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Convective acceleration is the spatial part $(\mathbf{u}\cdot \nabla )\mathbf{u}$ of the material derivative of fluid velocity. It can remain nonzero in steady flow, because a moving particle passes through positions with different velocities. The identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ relates it to vorticity.

Let $\mathbf{v}$ denote the velocity relative to the uniformly rotating frame, and write $r_{\perp }^2=x^2+y^2$. For rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, the Euler equations for an inviscid fluid in that frame areThe negative quadratic term is the centrifugal potential, whose negative gradient is the outward centrifugal acceleration per unit mass. In steady flow, dotting the equation with $\mathbf{v}$ eliminates the Coriolis force, since $\mathbf{v}\cdot (\mathbf{\Omega }\times \mathbf{v})=0$. The remaining acceleration term is $\mathbf{v}\cdot \nabla (|\mathbf{v}{|}^2/2)$. ThusBy part (b), the steady rotating-frame Bernoulli integral isSteadiness here is measured in the rotating frame, including the body-force potential used in that frame.

In steady flow, the Euler equations for an inviscid fluid and part (a) giveTaking the dot product with $\mathbf{u}$ yields $\mathbf{u}\cdot \nabla H=0$, because a cross product is perpendicular to its first factor. Part (b) therefore gives Bernoulli's quantity along streamlines:For rotational flow these constants can differ between streamlines, since $\mathbf{u}\times \mathbf{\omega }$ need not vanish and therefore $\nabla H$ need not be zero. Special rotational flows can still have one global constant when that cross product vanishes; the general conclusion is only constancy along individual streamlines.

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.