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I want to know how Reynolds Equation of Momentum is accord with the 3rd law of Newton, the law of action-reaction forces. For an ideal fluid, we have that

[tex]

\frac {\partial m \vec {v}} {\partial t} = \vec v [ \frac {\partial {\rho }} {\partial t} + \nabla [ \rho \vec v ] + \rho [\frac {\partial {\vec {v}} {\partial t} + ( \vec {v} \dot \nabla ) \vec v

[/itex]

\frac {\partial m \vec {v}} {\partial t} = \vec v [ \frac {\partial {\rho }} {\partial t} + \nabla [ \rho \vec v ] + \rho [\frac {\partial {\vec {v}} {\partial t} + ( \vec {v} \dot \nabla ) \vec v

Takint 2 points into the flow, I want to know if its posible to make that F12 = F21

Is this posible, or is a nonsense?

[tex]

\frac {\partial m \vec {v}} {\partial t} = \vec v [ \frac {\partial {\rho }} {\partial t} + \nabla [ \rho \vec v ] + \rho [\frac {\partial {\vec {v}} {\partial t} + ( \vec {v} \dot \nabla ) \vec v

[/itex]

\frac {\partial m \vec {v}} {\partial t} = \vec v [ \frac {\partial {\rho }} {\partial t} + \nabla [ \rho \vec v ] + \rho [\frac {\partial {\vec {v}} {\partial t} + ( \vec {v} \dot \nabla ) \vec v

Takint 2 points into the flow, I want to know if its posible to make that F12 = F21

Is this posible, or is a nonsense?

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