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Coriolis acceleration does no work

Physical scope

Pointwise algebraic consequence of the ideal rotating Boussinesq branch's Coriolis term in a right-handed local Cartesian model. No empirical motion or universal fluid behavior is asserted.

Example

In the adopted rotating Boussinesq model, let $u = (u_{1}, u_{2}, u_{3}) \in R^{3}$ be velocity in m s $^{- 1}$ , let $f \in R$ be the constant Coriolis parameter in s $^{- 1}$ , and let $e_{3} = (0, 0, 1)$ be the upward unit vector in the right-handed Cartesian coordinates. The momentum equation places $+ f e_{3} \times u$ on its left side, so its signed acceleration contribution on the right side is $- f e_{3} \times u$ . Its contribution to the instantaneous rate of change of specific kinetic energy $∣ u ∣^{2} / 2$ is the scalar product

$u \cdot (- f e_{3} \times u) = 0.$

Thus this ideal Coriolis term contributes zero specific power, in W kg $^{- 1}$ , at every point where the velocity is defined. This does not remove work by other forces or assert that an observed fluid follows the adopted model.

Example references

Geophysical approximation branches: adopted Coriolis acceleration convention and its physical scope.

Geoffrey K. Vallis, Fluid Dynamics of the Atmosphere and Ocean, University of Exeter ECMM719 full lecture notes (Autumn 2018), §1.2.2, equation (1.30) and the Coriolis no-work discussion, printed p. 6 (PDF page 10)

Facts & Assumptions

Given: The pointwise velocity vector and constant Coriolis coefficient in the adopted rotating Boussinesq branch.

[A1]

Geophysical approximation branches: the local Cartesian coordinates are right-handed, $e_{3} = (0, 0, 1)$ , and the rotating Boussinesq momentum equation places $+ f e_{3} \times u$ on its left side, with signed acceleration contribution $- f e_{3} \times u$ on the right.

[F1]

Geophysical approximation branches: the coordinate convention is $e_{3} \times u = (- u_{2}, u_{1}, 0)$ .

Verification

1.1givenF1algebra

By [F1], $f e_{3} \times u = f (- u_{2}, u_{1}, 0)$ .

2.1givenA1step 1.1algebra∎

Taking the Euclidean dot product with $u$ gives $u \cdot (- f e_{3} \times u) = f (u_{1} u_{2} - u_{2} u_{1}) = 0$ . The acceleration has units m s $^{- 2}$ , so the work rate has units $(m s^{- 1}) (m s^{- 2}) = m^{2} s^{- 3} = W k g^{- 1}$ and is zero.

Depends on

Geophysical approximation branches

Dependency tree · 2 levels

Sources

Geoffrey K. Vallis, Fluid Dynamics of the Atmosphere and Ocean, University of Exeter ECMM719 full lecture notes, Autumn 2018