The velocity \( \mathbf{u} \) of the fluid particle is obtained by time differentiation:
\[ \mathbf{u} = \frac{d\mathbf{x}}{dt} = \frac{d\mathbf{X}}{dt} + \frac{d\mathbf{x'}}{dt} = \mathbf{U} + \frac{d}{dt} \left( x'_1 \mathbf{e}'_1 + x'_2 \mathbf{e}'_2 + x'_3 \mathbf{e}'_3 \right) \] \[ = \mathbf{U} + {\frac{dx'_1}{dt} \mathbf{e}'_1 + \frac{dx'_2}{dt} \mathbf{e}'_2 + \frac{dx'_3}{dt} \mathbf{e}'_3} + {x'_1 \frac{d\mathbf{e}'_1}{dt} + x'_2 \frac{d\mathbf{e}'_2}{dt} + x'_3 \frac{d\mathbf{e}'_3}{dt}} \] \[ = \mathbf{U} + \mathbf{u'} + \mathbf{\Omega} \times \mathbf{x'} \]Take the time derivative to find the acceleration \( \mathbf{a} \) of a fluid particle at P
\[ \mathbf{a} = \frac{d \mathbf{u}}{dt} = \frac{d}{dt} \left( \mathbf{U} + \mathbf{u}' + \boldsymbol{\Omega} \times \mathbf{x}' \right) = \frac{d \mathbf{U}}{dt} + \mathbf{a}' + 2 \boldsymbol{\Omega} \times \mathbf{u}' + \frac{d \boldsymbol{\Omega}}{dt} \times \mathbf{x}' + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{x}') \]\(\because\) \( \frac{d\mathbf{x}'}{dt} = \mathbf{u}' + \boldsymbol{\Omega} \times \mathbf{x}' \), defined the velocity in the rotating frame, differentiate \( \mathbf{u} \) to find: \[ \mathbf{a} = \frac{d \mathbf{u}}{dt} = \frac{d \mathbf{U}}{dt} + \frac{d \mathbf{u}'}{dt} + \frac{d}{dt} \left( \boldsymbol{\Omega} \times \mathbf{x}' \right) \] Expanding each term \[ \mathbf{a} = \frac{d \mathbf{U}}{dt} + \frac{d}{dt} ( u'_1 \mathbf{e}'_1 + u'_2 \mathbf{e}'_2 + u'_3 \mathbf{e}'_3 ) + \frac{d \boldsymbol{\Omega}}{dt} \times \mathbf{x}' + \boldsymbol{\Omega} \times \frac{d \mathbf{x}'}{dt} \] \(\therefore\) \(\mathbf{a} = \frac{d \mathbf{U}}{dt} + \underbrace{\frac{d u'_1}{dt} \mathbf{e}'_1 + \frac{d u'_2}{dt} \mathbf{e}'_2 + \frac{d u'_3}{dt} \mathbf{e}'_3}_{\text{Fluid particle acceleration } \mathbf{a}' \text{ in non-inertial frame}} + \underbrace{u'_1 \frac{d \mathbf{e}'_1}{dt} + u'_2 \frac{d \mathbf{e}'_2}{dt} + u'_3 \frac{d \mathbf{e}'_3}{dt}}_{\text{Time derivatives of unit vectors}} + \frac{d \boldsymbol{\Omega}}{dt} \times \mathbf{x}' + \boldsymbol{\Omega} \times (\mathbf{u}' + \boldsymbol{\Omega} \times \mathbf{x}') \) \(\because\) The time derivatives of the unit vectors can be rewritten in terms of a cross product \[ u'_1 \frac{d \mathbf{e}'_1}{dt} + u'_2 \frac{d \mathbf{e}'_2}{dt} + u'_3 \frac{d \mathbf{e}'_3}{dt} = \boldsymbol{\Omega} \times \mathbf{u}' \] \(\therefore\) \(\mathbf{a} = \frac{d\mathbf{U}}{dt} + \mathbf{a}' + \boldsymbol{\Omega} \times \mathbf{u}' + \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}' + \boldsymbol{\Omega} \times \mathbf{u}' + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{x}')\) \( \boxed{\quad = \frac{d\mathbf{U}}{dt} + \mathbf{a}' + 2\boldsymbol{\Omega} \times \mathbf{u}' + \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}' + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{x}')}\)