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()
