Mass density. Equation of continuity.

Consider a sphere of infinitesimal radius centered at a point $P$ in the reference configuration. The material contained within the sphere has mass $\Delta M$. Let $\Delta V$ denote the volume of the sphere. The mass density $\rho_0$ at point $P$ in the reference configuration is defined as

$$\rho_0 = \lim_{\Delta V\rightarrow 0}\frac{\Delta M}{\Delta V}.$$ (3.11.1)

One of the assumptions in continuum mechanics is that the limit on the right-hand side of (3.11.1) exists at every point of the body. Since both $\Delta M$ and $\Delta V$ are positive, therefore, the mass density $\rho_0 (X_1,X_2,X_3)$ is positive. Note that from the point of view of Atomic Physics, the assumption that the right-hand side of (3.11.1) is well defined may not be justified. One can always envisage infinitesimal volume elements surrounding a point which contain no atomic particles at some time and hence will make the mass density at the point to be zero at that time. However, in continuum mechanics, we are concerned with gross effects or macroeffects of deformation, and lengths considered are much larger than the distance between adjacent atoms.

We now make the assumption that the mass of the material contained in every small volume element at $P$ is conserved. That is, the mass of the matter enclosed in the infinitesimal parallelepiped at $P$ equals the mass of the matter contained in the infinitesimal parallelepiped at $P^\prime$ into which the former is deformed. Denoting the mass density in the present configuration by $\rho$, we have

$$\rho dv = \rho_0dV.$$

Substituting for $dv$ from (3.10.2), we arrive at

$$\rho J = \rho_0.$$ (3.11.2)

or

$$\rho(X_1,X_2,X_3,t)J(X_1,X_2,X_3,t) = \rho_0(X_1,X_2,X_3).$$

Equation (3.11.2) which relates the mass density in the present configuration to the mass density in the reference configuration is the equation of continuity or the conservation of mass in the Lagrangian description.

To obtain the equation of continuity in the spatial description, we take the material derivative of (3.11.2) and thereby obtain

$$\dot\rho J + \rho \dot J = 0.$$ (3.11.3)

Since

$$J = \varepsilon_{ABC}F_{1A}F_{2B}F_{3C},$$

therefore,

$$\dot J = \varepsilon_{ABC}(\dot F_{1A}F_{2B}F_{3C} + F_{1A}\dot F_{2B}F_{3C} + F_{1A}F_{2B}\dot F_{3C}).$$ (3.11.4)

However,

$$\varepsilon_{ABC}\dot F_{1A}F_{2B}F_{3C} =\ \varepsilon_{ABC}\frac{\partial\dot x_1}{\partial X_A}F_{2B}F_{3C}\ ,$$

$$=\ \varepsilon_{ABC}\left(\frac{\partial v_1}{\partial x_1}\frac{\pa... ...rtial v_1}{\partial x_3}\frac{\partial x_3}{\partial X_A}\right)F_{2B}F_{3C}\ ,$$

$$=\ \varepsilon_{ABC}\frac{\partial v_1}{\partial x_1}F_{1A}F_{2B}F_{3C}+ 0 + 0\ ,$$

$$=\ J\frac{\partial v_1}{\partial x_1}.$$ (3.11.5)

Similarly, one can show that

$$\begin{split}&\varepsilon_{ABC}F_{1A}\dot F_{2B}F_{3C} = J\fr... ...\dot F_{3C} = J\frac{\partial v_3}{\partial x_3}\ . \end{split}$$ (3.11.6)

Substituting from (3.11.5) and (3.11.6) into (3.11.4) and then the result into (3.11.3) we get

$$\dot\rho J + \rho J\frac{\partial v_i}{\partial x_i} = 0$$

and thus conclude that

$$\dot\rho + \rho \frac{\partial v_i}{\partial x_i} = 0.$$ (3.11.7)

This equation can equivalently be written as

$$\frac{\partial\rho}{\partial t} + \frac{\partial}{\partial x_i}(\rho v_i) = 0,$$ (3.11.8)

which is the continuity equation in spatial description.

For an isochoric deformation

$$J = 1$$

and hence

$$\dot J = 0,\ \dot\rho = 0.$$ (3.11.9)

Equations (3.11.9) and (3.11.7) imply that for an isochoric deformation, the continuity equation assumes the form

$$\frac{\partial v_i}{\partial x_i} = 0.$$ (3.11.10)

Exercise: Show that

$$v_1 = -\frac{2x_1x_2x_3}{(x^2_1 + x^2_2)^2},\hspace{.2in} v_2 = \... ... - x^2_2)x_3}{(x^2_1 + x^2_2)^2},\hspace{.2in} v_3 = \frac{x_2}{x^2_1 + x^2_2},$$

are the components of a velocity field in an isochoric deformation.

Example: For the velocity field given by

$$v_i = \frac{x_i}{1+t},\ t \ge 0,$$

find the density of a material particle as a function of time.

Solution: For the given velocity field,

$$\frac{\partial v_i}{\partial x_i} = \frac{3}{1+t}.$$

Therefore, from the conservation of mass, we get

$$\dot\rho =\ -\rho\frac{\partial v_i}{\partial x_i} = - \frac{3\rho}{1 + t}.$$

$$\therefore\hspace*{.2in} \frac{d\rho}{\rho}=\ - \frac{3dt}{1 + t}.$$

Integration of this equation gives

$$\ell n\rho = -3\ell n(1 + t) + A\hspace{2in} (\ast)$$

where $A$ is a constant of integration. If $\rho = \rho_0$ at $t = 0$, then

$$\ell n\rho_0 = -3\ell n1 + A\ \ A = \ell n\rho_0.$$

Thus eqn. $(\ast )$ becomes

$$\rho = \rho_0/(1 + t)^3.$$

Exercise: Given the velocity field

$$v_i = x_1 t\delta_{i1} + x_2t\delta_{i2},$$

determine how the fluid density varies with time.

Exercise: In the spatial description the density of an incompressible fluid is given by $\rho = kx_2$. Find a permissible form for the velocity field, with $v_3 = 0$, in order that the conservation of mass equation be satisfied.