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For an axisymmetric field without azimuthal motion, the divergence in spherical coordinates isSubstituting the given axisymmetric stream function representation givesThus incompressible flow is automatic.

The velocity components are laboratory-frame components resolved about the instantaneous bubble center. The center itself translates with $U\hat{\mathbf{z}}=U\cos \theta \hat{\mathbf{r}}-U\sin \theta \hat{\mathbf{\theta }}$. The kinematic boundary condition for a spherical interface givesFor negligible internal dynamic viscosity, continuity of tangential traction imposes zero tangential shear stress, rather than a prescribed tangential velocity. In terms of the strain-rate tensor,Equivalently, ${\partial }_r{v}_{\theta }(a)=v_{\theta }(a)/a+U\sin \theta /a$. Axisymmetry gives $v_{\varphi }=0$, and the far-field boundary condition is $\mathbf{v}\to 0$. There is no no-slip boundary condition $v_{\theta }=-U\sin \theta$ on a clean bubble; the liquid can slip tangentially relative to the translating interface.

These are the conditions of the prescribed-sphere clean-bubble Stokes drag model. A genuinely free bubble also requires a normal stress balance. If one additionally insists on spatially uniform gas pressure, while removing both surface tension and the variation of hydrostatic pressure, that extra condition is incompatible with the translating spherical solution in part (c). The approximation must therefore be understood as computing drag on an imposed spherical interface, or as retaining the normal balancing effects when checking a free bubble.

Retain the clean-bubble approximation ${\eta }_a\ll {\eta }_w$ used in parts (b) and (c), with constant mass densities and diffusion coefficient. For a spherical bubble,ConsequentlyThe buoyancy minus bubble weight is $4\pi ({\rho }_w-{\rho }_a)g{a}^3/3$. Balancing it against the clean-bubble Stokes drag gives the instantaneous terminal velocitySet $z=0$ at the initial position. Combining this with the dissolution equation gives $da/dz=-D_a/(K{a}^3)$ and hence the dissolution-limited bubble rise lawThe expression applies until the bracket vanishes. The rise speed decreases continuously as the bubble shrinks. Equivalently, $z(t)=K(a_0^{2}t-D_a{t}^2)$, providing a direct check on the elimination of time.

The phrase “constant terminal velocity” must here mean rapid adjustment to a force balance at each radius. A literally time-independent speed is incompatible with that balance because $U\propto a^2$. If instead an external mechanism maintained a fixed $U_{\ast }$, the distinct result would be $a^2(z)=a_0^2-2D_{a}z/U_{\ast }$ and $z_{\mathrm{diss}}=U_{\ast }{a}_0^2/(2D_a)$. In particular, holding the initial terminal speed fixed doubles the predicted rise distance; it is not the freely rising shrinking-bubble model.

Using the supplied values and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ givesEquality means dissolution exactly on arrival at the surface; dissolution strictly before arrival needs a greater initial depth. This is a formal low-Reynolds number estimate. The same values predict $U(a_0)\simeq 0.033\mathrm{m}{\mathrm{s}}^{-1}$ and a radius-based Reynolds number ${\rho }_{w}U(a_0)a_0/{\eta }_w\simeq 3.3$; the diameter-based value is about $6.5$. Neither is $\ll 1$. Thus the numerical input does not consistently satisfy the Stokes flow assumption initially, and the $0.82\mathrm{m}$ result should not be presented as an accurate oceanic prediction. Within the stated ideal model, bubble compression, contaminants and changes in dissolved-gas concentration are also excluded.

Take the upward direction as ${\mathbf{e}}_z$, with polar angle $\theta$ measured from it. The liquid is quiescent at infinity. On a translating clean inviscid bubble, the kinematic boundary condition is $(\mathbf{u}-U{\mathbf{e}}_z)\cdot \mathbf{n}=0$, and the stress-free boundary condition is zero tangential traction. The normal traction must balance uniform gas pressure and any constant-curvature surface-tension jump.

For a vertical Stokeslet of strength $F$, the radial velocity is $u_r=F\cos \theta /(4\pi \mu r)$. Its traction on a concentric sphere is purely normal, so zero tangential traction is automatic. The kinematic condition at $r=a$ givesThe singularity is only a device for generating the exterior solution; there is no liquid point force inside the actual bubble. The integrated dynamic traction on the bubble is $-F{\mathbf{e}}_z$, giving the clean-bubble Stokes drag. The upward buoyancy is $4\pi \rho g{a}^3/3$, hence force balance gives

The normal stress also shows that surface tension is not required for this particular spherical solution. Let the ambient hydrostatic pressure at the bubble centre be $p_c$, so on its surface $p_h=p_c-\rho ga\cos \theta$. Adding the Stokeslet's dynamic normal traction giveswhere the terminal buoyancy value of $F$ cancels the dipole. This is the spherical clean bubble without surface tension balance: an initially spherical bubble can remain spherical in this ideal steady Stokes solution even when surface tension is zero. With uniform surface tension $\gamma$, choose gas pressure $p_b=p_c+2\gamma /a$; with $\gamma =0$, $p_b=p_c$. This proves compatibility of the spherical shape, not stability of arbitrary deformations or validity when inertia matters.

Write $D={\partial }_n{A}_n$ and use repeated-index summation. Because every $A_i$ is a harmonic function,andThus $\mu {\nabla }^2{u}_i-{\partial }_{i}p=0$, the incompressible Stokes equation. MoreoverThe antisymmetric first two terms cancel in the sum with ${\partial }_i{u}_j$, giving the Newtonian fluid Cauchy stress tensorFor $A_i=V_i(1-a/(2r))$, harmonic outside the origin,At $r=a$ this gives $\mathbf{u}=\frac{1}{2}[\mathbf{V}-(\mathbf{V}\cdot \mathbf{n})\mathbf{n}]$, where $\mathbf{n}=\mathbf{x}/a$, so the normal velocity is zero. AlsoHence on the surface,a purely normal traction, appropriate to a shear-free sphere rather than a no-slip solid sphere. Finally rotational symmetry gives ${\int }_{S^2}{n}_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$, so the fluid's force on the sphere isThis is the clean-bubble Stokes drag in the far-field velocity $\mathbf{V}$.

For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.