1. Originally Posted by Grey
The rock can and does move, though. Yes, it's bound to the Earth, but the Earth is moving as well, at the same acceleration. The bond between a rock on one side of the planet and a rock on the other side of the planet is anything but rigid; indeed, it's gravity that holds the solid parts of the Earth together, just like the fluid parts. Remember that at the scale we're talking about, the Earth actually behaves much more like a fluid than a rigid solid. That's why it's round, after all. It's true that the solid parts of the Earth do not move as far as the water (essentially, it behaves as a more viscous fluid) due to the tides(if it did move as far, we wouldn't see tides at all). But it's still the effects of the gravitational differential that we're seeing, not the direct acceleration.

And here's a cool way to see that.

By your calculations, the acceleration of a blob of water in the ocean due to the Moon is about 3 x 10-5 m/s2. Doing the same calculation for the effect of the Sun, we find that since the Sun is about 27 million times more massive than the Moon and about 400 times farther away, we get an acceleration of the same blob of water due to the Sun of about 170 times greater, or about 5 x 10-3 m/s2. That completely swamps the contribution from the Moon. If that acceleration were the primary driver of the tides, the Moon would be irrelevant to the tides; they would be completely driven by the position of the Sun. And we know that isn't the case.
See my post just before this one. It indeed appears that Profloater is erroneously using the absolute acceleration of the water and not allowing for the acceleration of the solid Earth beneath it. I stand by my reasoning as posted.

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3. Originally Posted by profloater
They are not all in free fall.
I should've added "in the moon's gravity". They are in free fall in the moon's gravity.
The fluids are in free fall but the solid surface is spinning and constrained by solid forces. A particle on the solid surface cannot adjust its speed of rotation except by elastic strain. You can view the whole body as in free fall but that is at moon orbit frequency not Earth spin frequency.

In a way this is similar to solid gyros and the way they respond to torques, it's the linear momentum of individual particles which keeps reversing due to spin but cannot move "sideways" due to solid tension forces.

So as I said just now, a solid particle feels the force from the moon on a tangent but cannot accelerate because it is bound into a solid and that binding means the net effect on spin is zero (I should say near zero because of solid tidal effects but they are much smaller when we consider the fluid tides.)

I had to really think about that free fall issue which you and Grey raised, but that is because I was thinking from the Earth frame of reference. Of course the actual tidal movements seen from an external reference is still spinning with the earth but locally slowing and speeding so that the tide momentum I described and calculated is relative to the Earth surface. When there is spin, the idea of free fall must be modified to include the preservation of angular momentum by internal tension forces, not just gravity. If there is a true solid no amount of gravity change can alter its angular momentum. But fluids are different.
Just watched this video:

Incredible. The guy slams every physics teacher for getting the tides wrong, says tides can't be like stretching taffy because the tidal force is too small, then says its like popping a pimple! If the tidal force is too small to stretch the ocean, it'll be too small to squeeze the ocean.

4. Originally Posted by Hornblower
See my post just before this one. It indeed appears that Profloater is erroneously using the absolute acceleration of the water and not allowing for the acceleration of the solid Earth beneath it. I stand by my reasoning as posted.
I agree that this seems to be precisely the error Profloater is making, and I believe that your reasoning is correct.

5. sackcloth and ashes; i have to apologise, I have been living a delusion. I can see it now from the figures, I do not know how I missed the odd million in the sun figures, I must have been hypnotised by an exponential wizard, and I cannot even blame Facebook. Thank you to Hornblower and Grey This has been a day of shame for me. But I have learned something. I should have known better. My typing may waver from the self flagellation. Now I will be quiet for a whole day.

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I think a good way to summarize the tides is to layer the complexity in three stages:
stage 1: starting from just the equipotential of the Earth's own gravity, you have an oblate spheroid that is axially symmetric and produces no tides as the Earth rotates. But introduce external perturbations from the Moon, and there is a forcing effect that varies on the timescale of the rotation of the Earth (as that alters the relative orientation to the Moon). This forcing effect is in turn modulated by the relative position of the Sun (spring and neap tides, on the timescale of the phases of the Moon), and even more importantly, the response to the forcing is complicated because the timescale the water would need to reach the equipotential shape is not much less than the timescale for the shape to rotate with the lunar driving (the daily timescale). However, the timescale for the land to participate in that movement is even longer, such that the land tide can be neglected and the contrast between the water and land movement is what we encounter on the beaches as waxing and waning tides.
Stage 2:
If we desire a more quantitative picture of how high the tides will be and when they will be high, but we also wish to retain simplicity, we must accept a poor approximation by imagining that the orientation of the Moon changes much more slowly than the timescale for the water to respond (i.e., a much longer day). Then the water will fit to the equipotential, creating a roughly (if exaggerated) "football" shape to the oceans. The rotation of that shape, relative to the more unchanging land shape, tells you both when the tide will be high, and how high it will be. Unfortunately, the actual timescale for the lunar driving is too fast for this to be a quantitatively good approximation. This fact is what motivated Gigabyte's point from the start.
Stage 3:
Accurately accounting for the slowness of the reaction of the water to the changing forcing of the roughly "football-shaped" equipotential requires including the coriolis effect. Also, the presence of obstacles in the form of continents, on large scales, and channels and bays, on smaller scales significantly alters the simple picture. As a result, the quantitative nature of tides is at times drastically altered from the simple stage 2 explanation. The diurnal periodicity of the driving is always there, but the phase and magnitude of the response in the tides at any given location varies so much from place to place that it is difficult to even find a prevailing correlation between the location of the Moon in the sky and the state of the tide that remains consistent from beach to beach, in stark contrast to the simple picture from stage 2. In particular, the combination of these significant complications introduces the concept of "amphidromic points," which are points in the ocean that exhibit little tidal variations in the height of the water, and are shown nicely at https://en.wikipedia.org/wiki/Amphid...onstituent.jpg. The height variations tend to sweep out a circular pattern around these tidal nodes like the minute hand of a clock, which show a location-dependent shift in phase from the equipotential variations that sweep through longitude lines. Hence, the situation for prediction of actual tides, both in terms of magnitude and phase, is quite complicated and sensitively dependent on the structure of the continents.

In fact, if we get the football picture for tides by imaging the forcing rate is very slow compared to the water response rate, we can also go to the opposite extreme and ask what tides would be like if the forcing rate were very fast compared to the water response. In that situation, we would only look at the first-order response of the water, which would then be simply the tangential motions of the water because there would only be time to set up a velocity field-- there would not be time to create much change in water height.

It seems to me this is the situation profloater was talking about, this opposite limit which is at some level an equally valid approximation to the actual timescales. The tidal effect is still inverse distance cubed, not inverse squared, because although the actual forces are inverse squared, the tilt of the equipotential tangent plane to the plane of the water introduces a small angle wherein we find the additional power of inverse distance (i.e., the lateral gradient contains an additional inverse distance), which is the part that was overlooked in his argument. But what remains true in his argument is that if we look at this timescale limit, we must focus on the tangential elements, where the effects of the tidal forcing is simply to introduce a velocity field into the water. There we find an important role for the coriolis effect and for the way continents force the velocity field to lie along the coastline. Then the directions of the velocity field do not at all follow the lateral gradients in the equipotential, any more than winds in the atmosphere follow the pressure gradients.

Indeed I suspect that including the coriolis effect on tides is analogous to how low pressure centers in the atmosphere produce cyclonic rotations (called "geostrophic flow"). If you imagine a pressure perturbation that peaks at some point in the atmosphere, you might expect the response of the air flow to also peak at that same point, but in fact the response is minimal at that point (the "eye of the hurricane")-- instead, the response proceeds in a circular fashion around that point, reminiscent of amphidromic points. Thus, in some sense, we might imagine that the tides on islands in the Pacific are more reminiscent of the weather patterns around atmospheric pressure fronts ruled by the relative location of various continents, rather than some direct connection to the equipotential. The varying equipotential then only sets the timescale for the tidal circulations and the overall scale of the height variations, not the details of where and when those variations peak, which brings us back to Gigabyte's argument.
Last edited by Ken G; 2018-Apr-15 at 07:18 AM.

7. So I must make a further admission, in my calculations I was not happy about the size of the momentum and now using the inverse cube that worry, which I did not air, is gone. My momentum was much too big for the actual tides observed and I did not see where the energy was dissipated. That is because of the resonance effect in the right sized oceans. I did indeed include Coriolis and again the momentum seemed too much. Now looking at the much reduced forcing, (100 million times less !) the momentum makes a lot more sense and the importance of the resonances can be put back into place. The North Atlantic for example can be seen in the GIF posted earlier to oscillate and even with the back force effect of the continental shelves this oscillation would be much greater with the kind of momentums I was previously calculating. That oscillation actually punches water right up to the arctic and the swirl effect of the Coriolis there creates very large warm currents. Thank you KenG for putting the context of the tangents in a better way.
Last edited by profloater; 2018-Apr-15 at 10:24 AM. Reason: typos

8. Originally Posted by grapes
I should've added "in the moon's gravity". They are in free fall in the moon's gravity.

Just watched this video:

Incredible. The guy slams every physics teacher for getting the tides wrong, says tides can't be like stretching taffy because the tidal force is too small, then says its like popping a pimple! If the tidal force is too small to stretch the ocean, it'll be too small to squeeze the ocean.
Thanks for that video, it's great.

9. Originally Posted by Ken G
Actually, believe it or not, a gravity that grows proportional to distance produces no tidal deformation at all, because the tidal forces would be spherically diverging about the center of the Earth. So the Earth would remain a sphere, squeezed equally in all directions, with no tides from gravity. Perhaps this shows why centrifugal forces are not a great way to see this, because centrifugal forces would deform the Earth into a slightly oblate spheroid, and given the 23 degree tilt of the rotation axis, would indeed produce tides. But if one just looks at the orbital plane, one can see that centrifugal forces and linearly increasing gravity produce no change at all to the equipotentials. One can explain it that all parts of the circular intersection with the orbital plane are happily in orbit without any internal forces from the Earth itself, given that a linearly increasing gravity produces orbits with all the same periods (a particularly simple version of Kepler's laws). But if one is happy with analyzing tidal forces all by themselves, it's more insightful just to say that a linearly increasing gravity produces spherically symmetric tidal forces, so no tides, just a slightly weaker Earth gravity.
We have to be careful when talking about centrifictional force, for two reasons. By saying that (linearly increasing) gravity produces orbits with all the same periods, you are ignoring the masses of the objects, since the actual orbit would be around the barycenter.

But it is tempting to analyze the system as a rotating solid (all orientations and distances constant) since the period would seem to be the same for all distances, but that induces a rotation in the orbiting body, with an effect from the rotation not strictly from the revolution.

10. Wow, I just answered a post from fifteen months ago! I've been wading through the posts the last week, one day at a time. I hope it's not too disjoint.

11. Originally Posted by grapes
Laplace does have it right, more or less.
https://en.wikipedia.org/wiki/Theory...idal_equations

U is the tide-forcing potential function that is "football shaped", a is earth radius, g is gravity, Ω is the angular rotation speed, D is depth, ς is vertical tidal elevation, and u and v are the velocity components in the latitude φ and longitude λ directions.

It looks to me like the first equation says that the rate of the change in the height of the tide depends upon the changes in the velocity times the depth. The next two equations are for the changes in the rates u and v, the negative term is the coriolis effect, and the only other term is the rate of change of the sum of the (negative) tide potential and the forcing potential. In other words, if the tide is parallel to the forcing, there will be no contribution from that last term.

Ken's tilted basin is a very good analogy, I've used it before myself.
Even worse, that last term is a little tricky, I've just realized. I borrowed the equations wholesale from wikipedia, but I'm pretty sure that the comment "u and v are the velocity components in the latitude φ and longitude λ directions" means that u is in the latitudinal direction, that is, the direction where λ changes.

You can see the similarities to the (linearized) wave equations here ( https://en.wikipedia.org/wiki/Shallo...servative_form ), the extra -bu and -bv terms represent viscous drag, otherwise the two sets line up equivalently (also, the cosine factor disappears, because x and y are a 2D perpendicular, unlike longitude and latitude, and by comparing them I noticed that I had a minus sign in the last equation above, when it should be a plus sign):

So, we can treat the tide as a gravity-forced wave, in that coordinate system, as Ken has pointed out. There are a couple differences--the speed of the wave, derived from that second set of equations, shouldn't really apply, since the equipotential surface doesn't move in the second set, but it moves with the wave in the first set of equations. A resonance in the ocean basins is one possibility--clearly the water cannot move from one side of the americas to the other.

Also, that helps answer Gigabyte's concerns about a 1000 mph wave crashing into shore--the surface of a wave that has conformed to the forcing equipotential does not have the energy to give up that such a wave in the second set of equations has.

12. Originally Posted by grapes
Yeah, going back through the thread, there are dozens of posts commenting on (and drawing conclusions from) illustrations of single component tidal results. M2 may be the strongest component, but it is a spherical degree two order two harmonic with axis parallel to the earth rotation axis--not at all the "football" which is a degree two order zero turned on its side.
I'd like to expand on this.

The football shape is a prolate spheroid, an ellipsoid with two equal shorter axes. It is still a degree two spherical harmonic, but its axis is perpendicular to the earth rotation axis. In tide theory, the components are broken up into harmonics whose axes are coincident with the earth rotation axis.

So, for a football on the equator (the moon or sun would be directly over the equator--that situation is rare but we can start there), it can be decomposed into two spherical harmonics. The first is an oblate spheroid, representing an average bulge/depression. The other component is M2-type, where the equipotential decreases from the equator by the square of the cosine of the latitude. It is modulated in the latitudinal direction by a simple cosine wave of two periods, the highest points are beneath the moon/sun, and antipodal to the moon/sun.

The solar tide is about half the lunar tide, and the resulting sum of the two is, first, an oblate spheroid that is just the
sum of their oblate spheroids and is constant on lines of latitude and so is not tide inducing as the earth rotates, and, second, a composite cosine wave that represents a phase shift. So, if my calculation is correct, that would mean when the sun is out of phase 120 degrees (60 degrees from the moon), the result is a M2-type tide forcing function that is shifted 30 degrees in phase, or 15 degrees away from the moon towards the sun. In other words, the actual shift on the surface of 2cos(2λ)+cos(2λ+2pi/3) = A cos(2λ+pi/6).

Illustrated at wolframalpha.com by plot(2*cos(2*λ)+cos(2*λ+2*pi/3);cos(2*λ+pi/6))

A is the square root of three (square root of 22+12+2*1*cos(2pi/3) ).

That analysis would pretty much work on any earth/moon/sun plane, in figuring out where the resultant bulge "points".

13. hesitatingly I comment, those three simultaneous partials need integrating to produce velocities. the integration will be in the form of a sinusoid from 0 to pi or a sin wt from 0 to 12 hours. At that time the direction reverses. Alternatively the forcing can be applied to a resonance set of equations with a damping factor. So again it gets complicated. The basin analogy seems just right but the scale means the propagation of the waves dominates the simple pendulum effect. One other analogy is a large bore hose in a U shape. If you observe such a hose and elevate one end, the level takes a characteristic time to drop, overshoot and stabilise. (A small bore hose is damped enough to just settle) if you then force the system with an oscillating up and down movement you can induce a resonant increase in amplitude. Not sure if that helps but a practical demonstration of a bowl, or a shallow baking tray would be better, with simulated land mass shapes would show basic tides when forced.

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The synchronous connection between the moon and the hump is the same as between the hump and the lamp post (next to your house).
Tidal waves are synchronized with the Earth's daily rotation, a 50-minute delay, according to the vortex theory of tides.
https://en.m.wikipedia.org/wiki/Lunitidal_interval shows the wave lag behind the moon.
https://en.m.wikipedia.org/wiki/Tide_clock shows the daily backlog of the wave.

15. Originally Posted by Fermer05
The synchronous connection between the moon and the hump is the same as between the hump and the lamp post (next to your house).
Tidal waves are synchronized with the Earth's daily rotation, a 50-minute delay, according to the vortex theory of tides.
https://en.m.wikipedia.org/wiki/Lunitidal_interval shows the wave lag behind the moon.
https://en.m.wikipedia.org/wiki/Tide_clock shows the daily backlog of the wave.

do not push your atm ideas here
to all: this will not be discussed in this thread

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Originally Posted by grapes
So, if my calculation is correct, that would mean when the sun is out of phase 120 degrees (60 degrees from the moon), the result is a M2-type tide forcing function that is shifted 30 degrees in phase, or 15 degrees away from the moon towards the sun.
I hesitate to comment on this, having not done the calculation. But it doesn't seem right, because it suggests a significant offset in the time of high tide, relative to the Moon, during what is known as a "neap" tide. What is normally claimed (for example, see https://oceanservice.noaa.gov/facts/springtide.html) is that neap tides are weaker because they are moderated by the opposite action of the Sun, not that the time of high tide is significantly shifted. So if what you are saying is correct, then almost everything written about neap tides is pure baloney. I don't say that's impossible, but it would be surprising.

17. Originally Posted by Ken G
I hesitate to comment on this, having not done the calculation. But it doesn't seem right, because it suggests a significant offset in the time of high tide, relative to the Moon, during what is known as a "neap" tide. What is normally claimed (for example, see https://oceanservice.noaa.gov/facts/springtide.html) is that neap tides are weaker because they are moderated by the opposite action of the Sun, not that the time of high tide is significantly shifted. So if what you are saying is correct, then almost everything written about neap tides is pure baloney. I don't say that's impossible, but it would be surprising.
Neap tides are when the sun is out of phase 180 degrees (90 degrees from the moon).

ETA: If the sun tidal effect were exactly equal to the moon's (instead of about half), then the sun 180 degrees out of phase would result in a lunar-solar tidal stress (as opposed to the strain that we are discussing in this thread) that would be similar to a centrifictional bulge. The earth, tilted with respect to their plane, could still rotate through differing tidal stresses.

EETA: The centrifictional bulge does have a small component of degree four. Its magnitude is much less than the degree two component, but it is comparable to the other geophysical components.

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