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# Kelvin's circulation theorem

In fluid mechanics, Kelvin's Circulation Theorem states "In an inviscid, barotropic flow with conservative body forces, the circulation around a closed curve moving with the fluid remains constant with time"[1]. The theorem was developed by William Thomson, 1st Baron Kelvin. Stated mathematically:

$\frac{\mathrm{d}\Gamma}{\mathrm{d}t} = 0$

where Γ is the circulation around a material contour C(t). Stated more simply this theorem says that if one observes a closed contour at one instant, and follows the contour over time (by following the motion of all of its fluid elements), the circulation over the two locations of this contour are equal.

This theorem does not hold in cases with viscous stresses, nonconservative body forces (for example a coriolis force) or non-barotropic pressure-density relations.

## Mathematical Proof

The circulation Γ around a closed material contour C(t) is defined by:

$\Gamma(t) = \oint_C \boldsymbol{u} \cdot \boldsymbol{ds}$

where u is the velocity vector, and ds is an element along the closed contour.

The governing equation for an inviscid fluid with a conservative body force is

$\frac{\mathrm{D} \boldsymbol{u}}{\mathrm{D} t} = - \frac{1}{\rho}\boldsymbol{\nabla}p + \boldsymbol{\nabla} \Phi$

where D/Dt is the convective derivative, ρ is the fluid density, p is the pressure and Φ is the potential for the body force. These are the Euler equations with a body force.

The condition of baratropicity implies that the density is a function only of the pressure, i.e. ρ = ρ(p).

Taking the time derivative of circulation gives

$\frac{\mathrm{d}\Gamma}{\mathrm{d} t} = \oint_C \frac{\mathrm{D} \boldsymbol{u}}{\mathrm{D}t} \cdot \boldsymbol{\mathrm{d}s} + \oint_C \boldsymbol{u} \cdot \frac{\mathrm{D} \boldsymbol{\mathrm{d}s}}{\mathrm{D}t}$

For the first term, we substitute from the governing equation, and then apply Stoke's theorem, thus:

$\oint_C \frac{\mathrm{D} \boldsymbol{u}}{\mathrm{D}t} \cdot \boldsymbol{ds} = \int_A \boldsymbol{\nabla} \times \left( -\frac{1}{\rho} \boldsymbol{\nabla} p + \boldsymbol{\nabla} \Phi \right) \cdot \boldsymbol{n} \, \mathrm{d}S = \int_A \frac{1}{\rho^2} \left( \boldsymbol{\nabla} \rho \times \boldsymbol{\nabla} p \right) \cdot \boldsymbol{n} \, \mathrm{d}S = 0$

The final equality arises since $\boldsymbol{\nabla} \rho \times \boldsymbol{\nabla} p=0$ owing to baratropicity.

For the second term, we note that evolution of the material line element is given by

$\frac{\mathrm{D} \boldsymbol{\mathrm{d}s}}{\mathrm{D}t} = \left( \boldsymbol{\mathrm{d}s} \cdot \boldsymbol{\nabla} \right) \boldsymbol{u}$

Hence

$\oint_C \boldsymbol{u} \cdot \frac{\mathrm{D} \boldsymbol{\mathrm{d}s}}{\mathrm{D}t} = \oint_C \boldsymbol{u} \cdot \left[ \left( \boldsymbol{\mathrm{d}s} \cdot \boldsymbol{\nabla} \right) \boldsymbol{u} \right] = \frac{1}{2} \oint_C \boldsymbol{\nabla} \left( |\boldsymbol{u}|^2 \right) \cdot \boldsymbol{\mathrm{d}s} = 0$

The last equality is obtained by applying Stokes theorem.

Since both terms are zero, we obtain the result

$\frac{\mathrm{d}\Gamma}{\mathrm{d}t} = 0$

## References

1. ^ Kundu, P and Cohen, I: "Fluid Mechanics", page 130. Academic Press 2002