A perfectly flat bucket of water?

[Wes Tausend]

...

I was thinking of a thought experiment concerning different systems of gravity-like acceleration.

I've noticed that using a stringline in construction is close, but not very accurate because the string always sags to some degree. I have sometimes fabricated a water level by attaching short clear extensions on the ends of a garden hose and assuming the water self-levels to natural flatness in the clear tubes. This is a much better method if long distance accuracy is important for subtle lot drainage.

So a water level is good. But how good is this, really? I live in the higher US latitudes, so, for me, it is probably more accurate east-west, than north-south, because of the equatorial bulge. The south end of my water level should always be too high by the tiniest fraction, and even worse in a tidal condition.

In another scenario, if I had a bucket of water sitting on the floor of my home, I might suppose that the surface is perfectly flat. This is likely the principle of manufacturing "float glass". Making float glass allows molten glass to float very level, and cool to hardness, on a bed of heavier molten metal.

I think the truth, though, is that my bucket of water and the floating glass both must have a very slight convex curvature to their top surface. The bucket water surface should be shaped in a slight convex curvature exactly like the ocean and so should the float glass. Now, admittedly, this is pretty flat for most purposes. On the same token, some mirrors are specifically spun, in a molten state, to produce a precision parabolic concave surface that may be used in Newtonian Reflector telescopes. I imagine this "reverse type" curvature effect must be taken into rpm selection for the very finest mirrors, and yet may not be entirely possible to perfect.

Another gravity-like acceleration is that of rotating space stations. In this case, I imagine the bucket, sitting against the inside circumference of the outer rim, to possibly have a concave surface, just the opposite of earth. As an example, if I were to not like the constantly curved floor for walking in artificial gravity, I could build a polygonal series of short flat floors entirely around the inner circumference so that I may walk on "flat" floors. Here though, I think that the series of floor "flats" would feel like they had a high spot in the middle of each flat polygonal side, and a dip as one walked into the joint angles. If I spilled water, it should run "downhill" into these dips. The dips are further from the hub center, and I imagine, the further out, the more acceleration. My bucket, sitting on such a floor, should do the same and develop some sort of natural concave curvature to it's surface.

The arbitrary "level seeking", of the spilled water, seems like it might, or might not, also cause flow to the sides of the floor "flats", since the outer edges are also further from hub dead center. Then, on the other hand, the floor edges are all perpendicularly equidistant to the hub axis which may only cause water curvature in the direction of rotation, not side-to-side. Which is correct? The latter seems like it may rule.

Considering the accuracy, or inaccuracy, of the above, can I theoretically slowly swing a water bucket around on earth, with a rope, in such a simple fashion as to carefully balance the "opposing(?) curve" acceleration forces into a perfectly flat surface? The idea of the original thought experiment was to achieve perfect natural flatness right here on mother earth, a process that can not apparently be easily done any other way. As an example, one might use a laser light to measure a float surface, but of course, gravity very slightly bends light too, so even that would not be good enough for perfection on Earth (unless the process was done by machining, and measured by light vertically).

The final attempt at flatness might be to simply slowly spin the bucket on a turntable, by calculated rpm, until the centrifugal force just barely flattens the natural curvature of earth. But this may not work either, if the spin opposing force is unevenly distributed in such a fashion as to tend to a parabolic curve, rather than equal-radius curve such as the shape of earth. This slight force difference may technically make any earthmade-spun parabolic mirror also just a shade imperfect.

It might just be, that the only way to get a perfectly flat bucket of water, or a perfect mirror, is to accelerate in linear fashion in a gravity-free environment. Then I would need a smooth rocket... and persnickety natural earthbound surface flatness is out.

Is there any other natural way to achieve real world flatness?

Thanks for your thoughts,
Wes
...

 

[mfb]

The Earth is not a disk, there is no single natural way to define "flat". What do you mean with that?

• same potential
  Water is exact here. Surfaces of the same potential are roughly oblate spheriods (=sphere with equatorial bulge), with smaller deviations due to local mass concentrations.
• Same distance from the geometric center of earth
  Well... who cares? Anyway, differential GPS can do this with a precision of a few centimeters, and astronomy should be able to do even better.
• Same relative height above an oblate spheroid
  Use differential GPS.
• A local area without curvature
  Note that this will not appear flat if it is large enough - the outer edge is higher above sea level than the inner side. Lasers will be able to do this with incredible precision (account for gravitational light deflection, if you care about picometers), but GPS works as well.
• Something else?

 

[Wes Tausend]

The Earth is not a disk, there is no single natural way to define "flat". What do you mean with that?

• same potential
  Water is exact here. Surfaces of the same potential are roughly oblate spheriods (=sphere with equatorial bulge), with smaller deviations due to local mass concentrations.
• Same distance from the geometric center of earth
  Well... who cares? Anyway, differential GPS can do this with a precision of a few centimeters, and astronomy should be able to do even better.
• Same relative height above an oblate spheroid
  Use differential GPS.
• A local area without curvature
  Note that this will not appear flat if it is large enough - the outer edge is higher above sea level than the inner side. Lasers will be able to do this with incredible precision (account for gravitational light deflection, if you care about picometers), but GPS works as well.
• Something else?

mfb,

I mean a flat, Euclidian, surface plane of water in a handheld, earthbound bucket no more than one foot across in diameter. It can be at any angle to the center of Earth if that will help. I don't believe a natural local area, the size of a water bucket, is technically flat on Earth unless we define it especially as so, or rather good enough, for the purpose of a particular, non-picky measurement.

But how may we simply cause said earthbound bucket to be perfectly flat in terms of a counter-acceleration to the natural curvature of earth? Is there a way? Perhaps not with an acceleration, but maybe with a carefully shaped magnetic force, I might suppose.

Thanks,
Wes
...

 

[mfb]

The water surface will have a curve radius of ~6370km. Within 30cm (~1 foot), this gives a deviation of a few nanometers - less than 100 layers of molecules. This is completely negligible relative to surface waves, evaporation, and other effects.

Telescope mirrors can be built with a similar precision. They are curved (to serve as telescope mirrors), but it would be possible to build them flat, too.In theory, you could probably counter the curvature of Earth with a very slow rotation - something like 1 revolution per hour. The remaining effects would be ~9 orders of magnitude below those few nanometers, way below the size of atoms.

 

[pervect]

mfb,

I mean a flat, Euclidian, surface plane of water in a handheld, earthbound bucket no more than one foot across in diameter. It can be at any angle to the center of Earth if that will help. I don't believe a natural local area, the size of a water bucket, is technically flat on Earth unless we define it especially as so, or rather good enough, for the purpose of a particular, non-picky measurement.

But how may we simply cause said earthbound bucket to be perfectly flat in terms of a counter-acceleration to the natural curvature of earth? Is there a way? Perhaps not with an acceleration, but maybe with a carefully shaped magnetic force, I might suppose.

Thanks,
Wes
...

The idea that comes to my mind is to try to use the same techniques Robert Forward suggest for gravity compensation in orbit.

In particular, his six-sphere compensator. See http://gravityresearchfoundation.org/pdf/awarded/1982/forward.pdf

Because tidal forces scale as GM/r^3, the amount of acceleration you'll have to compensate for will be greater (I estimate double, assuming LEO is 2000 km above the Earth's surface).

I can't say I've looked at in in great detail, but it looks like it might still be feasible.

I'm not sure about applying his other ideas, I don't think offhand they're needed.

 

[Wes Tausend]

The water surface will have a curve radius of ~6370km. Within 30cm (~1 foot), this gives a deviation of a few nanometers - less than 100 layers of molecules. This is completely negligible relative to surface waves, evaporation, and other effects.

Telescope mirrors can be built with a similar precision. They are curved (to serve as telescope mirrors), but it would be possible to build them flat, too.


In theory, you could probably counter the curvature of Earth with a very slow rotation - something like 1 revolution per hour. The remaining effects would be ~9 orders of magnitude below those few nanometers, way below the size of atoms.

mfb,

Thank you for some hard calculation. It does seem overly picky of me to be concerned about such an obscure principle, but there are instances where slight differences may matter as in the post following yours.

It seems that to, "counter the curvature of Earth with a very slow rotation", would not work all that perfect. Some high central spherical curvature might still exist in the rotating bucket when the outer edges have already become equal in height. I would surmise, that at that point, this would produce a circular surface standing wave-shape halfway between that of a sphere, and that of a parabolic ellipsoid. I think further rpm would likely produce ever higher edges before the center eventually comes close to any semblance of flattening. But perhaps I am mistaken.

Wes
...

 

[Wes Tausend]

The idea that comes to my mind is to try to use the same techniques Robert Forward suggest for gravity compensation in orbit.

In particular, his six-sphere compensator. See http://gravityresearchfoundation.org/pdf/awarded/1982/forward.pdf

Because tidal forces scale as GM/r^3, the amount of acceleration you'll have to compensate for will be greater (I estimate double, assuming LEO is 2000 km above the Earth's surface).

I can't say I've looked at in in great detail, but it looks like it might still be feasible.

I'm not sure about applying his other ideas, I don't think offhand they're needed.

pervect,

The link is a slightly different scenario, but a great example of nulling tiny forces in real life.

If it were possible, it seems having an identical sphere the size, mass and density of Earth just above the bucket on the ground would cancel all water surface curvatures for the extraordinary short time the experiment existed. But in LEO, a much smaller single oblate ellipsoid mass in close proximity, perhaps plus rotation, might suffice. Is there any simple way to accomplish flat spacetime on Earth's surface?

Your link to the described gravity cancelation experiment is reminiscent of the story of the fused quartz gyroscopes created for Gravity Probe B, that were the most nearly perfect spheres ever created. I am delighted to read about the success of such human endeavors. Thank you.

Thanks,
Wes
...

 

[pervect]

It's not as different as you think - making the gravity field of the Earth uniform requires basically the same corrections as it does in space to make it zero.

If you look at figure 1, in particular, you can see that the rate of change of gravity with height is 2s / meter , and that there is an inward compressive force of s / meter.

You can calculate the the vertical component on the surface of the Earth by taking $\frac{\partial}{\partial r} \left( \frac{GM}{r^2} \right) = -\frac{2GM}{r^3}$

Getting the horizontal component takes a bit more work, you need to include the direction of F as well as it's magnitude. There's a detailed but not particularly friendly derivation on the Wiki at http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=308246685 that tells you that the compressive force is 1/2 the stretching force.

This is exactly the same as Forward's figure 1.

The "ring of masses" produces the exact opposite effect, as you'll see in figure 2. (I haven't calculated it in detail myself)

So the physics is basically the same, regardless of whether you are zeroing the gravity in space, or making it uniform on the Earth's surface.

A detailed calculation would just add the potential fields of the six masses to that due to the Earth.

If you want to include rotational effects (I haven't), you'd need to use the effective potential , the same one the geophysicists use to compute the figure of the Earth.

 

[pervect]

Some design notes:

I analyzed the 6 mass configuration, the tidal forces are 6Gm/r^3 and 3Gm/r^3 in the two principal directions. Forward has a cryptic note to this effect on the diagram.

let M_e be the mass of the earth, r_e be the radius of the earth.

If the radius of the compensator circle is r, and the mass of the compensator masses is m, I get:

6Gm r^3 = 2GM_e r_e^3

so

m =(1/3) M_e (r/r_e)^3

The minimum required density will fit the mass m in a sphere of diameter < r/2

rho_min = 6m / ( pi r^3) = (2/pi)( M_e / r_e^3) , which is about 14 600 kg/m^3 or 14.6 gm/cm^3

Lead won't be quite dense enough, but gold, tungsten, or depleted uranium should work.

 

[Wes Tausend]

It's not as different as you think - making the gravity field of the Earth uniform requires basically the same corrections as it does in space to make it zero.

If you look at figure 1, in particular, you can see that the rate of change of gravity with height is 2s / meter , and that there is an inward compressive force of s / meter.

You can calculate the the vertical component on the surface of the Earth by taking $\frac{\partial}{\partial r} \left( \frac{GM}{r^2} \right) = -\frac{2GM}{r^3}$

Getting the horizontal component takes a bit more work, you need to include the direction of F as well as it's magnitude. There's a detailed but not particularly friendly derivation on the Wiki at http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=308246685 that tells you that the compressive force is 1/2 the stretching force.

This is exactly the same as Forward's figure 1.

The "ring of masses" produces the exact opposite effect, as you'll see in figure 2. (I haven't calculated it in detail myself)

So the physics is basically the same, regardless of whether you are zeroing the gravity in space, or making it uniform on the Earth's surface.

A detailed calculation would just add the potential fields of the six masses to that due to the Earth.

If you want to include rotational effects (I haven't), you'd need to use the effective potential , the same one the geophysicists use to compute the figure of the Earth.

Some design notes:

I analyzed the 6 mass configuration, the tidal forces are 6Gm/r^3 and 3Gm/r^3 in the two principal directions. Forward has a cryptic note to this effect on the diagram.

let M_e be the mass of the earth, r_e be the radius of the earth.

If the radius of the compensator circle is r, and the mass of the compensator masses is m, I get:

6Gm r^3 = 2GM_e r_e^3

so

m =(1/3) M_e (r/r_e)^3

The minimum required density will fit the mass m in a sphere of diameter < r/2

rho_min = 6m / ( pi r^3) = (2/pi)( M_e / r_e^3) , which is about 14 600 kg/m^3 or 14.6 gm/cm^3

Lead won't be quite dense enough, but gold, tungsten, or depleted uranium should work.

...

pervect,

I must admit I haven't done the math, and I appreciate you doing so. You have lost me, though.

I'm not sure we are talking about the same thing. It seems you are talking of removing tiny forces ("making the gravity field of the Earth uniform") that remain in tidal inequalities and self-gravity during free-fall while I am imagining flattening the effects of the entire, non-free-fall, full-field spacetime at ground level earth.

While I'm not accomplished in math, it does seem, from a geometric quality standpoint, that enough gravity to flatten (neutralize) the surface of a bucket of water sitting on the ground would require a mass either equal to earth, or the close proximity of marginally smaller masses of much higher density. However it wouldn't be the first time I would be surprised when the correct quantity of math results came in. In other words, it is easy for me to picture, or imagine, the supposed necessary qualities of a process, but I fall far short when it comes time to apply correct math to determine the proof of realworld quantities. Basically, I see the difference in the two scenarios is that the surface of Earth has far greater gravitational forces to overcome than low gravity residuals in LEO, or a free-fall condition.

As an example, suppose the correcting sphere, or spheres (to be used a few feet above earth) were one million times denser than Earth and we reduced itheir total mass/volume to 1/1,000,000 the size of earth. Then it seems reasonable to assume that they will still produce the same total gravity as earth, and identical to the surface of earth, at a distance far out in space, equal to normal Earth radius. That they equal Earth's mass might mean that we could use far smaller sized, but dense, spheres, or a sphere, very close to the top of the bucket to equal the pull of Earth's gravity so that the surface of the bucket assumed a neutral (flat) shape.

I earlier believed a single smaller sphere must also have the oblate ellipsoid shape (as alluded to in my above post) to evenly "spread" (think M&M shape) the spacetime attraction over the width of the bucket surface because of differences of radius between Earth and an equal-mass smaller/denser sphere, that arise when we use different sized spheres (therefore radius-angle to center that forms the original convex ocean-curve).

Since it seems there is no ordinary substance available to us that is many times as dense as earth, even a more reasonable artificial globular mass, or globes, would seem to end up being rather large. I'm also thinking any globe massive enough to equal the gravitational pull of Earth is going to cause a huge, crushing collision right after we have somehow placed it right above our earthbound bucket to momentarily flatten the surface of the water surface.

But I can not do the math. What size of sphere do you calculate would actually be needed for gold, tungsten, or depleted uranium? This is quite interesting that it may be much smaller than I think.

Thanks,
Wes
...

 

[nitsuj]

But I can not do the math. What size of sphere do you calculate would actually be needed for gold, tungsten, or depleted uranium? This is quite interesting that it may be much smaller than I think.

Thanks,
Wes
...

wiki says the Earth's mass is
5,972,000,000,000,000,000,000,000 kg

and the volume is
1,083,206,916,846 km cubed

Gravity is 9.8 m/s.

Gold is 19.3 grams per cubic cm.

converting the Earth's mass is
5,972,000,000,000,000,000,000,000,000 grams

and the volume is
1,083,206,916,846,000,000,000,000,000 cm cubed

Earth seems to calculate according to wiki values to 5.5 grams per cubic cm; again gold is 19.3 grams per cubic cm.

a quick google search to confirm...
5.52 g/cm(3)I don't think it'd work to simply "index" the density of gold, calculate the volume of the gold to have the same gravity (read mass) as Earth. Tidal effects I'd guess are really effected by volume / density.

But idealizing those difficulties away...19.3/5.5 = 3.5 times less volume

looks like this

EARTH 1,083,206,916,846 km cubed (1.1 trillion)
GOLD 309,430,051,813 km cubed (3.1 Billion) What a neat scenario to envision just how complicated tidal effects could be (are) and how complicated the math would be (is) to calculate what "real world" volume the gold sphere would have to be and where to place it in relation to the Earth and bucket.

 

[mfb]

@Wes Tausend: The non-homogeneous part is tiny compared to the homogeneous part (7 orders of magnitude smaller), you can cancel it with small masses close to the test volume.

 

[pervect]

It turns out there is a deep connection between the Riemann curvature tensor (one way of measuring the curvature of space-time) and the tidal forces we've been talking about. I'm not sure if it'd be good to try to go into any detail, my best guess is that it's too advanced for you (just a guess from the tone of your remarks, I don't know your math level).

The size of the required masses depends on the size of the compensated region. Let's work out the results for a 1 meter compensating circle. The compensation will be pretty good within 10cm of the center of this 1 meter circle.

So we compute, usig google calculator
m =(1/3) M_e (r/r_e)^3

M_e = 5.972E24 kg (Earth mass)
r = 1 meters
r_e = 6,371E3 m (Earth radius)

m = 7672 kg, 7+ metric tons

We need six of these masses (7+ metric tons each) , arranged in a hexagon in a 1 meter radius (2 meter diameter) circle, to build our compensator.

The density of gold is 19.3 gm/cm^3, or 19300 kg/m^3. Solving (4/3) rho pi r_s^3 = 7672 kg gives us r = .456 meters, so the two closest gold spheres won't quite be touching (but there will only be a small gap between them, with six masses around a 1 meter radius circle, the closest masses will be meter apart).

At about 50,000 dollars per kilo, that's a little less than 400 million per sphere.

Tungsten is about 1000x cheaper, so it's roughly 400,000 per sphere. The spheres will be very slightly bigger due to the very slightly lower density, but they still shouldn't quite touch.

If we want to scale up to a 2 meter compensating circle, the masses increase by a factor of 8, doubling in radius. And cost.

 

[nitsuj]

The size of the required masses depends on the size of the compensated region. Let's work out the results for a 1 meter compensating circle. The compensation will be pretty good within 10cm of the center of this 1 meter circle.If we want to scale up to a 2 meter compensating circle, the masses increase by a factor of 8, doubling in radius.

That's really neat!

 

[mfb]

Instead of a hexagon, I would expect that a ring (or ring-like structure) is better. It keeps the circular symmetry. It is harder to evaluate, however.

 

[pervect]

Instead of a hexagon, I would expect that a ring (or ring-like structure) is better. It keeps the circular symmetry. It is harder to evaluate, however.

I haven't tried to evaluate a ring (or torus) either. It might also reduce the required density a little bit - if it reduces it enough, you might be able to use something cheaper than tungsten. (You might be able to find something cheaper anyway, I didn't research the material possibilities in depth).

I'm not sure if it has the proper ratio of stretching / compressive tidal forces (because I haven't tried to evaluate it) however.

The six spherical masses have the advantage of being highly adjustable, though - to take into account local non-idealized variations.

They could also cause all sorts of havock rolling around the lab if they broke free of their mountings somehow (say an earthquake) :-)

 

[mfb]

Well, we would have to take masses elsewhere into account, too. Those balls need a good support, and local mass concentrations (apart from the balls) can influence the local field geometry significantly.

 

[Wes Tausend]

The water surface will have a curve radius of ~6370km. Within 30cm (~1 foot), this gives a deviation of a few nanometers - less than 100 layers of molecules. This is completely negligible relative to surface waves, evaporation, and other effects.

Telescope mirrors can be built with a similar precision. They are curved (to serve as telescope mirrors), but it would be possible to build them flat, too.


In theory, you could probably counter the curvature of Earth with a very slow rotation - something like 1 revolution per hour. The remaining effects would be ~9 orders of magnitude below those few nanometers, way below the size of atoms.

On second thought, I think you are correct that the parabolic rotation forces will very near cancel (to within atoms) the very low local curvature of Earth as represented by the water surface of the bucket.

In the early morning, just as I awake, my mind often becomes particularily lucid for a short time, and I have my answer. I bet I am not alone in this peculiar phenomenon.

What I did was picture a hand constructed graph with x as the horizontal axis and y as the vertical.

If I picture the earth, or rather any globular mass, set 30 cm in diameter, the horizontally curved edge of which is 1 meter above zero, centered on the + y axis, I can also construct an opposing parabolic curve, 1 meter in diameter, 30 cm below the x axis. By varying the shape of the parabolic curve, I can draw a series of coordinate points closely along the x-axis of what the average null line must be shaped as, that of average distance between them. The average line is meant to represent a cross section right through the center of the surface of our bucket of water which we are trying to straighten. In this extreme spherical case, I believe the most short condition of the line obtainable will still be quite distorted since the spherical globe has such a short radius. Ideally the line would be perfectly straight to represent our success at making the surface perfectly flat. This is the sort of faulty thinking I first imagined would objectionally distort a perfectly flat water surface on earth and I apologise for not catching it sooner.

I now more correctly picture the very large radius of Earth centered, say the edge of which is 1 meter above zero, on the + y axis, particularily a small, local slice of curved section the diameter of our bucket, perhaps only 30 cm across, I can also construct a variety of parabolic shapes, also 30 cm across, equidistant below the x axis. The two curves can again be compared along the x-axis to find their average as we seek neutral to form a flat water surface. Now the preset curvature-of-the-earth shaped section of spheroid is nearly straight along the x-axis to begin with. The shortest average line we can obtain between them, near the x axis, will be reached with a very minimal parabola. So minimal, I suspect you are perfectly correct that the deviation between curve types will be within atomic departure. Thank you for pointing this out to me.

I can picture a 3-D scene but not do well with the language of math which is definitely a severe shortcoming on this forum. It is usually so easy for me to picture complex geometry in my mind, I think it lead directly to my being lazy about having paid more attention to high school math when I had the chance.

I did take high school pre-engineering courses, but paid very little effort to retaining the math. At the time, I could see no reason to fully absorb these proper lessons other than for becoming an engineer, the mystery of whom I had never even knowingly met one. Upon question to my instructor, I was told that my most likely duty as a newly hired engineer would be that of draftsman, and, while I love geometry and detailed sketches on a napkin, I already highly disdained tedius inked drawings to say the least. Unfortunately I had no idea CAD would shortly make this soul-crushing task obsolete and free ones mind for more noble purposes of just what to draw.


Thanks again,
Wes
...

 

[Wes Tausend]

It turns out there is a deep connection between the Riemann curvature tensor (one way of measuring the curvature of space-time) and the tidal forces we've been talking about. I'm not sure if it'd be good to try to go into any detail, my best guess is that it's too advanced for you (just a guess from the tone of your remarks, I don't know your math level).

The size of the required masses depends on the size of the compensated region. Let's work out the results for a 1 meter compensating circle. The compensation will be pretty good within 10cm of the center of this 1 meter circle.

So we compute, usig google calculator
m =(1/3) M_e (r/r_e)^3

M_e = 5.972E24 kg (Earth mass)
r = 1 meters
r_e = 6,371E3 m (Earth radius)

m = 7672 kg, 7+ metric tons

We need six of these masses (7+ metric tons each) , arranged in a hexagon in a 1 meter radius (2 meter diameter) circle, to build our compensator.

The density of gold is 19.3 gm/cm^3, or 19300 kg/m^3. Solving (4/3) rho pi r_s^3 = 7672 kg gives us r = .456 meters, so the two closest gold spheres won't quite be touching (but there will only be a small gap between them, with six masses around a 1 meter radius circle, the closest masses will be meter apart).

At about 50,000 dollars per kilo, that's a little less than 400 million per sphere.

Tungsten is about 1000x cheaper, so it's roughly 400,000 per sphere. The spheres will be very slightly bigger due to the very slightly lower density, but they still shouldn't quite touch.

If we want to scale up to a 2 meter compensating circle, the masses increase by a factor of 8, doubling in radius. And cost.

pervect,

Something is still seriously wrong between our thoughts here. I know it is probably me, so I apologise in advance.

Can the surface curvature of the water on Earth be totally thought of as a tidal effect? Do the weighty spheres cause a dip in the shape of local gravity that flattens a small tide-free curvature area?

After some thought, I think I am wrong about placing an Earth sized mass directly above my bucket sitting on the ground to neutralize gravity. I think I am wrong besides the impossibilty of keeping the two masses separate long enough for observation.

I am wrong because I have been ignoring the tidal forces of placing a huge mass right above my bucket. The water surface of my bucket would likely not become flat in a neutral state of flat spacetime. I think the bucket, or at least the water, would immediately rise and develop a significantly sharper radius tidal bulge over and above the curvature of earth. One might even picture the upper portion of water floating up and then being drawn off like material is sometimes drawn off a star orbiting a companion black hole. At any rate, in spite of a plane of flat spacetime forming hopefully right at the water surface, it seems a severe tidal force would prevail that forms the water into an ellipsoid, draws it up and tears it apart. Am I wrong here?

On the other hand, I cannot help but still see you, pervect, as being concerned only with tidal forces. If the curvature of Earth is indeed a tidal effect, then this must be the correct manner of approach and I have a clinker in my thinker. When Dr. Forward spoke of canceling "earth gradients", I took it to mean that he has only redirected gradient lines of tidal force normally present in the thickness a disc of 30cm x 10 cm while it is in total free-fall. I believe he claims to do this with massive spheres strategically placed around the said disc to gravitationally draw local lines of force opposing that of horizontal side compressions by Earth tidal forces, so that all tidal forces cancel. See the compression lines (arrows) on Figure 2.1 at http://cws.unavco.org:8080/cws/straindata/Notesfrom2005class/tidenote.pdf. Some of the line orthogonal arrangements may be controversial since I think many regard some tidal effects to be purely centrifugal.

To cancel full vertical gravitational effects on the non-orbiting surface of earth, therefore in other than free-fall, it seems it would either take an Earth sized object directly above earth, or a denser, slightly smaller oblate spheroid held right above the singular thin plane of the waterbucket surface to create a full null flat spacetime, in said singular thin plane (unfortunately not throughout the bucket volume), in a similar 2-D graphical analysis that I attempted to describe in my previous post to mfb. I believe to further cancel horizontal self-gravitational effects would take the ring of which we have spoken.

I thought of it this way. If all forces of gravity were canceled so that the curvature of the bucket water surface were totally eliminated, we would have seemingly canceled planetary gravity altogether in a regional plane. Rather than having liquid water existing in the gravity-free region, we could assign the job to a rigid hockey puck placed horizontally with the flats facing our equal planetary masses. Would the hocky puck now float in this supposed gravity-free region with any stability? I believe the hocky puck would float, but in a state of instability, in the case of applying the massive Earth sized "compensator" just as it would "float" in gravity-free reality between the Earth and moon in just the right spot. But even between Earth and moon the likelyhood of eventual slow drift (falling) is likely to occur with a 50/50 chance towards either planetary mass. In addition, I think the puck is likely to orient itself to a condition of the rim facing the opposing planets, and possibly began to rotate continuously, because of the combination of tidal forces and the fact that it would be in orbit.

Does one become quite a bit lighter while walking the streets of New York amongst the tonnage of massive skyscrapers? Will a bucket of water lose it's curvature on the sidewalk below? What about a laser beam? Will it not curve like it does on the open prairie? Maybe. It would an interesting experiment.

Am I wrong here? If so, I apologise profusely for my naive misunderstanding.

Thanks,
Wes

P.S.
If we can cancel total gravity by the use of spheres here on earth, my family practitioner would be very interested to know this. Upon finding that he has a Physics Major, but prefers the salary of an MD, we joked that if I could ever understand GR well enough to build an anti-gravity machine, he wanted to know how to build it (He is also an avid pilot). I thought we both knew better under presently known laws.
...

 

[pervect]

We can't cancel total gravity here on Earth with available materials - we'd need something much denser than any known material that we can manipulate.

We know of materials dense enough actually (the electron degenerate matter in a white dwarf star, for instance, or the even more dense neutron degenerate matter at the core of a neutron star) - but we can't contain or manipulate them with existing technology.

Forward thought of total cancellation too - the structure you'd need is an overhead disk of such ultra-dense materials. I'm not sure I'd want to be under it :-).

But while we can't cancel gravity, I think we can make a bucket of water have a flat surface. That's a much less demanding task than nulling gravity altogether. All we need to do is straighten the field lines, which are a little bent.

As far as the math goes, the usual approach is to say that the surface of a bucket of water is an equipotential surface.

This makes the force always perpendicular to the surface. You might try http://web.ics.purdue.edu/~ecalais/teaching/eas450/Gravity2.pdf, I'm not sure if it's too advanced.

Hope this helps - if you want to go into equipotential surfaces more, maybe I can explain. If that seems like too much, I'm not sure where this conversation can go :-(.

[add]
As long as you deal with point masses and spheres, equipotential surfaces aren't hard to define:

The potential U
U = -GM/r1 - GM/r2 -GM/r3 ... -GM/rn

where r1, r2, r3...rn are the distances from some point where you want to calculate the potential to each of the point masses.

You'll need a bit of trig to compute the r's - mostly the pythagorean theorem.

Because it's 3d, graphing the planes of equal potential are harder. I'd suggest finding some advanced graphing or visualization software for that.

 

[Wes Tausend]

We can't cancel total gravity here on Earth with available materials - we'd need something much denser than any known material that we can manipulate.

We know of materials dense enough actually (the electron degenerate matter in a white dwarf star, for instance, or the even more dense neutron degenerate matter at the core of a neutron star) - but we can't contain or manipulate them with existing technology.

Forward thought of total cancellation too - the structure you'd need is an overhead disk of such ultra-dense materials. I'm not sure I'd want to be under it :-).

But while we can't cancel gravity, I think we can make a bucket of water have a flat surface. That's a much less demanding task than nulling gravity altogether. All we need to do is straighten the field lines, which are a little bent.

As far as the math goes, the usual approach is to say that the surface of a bucket of water is an equipotential surface.

This makes the force always perpendicular to the surface. You might try http://web.ics.purdue.edu/~ecalais/teaching/eas450/Gravity2.pdf, I'm not sure if it's too advanced.

Hope this helps - if you want to go into equipotential surfaces more, maybe I can explain. If that seems like too much, I'm not sure where this conversation can go :-(.

[add]
As long as you deal with point masses and spheres, equipotential surfaces aren't hard to define:

The potential U
U = -GM/r1 - GM/r2 -GM/r3 ... -GM/rn

where r1, r2, r3...rn are the distances from some point where you want to calculate the potential to each of the point masses.

You'll need a bit of trig to compute the r's - mostly the pythagorean theorem.

Because it's 3d, graphing the planes of equal potential are harder. I'd suggest finding some advanced graphing or visualization software for that.

pervect,

I looked at Gravitational Potential and I will have to think about this for a bit. I will be gone for the week-end but back again Monday.

I can see that down the road I need to review some math that I tended to blow off years ago. I know the equation must balance, but I don't recognise many of the terms and symbols used.

I can well imagine geometry, but the ratios must be real or it is all science fiction. Pictoral geometry just may not be enough anymore. I have some old, unopened, Simon & Shuster Pre-calculus/Calculus "placement" software that might allow me to brush up on quadratic functions and trig. I will first have to dust off an old Win ME/98, or Win 95 computer, the software has become that old, patiently waiting for the time-friendly atmosphere of retirement.

My surmising earlier, "Can the surface curvature of the water on Earth be totally thought of as a tidal effect? Do the weighty spheres cause a dip in the shape of local gravity that flattens a small tide-free curvature area?" might be close, more so as in realising even near-earth gravity tidal gradients are not all that formidable, similar to miniscule tidal effects in LEO.

The universe is a machine and I have a mechanical mind, but I feel like an exuberant pre-adolescent trying to join a group of experienced adults at a car show. While I have built soap-box racers, I am still not in the same discussion league until I've had my own broken pistons in a combustion engine; and can recite precisely how to prevent it. Meanwhile I can ask questions if I do not overly frustrate my benefactors. Retirement is scary; I am an old dog contemplating new tricks. Wish me luck.

Thanks for your patience pervect,
Wes
...

 

[pervect]

pervect,

I

My surmising earlier, "Can the surface curvature of the water on Earth be totally thought of as a tidal effect?

Let me see if I can rephrase what I'm saying - that's not quite right.

You can decompose gravity at the Earth's surface into a constant, uniform, vertical field, and a perturbing field This perturbing field is the tidal field, it's just what you get when you subtract a uniform field from the actual field.

If you can make the perturbing field zero, then all you constant vertical gravity.

If you have constant vertical gravity, water in it will be flat. And the surfaces of constant potential will be flat.

 

[Wes Tausend]

Let me see if I can rephrase what I'm saying - that's not quite right.

You can decompose gravity at the Earth's surface into a constant, uniform, vertical field, and a perturbing field This perturbing field is the tidal field, it's just what you get when you subtract a uniform field from the actual field.

If you can make the perturbing field zero, then all you constant vertical gravity.

If you have constant vertical gravity, water in it will be flat. And the surfaces of constant potential will be flat.

OK, pervect,

It is Monday and I am back as promised.

I think I finally get it. Your link on Gravitational Potential helped immensely. Thank you for your knowledge and patience.

I just couldn't see how the "rigid" lines (the field) could be easily bent. Perhaps my problem is that I can see the direction of vectors but, lacking a proper grasp and incorporation of math, I apparently have difficulty imagining the correct ratio of their magnitude.

This how I think I got past it:
The lines are not so rigid. Mfb has pointed out that a very slight rotation (he suggested about once an hour) would furnish enough tiny centrifugal force to bend the field, and I agreed when I finally realized how very close to flat, the local area of a small bucket diameter is. I now realize it would also take very little gravitational force, by the way of large, massive spheres, to similarly bend them.

For a bit I thought that the centers of the spheres would have to at least be slightly above the plane of water surface to work at all. But then being dead centered exactly on the plane now seems most correct. The idea is that both rotation and the massive spheres produce the same 90 degree angle of force to cause the field, therefore the water surface, to flatten. I did once unfortunately imagine that the problem was similar to swinging a weight around in a circle to such a degree that all vertical force was canceled by centrifugal force. This would be pretty much impossible, as the vertical force would always retain a small fraction of effect and some curvature would remain. This error lead me to believe that only an equal, but opposing huge gravitational field mass would fully cancel the curvature.

Your sphere masses, however, might produce an even better flatness than rotation. The question then remains, that if the spheres were too dense (too "powereful"), would they cause a concave water surface like excessive rotation? And if so, would the water start to bend in an inverted parabola, like does rotation, or would it retain a perfectly round, but inverted concave spheroid shape, the exact opposite of the geoid of a perfectly homogenous, non-rotating earth? I guess it doesn't really matter at this point, since I think I do understand most of the principle now.

One remaining effect not taken into account would be the miniscus around the edge of the bucket which would exceed all other effects as otherwise perfect flatness was approached, as it also would exceed the curvature in the very beginning.

Unless I'm still way off in my misunderstanding, I am done. Thank you all very much for your thoughts on this mental exercise.

Thanks,
Wes
...

 

[bahamagreen]

"Perfectly flat"?

In spite of all the analysis and calculations, at some point in order to confirm the perfect flatness you are going to have to take a close look at it. As you keep looking closer with whatever instruments and techniques, you are going to eventually notice that the surface is not planar... If you are measuring for a constant value of charge gradient, that gradient is going to be bumpy because of the molecules. Any other definition of "surface" you come up with that is not some kind of average is also going to be bumpy. Once you get down to the constituents of the molecules, the whole sense of a continuous surface is gone.

This is like the "How long is the coastline of Britain" problem... the greater resolution you employ, the greater the length... until it eventually become undefined because your resolution is finer than the individual particles that make up the coastline substance.

Ultimately, you might be able to define "perfectly flat" as indistinguishable from "flat +/- (1.616 × 10 −35 m) / 2" ...

 

[Wes Tausend]

...

True. Quantum. Which also might depend on the heat content defined at average sea level air pressure. Whatever average pressure really is...

Thanks,
Wes
...