How ambient pressure can affects my inner body? I’m a freediver. My lung shrinks when I getting down to the sea deeper and deeper by Boyle’s law. The law tells me that my volume of lung will shrink as 1/4 if I were in 4 bars of ambient pressure. I learned that, and I feel it when I get to 30m under the surface(=4 bars) of the sea.
But I actually don’t understand how the pressure can affect my inner body.
I know my outer body(skins, eyes, outer ears, ...) is not rigid material so all outer body will pressed 'a little'. But I feel it's not enough to shrink my lung that much.
So how come outer pressure affects to my inner body, especially my lung?
One thing that I think 'it might be' is, my belly is also pressed too. Maybe it's enough pressure to shrink my lung.
But I wish I can understand with better explanation.
 A: Here's why.  Let's think about the front of your chest as a kind of wall with your lungs on one side (inside your chest), and water on the other.  Now we'll work out what the force on that wall would be if your lungs were at atmospheric pressure and you were $30\,\mathrm{m}$ under water.
The pressure at depth $d$ under water is given by:
$$p = \rho g d + 1\,\mathrm{atm}$$
Where $\rho$ is the density of water and $g$ is the acceleration due to gravity, and $\mathrm{atm}$ is atmospheric pressure. (This is a slight approximation: it assumes $g$ is constant and water is incompressible: this is close enough to true here for this to be fine.)
At $30\,\mathrm{m}$ this is about $395\,\mathrm{kPa} = 395\,\mathrm{kN/m^2}$, assuming $1\,\mathrm{atm} = 101\,\mathrm{kPa}$.  (Sorry for working in pascals: I should be converting to bar, but I'm lazy).
Let's assume the area of the front wall of your chest is about $0.25\,\mathrm{m^2}$.  So, how much force would be exerted on it in this case?  Well, there's air at $1\,\mathrm{atm} =101\,\mathrm{kN/m^2}$ in your lungs pushing out, and water at $395\,\mathrm{kN/m^2}$ pushing out, and the answer is
$$(395-101)\,\mathrm{kN/m^2}\times 0.25\,\mathrm{m^2} = 73.5\,\mathrm{kN}$$
How big a force is this?  Well, a mass of $1\,\mathrm{kg}$ exerts a downward force of $9.8\,\mathrm{N}$ under gravity.  So this is about the same force that $7500\,\mathrm{kg}$ would exert if they were resting on your chest.
That's the equivalent of seven and a half tonnes resting on your chest.
Needless to say your chest can't withstand anything like this force.  So something gives, and what gives is that the air in your lungs compresses until the force it's exerting outwards approximately balances the force the water is exerting inwards.
A: What's really important is not absolute pressure, but pressure difference. When you start your dive, the air in your lungs is at 1 bar of pressure. At 30 m as you said the pressure is 4 bar, so there is a pressure difference between the lungs and the water. This is what determines the compression of your lungs.
The rest of your body is solid, so you won't feel any difference. (Actually, with a really high pressure you may feel a slight reduction in your volume, but that should be far outside of the boundaries of freediving.)
The fact that you are solid, but not rigid, is what determines the reduction in volume of your lungs. Behind the ribcage there is no resistance to pressure as long as there is a pressure difference, so the ribcage will "deform" (but not compress) to allow the compression of your lungs.
Compare that with what happens to, say, your legs. Your legs are subjected to the exact same pressure as your lungs and ribcage, but they are completely solid. This means that they won't feel any deformation.
On a side note, your lungs don't actually follow Boyle's law exactly, otherwise diving too deep (40 m? I'm not completely sure, sorry) would mean that your lungs get crushed. The freediving world record is below 200 m, so that's surely not the case. What actually happens is that the blood vessels in your lungs expand, getting full of blood, and thus helping reduce the total lungs' volume.
