Why Are Gravity Effects Neglected in Stokes Flow Analysis?

[Est120]

Homework Statement

find the appropiate Navier-Stokes equation in the low Re number limit

Relevant Equations

Navier Stokes equations

in the limit as Re→0 , we can neglect the material derivate of v ( Dv/Dt =0 ) but why in books they always make the gravity effects equal to 0?
i can't understand and no one really explains this stuff

 

[pasmith]

The density is being treated as constant, so the gravational force per unit mass is \mathbf{g} = -\nabla\left(-\int\mathbf{g} \cdot d\mathbf{x}\right).

You can therefore absorb it into the pressure term. It makes no difference to the mathematics if you work with p or p' = p - \int\mathbf{g} \cdot d\mathbf{x}.

 

[Est120]

The density is being treated as constant, so the gravational force per unit mass is \mathbf{g} = -\nabla\left(-\int\mathbf{g} \cdot d\mathbf{x}\right).

You can therefore absorb it into the pressure term. It makes no difference to the mathematics if you work with p or p' = p - \int\mathbf{g} \cdot d\mathbf{x}.

yeah, that's the so called "modified pressure" right? i hardly managed to get that but still almost no source of information talks about that...