# (Fluid Dynamics) Euler's equation including gravity

In fluid dynamics, we can write down the Euler's equation as

$$\dfrac{\partial \mathbf{v}}{\partial t} + ( \mathbf{v} \cdot \mathbf{\text{grad}} ) \mathbf{v} = - \dfrac{\mathbf{\text{grad}} \; p}{\rho}$$ .

If the fluid is in a gravitational field, we can add an extra term on the RHS, for example

$$\dfrac{\partial \mathbf{v}}{\partial t} + ( \mathbf{v} \cdot \mathbf{\text{grad}} ) \mathbf{v} = - \dfrac{\mathbf{\text{grad}} \; p}{\rho} + \mathbf{g}$$ .

My question is rather simple, in the above equation, does it imply that the direction of gravitational acceleration is in the same direction as the acceleration of the fluid? If so, what if the fluid is accelerating upwards due to a huge force $$- \mathbf{\text{grad}} \; p$$?

• if you'd like, you could use \nabla for $\nabla$. – user191954 Oct 13 '18 at 10:58

$$\mathbf{g}$$ in that equation is a generalized vector and its direction is undefined. It could be in any direction, depending what vector you choose to use. If you are using a standard cartesian reference frame, for $$\mathbf{g}$$ you would use (0, 0, -9.81).