# wine in tilted bottle - is surface horizontal or not & why

Discussion in 'Physics & Math' started by Billy T, Nov 16, 2015.

1. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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That is *not* equivalent to a manometer. The masses, in an upset condition, are not equal.

3. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I certainly am not and I said it was in unstable equlibrium when initially horizonatal so any small pertubation from horizontal will grow.

Because the wine molecules can move sideways from their higher position into a lower one, that problem, thanks to iceaura is resolved. Now I only want to discuss the problem where the movable masses are constrained in their motion.

The most clear example is the one system of two bricks on the ends of a rigid, massless, frictionless, equal arm, balance beam. Its center fulcrum is located 1cm from the surface of a 99cm radius sperical mass of neutron star matter to make the radial gravitational field strength very far from uninform in strength with radius. So lets confine future discussion to this completely rigid system.

I am of the opinion its stable equlibrium position is with one end (or bottom of one side, if beam is longer than a meter) of the balance beam resting on the surface of the neutron star. Do you agree?

Then, if yes, lets try to explain why the final equlibrium position of the mercury in the U shapped tube, is not with all the Hg on one side. I.e. when the Hg levels are equally distant from the center of the neutron star, that too is an unstable equlibrium position, much like the horizontal balance beam.

Last edited: Nov 16, 2015

5. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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See my last two posts: the system with two bricks is not related to the manometer because the bricks are equal in mass and the water columns are not.

7. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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They certainly can be if started (at t =0) in the unstable equlibrium position, like the balance beam was.

8. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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It's not an unstable equilibrium because any perturbation makes the masses in the manometer tubes unequal.

You're not even analyzing this correctly: if you want to know if an equilibrium is stable, you analyze the upset condition not the equilibrium state: in equilibrium, there is nothing to analyze!

I'm starting to wonder if you're really serious here. That would be out of character, but this thread is just bizarre.

9. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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That is true and on earth, as the extra weight of mercury in the higher level side would make greater pressure at the bottom of the U greater and that tends to restore the equal heights; but I am considering an extremely high gravity gradient near the surface of a small neutron star.

If raising the surface of the already taller Hg column 1 mm higher still against a much weaker downward force of gravity, requires less "lifting energy" in that much weaker field than the energy released by 1 mm fall of the surface of the surface in the shorter Hg column as that fall is thru a much stronger G field, then this "Earth based" reasoning does not apply. What is always true, is the levels will move so as to lower the total gravitational energy the single system stores.

It may be true that even in this extreme gravitational environment letting the taller column fall's surface fall 1 mm removes more stored energy than required to lift the shorter column 1 mm up, even though lifting that 1 mm tall equal mass must be done in a more intense G field. If that is the case, then yes the equal height columns is a stable equlibrium state. I.e. the pressure at the bottom of the U from both columns will become the same.

Evaluating these two energy changes associated with surface levels changing 1 mm, one up the other down, is a problem in integral calculus. That is why I suggested we consider it ONLY after reaching conclusions in the bricks on beam problem.

Again I ask if you agree that once a brick gets to be a little closer to the center of the neutron star if the force downward on it is not greater than the force downward on the other brick which is sightly more remote from the center of the neutron star? I. e. any small perturbation from the horizontal beam position, will grow.

Last edited: Nov 16, 2015
10. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Billy, your bricks on a balance model is fine, it just isn't related to a manometer. Draw a diagram (or grab one online). You'll see that the inequality you are describing does not exist - on any scale.

Whether you take a 1ft tall, 1 inch tall or 1mm tall element of water on the short side, the same element also exists at exactly the same height on the tall side. Always.

11. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Yes. that is a strong argument even for the manometer near surface of the neutron star. There is more net force downward on the taller side. Thanks for this obvious POV. It is more simple than trying to consider virtual changes in stored energy with level changes.

Thus only the balance beam problem remains. I claim when horizontal it is a case of unstable equlibrium. I.e. any small pertubatinon from a horizontal beam will grow until part of the beam contacts the neutron star. Do you agree?

12. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Yes, that's all fine.

13. ### iceauraValued Senior Member

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It is exactly the same as the wine bottle situation.

The 1 cm of rise is a net horizontal transfer - no penalty, potential energy the same. The 1 cm of fall is a lowering of potential energy in the field - the center of mass is now closer to the center of gravity of the star.

Neglecting the special circumstances of extra and extreme compression, etc., of course.