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Let $z$ increase downward. The slender viscous thread is locally in uniaxial extension. The transverse stress equals the ambient pressure, so the Trouton ratio gives the excess axial stress ${\sigma }_{zz}+p_e=3\mu w_z$. Mass conservation and axial force balance therefore give

In steady flow $Aw=Q$. Dividing the momentum equation by $A=Q/w$ givesWith $w=\hat{w}W$ and $z=\hat{z}Z$, chooseThe dimensionless equation becomesTreating $(W^{\prime }{)}^2$ as a function of $W$ and using an integrating factor yieldsBecause $A=Q/(\hat{w}W)$, the area sketches are the reciprocals of the functions $W$ found below. The dimensional vertical deviatoric stress is

For the uniaxial extensional flowthe spin tensor vanishes andThe flow is steady and homogeneous, so with $\xi =1$ the Jaumann derivative of $\mathbf{\alpha }$ vanishes. The structure equation yieldsWriting $k=c_2/c_1$ and using
$\dot{\mathbf{\gamma }}:\dot{\mathbf{\gamma }}=6{\dot{ϵ}}^2$ givesConsequentlyThe extensional viscosity is the tensile stress difference divided by $\dot{ϵ}$:For $b_2=0$ this is the constantThe factor three is the Trouton ratio associated with the effective zero-rate shear viscosity.

As $\dot{ϵ}\to 0$, $a\to 0$, $F\to 1$, andThe Trouton ratio is therefore three. This is also the small-rate Oldroyd-B result because finite extensibility is irrelevant while polymer deformation remains small.

A slender viscous thread has a cross-sectional scale much smaller than its axial variation scale. For an axisymmetric Newtonian thread of area $A$ and nearly uniform axial velocity $w$, mass conservation and the Trouton ratio givewhen inertia and surface tension are negligible and $z$ points downward.