A thinning jet has a radial acceleration. Its pressure therefore varies across a section, even when the leading axial velocity is uniform in . We now retain this pressure variation, together with the velocity and polymer-stress corrections required by the surface conditions.
The starting point is the leading-order derivation. We keep the same assumptions: an incompressible, axisymmetric liquid with no swirl, constant density , solvent viscosity and surface tension , negligible gas stress and no body force. Pressure is measured relative to the gas. Primes denote and dots denote .
1. Keep the radial ordering explicit
Let , where and are the radial and axial length scales. We use dimensional and throughout, with
The Taylor coefficients vary on the axial scale. Thus, and are successive relative corrections to the axial velocity. There are no additional factors multiplying the dimensional powers of .
The polymer stresses have the corresponding expansion
The equality of the leading radial and hoop stresses follows from axis regularity. Their coefficients need not be equal. We continue to write .
The even and odd powers follow from the parity of the fields. Under this regular slender expansion, the first relative correction to the leading equations is . All surface and reduced equations below are truncated at that accuracy unless identified as exact.
2. Collect the bulk equations
The and coefficients of axial momentum are
The radial Laplacian maps to and to . This is why enters (M2), even though its contribution to the velocity is fourth order. In the acceleration, the term from cancels the term from . The remaining material derivative in (M2) therefore has the form shown above. The radial derivatives are worked out in the Laplacian note.
To determine , we use the coefficient of in radial momentum. The radial acceleration at this order is
The polymer contribution is
Balancing these terms with pressure and solvent viscosity gives
Both the local and convective radial accelerations contribute. Dropping requires a separate time-scale argument; slenderness alone does not remove it.
3. Retain the correction to the volume flux
At , the kinematic condition is
Multiplying by gives
The same flux follows by integrating over the cross-section:
Thus, the centreline velocity and the mean velocity differ at second order.
4. Expand both surface tractions
The exact unit normal, tangent and curvature are
With this sign convention, a sphere has . Expanding for small slope gives
The last two terms are relative corrections to . The exact traction conditions are derived in the surface-stress note.
Normal traction
Solving for the centreline pressure gives
where
The first line accounts for curvature and the pressure difference between the axis and the surface. The second line is the solvent normal traction, including the change in the surface normal. The last line contains the three polymer corrections: the projection of the normal-stress difference, the projection of shear stress and the radial variation of .
Tangential traction
Multiplying by gives the exact surface condition
Let denote its leading contribution,
and the contribution smaller by two powers of slenderness,
At leading order we use . At second order, the condition is
This distinction determines the accuracy of the axial equation. In a perturbation expansion, the second-order correction to enters through and balances evaluated with the leading coefficients. Setting the two contributions to zero separately would remove that correction.
An alternative expression follows by using the leading condition to eliminate . The equivalent cubic contribution is
Indeed, . Their difference is beyond the retained order when (T) holds. The direct form (T2) avoids making this substitution.
5. Obtain the corrected axial equation
Using (T) to rearrange the viscous term in (M0) gives
Now substitute this expression and (N) into (M0). The explicit again cancels, leaving
Both final terms are second-order corrections. The pressure correction comes from normal traction, while changes the radial velocity profile through tangential traction. Polymer shear can still enter these corrections even though its explicit leading-order contribution has cancelled.
To use this hierarchy, retain (C), (M0), (M2), (R), (N) and (T), or replace (M0) and (N) by the equivalent equation (E2). In particular, (M2) supplies
Substituting this into (T2) removes as an independent unknown. It does not remove its contribution to the second-order balance. The polymer coefficients must be supplied by a constitutive equation expanded to the same accuracy, together with the appropriate initial and boundary conditions. The four leading closures alone do not constitute a complete second-order model.
6. Test a uniform filament
Take and impose homogeneous extension,
Assume spatially uniform, diagonal polymer stress, so all radial polymer coefficients and shear stresses vanish. The homogeneous velocity profile has , and (C) is satisfied exactly. Equation (R) gives
The terms from local and convective radial acceleration cancel. A radial pressure variation survives, but it has no axial gradient in this homogeneous ansatz. Consequently, it does not by itself balance axial acceleration: (M0) also requires if the stresses are spatially uniform and inertia is retained.
The same pressure variation does contribute to the force integrated over a section. We calculate that force, and the additional assumptions needed to use it as a local thinning model, in the next note.
