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Area of common surface of two bubbles

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Thread moved from the technical forums, so no Homework Template is shown
What will be the area of common surface of two identical bubbles of radius R , i know there common surface will be flat as the radius of curvature of comman surface will tends to Infinity , but how do i relate with area of flat surface
I tried to use
Energy = Surface tension * area
And then energy conservation
But failed
Any help will be appreciated
 
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Why would you want to use energy when it's just a problem in geometry? DRAW the two bubbles intersecting and go from there.
 
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Why would you want to use energy when it's just a problem in geometry? DRAW the two bubbles intersecting and go from there.

They will share a common flat surface where force on both sides is equal and we also don't have any other information so how we start
Otherwise thanks for replying
 
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They will share a common flat surface where force on both sides is equal and we also don't have any other information so how we start
Otherwise thanks for replying
I say again, and with added emphasis, DRAW the bubbles and the intersection and go from there.
 

haruspex

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I say again, and with added emphasis, DRAW the bubbles and the intersection and go from there.
To add a bit more of a hint, concentrate on a region where three films meet. Consider the force balance.
 
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To add a bit more of a hint, concentrate on a region where three films meet. Consider the force balance.
I think i had to find the angle between the three films to minimize the area considering Fnet to be zero
 

haruspex

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I think i had to find the angle between the three films to minimize the area considering Fnet to be zero
Rather than considering areas, it will be easier just to treat it as three forces pulling at a point. What can you say about the magnitudes of the three forces?
 
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Rather than considering areas, it will be easier just to treat it as three forces pulling at a point. What can you say about the magnitudes of the three forces?
Ohhh
The bubbles are identical so the force at a point will have same magnitudes but they have different direction , but Fnet = 0 so these three force vector must sum to zero only possible when they are at 120° hence Force net = 0 means potential energy = minimum hence area of circular film will be minimum if these two spherical bubbles intersect at 120°
Hence
Radius of circle = Cos 60 * Radius of bubble = R/2
Area = πr^2/4

But rather then using physics and geometry
If there is a way of using Calculus to Prove it
But anyways thanks for help
 
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How we can prof double bubble conjecture that is that three films of two soap bubbles intersect at 120° using only calculus
 

haruspex

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The bubbles are identical
It's not that the bubbles are identical. The films have the same surface tension, so exert the same force per unit length. Even if the bubbles were at different pressures the angles would still be 120°.
If there is a way of using Calculus to Prove it.
Yes. You could find the circle which minimises the total surface area for a given air content (moles). Bear in mind that the radius of the spheres is related to the pressure and thus to the volume.
 
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Th
It's not that the bubbles are identical. The films have the same surface tension, so exert the same force per unit length. Even if the bubbles were at different pressures the angles would still be 120°.

Yes. You could find the circle which minimises the total surface area for a given air content (moles). Bear in mind that the radius of the spheres is related to the pressure and thus to the volume.
But how can i apply the condition of fixed volume
If i make the area of circle as function of angle bw the bubbles then
Where can i apply condition of fixed air moles
Do i Missing something
 

Baluncore

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First reduce the problem from three surfaces to three lines.
We know that the three films are under tension and so of minimum area. Consider a section through the junction of the films, perpendicular to the line of the junction. Three lines meet at the common junction. You want to prove that those three lines meet at 120°.

Next reduce the problem from three lines to three points.
This is the same as minimising the length of road needed to join three roughly equally separated villages. There will be a common junction where roads from the villages meet. It is not surprising that the minimum total length of road is found when the position of the junction is such that the roads meet at 120°.

Consider three equally spaced points on a unit circle. Find a common point P(x,y) such that the sum of the distances from the three points to the common point is a minimum. If that point is at the origin then the angles will be 120°. You might use calculus to show that the total line length to the point has a minimum when the point is at the origin.
 
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I am unable to think how our problem is same as minimizing the length of road
 
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I am unable to think how our problem is same as minimizing the length of road
 

Baluncore

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I am unable to think how our problem is same as minimizing the length of road
That is because you are looking for difficulty, not elegant simplicity.

A bubble contains a volume of air. The internal pressure is very low. Where three liquid films meet at a line the liquid in the wall flows to balance the tension in the wall. A thin wall will draw liquid from a thick wall until the surface tension and wall thicknesses are the same. If a junction took the initial sectional form of a 'T' then the surface tension in the walls would pull the top of the 'T' into a 'Y'. That is reducing the area of the walls.

We can ignore the forces parallel with any three surface junction line and need only consider those forces in the film normal to the junction line. That is why it comes down to three roads meeting at a junction. Minimising the length of road is the same as making the thickest liquid film.
 
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That is because you are looking for difficulty, not elegant simplicity.

A bubble contains a volume of air. The internal pressure is very low. Where three liquid films meet at a line the liquid in the wall flows to balance the tension in the wall. A thin wall will draw liquid from a thick wall until the surface tension and wall thicknesses are the same. If a junction took the initial sectional form of a 'T' then the surface tension in the walls would pull the top of the 'T' into a 'Y'. That is reducing the area of the walls.

We can ignore the forces parallel with any three surface junction line and need only consider those forces in the film normal to the junction line. That is why it comes down to three roads meeting at a junction. Minimising the length of road is the same as making the thickest liquid film.


Okk i had understand answer pretty much good
But now what is the area of surface
 

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