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The exact question is

Two air bubbles in water

  1. attract each other
  2. repel each other
  3. do not exert any force on each other
  4. may attract or repel depending upon the distance between them.

The chapter is about gravitation. The given answer is

A lighter body inside a denser medium behaves like negative mass as far as gravitational force is considered. Two air bubbles i.e. two negative masses will attract each other.

What is negative mass in this context and how can it be applied to such macroscopic objects? How would it result in attraction?

My reasoning is:

enter image description here

Consider the bubble A in the above image. The air particles forming the bubble A would be attracted more to the left(away from B) as there are more dense particles towards that side-the air particles making up bubble B are less dense than the medium and they will attract the air from bubble A to a smaller extent than if the volume of bubble B was filled with the medium.

A similar case would apply to B due to lesser density of particles forming A and the bubbles would be (indirectly) repelled.

So what is happening in this case?

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  • $\begingroup$ Have you considered the water between the two bubbles? Which way does it tend to go? $\endgroup$ – Ali Jul 11 '13 at 10:49
  • $\begingroup$ IDK what you mean, but if B were not there, A would not have been attracted or repelled in any direction, the lighter volume of B relatively makes the attraction by the medium(now lesser of it) from the right lesser. $\endgroup$ – user80551 Jul 11 '13 at 10:51
  • $\begingroup$ I think you should only consider the water. The mass of air in the bubbles is negligible . for simplicity suppose the bubble is not air but vacuum. A sees less water in the direction of B and B less in the direction of A $\endgroup$ – anna v Jul 11 '13 at 10:54
  • $\begingroup$ Exactly, which means that the volume occupied by B is not attracting A to the same extent as it would be if that volume was occupied by water. $\endgroup$ – user80551 Jul 11 '13 at 10:55
  • $\begingroup$ I'm not considering it as vacuum as the air inside needs mass to be attracted. And of course the fact that it would collapse instantly. $\endgroup$ – user80551 Jul 11 '13 at 10:56
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Just consider the water, due to gravitational attraction(which I'm not sure how effective it is in this case); water molecules like to be as close to each other as possible. This means that they like to push the bubbles as close to each other as possible. Since the air has negligible mass, its gravitational forces can be neglected compared to the water ones.


The other way to go through this reasoning is by what has been suggested in the question, i.e. assuming the bubbles have negative mass. This solution has few steps, as following:

  • Say we have a huge spherical lump of water(with the radius $R$), without any bubbles inside. The gravity potential of this sphere is $\frac{-3GM^2}{5R}$, but we are not going to use that.

  • Now, say we don't remove a small spherical part(radius $r$ and mass $m$) of the water and replace it with a same-sized sphere with density $\rho'$(or mass $m'$) but rather add it to the current sphere(a ghost like sphere which can only interact through gravity with the world). To calculate the gravitational force acting on this sphere using Shell's theorem, we also need to know the distance from the center; assume it's $x$. Since the sphere had been in equilibrium before, the new net force will be(note the force should be proportional to the mass of each object):

$$ -\frac{GM'(x)}{x^2}m' \tag{1}$$ where $M'(x)$ is the mass of water inside a sphere with radius $x$.

  • The next step is to do the same with another small sphere of water. To make relations more simplified, I will put this one at $-x$. Now the net force on the first and second sphere will be:

$$F_1=-\frac{GM'(x)}{x^2} m' - \frac{G m' m'}{(2x)^2} \\ F_2=\frac{GM'(x)}{(x)^2} m' + \frac{G m' m'}{(2x)^2}$$

Or the accelerations:

$$a_1=-\frac{GM'(x)}{x^2} - \frac{G m' }{(2x)^2} \\ a_2=\frac{GM'(x)}{(x)^2} + \frac{G m' }{(2x)^2}$$

  • In the case of our problem $m' = -m$, therefore:

$$a_1=-\frac{GM'(x)}{x^2} + \frac{G m }{(2x)^2},$$

($M'(x) \gg m$)which is towards the center(and the other sphere). So it looks like the two spheres are attracting each other.


Now there are some ambiguities here:

  1. Do negative masses behave mathematically consistent(we can use $F=m a$), which I have assumed to be the case.
  2. Is this attraction duo to the big sphere of water or the other sphere? Looking at equation $(1)$, it seems to be the first case; unless my previous sign convention is wrong which simply means that the big sphere of water will try blow the bubbles away, although an attraction force between them(if they are closer than a certain distance they will attract).

Also I should point, if the bubbles get in touch; they will immediately collapse into a single bubble. This is due to the surface tension, not the gravitational effects for sure.

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  • $\begingroup$ What do you mean by negative mass? Its one of my questions. $\endgroup$ – user80551 Jul 11 '13 at 11:29
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    $\begingroup$ "water molecules like to be as close to each other as possible." Then why won't they push the bubbles away from each other and occupy the space between them? $\endgroup$ – user80551 Jul 11 '13 at 11:30
  • $\begingroup$ I will go into the details of it. $\endgroup$ – Ali Jul 11 '13 at 12:32
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Maybe this plot of the gravitational field will help: enter image description here

EDIT re the comment:

First of all you mean electric field lines, not magnetic. And yes, they do look the same because they are the same. The underlying field equations are identical (in the static nonrelativistic limit). The difference is that for gravity like charges attract whereas for electric forces like charges repel.

Maybe the plot is a little less than clear. Keep in mind this plot is the force on the fluid, not the bubbles. You were meant to take away that the fluid in between the bubbles flows out of the space and the bubbles get closer. ;)

To really do this problem properly you need some assumptions about the fluid: namely that it has surface tension to stabilise the bubbles and also that the flow is incompressible (otherwise you need to keep track of the density everywhere and it gets truly awful). You also need to put the system in a box (it can be a gigantic box - the size of it doesn't really matter in the end) just to avoid the ambiguities of having an infinite mass of fluid. Under these assumptions you can argue that the bubbles can't expand, and also any flow out of one region has to be balanced by a flow into another region.

I'm guessing you're not at the point of seeing field theory yet. Take this as an illustration that there are complicated ways of doing simple problems. The advantage of field theory is that it is much more general and powerful for other problems. But for this you don't really need it - the "negative mass" argument gives you the right answer. But this may give you confidence that the argument given about "negative mass" is correct. In fact, this is better because we don't need to invoke "negative mass" at all - we just talk about the fluid.

If you assume, based on the above, that the bubbles don't change shape or size, then you can treat the problem very simply. All you need to know is that the gravitational potential of a bubble is

$$ \phi(r) \propto \frac{1}{r}, $$

with a positive sign outside the bubble and

$$ \phi(r) = \text{a constant}, $$

inside the bubble. If you know Poisson's equation for the gravitational field you can derive this. You get the total potential by adding the potentials created by the two bubbles. You also need to know the gravitational energy a little parcel of fluid of volume $\Delta V$ is given by

$$ \Delta U = \phi(x,y,z) \rho(x,y,z) \Delta V, $$

where $\rho(x,y,z)$ is the density, which is constant everywhere outsite a bubble and zero anywhere inside. You add up the energy for every piece of fluid ("do an integral") to get the total energy, and see if it increases or decreases with respect to increasing the seperation between the bubbles. The system will naturally go in the direction that decreases the total energy.

So what does the energy look like? Here (energy units are arbitrary, length units are in bubble radii):

enter image description here

At seperations less than 2 radii the bubbles are intersecting so you can no longer trust the calculation - surely the bubbles start changing shape when they collide! But for larger seperations the calculation is ok and look, the energy increases. So it costs energy to pull the bubbles apart. The bubbles are attracted to each other!

The fact that nearly the same calculation done for electromagnetism gives charges which repel each other is an interesting and important clue to the difference between the two theories. If you are really super keen you might have a go at that problem one day.

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  • $\begingroup$ So they will move away? The plot looks similar to magnetic field lines produced by like charges. Also, could you provide the source for that plot? $\endgroup$ – user80551 Jul 11 '13 at 13:03
  • $\begingroup$ @user80551 See the edit... I made the plots in Mathematica. $\endgroup$ – Michael Brown Jul 11 '13 at 14:09
  • $\begingroup$ This is a great explanation, but unfortunately, I am accepting Ali's answer as it explains the negative masses. $\endgroup$ – user80551 Jul 11 '13 at 16:01
  • $\begingroup$ Can you explain the assumptions? IMO it's incompatible to have a finite amount of fluid AND an incompressible flow. Because the force diagram indicates that the water will flow away from the bubbles toward the outsides of the container, and by conservation of mass, it follows that the density near the bubbles would be lower than the density at the edge of the container, so the fluid must be getting compressed at some point. Also, you would need a container of water about the size of a planet to have any real gravitational effect on the bubbles. $\endgroup$ – Greg Jul 11 '13 at 16:27
  • $\begingroup$ @Greg Fluid has to flow in at the sides to fill the space the bubbles were in before they moved together. This could be arranged without any significant flow near the boundaries, since by assumption the size of the bubbles doesn't change. I'm imagining something like Stokes flow the velocity of which vanishes as $1/r$ from the bubble. You need the finite size to make Poisson's equation unambiguous. I forget how big you can make a water planet before it collapses into a black hole... it's pretty big though. :) $\endgroup$ – Michael Brown Jul 11 '13 at 16:38
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I think I see the argument:

One should go to the line joining A and B. And draw two vertical tangent on the right for A and on the left for B. There is more water volume to attract the water in the region on the left of A, so there will be a force towards B, and vice versa.

Anyway bubbles in liquids are not a simple matter like your problem states.

see : http://www.tandfonline.com/doi/pdf/10.1080/18811248.2001.9715037

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What is negative mass in this context

Consider a single bubble in the middle of a glass of water on your desk.

Normally, gravity pulls air above the desk downwards towards the desk surface.

However, in this case the air in the bubble does not move downwards under the influence of gravity. It moves in the opposite direction. It is as if the air in the bubble has negative mass and is repelled by gravity.

In fact the air does not have negative mass and is not repelled but the real situation is more complex to describe.

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  • $\begingroup$ "does not move downwards under the influence of gravity." Because the buoyant force on it is greater than the gravitational force pulling it down. Gravity tries to pull it down. $\endgroup$ – user80551 Jul 11 '13 at 13:02
  • $\begingroup$ @user80551: Yes, that is part of "the real situation is more complex" which I was deliberately avoiding describing. $\endgroup$ – RedGrittyBrick Jul 11 '13 at 13:24
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Here's how I look at it:

When you 'release' the air bubbles, they tend to rise due to the buoyant force, $\rho_wgv$. Thus, they are in motion relative to the static fluid. Assuming that the air bubbles are spherical, then in their motion they change the pressure of the water (i.e. the streamlines of water are closer together). Now since we have 2 air bubbles, the area in between them is affected by the motion of both bubbles, and the streamlines are further compressed. This results in lower pressure in the area between the bubbles relative to the pressure on the left of A and the right of B. Thus the air bubbles are 'pulled' toward each other. This is similar to the Venturi Effect.


I'm not sure whether we are to assume that the bubbles are in motion or if they are being held fixed, but if they are indeed fixed, then here is an experiment to try:

Take 2 ping-pong balls, and connect each one to a seperate string. Place the ping-pong balls in a container filled with water, and connect the strings to the bottom of the container (right underneath each ping-pong ball). Stabilize the ping-pong balls so that each string is perpendicular to the bottom of the container. Then, let them go and observe what happens.

I got it:

Imagine a(n) (infinite) universe that is completely filled with water. Water as far as the eye can see. You can intuitvely guess that the force on any water molecule is 0, because it is drawn toward every other water molecule (in every direction) in the this universe. Now, let's say there is a void in this water universe. A spherical region where no water exists. Now imagine a 'test particle,' with mass $M$, and calculate the gravitational force in the direction away from the center of the circle. Assuming that there is a linear density in the water, such that each unit of distance $r$ contains a mass $m$, the force in that direction is: $$GMm\sum\limits_{i=0}^\infty \frac{1}{(0.5+i)^2}\approx\left(GMm\right)*4.9348$$

enter image description here

Because of the void in the opposite direction, the force in that direction is less than what we calculated above. So the net force is away from the void. In fact, a test particle placed anywhere near this void will experience a net force away from it!

What that implies is that the water is pushed away from the void in all directions. This leads me to believe that the void (AKA bubble) would actually expand to occupy the space left behind by the water.

Now imagine two voids. The force field looks like what Michael Brown has posted. Since the water is pushed away from the two bubbles, the question is what does it leave behind? It looks as though the two bubbles expand until they become one big bubble (which is also expanding). Any way you look at it, the bubbles combine into one.

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  • $\begingroup$ Note that the question is about the mutual gravitational attraction of two bubbles, not their motion under an external gravitational field. $\endgroup$ – Michael Brown Jul 11 '13 at 14:50
  • $\begingroup$ Comment on the edit: the bubbles are still attracted even if there is surface tension to stabilise them. $\endgroup$ – Michael Brown Jul 12 '13 at 8:17
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In the water when there are two bubbles the surface are of the water (which is the surface are of the bubble) is more when the bubbles are collapsed. As the water tries to make the surface area least they will come closer and and collapse. The process is happening due to the force of water between the bubbles on the bubbles.

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