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The bit that makes sense – tidal forces

My physics teacher explained that most tidal effect is caused by the moon rotating around the Earth, and some also by the Sun.

They said that in the Earth and Moon system, the bodies are are in free fall about each other. But that points on the surface of Earth, not being at Earth's centre of gravity, experience a slightly different pulls towards the moon.

The pull is a little greater if they are on the moon's side, and slightly lesser if they are on the side away from the moon. Once free fall is removed, on the moon side this feels like a pull towards the moon and on the the opposite side it feels like a repulsion from the moon.

This makes sense to me, and is backed up by other questions and answers here, like this and also this Phys.SE question.

The bit that doesn't make sense – tidal bulges

They also said that there are "tidal bulges" on opposite sides of the Earth caused by these forces. The bulges stay still relative to the moon and Earth rotating through the bulges explains why we get two tides a day. They drew a picture like this one…

Tidal Bulges

An image search for tidal bulges finds hundreds of similar examples.

…But, if there is a tidal bulge on both sides of Earth, like a big wave with two peaks going round and around, how can an island, like Great Britain where I live, simultaneously have a high tide on one side and a low tide on the other?

For example:

Two ports with tides 6 hours, or 180º apart. It's high tide at one while low tide at the other. But only about 240 miles by road.

Great Britain is much smaller than Earth. It's probably not even as big as the letter "A" in the word "TIDAL" in that picture.


To prove this isn't just Britain being a crazy anomaly, here is another example from New Zealand:

Two ports that are 180º (6 hours) apart, but separated by just 200 delightful miles through a national park. New Zealand, unlike the UK, is in fairly open ocean.

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Tides in the British Isles are very complicated. They're strongly affected by water having to flow through the Irish Sea, the English Channel and the rather shallow North Sea. Tides in Liverpool and the Severn Estuary are compounded by resonance; Southampton has a weird (unique?) double-peaked tide because of interactions around the Isle of Wight. That in no way invalidates the question (+1!); it just means that the UK isn't a great example to use. –  David Richerby Jun 26 at 20:02
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The "bulge" as drawn is an equipotential surface. How water moves in response to this (where there is a gradient in the surface there will be a force on the water) depends on the geography of the "container" of the water (depth as well as coast lines). "It's complicated". –  Floris Jun 27 at 20:32
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Hi @Floris, I think the surface (greatly exaggerated) might look like that. That's not the explanation that is frequently given, however. A web search, and answers here, confirms this, again and again. I believe many people operate under the idea that the tidal bulge theory is about right, with just a few tweaks needed for continents and such – I certainly used to. The reality is: it is nonsense. It couldn't be more comprehensively rejected by the evidence in measurements. In my opinion it simply causes confusion and should be rejected entirely, even as a expositional tool. –  Benjohn Jun 28 at 18:43
    
The complexity, IMO, isn't all that great, provided you're not after a complete fluid dynamics model of the entire thing. You've got a driving function wobbling lots of water, so the major period of tidal waves matches the period of the driving function. Shaking a bowl of water with rocks in it might provide a compelling model :-) –  Benjohn Jun 28 at 18:47
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@DavidRicherby I've added another example from New Zealand. Thanks. –  Benjohn Jul 5 at 23:05

7 Answers 7

up vote 108 down vote accepted

There is no tidal bulge.

This was one of Newton's few mistakes. Newton did get the tidal forcing function correct, but the response to that forcing in the oceans: completely wrong.

Newton's equilibrium theory of the tides with its two tidal bulges is falsified by observation. If this hypothesis was correct, high tide would occur when the Moon is at zenith and at nadir. Most places on the Earth's oceans do have a high tide every 12.421 hours, but whether those high tides occur at zenith and nadir is sheer luck. In most places, there's a predictable offset from the Moon's zenith/nadir and the time of high tide, and that offset is not zero.

One of the most confounding places with regard to the tides is Newton's back yard. If Newton's equilibrium theory was correct, high tide would occur at more or less the same time across the North Sea. That is not what is observed. At any time of day, one can always find a place in the North Sea that is experiencing high tide, and another that is simultaneously experiencing low tide.

Why isn't there a bulge?

Beyond the evidence, there are a number of reasons a tidal bulge cannot exist in the oceans.

The tidal bulge cannot exist because the way water waves propagate. If the tidal bulge did exist, it would form a wave with a wavelength of half the Earth's circumference. That wavelength is much greater than the depth of the ocean, which means the wave would be a shallow wave. The speed of a shallow wave at some location is approximately $\sqrt{gd}$, where $d$ is the depth of the ocean at that location. This tidal wave could only move at 330 m/s over even the deepest oceanic trench, 205 m/s over the mean depth of 4267 m, and less than that in shallow waters. Compare with the 465 m/s rotational velocity at the equator. The shallow tidal wave cannot keep up with the Earth's rotation.

The tidal bulge cannot exist because the Earth isn't completely covered by water. There are two huge north-south barriers to Newton's tidal bulge, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. The tides on the Panama's Pacific coast are very, very different from the tides just 100 kilometers away on Panama's Caribbean coast.

A third reason the tidal bulge cannot exist is the Coriolis effect. That the Earth is rotating at a rate different from the Moon's orbital rate means that the Coriolis effect would act to sheer the tidal wave apart even if the Earth was completely covered by a very deep ocean.

What is the right model?

What Newton got wrong, Laplace got right.

Laplace's dynamic theory of the tides accounts for the problems mentioned above. It explains why it's always high tide somewhere in the North Sea (and Patagonia, and the coast of New Zealand, and a few other places on the Earth where tides are just completely whacko). The tidal forcing functions combined with oceanic basin depths and outlines results in amphidromic systems. There are points on the surface, "amphidromic points", that experience no tides, at least with respect to one of the many forcing functions of the tides. The tidal responses rotate about these amphidromic points.

There are a large number of frequency responses to the overall tidal forcing functions. The Moon is the dominant force with regard to the tides. It helps to look at things from the perspective of the frequency domain. From this perspective, the dominant frequency on most places on the Earth is 1 cycle per 12.421 hours, the M2 tidal frequency. The second largest is the 1 cycle per 12 hours due to the Sun, the S2 tidal frequency. Since the forcing function is not quite symmetric, there are also 1 cycle per 24.841 hours responses (the M1 tidal frequency), 1 cycle per 24 hours responses (the S1 tidal frequency), and a slew of others. Each of these has its own amphidromic system.

With regard to the North Sea, there are three M2 tidal amphidromic points in the neighborhood of the North Sea. This nicely explains why the tides are so very goofy in the North Sea.

Images

For those who like imagery, here are a few key images. I'm hoping that the owners of these images won't rearrange their websites.

The tidal force


Source: http://physics.mercer.edu/hpage/tidal%20asymmetry/asymmetry.html

This is what Newton did get right. The tidal force is away from the center of the Earth when the Moon (or Sun) is at zenith or nadir, inward when the Moon (or Sun) is on the horizon. The vertical component is the driving force behind the response of the Earth as a whole to these tidal forces. This question isn't about the Earth tides. The question is about the oceanic tides, and there it's the horizontal component that is the driving force.

The global M2 tidal response


Source: http://en.wikipedia.org/wiki/File:M2_tidal_constituent.jpg


Source: http://volkov.oce.orst.edu/tides/global.html

The M2 constituent of the tides is the roughly twice per day response to the tidal forcing function that results from the Moon. This is the dominant component of the tides in many parts of the world. The first image shows the M2 amphidromic points, points where there is no M2 component of the tides. Even though these points have zero response to this component, these amphidromic points are nonetheless critical in modeling the tidal response. The second image, an animated gif, shows the response over time.

The M2 tidal response in the North Sea


Source: http://www.geog.ucsb.edu/~dylan/ocean.html

I mentioned the North Sea multiple times in my response. The North Atlantic is where 40% of the M2 tidal dissipation occurs, and the North Sea is the hub of this dissipation.

Energy flow of the semi-diurnal, lunar tidal wave (M2) (June 27 update)


Source: http://earth.eo.esa.int/brat/html/appli/ocean/tides_en.html

I added this image in response to Ralph's comment. The above image displays transfer of energy from places where tidal energy is created to places where it is dissipated. This energy transfer explains the weird tides in Patagonia, one of the places on the Earth where tides are highest and most counterintuitive. Those Patagonian tides are largely a result of energy transfer from the Pacific to the Atlantic. It also shows the huge transfer of energy to the North Atlantic, which is where 40% of the M2 tidal dissipation occurs.

Note that this energy transfer is generally eastward. You can think of this as a representing "net tidal bulge". Or not. I prefer "or not".

Another update (September 9), based on recent comments

Isn't a Tsunami a shallow water wave as well as compared to the ocean basins? I know the wavelength is smaller but it is still a shallow water wave and hence would propagate at the same speed. Why dont don't they suffer from what you mentioned regarding the rotational velocity of the earth.

Firstly, there's a big difference between a tsunami and the tides. A tsunami is the the result of a non-linear damped harmonic oscillator (the Earth's oceans) to an impulse (an earthquake). The tides are the response to a cyclical driving force. That said,

  • As is the case with any harmonic oscillator, the impulse response is informative of the response to a cyclical driving force.
  • Tsunamis are subject to the Coriolis effect. The effect is small, but present. The reason it is small is because tsunami are, for the most part, short term events relative to the Earth's rotation rate. The Coriolis effect becomes apparent in the long-term response of the oceans to a tsunami. Topography is much more important for a tsunami.

The link that follows provides an animation of the 2004 Indonesian earthquake tsunami.

References for the above:

Dao, M. H., & Tkalich, P. (2007). Tsunami propagation modelling? a sensitivity study. Natural Hazards and Earth System Science, 7(6), 741-754.

Eze, C. L., Uko, D. E., Gobo, A. E., Sigalo, F. B., & Israel-Cookey, C. (2009). Mathematical Modelling of Tsunami Propagation. Journal of Applied Sciences and Environmental Management, 13(3).

Kowalik, Z., Knight, W., Logan, T., & Whitmore, P. (2005). Numerical modeling of the global tsunami: Indonesian tsunami of 26 December 2004. Science of Tsunami Hazards, 23(1), 40-56.

This is an interesting answer full of cool facts and diagrams, but I think it's a little overstated. Newton's explanation wasn't wrong, it was an approximation. He knew it was an approximation -- obviously he was aware that the earth had land as well as water, that tides were of different heights in different places, and so on. I don't think it's a coincidence that the height of the bulge in the equipotential is of very nearly the right size to explain the observed heights of the tides.

Newton's analysis was a good start. Newton certainly did describe the tidal force properly. He didn't have the mathematical tools to do any better than what he did. Fourier analysis, proper treatment of non-inertial frames, and fluid dynamics all post-date Newton by about a century.

Besides the issues cited above, Newton ignored the horizontal component of the tidal force and only looked at the vertical component. The horizontal component wouldn't be important if the Earth was tidally locked to the Moon. The dynamical theory of the tides essentially ignores the vertical component and only looks at the horizontal component. This gives a very different picture of the tides.

I'm far from alone in saying the tidal bulge doesn't exist. For example, from this lecture, the page on dynamic tides rhetorically asks "But how can water confined to a basin engage in wave motion at all like the “tidal bulges” that supposedly sweep around the globe as depicted in equilibrium theory?" and immediately responds (emphasis mine) "The answer is – it can’t."

In Affholder, M., & Valiron, F. (2001). Descriptive Physical Oceanography. CRC Press, the authors introduce Newton's equilibrium tide but then write (emphasis mine) "For the tidal wave to move at this enormous speed of 1600 km/h, the ideal ocean depth would have to be 22 km. Taking the average depth of the ocean as 3.9 km, the speed of the tidal elevations can only be 700 km/h. Therefore the equilibrium position at any instant required by this theory cannot be established."

Oceanographers still teach Newton's equilibrium tide theory for a number of reasons. It does give a proper picture of the tidal forcing function. Moreover, many students do not understand how many places can have two tides a day. For that matter, most oceanography instructors and textbook authors don't understand! Many oceanographers and their texts still hold that the inner bulge is a consequence of gravity but the other bulge is a consequence of a so-called centrifugal force. This drives geophysicists and geodocists absolutely nuts. That's starting to change; in the last ten years or so, some oceanography texts have finally started teaching that the only force that is needed to explain the tides is gravitation.

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@annav - This question was about the ocean tides, not the Earth tides. It's the Earth tides that the LHC takes into account. The Earth tides have an observable affect the timing of atomic clocks, the pointing of milliarcsecond telescopes, the orbits of satellites in low Earth orbit, etc. The behavior of the Earth tides is close to Newtonian. But not the ocean tides. –  David Hammen Jun 26 at 11:25
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Much of what you say makes sense, but I don't agree with that part starting "If the tidal bulge did exist, it would form a wave with a wavelength of half the Earth's circumference". Arguing that the speed of displacement of a "tidal bulge" does not match the speed of wave propagation just shows that the system cannot resonate at the frequency that the Moon drives it at. Which is very fortunate! In a spherically symmetric system (no continents) there would be bulges, they would just don't propagate as waves. A system will oscillate when driven at any frequency, just not very strongly. –  Marc van Leeuwen Jun 27 at 5:02
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@Ralph: Wikipedia also has an article on Lies to children. The tidal bugle explanation is a lie to children. One explanation of tidal acceleration: Conservation of angular momentum. Is a detailed mechanism truly needed? It certainly exists, but who cares! One of the niceties of invoking the conservation laws is that they sweep the need for a mechanism under the rug. Admittedly, that's not all that satisfying, but then again, neither are lies to children. (Continued) –  David Hammen Jun 27 at 19:29
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If you don't want a "lie to children", the answer is going to be rather complex and also more than a bit empirical. If you look at the images (I added a new one on energy transfer just to answer your question), you'll see asymmetries in those amphidromic systems, and you'll also see energy transfer (aka water flow) amongst amphidromic systems. You can interpret those asymmetries and energy transfers as representing a "net tidal bulge". Or not. You can just say a mechanism exists thanks to conservation of angular momentum. (Continued) –  David Hammen Jun 27 at 19:30
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Researchers in the field don't particularly care about a detailed mechanism that explains transfer of angular momentum from the Earth to the Moon. They are more concerned with what observations of the Moon's recession tells us about tidal dissipation, and what paleo observations of length of day tells us about dissipation (and some other interesting things, such as whether Newton's gravitation constant and the speed of light are indeed constant). –  David Hammen Jun 27 at 19:30

The picture of high tides on opposite sides of the Earth with a period of about 12 hours (actually 12 hours 25 minutes, due to the rotation of the Earth) is an oversimplification. It's just a starting point. Tides would behave this way in the limit of an all-water Earth with ocean depth so great that it had no effect on the surface wave.

But the Earth has continents , peninsulas, bays, estuaries and the like, and the ocean has a finite depth causing frictional effects on ocean waves, and characteristic frequencies of the ocean basins. All of these factors, plus the Coriolis effect due to the Earth's rotation affect the boundary conditions of the variation in ocean height due to tides. In turn, depending on local coastal geography, local basins can have characteristic resonant frequencies leading to local constructive or destructive interference with the tides.

All of these effects lead to higher order harmonics in the tides on top of the 12 hour 25 minute primary tide. By higher order, I mean that these components of the tides have higher frequencies (shorter periods). And they can be locally important.

It's these short period effects (periods of a few hours, not 12+ ) that would explain what's going on in locations such as the two places in England.

The Wikipedia article on the theory of tides has several links to papers about harmonic analysis of tides done by George Darwin (Charles's son) and others in the early 1900s. Nowadays this work is done with numerical simulations, but that work builds on the work done eariler.

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I don't think that what you say about harmonics holds up to evidence. While harmonics of the driving effect make the situation even more complex, just the first Fourier component at the fundamental period of the moon fails to show evidence of the tidal bulges. An animation showing the moon fundamental without any other harmonics can be found here. –  Benjohn Jun 26 at 22:58
    
With respect, I disagree. See the references cited in the Wikipedia article I added to my answer as a link. –  paisanco Jun 27 at 2:15
    
Thank you. As I understand, G. Darwin developed the harmonic approach to prediction, using numerical analysis in Fourier's frequency domain. His mnemonics for the forcing periods remain in use. M2 is the wave that has a frequency of two cycles a day. The hypothetical bulge shaped response has precisely the M2 frequency so must exist only in the M2 component. Bulges would appear in M2 as latitude constant magnitude and phase = 2 x longitude. Global M2 data does not show this. Instead the M2 responses are localised, circulating around single oceans or countries, not the Earth. –  Benjohn Jun 27 at 8:37
    
Someone much more proficient with numeric analysis tools than me could compute the total "bulge like feature"'s energy present in the M2 response of the measured global tidal data. I suspect it would be a tiny fraction with no more evidential support than assorted other smooth patterns fitted to the data. –  Benjohn Jun 27 at 8:43

Unlike sea tide, which is quite complex, as other answers explain, the solid (not-so-solid for this part) Earth tide tends to be simple and the first-order picture can be reasonably approximated by the "bulges" metaphor mentioned in the question.

Solid earth tide has an amplitude of ~1 ft typically and it can be safely ignored in most situations, including common surveying. This is probably the reason why most people are not aware of this phenomenon. Yet is is interesting to realize that your house moves up and down some thirty centemeters twice a day.

The wikipedia entry has a good explanation and referece further refernce.

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While this provides some good information on the topic (+1), it doesn't actually answer the question: "[...] simultaneously have a high tide on one side and a low tide on the other?" (-1). So a net +0 from me. –  Kyle Jun 26 at 23:30
    
This is really interesting information beyond the question, I'm glad you added it, thanks (+1). –  Benjohn Jun 29 at 7:45

The specific problem of your location is answered partially in the comments. I suppose it is the six hours that is problematic for you.

Edit after reading main answer that there are lots of bulges due to the ocean landscape boundary conditions and fluid mechanics.

What does it mean that a bulge, a high twelve foot tide, comes from the west , lets say at 12:00 hours ? Water is sucked up into the bulge. The bulge on the ocean side as it reaches the UK will start sucking the water all around the island. There is not as much water as on the ocean side and the water on the east gets low while the water or the west rises. As the pull of the moon/sun goes over the UK part and moves towards the continent there will be a small lift of the water in the channel but at the same time the water that was making the bulge on the west is filling up back to equilibrium in the channel and there will be a small not really visible effect at less than an hour while the bulge is passing over to the continent.

It is interesting to know that even the solid earth has bulges, called earth tides of order 40 to 50 cms height. They have to be taken into account to keep the beams stable at LHC.

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Yes, the earth has high tides on opposite sides. That is why high tides come about 12 hours apart.

The timing of tides at nearby places is very dependent on local landforms. You can probably see a nice gradation of tide time if you look at the towns between Holyhead and Whitby. The delay may be different for high and low tides. The tides get very complicated when the geography is difficult.

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But Holyhead would need to be at the top of the bulge while Whitby is at the bottom. That would place them 10,000 km apart (90º around the world's equator, thought Britain isn't at the equator), but Google Maps says the distance is only 383 km by car. –  Benjohn Jun 25 at 22:20
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You have a very long period water wave, which has to flow through the channels and around the island. The bulges you see would be valid for a planet wholly covered by water. In fact the wave piles up on the west coast of an ocean, leading to much higher tides on one side of an ocean than another. As the tide goes up rivers it can locally be delayed by much more than 12 hours. –  Ross Millikan Jun 25 at 22:58
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Standing near a narrow spot connecting two substantial bodies of water can be instructive. I watched the water rushing though a narrow channel between two islands near the mouth of the Puget sound not so long ago. Thrilling. And it makes it obvious that the topology can induce substantial delays in the tides. –  dmckee Jun 26 at 2:23

Tidal movement is various at various geo-oceanic points on our globe. Ocean is water which is fluid, as distinct from the land exposed surfaces on earth which are of course rigid; to extent of not being possible for moon effect to draw the land surface to a bulge. Therefore, ocean depth will bulge in direct response to moon position relative to Earth. Notwithstanding the above, the ocean depth will vary with some response to other effects of the physics of ocean temperature, due to oceanic fissures releasing volcanic conditions causing ocean temperature to rise. Also tectonic plate movements (earthquakes) will cause varying effects on oceanic depths, which may be interpreted to be lunar caused tidal variation at some locations.

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Welcome to Physics SE! You provided interesting information, but did not answer the question "Does Earth really have two high-tide bulges on opposite sides?" You guessed it has something to do with tectonitc plate movements, but did not give a reference to underline it. Anyway welcome :) –  Stefan Bischof Jun 29 at 12:53

The answer starts with realizing that no water molecules move as quickly as the moon overhead moves around the earth. Even if there were no land masses in the way and the planet was all water, there is no way that a single water molecule could move around the earth fast enough to keep up with the moon (earth's circumference = 24,000 miles in 12 hours, which is 2000 miles per hour!).

My explanation -

All a single water molecule can do is sort of drift towards the moon. However, it cannot drift towards the moon when it is in the Far Tidal Bulge (earth is in the way), or in the Near Tidal Bulge (oceans rarely fly).

However, the water molecules on the sides of the earth (in the diagram above) CAN move towards the moon. They may not move far (my guess is some number of miles/kilometers), but they move far enough to stretch the water on the sides over into the Near Tidal Bulge. This is why the Near Tidal Bulge is bigger than the Far Tidal Bulge.

It may be helpful to think what would happen if the moon moved extremely slowly. In that case the water could keep up to the moon and, I think, the Far Tidal Bulge would be small or even disappear. Or, if the moon moved extremely quickly then water would have no time move significantly between orbits so there would be no tides. Right now we have a case between those two extremes.

Excellent question! I remember thinking the books were wrong when I first saw this.

Edit - On the far side, although gravity is weaker there due to bigger R in Fg=G m1 m2 / R^2, I find that explanation not convincing. My explanation works even if the gravitational field is constant, and the 1/R^2 explanation fails in the case of an extremely slowly rotating moon. Guess I'll need to crunch numbers to prove this, since this is not the official answer.

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