# Conservation of quasimomentum and added-mass: help needed to understand a derivation in G. Falkovich's fluid mechanics book

## Context

Starting at page 30 of Falkovich's Fluid Mechanics, there is an interesting discussion about the relationship between the conservation of quasimomentum in an isentropic ideal fluid and the added-mass force that a submerged body experiments when it is accelerated by external forces. With some difficulty I have been able to work my way through it up to page 33, where the final sequence of equations leads to the relation

$$\dot{K}_i = -\int \frac{\partial p}{\partial r_i}\, \mathrm d \mathbf{r} \qquad \tag{1}\label{Eq:Quasimomentum}$$

where $$K_i$$ is the component i of the quasimomentum, $$p$$ is the pressure and $$\mathbf{r}$$ is the position vector of the fluid particles; i.e., $$\mathbf{r}=\mathbf{r}(\mathbf{R},t)$$, where $$\mathbf{R}$$ is the reference (initial) positions of the fluid particles. The time derivative in the equation above, denoted with an overhead dot, must be interpreted as the material derivative, that is, following the motion of the fluid particles. Note that the RHS above is (minus) the $$i$$-th component of the force applied by the fluid on the particle. The integration domain is the whole fluid, which is assumed infinite. The definition of the quasimomentum is (p. 33, Eq. 1.42)

$$K_i = -\rho_0 \int v_j \left ( \delta_{ij} - \frac{\partial r_j}{\partial R_i}\right ) \, \mathrm d \mathbf{R}$$

where $$\rho_0$$ is the initial density of the fluid, $$v_i$$ is the $$i$$-th component of the fluid velocity and $$\delta_{ij}$$ is Kronecker's delta.

## Where I got lost

My problem is the last step in the sequence of equalities that lead to the equation above (p. 33, Eq. 1.44). There the equation of motion $$\rho \dot{\mathbf{v}} = - \partial p / \partial \mathbf{r}$$, with $$\rho(R,t)=\rho_0 \det(\left [ \partial r_i/\partial R_j \right ])$$, is introduced in the previous equations to obtain (all straight-forward substitutions)

$$\dot{K}_i= - \int{\frac{\partial p}{\partial r_i} \, \mathrm{d}\mathbf{r}} + \int{\frac{\partial }{\partial R_i} \left ( \rho_0W - \frac{\rho_0v^2}{2} \right )\, \mathrm{d}\mathbf{R}} \qquad \tag{2}\label{Eq:DerivativeQuasimomentum}$$

where I have corrected what I believe must be a typo in the original text (the density should multiply the whole parenthesis in the second term, and not just the $$v^2/2$$ term). The quantity $$W$$ represents the enthalpy density, defined as $$W=E+p/\rho$$, with $$E$$ the internal energy of the fluid (see p.32). To fix ideas, the Hamiltonian for the same flow is defined (p.32, Eq. 1.41) as

$$\mathcal{H} = \int{\rho_0 \left ( W + \frac{v^2}{2} \right )\, \mathrm{d}\mathbf{R}}.$$

Now, I do not understand the following sentence, which is the only reason given in the book for neglecting the second term in Eq. \ref{Eq:DerivativeQuasimomentum} to obtain Eq. \ref{Eq:Quasimomentum}:

In the fourth line [i.e., our Eq. \ref{Eq:DerivativeQuasimomentum}], the integral over the reference space R of the total derivative in the second term is identical to zero, while the integral over r in the first term excludes the volume of the body, so that the boundary term remains, which is minus the force acting on the body.

Why is the second term in Eq. \ref{Eq:DerivativeQuasimomentum} equal to zero? Why does he mention the fact that the volume of the body is excluded in the integral of the first term only (should one not exclude its volume at the initial time in the second term too?). Why does he call the derivative a total derivative?

Can anyone give me a hint?

I think the reference space $$\mathbf{R}$$ used to define quasimomentum is, by definition, one in which the fluid has constant density $$\rho_0$$ over an entire Euclidean space. The volume occupied by the body is mapped to a single point, say $$\mathbf{R} = \mathbf{0}$$, at which fields can be discontinuous. The book's description of $$\mathbf{R}$$ as the "initial" position of fluid particles is confusing and should be taken in a conceptual sense, as if we started with a uniform fluid in which the body was then "inflated" from zero volume, whether or not this actually happened. Thus, $$\mathbf{q} = \mathbf{r} - \mathbf{R}$$ is the fluid displacement attributable to the body's presence.
For example, consider an incompressible fluid containing a spherical body. If we choose a time $$t_0$$ at which the body is centered at say $$\mathbf{r} = \mathbf{0}$$ and has radius $$a$$, a possible relation defining a reference space is $$\mathbf{r}(\mathbf{R},t_0) = (R^3 + a^3)^{1/3} \mathbf{R}/R$$. This satisfies the book's condition that $$\partial r_j/\partial R_i \to \delta_{ij}$$ for $$R \to \infty$$.
The quasimomentum appears not to be uniquely defined, because we can perform an anisotropic volume-preserving transformation of $$\mathbf{R}$$ that maintains that asymptotic condition.
• Thank you. At some point I had imagined something along the lines could be going on, since Falkovich writes (p. 47) '$\rho_0$ is the density in the reference (initial) state, which can always be chosen to be uniform' (I never got this). Your solution is also coherent with ignoring the interphase surface for the current deformed) coordinates as mentioned in my questions. Still, I don't think I would have been able to resolve this without your help. There is very little out there about quasimimentum for continuum mechanics! – Guillermo BCN Mar 11 '19 at 9:03