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

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

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

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Last night at a party where all were speaking protuguese at once and I could not follow any, there happen to be a 2/3 full bottle of wine in ice bucket that was quite tilted. (too much ice in bottom had not yet melted.) I assumed the surface was horizontal to achieve the least energy state, but then noted the surface was an ellipse. Also that the minor axis of that ellipse was of length equal to the ID of the assumed circular bottle.

I begain to imagine another plain passing thru that minor axis and diameter not quite horzontal. That would make a curved wedge shape above the liquid surface, exactly the same shape and volume as the part of that plan passing under the liquid surface. In fact if the new plane were the surface, the gravitational energy would be unchange - so why is the real surface horizontal?

Quickly I supplied my self an answer: the new, not horizontal surface area is greater and that requires more energy even if the gravitational energy remains the same. At that point I should have drunk the wine but I realized if the surface were circular (not horizontal) same as when the bottle was vertical, the gravitational energy would still be the same, but then the surface energy required to make that smaller surface would be less - so the horizonal surface is NOT the least energy configuration, yet is seems to be the one the wine surface assumed - why?

The only good thing was once again I was made aware of why I am never bored - mind constantly thinking even when for hours there is nothing I can do but sit and smile. I gave up on this problem after it entertained me for more than an hour. Then I evaluated the looks of the younger ladies at the party and drank more wine.

Last edited: Nov 16, 2015

3. ### danshawenValued Senior Member

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So, you've never pondered the same question about an ordinary bubble level?

A bubble has surface tension which explains the shape and characteristics of the meniscus. Like most problems in physics, an equilibrium is reached between two, or in this case, three forces. Besidea gravity and surface tension, there is also the pressure of the trapped vapor in the bubble.

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

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Last night? And you're still drunk? Must have been some party!

7. ### iceauraValued Senior Member

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In the tilted surface condition, a wedge of wine is being supported above a wedge shaped gap closer to the center of gravity - and it flows downhill, naturally, since nothing prevents it from doing so.

It might become clear if one imagines the exact wine molecules in the uphill wedge flowing into the downhill hole , rather than displacing the molecules directly under them - whose potential energy does not change, upon being transferred horizontally, and so can be neglected.

Last edited: Nov 16, 2015
8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Thanks. I had had quite a few glasses of wine before they brought the ice bucket with excessive ice in it out and set it down in front of me. So much ice in the bucket that the bottle was tilted at nearly 45 degrees to the vertical, and resting on the rim of the ice bucet.

I was not aware of my tacit and limiting assumption that the wine molecules could only move vertically; however, that is the case for mercury molecues in a U shapped manometer tube. (Both ends open to the same air pressure.) If the surface in the right arm is 2 cm higher than in the left arm, we all know that the right surface will fall down 1 cm and the left will rise 1cm so both surfaces will be the same distance from the center of the Earth. Understanding this is not so easy. Lets make it even tougher:

Assume this mercury manometer is not on Earth but one meter from the center of very cold* spherical piece of a neuron star of radius 99cm. I. e. the bottom of the U shaped glass tube is only 1 cm above the star's surface. Now the work done on the lower left side Hg as it rises the first mm is much greater than the gravitational energy released as the higher right side Hg falls 1mm as the left side surface is rising in a much stronger gravitational field than the right side is falling down through. I. e. It seems to require net energy for the two sides to have their surfaces become equally distant from the center of the neutron star due to fact gravity is an inverse square field.

Can you (anyone) explain why the two surfaces become equally distant from the center of the neutron star?

* I don't know if a small sphere of dense neutron star mater is possible but if cold enough, it would seem to be.

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

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Billy, 99% of drunk/high musings are complete garbage, as is this one. When you sober up, you have to toss them in the trash and re-ask yourself the original question from scratch, not continue-on as if the drunk/high musing was something profound.

There is nothing here beyond what you learned in junior high.

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

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Then you should be able to explain how the mercury in the arm of the U shaped manometer with the lower surface rises in a stronger gravitational field the same amount as the the Hg in the other arm falls down thru a weaker gravitational field without some energy input suppling the net increase in stored gravational energy.

PS thanks to the mod who added the missing "t" in the thread's title.

Last edited: Nov 16, 2015
11. ### exchemistValued Senior Member

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Are you still pissed?

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

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No. Puzzeled, yes.

Also puzzeled as to why no one tries to answer the U shapped manometer near small neutron star question.

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

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Billy, the manometer is a symmetrical tube! The two sides have different amounts of mercury in them!

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

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Not normally when both tops are open to same air pressure.

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

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Probably because they are flabbergasted as to why this would be confusing you. Aren't you a PhD physicist? Honestly, if you aren't on a multi-day bender, you may want to consider seeing a doctor!

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

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I don't know what you are trying to say there: You specified a tube with an uneven height of mercury in it.

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

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Yes as the initial or t=0 condition, not as an equlibrium condition.

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

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well that would cost money and his (her?) understanding of the physical forces acting would be very inferior to mine.

What I seek is rational explanation, for what most of us think will happen - I.e. the two surfaces will become equally distant from the center of the small neutron star.

Not that all the Hg will flow into one side of the U as the Hg sinking to bottom of the U is in stronger gravity field and gives more energy release than lifting that volume of Hg up in the weaker G field requires.

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

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The higher surface has a larger gravitational potential energy and so it falls.

Or, if the lower surface is at height he and the higher surface h2, the mercury at height h1 on the high side has h2-h1 mercury sitting on top of it, applying a force!

Your attempted confounding factor of an uneven gravitational field actually isn't.
Any doctor worth his salt should be able to explain it to you, but that isn't why you'd be going to see him.

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

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Yes, but to do so it must push up the Hg an equal height in the other side of the U tube. Once past the equal height position with some KE momentum, the greater strength gravity field it is then falling thru, supplies more than enough energy to lift that same falling volume up in the other arm of the U by the lenght of the fall as the added and equal increase in the height of the rising Hg column is work done against a weaker G field.

Crudely, my quetion is:
Why does not all the Hg flow into one side of the U, in this very non-uniform strength G field?

Last edited: Nov 16, 2015
21. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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To state the problem in other terms, imagine a rigid, massless frictionless equal arm length balance beam with two equal mass bricks, one at each end of the horizontal beam. That is an unstable equlibrium condition as if one brick descends into a stronger gravity field*, the other brick is being lifted in a weaker G field. I.e. net Gravity energy is being released - same reason why water runs down hill.

* Still assuming the balance beam is only 1cm above the surface of a 99cm radius small neutron star, so the inverse square decrease of gravity is very important. I. e. the gravitational force acting on the descending brick is larger than the gravitational force acting on the rising brick.

PS I hear that the theoretical physics supply store has a 20% discount sale on rigid, massless frictionless, equal-arm-length balance beams until the end of November.

Last edited: Nov 16, 2015
22. ### OnlyMeValued Senior Member

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Billy, you cannot treat the two sides of the tube or the balance beam as separate systems. They are both parts of a single system and as long as each is symmetrically situated relative to the gravitational source, both systems will end in a state of equilibrium.., or balance...

The issue with the wine in the bottle, I believe was earlier answered... The variable you were missing was the surface tension betwee the liquid and the bottle which because of the shape of the bottle and that it is laid sideways returns asymmetric results relative to the gravitational filed. The surface of a liquid in any tube is concave. It is just usually only noticeable in small diameter containers-tubes.., or when drunk.., affected by other aspects of the liquid in question.

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

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You appear to be modeling this as two equal masses, one on each side of the tube. It isn't. The high side has an equal amount as the low side -- plus some more.

Maybe if you drew a diagram it would be more obvious to you.