to the pressure wave itself, but due to the overall displacement of air in the forward direction caused by the passage of the pressure wave.

 

[Frabjous]

At a sufficient distance from the origin (which may admittedly rule out NZ and Australia), I would expect the pressure wave itself to be barely detectable on a scale of minutes because of being too rapid to be recorded by weather instruments, and the average deviation due to the pressure wave itself would be zero if it was moving with a straight wavefront. The effect I'm expecting to see is not due to the pressure wave itself, but due to the overall displacement of air in the forward direction caused by the passage of the pressure wave.

At large distances, it is locally going to look like a plane wave so I do not understand why the sign changes.

 

[Orthoceras]

The Bureau of Meteorology, Western Australia twittered the following pressure recordings. Curiously, the positive peak and the negative peak do not have the same propagation speed.

s

 

[Jonathan Scott]

At large distances, it is locally going to look like a plane wave so I do not understand why the sign changes.

The observation was 4/5 of the way to the other side, where the radius of the ring changes rapidly as the wave propagates. By then I would expect the pressure wave itself to barely register, but the wave will displace a large amount of air forwards as it passes, leaving it some distance from its original position. If it displaces it in an inwards direction, reducing the circumference, I would expect the wave to be followed by a temporary period of excess pressure, and if it displaces it in an outwards direction, I would expect it to be followed by a temporary period of lower pressure.

 

[Jonathan Scott]

The Bureau of Meteorology, Western Australia twittered the following pressure recordings. Curiously, the positive peak and the negative peak do not have the same propagation speed.

That's very interesting, thanks. Clearly at that point the excess pressure lasted long enough to register directly, and we then see the rebound negative pressure. The overall effect is gradually getting smoothed and spread out.

However, the effects look symmetrical here, without an overall temporary higher or lower pressure, so I suspect that this close to the origin the displacement effect which I have suggested above is hidden by the direct effects of the pressure wave itself.

 

[Orthoceras]

Curiously, the positive peak and the negative peak do not have the same propagation speed.

When extrapolating it back to the eruption of the vulcano, the negative peak occurred almost at the same time as the positive peak (less than 5 minutes later).

 

[boneh3ad]

The Bureau of Meteorology, Western Australia twittered the following pressure recordings. Curiously, the positive peak and the negative peak do not have the same propagation speed.

View attachment 295685s

As I previously mentioned, shocks travel faster than the speed of sound. If the initial rise is still strong enough to be a shock, it will be traveling faster than the negative portion that follows, which travels at the speed of sound.

 

[Borek]

I would not assume constant speed, it is a function of air density, which in turn depends on temperature, pressure and humidity.

 

[boneh3ad]

I would not assume constant speed, it is a function of air density, which in turn depends on temperature, pressure and humidity.

In an ideal gas, it's really just a function of temperature thanks to the simple equation of state. The atmosphere is approximated really well by an ideal gas model. The speed of sound, $a$, is
a = \sqrt{\gamma R T},
where $\gamma$ is the ratio of specific heats, $R$ is the specific gas constant, and $T$ is the local temperature.

 

[cjl]

Yep, though in the atmosphere the humidity also creates an effect since that's changing the gas composition (and thus the specific heat ratio and the gas constant).

 

[256bits]

View attachment 295659

I think that is basically the correct idea.

A huge volume of gas rises from the ocean displacing a volume of air already there.
Subsequently a high pressure is produced.
The hot gas rises ( 60000 feet I believe ) and surrounding air backflows into the region.
Subsequently a low pressure is produced.

The high and low pressures travel outwards from the volcano area around the world.
It's a tsunami, but this time not in the water but in the air, both water and air being fluids.

 

[Borek]

I think that is basically the correct idea.

A huge volume of gas rises from the ocean displacing a volume of air already there.
Subsequently a high pressure is produced.
The hot gas rises ( 60000 feet I believe ) and surrounding air backflows into the region.
Subsequently a low pressure is produced.

The high and low pressures travel outwards from the volcano area around the world.
It's a tsunami, but this time not in the water but in the air, both water and air being fluids.

That's not what I am asking. Fact that the shockwave is immediately followed by the drop in the pressure is obvious. But why the same wave after getting around the globe and hitting me several hour later shows as a drop? That's something completely different.

 

[Borek]

In an ideal gas, it's really just a function of temperature thanks to the simple equation of state. The atmosphere is approximated really well by an ideal gas model. The speed of sound, $a$, is
a = \sqrt{\gamma R T},

OK - but even if it is a good approximation it still says "don't assume constant speed of sound", as the wave travels around the world, through areas with a very different weather.

 

[Orthoceras]

But why the same wave after getting around the globe and hitting me several hour later shows as a drop? That's something completely different.

Again I wonder whether the shape of the pressure wave is distorted at the antinode. The pressure wave inevitably is a vortex ring, it must have a shear layer above sea or land, because of friction. The vortex ring should have a problem propagating through the antinode.

OK - but even if it is a good approximation it still says "don't assume constant speed of sound", as the wave travels around the world, through areas with a very different weather.

You might expect slow propagation over the poles because of the cold. That should have been visible in the isochrones of the pressure wave of the Krakatoa, especially in the center of figure 2B, the return of the pressure wave to Krakatoa after a full cycle around the earth. The isochrone in the center has lobes that depict the slowest arrival directions. Figure 2B shows that the slowest arrival directions are not those from the south pole and the north pole. Instead, the figure shows that the eastward travel around the world was 1 hour slower than the westward travel. This could be explained by the Trade Winds.

 

[Borek]

Again I wonder whether the shape of the pressure wave is distorted at the antinode.

That would be a logical place for something to happen, still, the question is: what it is that is happening there?

 

[Jonathan Scott]

That's not what I am asking. Fact that the shockwave is immediately followed by the drop in the pressure is obvious. But why the same wave after getting around the globe and hitting me several hour later shows as a drop? That's something completely different.

I still think this was covered by my explanation, unless anyone can find a problem with it. (I'm not guaranteeing that there isn't one, but so far I've not seen anything against it!)

The eruption creates a huge volume of gas suddenly, displacing the surrounding atmosphere outwards by several km and creating a shock wave with much higher density than normal, which then propagates outwards in an approximate circle. The wave gradually smooths out as it loses energy, and continues to travel at around the speed of sound. As the wave of higher density passes any given point, the overall effect of its passage is that air is displaced in the forward direction. This is like a ridge in a rug; if you smooth it out towards the edge, you are simply moving the affected part forward by the amount of slack in the ridge.

The pressure wave averages out in the sense that the forces as it arrives and departs must balance, as it leaves the atmosphere approximately at rest as before. As it smooths out, the overall rise then dip becomes less significant. However, if the wave is circular and expanding or contracting, the resulting displacement will temporarily affect the density because of the change of circumference. This means that even when the intensity of the pressure wave itself is low, the overall displacement of air due to its passage can temporarily change the pressure overall in the positive or negative direction.

I don't think any special interaction would occur near the opposite point because relatively weak pressure waves would simply add linearly and pass across each other, although of course the local intensity near that point might be interestingly high from an observer point of view.