Why is the total force at a free surface zero? I am looking into waves on a free surface for which their are two main conditions:


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*Kinematic condition: Particles on the surface remain on the surface.

*Dynamic condition: Forces acting on the surface are in equilibrium.


Where do these conditions come from?  
Reference


*

*Here (link to Google Books)

*And here
 A: The so called kinematic condition is essentially an equivalence between Eulerian and Lagrangian descriptions of a fluid. Consider a particle on the surface located at $\vec{x}$. Furthermore, for an irrotaional inviscid fluid the body of the fluid obeys Laplace's equation which means there exists a velocity potential $\phi$. The first condition is: 
$$\frac{dz}{dt} = \phi_z \quad @z=\eta$$
where $d/dt$ is the total derivative, $\phi$ is the velocity potential. Now, if we parametrize all points on the surface as a function $\eta(\vec{x}(t),t)$ then by definition 
$$\frac{d\eta}{dt}  = \frac{\partial \eta}{\partial t}+\nabla\phi\cdot \nabla \eta = \phi_z$$ 
evaluated at $z=\eta$. 
Secondly, one has the dynamic boundary condition (Let's just do this for gravity waves and ignore things like capillarity). This is the statement that pressure is continuous across the interface (do you know why we do this?). To get this, we begin with Euler's equations: 
$$\frac{\partial \vec{u}}{\partial t}+\vec{u}\nabla\vec{u} =-\nabla p +g\hat{z}.$$
The advective term can be rewritten as 
$$\vec{u}\nabla\vec{u}= \frac{1}{2}\nabla (\vec{u}\cdot \vec{u}) -\vec{u}\times \vec{\omega}$$
with $\vec{\omega} = \nabla \times \vec{u}$ the vorticity, identically zero for our flow. Therefore one can rewrite the Euler equations at the surface as
$$\nabla ( \phi_t+\frac{1}{2}(\nabla \phi)^2 +g z+p)=0$$
as $z=\eta$
Hence, at the surface ($z=\eta)$ we have 
$$\phi_t+\frac{1}{2}(\nabla \phi)^2 +g z+p=B(t)$$
for some constant $B(t)$. Usually, in the absence of forcing we set $p=0$ at the surface. 
