Circulation can produce a net force perpendicular to a uniform free stream even when the surrounding flow is modeled as inviscid and irrotational outside the body or vortex core.

For a steady two-dimensional incompressible flow, let the far-field velocity be

$$\mathbf U_\infty=U_\infty\hat{\mathbf x},$$

and define positive circulation $\Gamma>0$ as counterclockwise when viewed from the positive $z$ direction.

The Kutta-Joukowski theorem gives the force per unit span associated with circulation as

$$\boxed{\mathbf F'=\rho,\mathbf U_\infty\times\boldsymbol\Gamma},$$

where

$$\boldsymbol\Gamma=\Gamma\hat{\mathbf z}.$$

For the chosen axes,

$$\mathbf U_\infty\times\boldsymbol\Gamma =U_\infty\Gamma (\hat{\mathbf x}\times\hat{\mathbf z}) =-U_\infty\Gamma\hat{\mathbf y},$$

so

$$\boxed{F'y=-\rho U\infty\Gamma}.$$

Thus, under this sign convention,

• counterclockwise circulation $(\Gamma>0)$ produces downward lift;
• clockwise circulation $(\Gamma<0)$ produces upward lift.

The magnitude is

$$\boxed{|L'|=\rho U_\infty|\Gamma|}.$$

Derivation from the circulating cylinder

The circulating potential flow around a cylinder of radius $a$ has surface tangential velocity

$$v_\theta(a,\theta) =-2U_\infty\sin\theta+rac{\Gamma}{2\pi a}.$$

Define

$$\lambda=\frac{\Gamma}{2\pi aU_\infty}.$$

Then

$$\frac{v_\theta}{U_\infty}=-2\sin\theta+\lambda.$$

Steady Bernoulli gives the surface pressure coefficient

$$C_p =1-\left(-2\sin\theta+\lambda\right)^2.$$

For a surface element per unit span,

$$dA'=a,d\theta.$$

Pressure acts inward on the body, so the vertical force element is

$$dF'y =-(p-p\infty)\sin\theta,a,d\theta.$$

Using

$$p-p_\infty=\frac12\rho U_\infty^2C_p,$$

we obtain

$$F'y =-\frac12\rho U\infty^2a \int_0^{2\pi}C_p\sin\theta,d\theta.$$

Expand

$$C_p =1-4\sin^2\theta +4\lambda\sin\theta -\lambda^2.$$

When multiplied by $\sin\theta$ and integrated over a full period, the terms proportional to

$$\sin\theta$$

and

$$\sin^3\theta$$

vanish. Only the cross term remains:

$$\int_0^{2\pi}C_p\sin\theta,d\theta =4\lambda\int_0^{2\pi}\sin^2\theta,d\theta =4\pi\lambda.$$

Therefore

$$F'y =-\frac12\rho U\infty^2a(4\pi\lambda).$$

Substituting

$$\lambda=\frac{\Gamma}{2\pi aU_\infty}$$

gives

$$\boxed{F'y=-\rho U\infty\Gamma}.$$

The horizontal pressure-force integral remains zero. Circulation breaks the top-bottom pressure symmetry and creates lift while the ideal cylinder still has zero drag.

Why the theorem is more general than the cylinder

The cylinder provides an exactly solvable demonstration, but the Kutta-Joukowski relation is not a special cylinder formula. For an appropriate two-dimensional inviscid outer flow with uniform far field, the net transverse force per unit span depends on the total circulation around the body rather than the detailed body shape:

$$\boxed{\mathbf F'=\rho\mathbf U_\infty\times\boldsymbol\Gamma}.$$

The body shape determines which velocity and pressure distribution realizes that circulation; the theorem compresses the resulting integrated lift into one global quantity.

Worked example

An airfoil section moves through air approximated by

$$\rho=1.20,\mathrm{kg/m^3}$$

with free-stream speed

$$U_\infty=30,\mathrm{m/s}.$$

Suppose its outer flow has clockwise circulation

$$\Gamma=-4.0,\mathrm{m^2/s}.$$

Then

$$F'y=-\rho U\infty\Gamma$$

$$=-(1.20)(30)(-4.0) =\boxed{144,\mathrm{N/m}}.$$

The positive sign means the lift is upward under the chosen coordinate convention.

What the theorem does not determine

Kutta-Joukowski answers:

If the circulation is $\Gamma$, what lift does the ideal two-dimensional outer flow produce?

It does not by itself determine the value of $\Gamma$ around an airfoil.

For a sharp trailing edge, potential flow admits many mathematical circulation values that satisfy no penetration. The additional physical rule used to select the relevant circulation is the Kutta condition.

This distinction is fundamental: circulation selection and force from circulation are separate problems.