HR problem deals with a spherical fluid viscous drop falling in a different fluid under influence of gravity. The outer fluid is of uniform speed $U$ in direction of gravity, far away from the drop. The boundary solutions posed are


1. zero velocity of both flows on drop boundary
2. continuity of tangential velocities
3. continuity of tangential stress


Working with Stokes stream functions in spherical ccordinates we have solutions


$$\psi\_{drop}=\dfrac{Ur^2\sin^2 \theta}{4(1+\kappa)}\left(1-\frac
{r^2}{a^2}\right)$$


$$\psi\_{fluid}=-\dfrac{Ur^2\sin^2\theta}{2} \left(1-\dfrac{a(2+3\kappa)}{2r(1+\kappa)} + \dfrac{\kappa a^3}{2r^3(1+\kappa)} \right)$$


where $a$ is the radius of the drop and $\kappa=\frac
{\mu\_{drop}}{\mu\_{fluid}}$.


In the above we do not pose the condition of continuity of normal stresses but apriori claim that the spherical shape is preserved by action of surface tension. However, Batchelor claims that, after checking the difference of of normal stresses for above stream functions, we get that we do not need surface tension since 
$$p\_{drop} - 2 \mu\_{drop} \sigma\_{drop,rr}=p\_{fluid} - 2 \mu\_{fluid} \sigma\_{fluid,rr}$$
after neglecting constants. Here $p$ is modified pressure and $\sigma\_{rr}=\frac{\partial u\_r}{\partial r}$.
However, I computed these expressions several times and I do not get equality.


My question is: Did I missunderstand Batchelor or there is some mistake? Is the spherical shape consistent with other assumptions or it is consequence of high surface tension?



Hadamard-Rybczynski problem
Yes, I partially missunderstood Batchelor, leading myself by some other literature. Spherical form of drop is due to conitnuity of normal stresses but in form $-p\_{abs}+2\mu \sigma\_{rr}$ where $p\_{abs}=p+\rho \mathbf{g}\cdot \mathbf{x}$ and this in case when $U$ is the velocity of free falling $U=-\frac{2}{3}\frac{a^2 g \rho}{\mu} \left(\frac{\bar{\rho}}{\rho} -1 \right)\dfrac{\kappa +1}{3\kappa +2}.$


