Analogous to the tides of Earths oceans, do the Moon and Sun cause our atmosphere to bulge in what could be described as a low and high tide?
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3possible duplicate of Does the moon affect the Earth's climate? – Aug 22 '15 at 20:26
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8@ChrisWhite The answers posted to that question are vague and completely lacking in physical analysis. I would like to hear what a true expert has to teach us about this subject. Also, that question is improperly titled or asks more than one item. Mine is focused and will be search engine friendly. – Alex Aug 22 '15 at 20:51
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Very nice question Alex, best of luck with it, does a solar eclipse enhance the effect? – Aug 22 '15 at 21:05
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4This is not a duplicate question. The supposed duplicate asks about the Moon; this question asks about the Moon and the Sun. The Sun is dominates over the Moon by more than an order of magnitude in terms of contribution to atmospheric tides. – David Hammen Aug 22 '15 at 22:45
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The differential force of gravity on the atmosphere works the same as it does for the rest of the earth (the oceans etc). However, moving the equipotential surface by a few m will be almost undetectable on the atmosphere, since the density of the atmosphere decreases so gradually – over many km. Contrast this with the surface of the ocean, which is crisp.
So while it might be theoretically possible to look for small changes in the height of an isobar of, say, $10^4\,\mathrm{Pa}$, I don't think that it will be possible to measure such a change in practice.
See for example this graph from the Australian weather service showing pressure changes over four days. The units on the left are $\mathrm{hPa}$ – you expect tidal variations to be much smaller. It may take a while (many cycles) to pick out the lunar variations - although I am sure it has been done.
There is a thing called "lunar atmospheric tides" - see Wikipedia which describes the math behind this. And it describes it as "weak".
So the short answer is "yes".
For a good (27 page) review of the subject, see this 1979 article by Lindzen
The introduction of that article states:
1 INTRODUCTION
Atmospheric tides refer to those oscillations in the atmosphere whose periods are integral fractions of a lunar or solar day. The 24-hour Fourier component is referred to as a diurnal tide, the 12-hour component as a semidiurmal tide. The total tidal variation is referrred to as the daily variation. Although atmospheric tides are, in small measure, gravitationally forced, they are primarily forced by daily variations in solar insolation.
So – the main cause of daily variation is solar heating. There is a (much) smaller component due to gravity:
... atmospheric tides are, in small measure, gravitationally forced...
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5I don't own the latest gadgets, but according to Randall Munroe, "If your phone has a barometer in it, as a lot of new Android phones do, you can download an app and actually see the pressure difference between your head and your feet." So apparently ~20 Pa sensitivity has become commonplace. – Aug 22 '15 at 21:09
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6@ChrisWhite of course you can measure a tiny pressure change - the question is how you distinguish lunar variations from all the other factors that influence pressure - there is a lot of churn in the atmosphere. – Floris Aug 22 '15 at 21:17
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Well, you could take data for a couple millenia and extract a periodic signal at the end ;) What's interesting is you'll find lots of people advocating barometers for finding altitude to within ~5 m, as long as they are calibrated that day. But if I did the math right your chart shows the readings could drift by ~30+ m each hour, and the trend could continue unabated all day. – Aug 22 '15 at 21:28
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@ChrisWhite - while it's possible to measure the pressure difference due to a 5 m change in altitude, you absolutely need to know the reference pressure at that time (not just "that day") if you want to reach that kind of accuracy. Your local airport might publish hourly pressures - see for instance http://w1.weather.gov/obhistory/KALB.html . This afternoon, with weather "unchanging", pressure was dropping by a leisurely 0.3 mbar / hour. You'd lose your 5 m accuracy within a couple of hours. – Floris Aug 22 '15 at 21:44
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3I would still suspect that atmospheric motion caused by the moons gravity differential would be much larger (in volume) than that of the oceans. But is it clear that such motion would necessarily show as a pressure change? After all, the extra air above a given point on the surface is partially "supported" by Moon's gravitational pull. – Jyrki Lahtonen Aug 22 '15 at 21:59
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1The very first line of your answer seems wrong to me. Since water is a far denser fluid than air, the tide height difference of air shouldn't be a few meters but far more. And if the equipotential surface in atmosphere is not moving significantly enough, then simply the tide is not caused by the moon. – Jul 28 '16 at 04:59
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@displayName - that is simply not true. The surface of either a light or a heavy liquid will follow the same equipotential surface, as they are subject to the same laws of physics (gravity, inertia). Density doesn't come into it. Of course the actual motion of the oceans doesn't really follow the equipotential surface because of obstacles (continents) that limit free flow. But the principle is the same, and the distortion of the EPS is not very large. And it's the distortion multiplied by the density and gravity that gives the pressure variations. – Floris Jul 28 '16 at 21:04
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1@Floris: I didn't say anywhere that EPS is subjected to different laws. Can you please elaborate though on how the EPS distortion of water and air will be comparable in presence of same gravitational force? – Jul 29 '16 at 00:29
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@displayName An equipotential surface should be the surface to which a fluid will come to equilibrium if allowed. The shape of the EPS is the same for water and air - so why would air, though less dense, have "bigger" tides? – Floris Jul 29 '16 at 02:13
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1@Floris: That is what I have asked you. Are you saying that the tides in water and air will be of same height when they are subjected to same force? – Jul 29 '16 at 02:16
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1@Floris: Moon's gravity will subject the Earth's atmosphere and ocean to same gravitational force, right? If you say No please clarify a bit further. Second, how did you conclude that tide in air and water will be of same height? – Jul 29 '16 at 02:39
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No - force is gravitational field times mass. The mass of the atmosphere is much smaller - so it will experience a smaller force. This is why a heavy object weighs more than a light object - they are both attracted by the earth yet the force is different. Same thing with moon and tides. – Floris Jul 29 '16 at 02:41
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@Floris: Is the presence of Moon's gravitational field significant enough to cause movement on an object on Earth? – Jul 29 '16 at 02:50
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@displayName it is enough to be measurable. It causes small shifts in equilibrium conditions. On a macro scale that does lead to motion. – Floris Jul 29 '16 at 02:56
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@Floris: Measurability is because of the capability of the measuring instrument. What's is the Moon's field's value? What is Earth's field's value at the location of that same object? How does the small shift in equilibrium conditions result in a macro scale movement? Is it proved or just a hypothesis? If it is proved then same phenomenon should be present in river's water too. Is there a measurable difference in flow of river water on a full moon vs a no moon night? – Jul 29 '16 at 03:15
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Where I live (over 100 miles from the ocean) there are measurable tidal effects in the (Hudson) river - does that count? There are quite a few questions on this site about tides - I suggest you search them to see if it clear things up. A long comment thread is not the way to satisfy your curiosity. – Floris Jul 29 '16 at 04:07
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1@Floris: You can append the response to my comment directly to your answer. It will make the answer more useful to future users. Can't force you though. It's your wish in the end. Secondly, my question stays put. Is the tidal effect in Hudson river provably due to Moon's gravity? This even says that tides in Hudson are not because of direct action of Moon (and that too is wrong). Lastly, you did not respond to the question about Moon's vs Earth's field on an object on Earth's surface. – Jul 29 '16 at 04:30
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@Floris: If you don't want to discuss further, please let me know. I can continue as long as we don't reach the right conclusion. – Jul 29 '16 at 04:32
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@Floris: Not trying to engage you in a discussion, but on Notes and References of the same Wikipedia page linked by you in your answer, the 4th link opens the news when Atmospheric tides were discovered in 1947. They are 40 miles high. – Aug 04 '16 at 03:28
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The link about the Hudson shows that there is an excitation of a wave due to the lunar attraction (when moon and sun cooperate it is bigger - hence the difference between spring tide and neap tides; this demonstrates the moon is playing a role). The shape of the river acts as an amplifier so you can measure the (amplified) effect quite far away. – Floris Aug 04 '16 at 11:06
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Here is a recent link showing the pressure change due to the moon is tiny - but studying precipitation patterns over many years it gives a measurable influence on precipitation: https://www.sciencenews.org/article/atmospheric-tides-alter-rainfall-rate – Floris Aug 04 '16 at 11:11
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And here is a link to an excellent paper describing more than you ever wanted to know: http://www-eaps.mit.edu/faculty/lindzen/29_Atmos_Tides.pdf - I am still digesting it but it shows that long series of pressure observations do indeed show a small barometric variation that corresponds to the lunar tide. Laplace tried and failed ("need more data!"), as did many others. But these days it is well established. And the (pressure) amplitude is tiny (10's of microbars - see e.g. table 2L2 on page 72 in the above) – Floris Aug 04 '16 at 15:46
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@Floris: Please tag me by my user handle so that I know you have replied. I have not explored completely what you mentioned about the Hudson river. Will get to you regarding that. For now, the sciencenews paper that you linked to says - Air gathers on Earth’s moon-facing side and on the opposite end of the globe. If the pull is due to Moon, why is there a pull on the opposite side? Moon should be pulling air only towards it, hence the accumulation should be only on one side. No? – Aug 08 '16 at 17:39
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@displayName - the "two bulges" for tides is the subject of many questions on this site; probably the best one (which explains that the oceans don't really have these bulges because of many effects) is http://physics.stackexchange.com/q/121830/26969 – Floris Aug 08 '16 at 19:33
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@Floris: On the same question, please go to the image titled "The tidal force". It shows the tidal force in two directions pointed as n and z and both those directions are opposite to each other. However, if you are saying that the two bulges or tides are not present, you are essentially refuting the same link that you are giving to me. Did I misunderstand you? – Aug 08 '16 at 19:39
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@displayName You need to read the answers carefully. The equipotential surface has two bulges, for reasons discussed in many of the answers (and the question). The fact that the ocean doesn't have such an obvious "bulge" is a consequence of the shape of the oceans and continents. This results in constraints on the flow of the water that prevents "bulges" in real life; but the (apparent) forces that would cause an Earth made entirely from water to take on the shape shown - those forces are real (in the rotating frame of reference). – Floris Aug 08 '16 at 19:44
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@Floris: I don't see any reason for two bulges/forces in opposite directions and the resultant distortion of the EPS. The force of gravitation is only attractive, and not attractive on one side and repulsive on the other side. – Aug 08 '16 at 19:59
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@displayName there are two common explanations. One invokes the centrifugal force in the rotating frame of reference which is greater as you move further from the barycenter; the other considers the earth plus water in free fall in the moons gravitational field - one side being pulled more, the other less. Both lead to "two bulges" in the EPS. See also this answer for detailed mathematical treatment. – Floris Aug 08 '16 at 20:59
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@Floris: If we go by the first common explanation that Earth's centrifugal force in the rotating frame of reference is the cause for the bulge/force then it means that tide is due to Earth's phenomenon and not due to presence of Moon. (contd..) – Aug 08 '16 at 21:25
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@Floris: I'm not able to go by second common explanation at all. Why? Because the universe is not in free fall. Rather, the universe is expanding. – Aug 08 '16 at 21:27
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Unless your barycenter is off center (mutual center of mass of moon and earth) there would be no explanation. It's the moon. Believe it. There are MANY questions / answers in the topic already on this site. – Floris Aug 08 '16 at 21:28
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Any satellite orbiting another object is in effect in free fall! And the "far side" falls less quickly, because it experiences less attraction. – Floris Aug 08 '16 at 21:30
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@Floris: Finally, if you pay close attention to the mathematical treatment in the answer you have linked you will see that it only uses equations to mathematically describe how tides would be, if they were following the assumptions that are there in that answer and that Moon indeed is the reason for tides. It does not prove anywhere that that mathematical description explains the reality on Earth. – Aug 08 '16 at 21:30
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If an assumption leads to a prediction that matches observation, that is usually considered support of that assumption. – Floris Aug 08 '16 at 21:33
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@Floris: Where is the answer saying that the prediction has matched the observations? – Aug 08 '16 at 22:28
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@Floris: There are many points in multiple comments above and in your linked answer, which I would like to explore further. For one, going to the answer you have linked right above, you mention the tide producing forces of the sun and moon. The force from Moon is 10^-5 from my calculations. Other people on PSE have given an even lower order of magnitude. May I know how do these small forces generate a tide? – Aug 09 '16 at 02:14
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A small (differential) force acting for a long time will move large bodies of water etc - since the water will attempt to distribute according to the equipotential surface. A force equal to 10$^{-5}$ of the force of gravity will give an acceleration of 0.0001 m/s^2. Keep that going for 3 hours (~10,000 seconds) and you could have a velocity of 1 m/s and a displacement of hundreds of meters. But of course once there is a small slope in the surface of the water there will be a restoring force. This is why (in open water) tides don't look like they have much "slope" to them. But tides exist. – Floris Aug 09 '16 at 03:33
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@Floris: The force I calculated was not 10^-5 of g. It was itself 10^-5. This force was in the direction of Moon's center. It horizontal component was itself times Cos(~89 degrees). Also this force is cyclic. So before it is applied in one direction for a few hours, it has been applied in the opposite direction for the same number of hours. The forces are barely causing a few meters of water displacement. Is it an acceptable explanation for huge tides? – Aug 09 '16 at 04:46
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A force has units (N) and applies to a certain mass. What acceleration did you calculate? Note also that "huge" tides are usually an interplay between water and sea bottom (like a breaking wave on the shore). For instance in Bay of Fundy the shape of shore line and rising sea bottom act as an amplifier of the tidal motion - resulting in spectacular phenomena (look it up). – Floris Aug 09 '16 at 12:39
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@Floris, tides on the Hudson can't be caused by the moon, since it cannot produce a tide even in the Mediterranean. The differential pull is half a micron: .5 *10 ^-7 $g$ (= 9.8) so, $g$ beneath the moon is $9.799 999 95$ m/s^2. How can a half-a-micron-difference rise a tide on the Hudson or 1 meter on the ocean? – Aug 10 '16 at 04:32
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@user104372 I suggest you turn all your doubts into a well formulated question - this discussion is not productive. An important factor in all this is tidal resonance - the interaction between a small excitation and the container in which a wave can build up over many cycles. See https://en.m.wikipedia.org/wiki/Tidal_resonance and study Bay of Fundy. – Floris Aug 10 '16 at 12:25
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Analogous to the tides of Earths oceans, do the Moon and Sun cause our atmosphere to bulge in what could be described as a low and high tide?
The answer is yes, if you generalize beyond gravitation. Sunlight heats the atmosphere, and this causes atmospheric tides. The two dominant effects are absorption of visible and near infrared sunlight by water vapor in the troposphere and absorption of ultraviolet by ozone in the stratosphere.
The article by Lindzen cited in Floris' answer says exactly that. Atmospheric tides are caused primarily by solar heating rather than by gravitation. They are still "tides", however.
David Hammen
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1Absolutely correct that variation due to solar heating exists - not sure that is a "tide" in the sense the OP intended. The article I linked mentions that gravitational tides exist (with a roughly 12-hour period), but they are MUCH weaker than the (diurnal - 24 hour cycle) variation due to solar heating. I pulled out the relevant quote and added it to my answer – Floris Aug 23 '15 at 02:34
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@Floris You would think that, being forced by solar heating, the atmospheric tides would be diurnal. But they are semidurnal. The solar heating signal is basically a truncated sinusoid, being zero all night, and thus has a large semidurnal component. And in terms of the effect on atmospheric pressure, especially in the tropics, this semidiurnal component dominates. – Ben51 Jan 23 '18 at 18:26
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@Ben51 are you referring to the fact that from a Fourier transform perspective a sinusoid has a second harmonic? – Floris Jan 24 '18 at 00:00
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@Floris Yes! (Well, not a full sinusoid, of course, but rather a truncated sinusoid, with all the negative lobes chopped off). There is quite a lot of energy in the 12-hour harmonic, and that is what the tropical atmosphere responds to. Fair-weather barometric pressure records from the tropics are mostly just clean 12-hour oscillations. – Ben51 Jan 24 '18 at 03:44
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This isn't really an answer, and I will take it down shortly, just want to put it here as a demonstration of the clear dominance of the 12-hour constituent in the atmospheric tides in the tropics (can't put images in comments). The pressure record comes from last year in the Eastern tropical Pacific. There's a fair amount of variability on weekly timescales, and the 12-hour oscillation stands out, but you can't really make out a 24-hour component at all.
So even though the tides are driven by solar heating, which certainly sounds like it would mean they'd follow a daily cycle, they actually have a half-day cycle.
Ben51
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This is very surprising. Do you have a link to the source of this material? For one thing, the lunar tidal force is stronger than the solar one - yet the cycles shown here are exactly 12 hours long, not 12 hours and 25 minutes. Does a rapid cooling/heating at sunrise/sunset account for this (changing the density of the atmosphere due to humidity?) – Floris Jan 24 '18 at 04:19
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Honestly I don't fully understand the mechanism. I had it beaten into my head that it was a heating effect, not a gravitational one, which I don't question. As for why there's a stronger response to the 12-hour forcing than the 24-hour, I always figured it was a resonance thing. The data is mine, there's no link to it, but I have other examples, and it shouldn't be hard to find comparable ones online. – Ben51 Jan 24 '18 at 04:24
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Does the wind pick up mornings and evenings? The flatness of the over all curve (total pressure range) is astonishing. Where did you measure this? I know many islands have wind in the morning and evening (because of the heating/cooling effect of the water vs the land) - Bernoulli provides the corresponding pressure change. – Floris Jan 24 '18 at 05:14
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This is at 10 N, 125 W, in the middle of the ocean sort of halfway between Hawaii and Central America. There are maybe barely discernible peaks in the spectrum of wind speed at day and half day periods, but you certainly don't see anything jump out at you when you look at a timeseries. I chose a pretty quiet period to emphasize the tides. There are bigger pressure fluctuations sometimes. – Ben51 Jan 24 '18 at 05:20
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OK - so no effects due to differential heating as there is no land nearby?. Interesting - I can’t explain it. – Floris Jan 24 '18 at 05:31
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There's a book about it by Chapman and Lindzen. I haven't read through enough to see if there's a definitive answer, but there's a great quote from Lord Kelvin, which I just pasted in above. – Ben51 Jan 24 '18 at 05:40
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That is a fascinating quote - speculative, but it makes sense. He clearly concludes this is not a gravitational effect - but whether there is in fact an excited resonance is left unanswered. I believe this merits a follow up question “what is the mechanism of the semidiurnal variation of atmospheric pressure?”. Since this is not “tides” (in the gravitational sense), I think it is distinct from the present post. – Floris Jan 24 '18 at 13:45