Why

A smooth stress tensor has a unique value at the axis, even though the radial and azimuthal basis vectors depend on the direction of approach. Rotational symmetry therefore makes the radial and hoop stresses equal at . Smoothness also determines how quickly their difference can grow away from the axis.

We assume a smooth, symmetric stress tensor whose Cartesian components admit a Taylor expansion near the axis. The field is axisymmetric: rotating the position about rotates its vector and tensor components accordingly. All powers of below are dimensional Taylor powers.

1. Establish parity on a line through the axis

Take the line and rotate it by about . This maps to . A scalar and an axial vector component are unchanged by this rotation, whereas a transverse vector component changes sign. Thus, for an axisymmetric velocity and pressure,

Here, even and odd parity refer to smooth continuation along a signed transverse coordinate through the axis; the cylindrical radius itself is nonnegative. Continuity then relates the radial-velocity coefficients to axial derivatives, giving and .

2. Use rotational symmetry at the axis

At , a symmetric transverse stress tensor invariant under every rotation must be proportional to the identity. Hence

or, in cylindrical components,

The mixed transverse-axial components also vanish at the axis because no nonzero transverse vector is invariant under all rotations.

3. Determine the first allowed difference

On the positive -axis, and . A rotation by changes the sign of each transverse basis vector, so the two signs cancel in a transverse rank-two tensor component. Both and are therefore even functions of .

Their Taylor series have equal constant terms and no linear terms. We can write

Subtracting gives

The difference may vanish faster, but it cannot start at or under these smoothness assumptions. Continuity at the axis alone would not establish the quadratic ordering.

4. Check the Cartesian representation

For the meridionally reflection-symmetric stress used in the slender-jet model, . Its transverse block is

where and . Therefore,

Substituting the quadratic difference gives

which is smooth at the origin. The directional factor is thus cancelled by the radial dependence established in step 3.

The reflection assumption removes azimuthal shear away from the axis. Axisymmetry and a swirl-free velocity alone do not impose that condition on an arbitrary prescribed polymer stress. The regularity of the remaining polymer components, and their contribution to momentum, are developed in the polymer-stress note.