Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This seems natural, but I can't wrap my head around it when I think about it.

When I squeeze an open bottle filled with water, the water will spill out. When I squeeze a bottle, the material collapses where I squeeze it, but expands in other areas, resulting in a constant volume. If the volume is constant, then I would think that the water shouldn't spill out.

If I were to guess, there is something related to the pressure my hand is creating inside the bottle, but I'm not entirely sure.

share|cite|improve this question
If the volume is constant, why can you not squeeze a bottle as much with the cap screwed on? – Jon Hanna May 1 '14 at 14:04
"material collapses where I squeeze it, but expands in other areas, resulting in a constant volume" sounds to me like "I bet 100\$ on horses, and got 5\$ back at the cage, resulting in a constant wallet volume" – Spork May 1 '14 at 15:22
up vote 13 down vote accepted

The surface area of the bottle is conserved, but the volume is not. Squeezing the bottle deforms it into a shape whose volume to surface area ratio is lower than it was previously.

As an example consider a bottle whose cross-section is initially a circle. The volume of the bottle will be $V_0=\pi r^2h$ where $h$ is the height of the bottle, and the surface area will be $A_0=2\pi rh$.

As we squeeze the bottle we deform it into some kind of ellipse. If we choose the major and minor axes of the ellipse to be $a=xr$ and $b=\sqrt{2-x^2}r$, then the new surface area of the bottle is given approximately by $$ A_1\approx h 2\pi\sqrt{\frac12(a^2+b^2)}=2\pi rh\quad\Rightarrow\quad\frac{A_1}{A_0}=1, $$ where we have neglected the end caps. So, the surface area of the bottle is conserved. The new volume is given by $$ V_1=\pi abh=\pi r^2h x\sqrt{2-x^2}\quad\Rightarrow\quad\frac{V_1}{V_0}=x\sqrt{2-x^2}, $$ where we have again neglected the end caps. So you can see that even though the surface area is conserved, the volume is maximum only when $x=1$. A plot of the last term, $x\sqrt{2-x^2}$ is shown below to confirm this fact.

enter image description here

share|cite|improve this answer

Squeezing the bottle does decrease its volume. Rather than a bottle, it may be more helpful to think of a full toothpaste tube; the mechanics will be the same.

If you squeeze the middle of the tube, the middle will collapse, the back will expand, and the front will expand and squirt out some toothpaste. Treating the toothpaste in the tube (or the water in the bottle) as incompressible, if the tube is full then the volume of the inside of the tube = the volume of toothpaste inside the tube. When you squeeze, some of the toothpaste comes out, indicating there is now less toothpaste inside the tube. But the tube is still full of toothpaste. Therefore, the tube volume is decreased by the amount of toothpaste that came out.

Likewise, although it may look like the collapsing middle of the water bottle is offset by the expanding top and/or bottom of the bottle, it doesn't quite do it. Assuming the bottle is completely full of water, the decrease in the bottle's volume will be easily measured by capturing the water that spills out and measuring its volume.

share|cite|improve this answer
Where does that volume go? Is that too much of a broad question to ask? – tambykojak May 1 '14 at 5:55
Tsk! Squeeze the toothpaste tube from the end! – innisfree May 1 '14 at 5:58
The volume of water goes all over the floor. As for the container, its "volume" is really capacity. In the extreme case, what if you flatten the bottle completely. It will have zero water in it, and have a capacity of zero. The material volume of the bottle is the volume of a small amount of plastic, and won't change when you squeeze the bottle. – jdj081 May 1 '14 at 5:59
@jdj081 I think you should incorporate the volume/capacity comment into the answer, it's a vital point. – Angew May 1 '14 at 11:46
"Where does the volume go": volume isn't matter or even space; it's not conserved. It's actually just a concept: a measurement of a portion of space. Is it surprising that the same length of string that can form a circle with area $\pi$ (that is, length $2\pi$) can also be squashed into a doubled-back line (of length $\pi$ covered twice) with no interior? – Ryan Reich May 1 '14 at 16:35

jdj081's answer is good. I just want to address where I think you originally went wrong.

Your confusion lies in using the word "volume" in two different ways. You should differentiate between volume of the container (capacity is a better term, as jdj081 states) and volume of the liquid.

The liquid's volume doesn't change. The container's volume does. Since not all of the liquid can fit, it spills out of the container. Yes, the liquid's volume is still the same, but this does not mean the volume's container is also the same.

share|cite|improve this answer
Yes I was forgetting that water doesn't compress easily and the container compresses very easy. – tambykojak May 2 '14 at 5:17

Just another approach to visualize it. (two dimensional) If you look on a bottle from above, you can see the round shape. A circle is the optimal shape, where you can have the biggest amount of content per package. If you start to squeeze the circle (no matter how), the amount of stuff you can put into is decreasing. Cause the amount of package is not increasing, where should it come from. (If you have an inelastic package of course.)

share|cite|improve this answer

protected by Qmechanic Apr 20 at 11:53

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.