# Scuba diving in free fall, do I need to worry about the Bends?

The Integral Trees is a 1984 science fiction novel by Larry Niven with an entire ecological system of oxygen breathing people and creatures, in free fall, in a gas torus. Water forms ponds, lakes, and puddles, in spheres as you would expect water to do in free fall.

On Earth when diving in water the pressure increases by 1 atmosphere about every 10 meters or 33 feet of depth. One of the complications of this is the bends. In fact the ISS is maintained at 1 atmosphere for similar reasons

Assuming I have located a large body of water, like a lake or even a small sea, in an actual Goldblatt's World. It might even be a lake in space station. Would I need to worry about increased pressure as I dove deeper into it? What similarities or differences would I encounter diving in free fall as opposed to Earth?

• Way worse problem would be "surfacing" - leaving the water. Surface tension would make a big blob of water to keep sticking to your body, distributing itself semi-uniformly, including your face. You'd need some means of separating it - a minature centrifuge (even a spinning bar to grip and turn around), a strong current of air, or some really big towels. You just can't count on the water dripping off you. – SF. Jun 28 '16 at 11:38
• Since this question originated in a sf story, I feel ok with pointing out that there is a very good sf story about just this kind of swimming pool: John Varley's "Blue Champagne". Although he does use hand-wavey "fields" for some of his effects. isfdb.org/cgi-bin/title.cgi?41594 – Organic Marble Jun 29 '16 at 13:46
• Here are two other, different questions that are related to swimming on the surface of a body in substantially reduced gravity on Enceladus and on Mars. They both have some good answers well worth reading. – uhoh Jun 29 '16 at 15:47

tl;dr: The pressure increase due to self-gravity reaches one atmosphere for a sphere of water with a radius of about 1000 km. Since water is still viscous but the average buoyancy along trip would be a tiny fraction of what you'd experience on Earth, it would take a very long time to travel that distance. If you used a winch to pull yourself through thousands of kilometers of water, you might be able to get the bends, but you'd have to work at it. But the bends is a complicated function of pressure difference, rate of change, and total time. Here is the math to calculate the pressure profile at least.

update: As an fyi, in this answer for a different low-gravity swimming question, several thresholds for different problems besides the bends (if I understand correctly) are also listed. The the first falls at about 30m depth in earth gravity, where every 10m is roughly one additional atmosphere.

If you assume the density of the water $\rho$ is constant, then as you move away from the center of the sphere, the pressure in each shell of water of thickness $dr$ increases by $dP$ as:

$$\frac{dP_g}{dr} = -\rho g(r) .$$

From Equation (3) in the derivation of the Adams Williamson equation, where $g(r)$ is the gravitational acceleration inside the sphere. Again assuming the density is constant:

$$g(r)=\frac{4}{3} \pi G r \rho$$

which comes from Gravity of Earth (depth) and can be derived from the historically significant Shell Theorem of Newton.

Integrating $\frac{dP_g}{dr}$ out from $r=0$ and adding a constant to make it zero at the surface $r=R_0$, the pressure due to self-gravity becomes:

$$P_g(r)=\frac{4}{3} \pi G \rho \frac{{R_0}^2-r^2}{2} ,$$

and the maximum pressure due to gravity in the center:

$$P_{g0}=\frac{2}{3} \pi G \rho {R_0}^2 .$$

There is also a uniform contribution to the pressure due to the surface tension, which tends to maintain its spherical shape and therefore minimal surface area. For a sphere, the Young-Laplace equation gives the pressure difference across an interface as:

$$\Delta P_s = \gamma \frac{2}{R_0} .$$

Remembering to include the ambient pressure $P_a$, the total pressure inside the water sphere (maintaining assumption of uniform density) is:

$$P_{tot} = P_g + \Delta P_s + P_a$$

$$P_{tot}(r) =\frac{4}{3} \pi G \rho \frac{{R_0}^2-r^2}{2} + \gamma \frac{2}{R_0} + P_a$$

In the plot below, I've left out the ambient pressure. Atmospheric pressure is about $10^5$ Newtons per square meters (Pascal). The pressure increase at the center of a sphere of water floating in air reaches one atmosphere when the radius reaches one micron (due to dominant surface tension) and when it reaches one thousand kilometers (due to self-gravity).

I used Python to make the plot:

def g(r, rho):

return (4./3.) * pi * G * r * rho

def Pgrav(r, R0):

# dP/dr = -rho*g(r) just integrate

return (4./3.) * pi * G * rho * (R0**2 - r**2) / 2.

def dPsurf(R0):

return 2. * gamma / R0

import numpy as np
import matplotlib.pyplot as plt

pi    = np.pi
G     = 6.674E-11       # N m^2/kg^2
rho   = 1000.         # kg/m^3 water roughly
gamma = 73. *1E-03  #  N/m  against air, at 20C, roughly

R0    = 100000.          # m
r  = np.linspace(0, R0, 1001)

Pg  = Pgrav(r, R0)
dPs = dPsurf(R0) * np.ones_like(r)
Pa  = 1E+05 * np.ones_like(r)    # N/m^2  roughly

Ptot = Pg + dPs + Pa
Pgs  = Pg + dPs

plt.figure()
plt.plot(r, Pgs, '-k')
plt.plot(r, Pg)
plt.plot(r, dPs)
plt.show()

R0  = np.logspace(-6, 6, 1001)

Pg  = Pgrav(0, R0)
dPs = dPsurf(R0)

Pgs = Pg + dPs

plt.figure()
plt.plot(R0, Pgs, '-k')
plt.plot(R0, Pg)
plt.plot(R0, dPs)
plt.yscale('log')
plt.xscale('log')
plt.xlabel('R0 (meters)', fontsize=18)
plt.ylabel('Pressure (Pa = N/m^2)', fontsize=18)
plt.show()

• I'm starting to double check the math now - help is always welcome! – uhoh Jun 28 '16 at 3:28

The increase in pressure with depth is solely due to gravitation (rho x g x h) == (density times gravity times depth). If the body of water is in free fall, there would be no such increase, since g = 0.

• The water blob is held together by surface tension; does that cause any higher pressure at core than at surface? (i.e. Is rho x g x h really all the source of pressure, or just the dominant source?) – Russell Borogove Jun 27 '16 at 22:47
• Good question; it was always negligible for anything I did. – Organic Marble Jun 27 '16 at 23:12
• This paper (which I didn't totally follow) seems to imply that internal pressure due to surface tension drops off dramatically as the size of the droplet increases. eng.utah.edu/~lzang/images/lecture-8.pdf – Organic Marble Jun 27 '16 at 23:18
• However, there might be uniformly distributed pressure due to the container the lake is in... – jpaugh Jun 27 '16 at 23:25
• Container? The question is about spherical "lakes" (big droplets) in free fall. From the wikipedia article "The Smoke Ring contains numerous "ponds," globs of water of various sizes which float free like everything else." – Organic Marble Jun 27 '16 at 23:25