# Why does the 'vortex stretching term' not appear in Kelvin Circulation theorem?

The Kelvins circulation theorem states that "In a inviscid, barotropic fluid with conservative body forces, the circulation around a closed curve moving with the fluid remains constant with time; if the motion is observed from an inertial frame of reference". This means that for an ideal fluid (zero viscosity) with constant density (hence barotropic) in presence of only gravitational force (conservative) observed from an inertial frame of reference, an irrotational flow shall remain irrotational. Very well.

Now, consider the vorticity transport equation for an ideal, constant density fluid with only gravitational force observed from an inertial frame of reference:

$$\frac{D\vec{\omega}}{Dt}=(\vec{\omega}.\nabla)\vec{U}$$

As per this equation, the term on the RHS (vortex stretching term) is an agent for change in vorticity following a fluid particle. So, even under conditions required for the Kelvins circulation theorem (inviscid, barotropic and only conservative forces), the vorticity transport equation shows that vorticity following a fluid particle can change due to the mechanism of vortex stretching; so an initially irrotational flow may become rotational due to vortex stretching. Is this correct?

How can we reconcile the absence of vortex stretching effect in Kelvins circulation theorem? Thanks and with regards.

The vorticity transport equation views the vorticity as vector value 0-form. However, Kelvin’s theorem views it as a 2-form. In the former formalism, it is nit conserved due to vortex stretching while in the latter it is.

More physically, if a point follows the fluid, it will see the vorticity change. However, if a surface follows the fluid, the the vorticity flux is conserved. It is not the same quantities that are at stake so the conservation of one does not imply the conservation of the other.

Mathematically, the convective part: $$L_vf=(v\cdot\nabla)f$$ is the Lie derivative of the 0-form $$f$$. The analogue for a general 2-form is: $$L_v\omega=\nabla\times (\omega\times v)+(\nabla\cdot \omega)v$$ Using incompressibility $$\nabla\cdot v=0$$ and the incompressibility of vorticity (from the curl) $$\nabla\cdot \omega=0$$, the vorticity transport is equivalent to: $$L_v\omega=0$$

Btw, you can similarly interpret Euler’s equation in velocity as the transport of velocity as a 1-form and also get Kelvin’s theorem.

Hope this helps.

• Thanks for bringing out the important distinction between 'conservation of vorticity' and the 'conservation of the flux of vorticity'. So, based on the former, using the vorticity transport equation, can we say that an initially irrotational flow may become rotational due to vortex-stretching term as the flow evolves; given other conditions like inviscid, barotropic and only conservative body forces?? Commented Sep 3, 2023 at 11:33
• An initially irrotational flow stays irrotational according to to Kelvin’s theorem, i.e. looking at vorticity as a 2-form. It’s just less obvious when you consider vorticity as a $0$ form and proving it directly would be harder (and useless?). Intuitively, since the vortex stretching is proportional to vorticity, it will never contribute if it starts at zero.
– LPZ
Commented Sep 3, 2023 at 14:53
• Thanks for clarifying that an irrotational flow will remain irrotational. Yes, proving from the vorticity equation is difficult and I didn't find in any books. Intuitively, it does indeed seem that if the fluid is starting from zero vorticity, the RHS is initially zero. Considering that there are no other vorticity generating mechanisms, it continues to be zero. Hence its rate of change following a fluid particle is zero, i.e. the flow continues to be irrotational. Commented Sep 3, 2023 at 15:38