of interface with the orientation n , so that the integral over all orientations gives the total amount of interface per unit volume. Therefore, for the state variables Q and q one can write the following definitions where 1 is the unit tensor, and ∫ d $^{2}$n denotes an integration over the unit sphere. Introduc- tion of the distribution function and further averaging over all possible orientations imply the loss of information of the detailed morphology of the system. This coarse-grained model keeps only the information about the average amount of interfacial area per unit volume and its orientation. However, the detailed information about the morphology, i.e., the explicit interfacial position and orientation, is lost. The time evolution of the new configurational variables Q and q is determined by two factors: the external flow field which orients and enlarges the interface, and the interfacial tension which has the opposite effect and governs the relaxation of the interface. Therefore, the time evolution can be separated into two parts: The convective part of the time evolution can be found by considering the convection be- havior of the unit vector n normal to the interface under affine deformations and volume preservation assumptions. Then, the convection of the state variables is derived using the above definitions (1.2) where ˙ γ = ∇ v + ( ∇ v ) T is the symmetrized velocity gradient tensor [33]. In the equation for the interfacial shape (1.4b), the fourth moment of the interface normal n appears, which must be expressed in terms of the state variables through an appropriate closure approximation ∂Q ∂t ∣ convection = −∇ · ( Q v ) − ( ∇ v ) T : q + 2 3 Q 1 : ( ∇ v ) T , (1.4a) ∂ q ∂t ∣ convection = − ( ∇ v ) · q − q · ( ∇ v ) T − 1 3 Q ˙ γ − v · ∇ q + 1 3 1 ( ∇ v ) : q + [ q + 1 9 Q 1 ] Tr ( ∇ v ) + nnnn : ( ∇ v ) T . (1.4b) ∂Q ∂t = ∂Q ∂t ∣ convection + ∂Q ∂t ∣ relaxation , (1.3a) ∂ q ∂t = ∂ q ∂t ∣ convection + ∂ q ∂t ∣ relaxation . (1.3b) Q = ∫ f ( n ) d $^{2}$n, (1.2a) q = ( nn − 1 3 1 ) f ( n ) d $^{2}$n = nn − 1 3 Q 1 , (1.2b) 4