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Thread: Question about Earth tides

  1. #211
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    Quote Originally Posted by Gigabyte View Post
    Yes, but guess what? I checked the http://volkov.oce.orst.edu/tides/ page and they have the 24 hour tide animation for April 3 2018 up!

    http://volkov.oce.orst.edu/tides/pic/tpxo9.gif

    I don't know if they are updating daily now, but it's wicked cool. And as always, it shows the actual tidal behavior of the planet.
    What is "wicked cool" about it? Apart from being current, it just seems to show the same stuff happening that we know happens.

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    The vortex theory of tides can be easily verified by the connection between the height of the tidal wave and the rotation speed of the whirlpools.

    The list of seas with an average swirl speed of more than 0.5 km / h, and an average tidal wave height of more than 5 cm:
    Irish Sea, North Sea, Barents Sea, Baffin Sea, White Sea, Bering Sea, Sea of Okhotsk, Arabian Sea, Sargasso Sea, Hudson Bay, Maine Bay, Gulf of Alaska. and so on.
    The list of seas with an average swirl speed of less than 0.5 km / h, and an average tidal wave height of less than 5 cm:
    The Baltic Sea, the Greenland Sea, the Black Sea, the Sea of Azov, the Caspian Sea, the Chukchi Sea, the Kara Sea, the Laptev Sea, the Red Sea, the Marmara Sea, the Caribbean Sea, the Sea of Japan, the Gulf of Mexico, etc.
    Note: The height of the tidal wave (soliton) and the amplitude of the tides is not the same.
    Last edited by Fermer05; 2018-Apr-04 at 06:26 PM.

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    Quote Originally Posted by Fermer05 View Post
    The vortex theory of tides can be easily verified ...
    Fermer05, did you miss this post?

    Quote Originally Posted by PetersCreek View Post
    Fermer05,

    Your ideas about tides are very much against the mainstream (ATM) and may not be advocated outside of the ATM forum. If you wish to pursue your claims, you must do so there. Please read our rules and ATM advice, both linked in my signature line below.
    To be clear, the "vortex theory" is ATM and will not be advocated outside of the ATM forum. If you want to pursue it, start a thread in ATM.

    You are receiving an infraction for ATM outside of ATM.
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  4. 2018-Apr-04, 04:14 PM
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  5. 2018-Apr-04, 05:04 PM
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  6. #214
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    Quote Originally Posted by Gigabyte View Post
    Quote Originally Posted by grapes
    By observation, do you mean the animation here: https://www.youtube.com/watch?v=5zi7N06JXD4
    Quote Originally Posted by grapes
    What I meant by this was, I seem to see a mid-ocean near-equatorial bulge that progresses across the pacific at about the speed of earth rotation. We don't know whether the bulge is lagging behind the moon or not, but the bulge does seem to be there in that animation.
    No, there is no bulge moving at 1000 mph
    I seem to see one. They start at the right of the image, progress to the left, and the time taken is around a day.

    The bulge is obscured by its interaction with the continents, because of the lack of water, but it does appear. I'm still not sure what the lunar/solar phase is, especially since that's a simulation. Ken was convinced that there would be small correlation, but I'm not convinced by his argument.

    Physically, the bulge exists in the sense that the earth's shape is defined as the geoid, the time-averaged equipotential surface, and the moon's (and sun's) gravity modifies that equipotential surface.

    Ken made that point early on in the thread:
    Quote Originally Posted by Ken G View Post
    Yes, we can certainly say that understanding tides starts with the shape of the equipotential of the effective gravity, which includes corrections to Earth gravity from the Moon's gravity and the Earth's orbital motion in the Earth-Moon system. That equipotential looks like two bulges of very small amplitude (the bulge only lifts the water about a foot above the average height), and that is pretty much what you find in the open ocean (though currents from the continents and the ocean floor affect that too). But that cannot end the discussion, because obviously tides on beaches have a much higher amplitude, and in inlets (like the Bay of Fundy) can be 10-20 times higher than that There can also be one tide per day, for the reason Grant gave (the bulges aren't on the equator), and due to more complicated resonances in the current flows along coastlines of various shapes. But all of this is caused by that double-bulge equipotential, no one could ever understand the first thing about tides without understanding those two bulges.

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    Quote Originally Posted by grapes View Post
    I seem to see one. They start at the right of the image, progress to the left, and the time taken is around a day.

    The bulge is obscured by its interaction with the continents, because of the lack of water, but it does appear. I'm still not sure what the lunar/solar phase is, especially since that's a simulation. Ken was convinced that there would be small correlation, but I'm not convinced by his argument.

    Physically, the bulge exists in the sense that the earth's shape is defined as the geoid, the time-averaged equipotential surface, and the moon's (and sun's) gravity modifies that equipotential surface.

    Ken made that point early on in the thread:
    I still do not agree even with Ken, the major factor driving the tides is the tangential acceleration at the edges. Not the equipotentials the fluids accelerate and then reverse. Huge tangential momentums develop, then the contours cause the rising water. This is orders bigger than the bulges which are radial displacements. Euler was not wrong about this but many eminent people do not understand.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Let me clarify the use of the term "bulge", which does sound like a radial displacement in the water, but actually it's not that-- it's a bulge in the equipotential. Normally Earth's equipotential is a sphere (ignoring oblateness), but tidal forces turn that into a football (to exaggerate). That's all that tides come from, without that "bulge" there's no tides. But I completely agree that the water does not fit to the shape of those bulges, because the timescale for the water to reach the bulge is comparable for the timescale for the bulge to not be there any more. In short, it is very important that the bulge is rotating, and doing so on a relatively fast timescale from the point of view of the water. This makes everything much more complicated, even before we introduce coastlines and channels and all that. The tendency for the complexities to produce special points that have essentially no tidal effects, with tidal rises rotating around those points, is a very different behavior than following the bulge of a football, and I confess I did not appreciate that at the start of this thread. Nevertheless, it's all caused by that football-- a football in equipotential, not water. I certainly agree, however, that coriolis forces must not be neglected, that's clear from those tidal movies.
    Last edited by Ken G; 2018-Apr-10 at 09:16 PM.

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    Fermer has been busy...

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    Quote Originally Posted by profloater View Post
    I still do not agree even with Ken, the major factor driving the tides is the tangential acceleration at the edges. Not the equipotentials the fluids accelerate and then reverse. Huge tangential momentums develop, then the contours cause the rising water. This is orders bigger than the bulges which are radial displacements. Euler was not wrong about this but many eminent people do not understand.
    The driving factor in the dynamic theory of Laplace is the equipotential. What do you mean about Euler?
    Quote Originally Posted by Ken G View Post
    Let me clarify the use of the term "bulge", which does sound like a radial displacement in the water, but actually it's not that-- it's a bulge in the equipotential.
    Yes. This thread is mainly concerned about the expression of that bulge in the actual tide, I think.
    Normally Earth's equipotential is a sphere (ignoring oblateness), but tidal forces turn that into a football (to exaggerate).
    Looking back through the thread, where the resultant moon/sun effect is discussed, I'd just like to point out that if the tidal effect of the sun were actually equal to the moon's, then the bulge at neap tide would be an oblate spheroid, not a football (prolate spheroid).
    That's all that tides come from, without that "bulge" there's no tides. But I completely agree that the water does not fit to the shape of those bulges, because the timescale for the water to reach the bulge is comparable for the timescale for the bulge to not be there any more. In short, it is very important that the bulge is rotating, and doing so on a relatively fast timescale from the point of view of the water. This makes everything much more complicated, even before we introduce coastlines and channels and all that. The tendency for the complexities to produce special points that have essentially no tidal effects, with tidal rises rotating around those points, is a very different behavior than following the bulge of a football, and I confess I did not appreciate that at the start of this thread. Nevertheless, it's all caused by that football-- a football in equipotential, not water. I certainly agree, however, that coriolis forces must not be neglected, that's clear from those tidal movies.

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    Quote Originally Posted by grapes View Post
    Looking back through the thread, where the resultant moon/sun effect is discussed, I'd just like to point out that if the tidal effect of the sun were actually equal to the moon's, then the bulge at neap tide would be an oblate spheroid, not a football (prolate spheroid).
    Yes good point, the football focuses only on the Moon. The general shape is like an ellipsoid but with three different elongations along the three different axes, so it's not even an oblate spheroid (an oblate spheroid has the two larger elongations the same, a football has the two smaller elongations the same, but this shape has none of the three the same in general).
    Last edited by Ken G; 2018-Apr-11 at 08:31 PM.

  12. #220
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    And the axes are not orthogonal, except on rare occasions.

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    Quote Originally Posted by grant hutchison View Post
    And the axes are not orthogonal, except on rare occasions.
    It seems to me they must always be orthogonal. Regardless of the complexity of the mass distribution of their sources, the tidal accelerations (being small) can be thought of as arising from a Jacobian matrix with entries like [dg_x/dy] etc., multiplying a displacement vector like (dx, dy, dz) from the center of mass of the Earth. Since the accelerations come from a gradient of a potential function U, we have g_x =- dU/dx and so on. This means the xy entry in the Jacobian matrix looks like -d^2 U / dxdy, and the yx entry looks like -d^2 U / dydx. But that's the same-- so the matrix is symmetric. I strongly suspect that in your widespread experience you have encountered the extreme significance of symmetric matrices, so you know that they can always be diagonalized by a set of perpendicular axes. Along those directions, the tidal squeeze is aligned along those directions (as can be seen by multiplying out the diagonalized matrix), so the surface of the equipotential must cross those axes perpendicular to those axes. Hence they are principal axes of the deformation, and this would be true even if the Earth orbited a triple star and had a distant moon of equal mass to itself. You have no doubt assimilated the moral: there are only so many things a small gravitational perturbation is capable of doing!
    Last edited by Ken G; 2018-Apr-12 at 02:33 PM.

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    I think Grant was talking about the earth-moon axis and the earth-sun axis. Even at neap tide each month they are only approx orthogonal.

    ETA: but yeah, to agree with your comment, the result is a tri-axial spheroid then, a flattened (American) football then, when the sun is involved. And, it wouldn't necessarily "point" at the moon.

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    I actually meant that the axes are only very rarely orthogonal to the Earth's rotation axis - the triaxial tidal ellipsoid is lolloping around in a very complicated dance relative to the principle axes of the figure of the Earth, as well as continuously varying in its axial ratios. So any glimpse of a single day's worth of tidal data is going to be unrepresentative of the long-term average behaviour.
    But it was one of those occasions when what came out of my fingers bore only a passing relationship to what was going on in my head.

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    Quote Originally Posted by grant hutchison View Post
    I actually meant that the axes are only very rarely orthogonal to the Earth's rotation axis - the triaxial tidal ellipsoid is lolloping around in a very complicated dance relative to the principle axes of the figure of the Earth, as well as continuously varying in its axial ratios. So any glimpse of a single day's worth of tidal data is going to be unrepresentative of the long-term average behaviour.
    But it was one of those occasions when what came out of my fingers bore only a passing relationship to what was going on in my head.
    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.

  17. #225
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    Quote Originally Posted by grapes View Post
    The driving factor in the dynamic theory of Laplace is the equipotential. What do you mean about Euler?

    Yes. This thread is mainly concerned about the expression of that bulge in the actual tide, I think.

    Looking back through the thread, where the resultant moon/sun effect is discussed, I'd just like to point out that if the tidal effect of the sun were actually equal to the moon's, then the bulge at neap tide would be an oblate spheroid, not a football (prolate spheroid).
    Euler was the first to challenge Newton and pointed out the significance of the tides of fluids on the surface.
    It is easy to check.
    if we consider a kilogram of water in an ocean, choosing a point on the equator say, but choosing a time in the rotation of the Earth when our kg "sees" the moon and sun on the horizon, and not at the zenith.

    That kg of water experiences a tangential pull from those bodies which can be calculated using Newton, and the inverse square distance. Being in a fluid it is accelerating toward the moon and so are all the kg of fluids in that region.

    That situation continues for a few hours. In a simple way we could say it lasts for about 6 hours from the +45 to - 45 degree positions. Then for about six hours the chosen point is on the backside or frontside, before another six hour period where the force is reversed. Really its a sinusoidal integral over 12 hours from moon overhead.

    In that time period the small acceleration is integrated into a velocity and the velocity times the mass of water (or air) is a momentum tangential at at near constant earth radius. Without land contour it would not amount to more than sideways currents that reverse every 12 hours. In addition and much smaller, the equipotential bulge is rotating and trying the lift the water but this is a very small effect.

    The moving water, whole oceans of it, is then affected by the sea bed, the continents, the forced changes in lattitude, hence Coriolis and the natural sloshing frequencies of various trapped waters. It is a rythmic driving maximised when the sun and moon are aligned and minimised when the sun is the opposite side every couple of weeks later. It turns out the gravity pull of the moon is about double that of the sun.

    Any solid planet would not see those changes of course, there the tidal calculation is the inverse cube of the distance and is a strain not a sliding acceleration.

    Anyone can plug in the distances and masses to confirm the significant basic velocity of a free liquid or the atmosphere. I feel the established solid body tidal force equations have allowed people to forget that sideways currents under simple gravity face hardly any resistance until the boundaries change in shape. The actual movements are certainly complex and there are countercurrents due to resonances and the moving situation between sun and moon.
    Last edited by profloater; 2018-Apr-12 at 05:36 PM. Reason: typo
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    here is a simplified example using the moon and force = G m m / d^2
    I will work to one significant figure to make it easy to follow:

    Moon mass 7 . 10^22 kg
    moon distance 4 . 10^8 m
    G = 7 . 10^-11

    acceleration sideways due to that force on 1 kg = 7. 7 /16 . 10^-5 = 3. 10^-5 m/s/s

    using the 6 hour approximation or about 2 . 10^4 seconds, and speed = accn x time

    the speed implied is 0.6 m/s

    It's not a huge figure, about 2 mph, but then you work out the momentum implied for all the water in the region: that is huge.
    Last edited by profloater; 2018-Apr-12 at 05:57 PM. Reason: typo again
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    Euler was the first to challenge Newton and pointed out the significance of the tides of fluids on the surface.
    It is easy to check.
    if we consider a kilogram of water in an ocean, choosing a point on the equator say, but choosing a time in the rotation of the Earth when our kg "sees" the moon and sun on the horizon, and not at the zenith.

    That kg of water experiences a tangential pull from those bodies which can be calculated using Newton, and the inverse square distance. Being in a fluid it is accelerating toward the moon and so are all the kg of fluids in that region.
    So are all the solid kg in that region, the net result relative to the earth is that there is no tangential tidal force there.
    That situation continues for a few hours. In a simple way we could say it lasts for about 6 hours from the +45 to - 45 degree positions. Then for about six hours the chosen point is on the backside or frontside, before another six hour period where the force is reversed. Really its a sinusoidal integral over 12 hours from moon overhead.

    In that time period the small acceleration is integrated into a velocity and the velocity times the mass of water (or air) is a momentum tangential at at near constant earth radius. Without land contour it would not amount to more than sideways currents that reverse every 12 hours. In addition and much smaller, the equipotential bulge is rotating and trying the lift the water but this is a very small effect.

    The moving water, whole oceans of it, is then affected by the sea bed, the continents, the forced changes in lattitude, hence Coriolis and the natural sloshing frequencies of various trapped waters. It is a rythmic driving maximised when the sun and moon are aligned and minimised when the sun is the opposite side every couple of weeks later. It turns out the gravity pull of the moon is about double that of the sun.

    Any solid planet would not see those changes of course, there the tidal calculation is the inverse cube of the distance and is a strain not a sliding acceleration.

    Anyone can plug in the distances and masses to confirm the significant basic velocity of a free liquid or the atmosphere. I feel the established solid body tidal force equations have allowed people to forget that sideways currents under simple gravity face hardly any resistance until the boundaries change in shape. The actual movements are certainly complex and there are countercurrents due to resonances and the moving situation between sun and moon.

  20. #228
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    Quote Originally Posted by grapes View Post
    So are all the solid kg in that region, the net result relative to the earth is that there is no tangential tidal force there.
    Yes of course but the force is trivial for solids. Fluids can and will move. When you consider that all the water in that region is accelerated at the same time, there is no friction of viscous resistance. The water just starts to move and as calculated will reach significant speed as a current. We are talking about millions of tons of water moving sideways as a current, not just my test kg. This is a very important issue if anyone wishes to understand the tides.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    This is mainstream gravity. A unit mass in that position has a radial force due to the Earth of about 10 Newtons and a force toward the moon of 3. 10^-5 Newtons. Obviously the moon force is trivial for a restrained solid, but a kg of water or air will slide sideways because there is nothing to stop it. That acceleration due to that force goes on for hours before it reverses.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    Yes of course but the force is trivial for solids. Fluids can and will move. When you consider that all the water in that region is accelerated at the same time, there is no friction of viscous resistance. The water just starts to move and as calculated will reach significant speed as a current. We are talking about millions of tons of water moving sideways as a current, not just my test kg. This is a very important issue if anyone wishes to understand the tides.
    No, the point that grapes is making (I think) is that the acceleration due to the Moon that you calculate for a blob of water in the ocean is exactly the same as the acceleration due to the Moon of the entire Earth. Which also experiences no net friction, because it's all moving together. The whole Earth accelerates toward the Moon because of that force, not just the blob of water. So the net motion of the water relative to the Earth is zero. There is no "restrained solid", because the gravity from the Moon affects the entire Earth. The only thing relevant for the tides is the differential gravity between various points on the Earth, because the Earth is an extended object rather than a single point.
    Conserve energy. Commute with the Hamiltonian.

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    This paper
    https://www.tandfonline.com/doi/pdf/10.3137/OC311.2009
    is about modelling the continental shelf effect on the deep ocean tidal resonances. It's complicated. However it confirms the idea that the continental shelves reduce the amplitude of the ocean resonances by what they call a back effect. This can be visualised as a rise in water level at the shelf as the deep water momentum forces it to rise, and that rise is a driver by pressure to hold back the tide. The large whole ocean resonance would be driven by the vertical rises at each side like water in a basin, but the tidal driving force is caused by the momentum generated every cycle.

    A big question therefore is not so much what causes the tides but why they are not even higher due to resonance. Modelling like that described in the paper may help model the tides during ice ages and when the land masses were differently disposed in the past.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by Grey View Post
    No, the point that grapes is making (I think) is that the acceleration due to the Moon that you calculate for a blob of water in the ocean is exactly the same as the acceleration due to the Moon of the entire Earth. Which also experiences no net friction, because it's all moving together. The whole Earth accelerates toward the Moon because of that force, not just the blob of water. So the net motion of the water relative to the Earth is zero. There is no "restrained solid", because the gravity from the Moon affects the entire Earth. The only thing relevant for the tides is the differential gravity between various points on the Earth, because the Earth is an extended object rather than a single point.
    OK that is a different point but easily countered.

    The Earth is spinning and any effect from the moon on its orbit would be at moon orbit frequency. The water on the surface has no equivalent inertia, the kg test mass and all the mas of fluid is completely free to move at no radial change, sliding at constant radius. The waters at the opposites sides of the earth also move towards the moon of course so that if the earth were not spinning we would see the equipotential bulge as the moon proceeds around the earth, travelling around phase locked to the moon.

    But the earth does rotate and those sliding velocities get reversed every 12 hours (approx) The integrating effect is very real. The change in gravity potential by the moon has that same tiny number but vertically the lift of the water would be too small to measure. The 0.6 m/s current I just calculated above is very significant.

    So the tangential force on a kg of rock and a kg of water are of course the same but the rock cannot move.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by Grey View Post
    No, the point that grapes is making (I think) is that the acceleration due to the Moon that you calculate for a blob of water in the ocean is exactly the same as the acceleration due to the Moon of the entire Earth. Which also experiences no net friction, because it's all moving together. The whole Earth accelerates toward the Moon because of that force, not just the blob of water. So the net motion of the water relative to the Earth is zero. There is no "restrained solid", because the gravity from the Moon affects the entire Earth. The only thing relevant for the tides is the differential gravity between various points on the Earth, because the Earth is an extended object rather than a single point.
    I feel I did not answer well. If we consider the earth as a rigid solid it accelerates with the water on it toward the moon. But the tidal forces on the nearest point and furthest point are different and that is found in the inverse cube relationship. If we consider just one point it experiences a reversing or sinusoidal force which the centre of the Earth does not feel. The Earth clearly does not move as a body at its spin rate. The reversing forces at the tangents are not large enough to break pieces off so we discount them, but the fluids respond to the inverse square. So while the Earth responds to the moon orbit period, the fluids respond to the spin rate.

    I will confess I am unsettled so I thank you for your comment. Maybe I will spend some more time on this.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    So are all the solid kg in that region, the net result relative to the earth is that there is no tangential tidal force there.
    Yes of course but the force is trivial for solids. Fluids can and will move.
    Not if the net force is zero, that's more or less what is correctly handled by the dynamic theory of Laplace. The tidal force itself is not important.
    When you consider that all the water in that region is accelerated at the same time, there is no friction of viscous resistance. The water just starts to move and as calculated will reach significant speed as a current. We are talking about millions of tons of water moving sideways as a current, not just my test kg. This is a very important issue if anyone wishes to understand the tides.

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    Quote Originally Posted by grapes View Post
    Not if the net force is zero, that's more or less what is correctly handled by the dynamic theory of Laplace. The tidal force itself is not important.
    Let me add further notes.
    The angular momentum of the spinning Earth is represented by all the individual particles linear momentum each held into a circle by centripetal force. That restraining force for a rock is inter atomic, inter molecular as we know as a solid.

    So the rock in the Atlantic which at one moment is travelling away from the moon experiences a few microNewtons of force trying to decelerate it but it is bound to its opposite rock in the pacific which at that time is experiencing an equal and opposite moon force trying to accelerate it. Same for all the solid that makes up the great flywheel that is the spinning Earth so its spin is not affected by the moon. Those solid tensions are orders greater than gravity.

    However the water and the air above those rocks are not bound together at all. They are held in place by gravity only. So the kg of water in the Atlantic is being decelerated and the kg of water in the pacific is being accelerated by those small tangential forces from the moon.

    In fact the water and air see a 24 hour sinusoidal forcing. When under the moon they have a small decrease in their local gravity, but if they try to rise they are directly still constrained by that gravity. However when at the tangent so that the moon force is from the side, they have no constraint equivalent to the solid forces. Not even viscous effects (which are responsible for driving the water araound with the solid earth) because all their neighbours are feeling the same force.

    So relative to the solid surface the fluids are free to move sideways and that acceleration causes local relative velocity.

    I am grateful for the chance to explain that more fully because it is true that all particles feel the moon force but the surface feels it as a sinusoidal force. I simplified it to six hours of sideways acceleration but the principle is the same.

    I know it is also a simplification to call the Earth a solid but in relative terms the surface rocks are tied together while the fluids have no such tensions and respond to every force by moving.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

  28. #236
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    Quote Originally Posted by profloater View Post
    Let me add further notes.
    The angular momentum of the spinning Earth is represented by all the individual particles linear momentum each held into a circle by centripetal force. That restraining force for a rock is inter atomic, inter molecular as we know as a solid.

    So the rock in the Atlantic which at one moment is travelling away from the moon experiences a few microNewtons of force trying to decelerate it but it is bound to its opposite rock in the pacific which at that time is experiencing an equal and opposite moon force trying to accelerate it. Same for all the solid that makes up the great flywheel that is the spinning Earth so its spin is not affected by the moon. Those solid tensions are orders greater than gravity.

    However the water and the air above those rocks are not bound together at all. They are held in place by gravity only. So the kg of water in the Atlantic is being decelerated and the kg of water in the pacific is being accelerated by those small tangential forces from the moon.

    In fact the water and air see a 24 hour sinusoidal forcing. When under the moon they have a small decrease in their local gravity, but if they try to rise they are directly still constrained by that gravity. However when at the tangent so that the moon force is from the side, they have no constraint equivalent to the solid forces. Not even viscous effects (which are responsible for driving the water araound with the solid earth) because all their neighbours are feeling the same force.
    Not only their neighbors, but the entire earth. They are in free fall. That's why there is no tangential tidal force there.

    It would be like liquids on the ISS -- they would appear to float, maybe bead up, there is no significant acceleration relative to the ISS.
    So relative to the solid surface the fluids are free to move sideways and that acceleration causes local relative velocity.

    I am grateful for the chance to explain that more fully because it is true that all particles feel the moon force but the surface feels it as a sinusoidal force. I simplified it to six hours of sideways acceleration but the principle is the same.

    I know it is also a simplification to call the Earth a solid but in relative terms the surface rocks are tied together while the fluids have no such tensions and respond to every force by moving.

  29. #237
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    Quote Originally Posted by grapes View Post
    Not only their neighbors, but the entire earth. They are in free fall. That's why there is no tangential tidal force there.

    It would be like liquids on the ISS -- they would appear to float, maybe bead up, there is no significant acceleration relative to the ISS.
    They are not all in free fall. 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.
    Last edited by profloater; 2018-Apr-13 at 07:54 AM. Reason: typo
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    In the space station or indeed any such thing, if it is basically a spinning solid, then any liquid would still feel the tides in the same way. The spin would cause linear momentum to by constantly modified by the solid surfaces and the fluid would be a film on the surface but assuming it's thick enough to be free of boundary layer ie viscous effects, then tides would happen at spin frequency. The ISS is not spinning so the reversal of momentum is not there in the solid structure.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

  31. #239
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    Quote Originally Posted by profloater View Post
    So the tangential force on a kg of rock and a kg of water are of course the same but the rock cannot move.
    Quote Originally Posted by profloater View Post
    I know it is also a simplification to call the Earth a solid but in relative terms the surface rocks are tied together while the fluids have no such tensions and respond to every force by moving.
    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.
    Conserve energy. Commute with the Hamiltonian.

  32. #240
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    Quote Originally Posted by profloater View Post
    another thread about tides? And who is recalling Euler? yes the equipotential bulges are there but they are tiny compared to the lateral integrated accelerations caused by the tangential force on the fluids. Then the velocities accumulated represent huge sideways momentums which get channeled by the contours and stirred up by Coriolis as they change lattitude. There is this confusion between tidal forces on a solid planet or moon (inverse distance cubed) and the tangential forces on fluids (inverse distance squared) which are free to move sideways at constant radius. the latter are the driving force for the tides we experience on Earth.
    A novice could glance at this post and think you are arguing that the motion of the ocean water will go as inverse square rather than inverse cube of the distance to the perturbing body. If you are arguing such it is my opinion that you are mistaken.

    Let us start at a point 90 degrees from the sublunar point, so we are abreast of the center of the Earth. The acceleration of a sample of water here toward the Moon is indeed inversely proportional
    to the square of the distance R to the Moon, but the underlying solid body is accelerating at the same rate, so there is no motion relative to the bottom. Now let's go a distance r toward the Moon, and for the purpose of argument assume that the solid body is not deformable while the water can move sideways with no resistance. Let G and M be unity to reduce visual clutter in the presentation.

    Let A be the local acceleration of the water while a is the acceleration of the solid body.

    A = 1/(R-r)2

    a = 1/R2

    The acceleration of the water relative to the solid body is the difference between the two, which I will call dA.

    dA = 1/(R-r)2 - 1/R2

    = [R2 - (R-r)2]/[R2(R-r)2]

    = [2Rr - r2]/[R2(R-r)2]

    For R>>r, a very good approximation follows:

    dA ~ 2Rr/R4 ~ 2r/R3

    Thus the acceleration of the water relative to the bottom is closely approximated by inverse cube of R, not inverse square.

    Nobody can deny that nevertheless the vast tonnage of water involved packs a wallop when it piles up in places like the Bay of Fundy.

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