Analysis of motion

Our discussion of kinematics thus far has focused on the deformation of the initial placement of a body to its current placement. We will now consider the time evolution of a body as it moves and deforms in three dimensional space. We will begin our discussion by focusing on the evolution of a single material point, and subsequently extend it to the time evolution of the deformation gradient. We will, in particular, derive the mass continuity equation as an illustration of working with time derivatives. Finally, we will have a look at the important topic of taking the time derivatives of integrals whose domain and integrands both depend on time.

Basic definitions

Suppose that $(\chi_t)_{t \in I}$ is a motion of a material body $\mathcal{B}$ in three dimensional space, as observed by an observer. At the initial instant of time, chosen without loss of generality as $t = 0$, the body occupies the open subset $B_0 \subseteq \mathbb{R}^3$ of three dimensional space. Let us now focus attention on a fixed point $\mathsf{X} \in B_0$ and track its evolution as the body moves and deforms in space over the time interval $I$. The path followed by the body, given by $$\{\varphi_{\mathsf{X}}:I \to \mathbb{R}^3 \,|\, \text{ for any } t \in I, \; \varphi_{\mathsf{X}}(t) = \varphi(\mathsf{X},t)\},$$ is called the pathline of the point $\mathsf{X} \in B_0$. Different particles will have different pathlines, but they never intersect owing to the fact that $J(\mathsf{X},t) > 0$.

Remark

Pathlines are especially convenient when working with solids, where the correlation between the instantaneous relative positions of the particles remain similar over typical time intervals of interest. In fluids, on the other hand, the chaotic nature of the motion often results in the pathlines being intricately mixed and knotted. In such cases, it is more helpful to work with the streamlines and streaklines. A streamline is a line, defined at every instant of time, such that the tangent vector at any point on the line coincides with the velocity of the particle occupying that position at that specific instant of time. Note that a streamline is globally computed at a given time instant, and can vary over time. A streakline, on the other hand, is the trace of the current position of every particle that occupied a particular fixed point at some previous instant of time. A streakline is what is observed when a dye is injected in a fluid flow a fixed point.

Material velocity and acceleration

The material velocity of the particle initially at $\mathsf{X} \in B_0$ is defined as the map $\mathsf{V}_{\mathsf{X}}:I \to T\mathbb{R}^3$ such that, for any $t \in I$, $$\mathsf{V}_{\mathsf{X}}(t) = \frac{\partial \varphi(\mathsf{X},t)}{\partial t}.$$ We will also use the notation $\mathsf{V}(\mathsf{X},t)$ in place of $\mathsf{V}_{\mathsf{X}}(t)$, when convenient. If the observer chooses a global basis $(\mathsf{e}_i)$ for $\mathbb{R}^3$, then the components of the material velocity can be written as $$\mathsf{V}(\mathsf{X},t) = \sum V_i(\mathsf{X},t) \mathsf{e}_i.$$ Note that we use lowercase subscript indices to indicate the components of $\mathsf{V}_{\mathsf{X}}(t)$. This is because of the fact that though the spatial reference of the point whose velocity we compute is $\mathsf{X} \in B_0$, the velocity vector at time (t) is itself not in $T_{\mathsf{X}}B_0$, but rather in $T_{\varphi_t(\mathsf{X})}\mathbb{R}^3$.

Remark

Note also that if we focus on a particular instant of time $t$, $\mathsf{V}(\mathsf{X},t)$ is not necessarily in $T_{\varphi_t(\mathsf{X})}B_t$, but rather in $T_{\varphi_t(\mathsf{X})}\mathbb{R}^3$. This is due to the fact that if the body were a surface or a curve in three dimensional space, then $T_{\mathsf{x}}B_t \subsetneq T_{\mathsf{x}}\mathbb{R}^3$ for every $\mathsf{x} \in B_t$. Since we only deal with cases where the placement of the body at any instant of time is an open subset of $\mathbb{R}^3$, thereby excluding surface-like and line-like bodies from our discussion, this distinction is not important for our purposes. It is nevertheless a good idea to keep this distinction in mind for the sake of conceptual clarity.

The material acceleration of the particle initially at $\mathsf{X} \in B_0$ is defined as the map $\mathsf{A}_{\mathsf{X}}:I \to T\mathbb{R}^3$ such that, for any instant of time $t \in I$, $$\mathsf{A}_{\mathsf{X}}(t) = \frac{\partial \mathsf{V}(\mathsf{X},t)}{\partial t} = \frac{\partial^2 \varphi(\mathsf{X},t)}{\partial t}.$$ As before, we will use the notation $\mathsf{A}(\mathsf{X},t)$ in place of $\mathsf{A}_{\mathsf{X}}(t)$ when required. Given a global coordinate system, the components of the material acceleration can be computed just as in the case of the material velocity.

Remark

Techincally speaking, the material acceleration $\mathsf{A}_{\mathsf{X}}(t)$ of a particle intially at $\mathsf{X} \in B_0$ at time $t$ lives in the linear space $T_{\mathsf{x}}(T_{\mathsf{x}}\mathbb{R}^3)$, where $\mathsf{x} = \varphi_t(\mathsf{X})$. However, since the tangent space at a point is always a linear space, we will use the more relaxed notion that $\mathsf{A}_{\mathsf{X}}(t) \in T_{\mathsf{x}}\mathbb{R}^3$.

Spatial velocity and acceleration

The material version of the velocity and acceleration are quite convenient when studying the behavior of solids. When working with fluids, it is more convenient to focus on the nature of the particles occupying a given point in space at a given instant of time, rather than tracking individual particles (which is often infeasible). Towards this end, we will develop the spatial versions of the velocities and accelerations.

Suppose that the body occupies the region $B_t \subseteq \mathbb{R}^3$ at the instant of time $t \in I$. Let us focus on a point $\mathsf{x} \in B_t$ and denote by $\mathsf{v}(\mathsf{x},t) \in T_{\mathsf{x}}\mathbb{R}^3$ the velocity of the particle that occupies the position $\mathsf{x} \in B_t$ at the instant of time $t$. Since this is nothing but the velocity at time $t$ of the particle that was intially at $\varphi_t^{-1}(\mathsf{x})$, we have the following identity: $$\mathsf{v}(\mathsf{x},t) = \mathsf{V}(\varphi_t^{-1}(\mathsf{x}), t).$$ The spatial velocity of the particle at $\mathsf{x} \in \mathbb{R}^3$ is defined as the map $\mathsf{v}_{\mathsf{x}}:I \to T\mathbb{R}^3$ such that $\mathsf{v}_{\mathsf{x}}(t) = \mathsf{v}(\mathsf{x},t)$, for any $t \in I$.

Remark

By defining, for any $t \in I$, $\mathsf{v}_t:B_t \to T\mathbb{R}^3$ and $\mathsf{V}_t:B_0 \to T\mathbb{R}^3$ in the usual manner, we see that $\mathsf{v}_t = \mathsf{V}_t \circ \varphi_t^{-1}$.

The spatial acceleration, $\mathsf{a}_{\mathsf{x}}:I \to T\mathbb {R}^3$, of the particle at $\mathsf{x} \in \mathbb{R}^3$ is defined similarly as the acceleration of the particle that was initially at $\varphi_t^{-1}(\mathsf{x})$: for any $t \in I$, $$\mathsf{a}_{\mathsf{x}}(t) = \mathsf{A}(\varphi_t^{-1}(\mathsf{x}), t).$$ As before, we will at times use the notation $\mathsf{a}(\mathsf{x},t)$ in place of $\mathsf{a}_{\mathsf{x}}(t)$.

If the observer chooses a global basis $(\mathsf{e}_i)$ for $\mathbb{R}^3$, the components of the spatial velocity can be expressed as $$\mathsf{v}(\mathsf{x},t) = \sum v_i(\mathsf{x},t)\mathsf{e}_i.$$ The coordinate representation of the spatial acceleration is computed likewise.

Material time derivatives

Recall that the material acceleration is the time derivative of the material velocity. Let us now look at the relation between the spatial velocity and the spatial acceleration. For any $\mathsf{X} \in B_0$ and $\mathsf{x} = \varphi_t(\mathsf{X}) \in B_t$, we see that $$\begin{split} \mathsf{a}(\mathsf{x},t) &= \mathsf{A}(\mathsf{X},t)\\ &= \frac{\partial \mathsf{V}(\mathsf{X},t)}{\partial t}\\ &= \frac{\partial \mathsf{v}(\varphi(\mathsf{X},t), t)}{\partial t}\\ &= \frac{\partial \mathsf{v}(\mathsf{x},t)}{\partial t} + \text{grad }\mathsf{v}(\mathsf{x},t) \cdot \frac{\partial \varphi(\mathsf{X},t)}{\partial t}\\ &= \frac{\partial \mathsf{v}(\mathsf{x},t)}{\partial t} + \text{grad }\mathsf{v}(\mathsf{x},t) \cdot \mathsf{v}(\mathsf{x},t). \end{split}$$ Let us now introduce the material time derivative, $D\mathsf{v}(\mathsf{x},t)/Dt$, of the spatial velocity as follows: $$\frac{D\mathsf{v}(\mathsf{x},t)}{Dt} = \frac{\partial \mathsf{v}(\mathsf{x},t)}{\partial t} + \text{grad }\mathsf{v}(\mathsf{x},t) \cdot \mathsf{v}(\mathsf{x},t).$$

Remark

We have introduce the notation $\text{grad }\cdot$ to denote the gradient of a field variable that depends on $\mathsf{x} \in \mathbb{R}^3$. Note that this is different from the gradient operation $\text{GRAD }\cdot$ that is defined only for fields defined on $B_0$.

We thus obtain the following relation between the spatial velocity and the spatial acceleration: $$\mathsf{a}(\mathsf{x},t) = \frac{D\mathsf{v}(\mathsf{x},t)}{Dt}.$$ Unlike the relation between the material velocity and the material acceleration, the spatial acceleration is not the (partial) time derivative of the spatial velocity; in addition to the partial time derivative of the spatial velocity with respect to time, it also has the the convective term $\text{grad }\mathsf{v}(\mathsf{x},t) \cdot \mathsf{v}(\mathsf{x},t)$. The physical intuition behind this expression for the spatial acceleration is that the velocity at a given point in space changes as a consequence of both a local change in velocity, and because of material convected to the considered point due to the motion of the body.

Let us quickly work out the expression for the material time derivative of the velocity according to a global basis $\mathsf{e}_i$ of $\mathbb{R}^3$ chosen by an observer. It follows from a simple calculation that $$a_i(\mathsf{x},t) = \frac{\partial v_i(\mathsf{x},t)}{\partial t} + \sum \frac{\partial v_i(\mathsf{x},t)}{\partial x_j}v_j(\mathsf{x},t).$$

Note

It is important to note that $\text{grad }\mathsf{v} \cdot \mathsf{v} \neq \mathsf{v} \cdot \text{grad }\mathsf{v}$ in general. Many authors write the convective term of the material time derivative of the spatial velocity as $\mathsf{v} \cdot \nabla \mathsf{v}$ in place of what we write as $\text{grad }\mathsf{v} \cdot \mathsf{v}$. To avoid ambiguity, we will always use the notation $\text{grad }\mathsf{v} \cdot \mathsf{v}$.

For completeness, let us also note that the material time derivative of the material velocity is defined simply as the partial time derivative: $$\frac{D\mathsf{V}(\mathsf{X},t)}{Dt} = \frac{\partial \mathsf{V}(\mathsf{X},t)}{\partial t}.$$ Notice how the same term material time derivative is used to describe the time derivative as we follow a particle for both material and spatial velocity fields. Though this terminology can be confusing, it does not cause any problems in practice since the meaning is always evident from the context.

Remark

Many authors use the notation $\dot{\mathsf{v}}(\mathsf{x},t)$ for the material time derivative. We will also adopt this convention at places. In terms of this notation, the relation between the spatial velocity and spatial acceleration can be written succinctly as $\mathsf{a}(\mathsf{x},t) = \dot{\mathsf{v}}(\mathsf{x},t)$. Note that in the case of the material velocity, the dot notation is identical to the partial derivative with respect to time.

The foregoing definition of the material time derivative can be extended to both scalar and tensor fields. For instance, if $\tau(\mathsf{x},t) \in \otimes^k T_{\mathsf{x}}\mathbb{R}^3$ denotes a tensor field of order $k$, then its material time derivative is computed as $$\frac{D\tau(\mathsf{x},t)}{Dt} = \frac{\partial \tau(\mathsf{x},t)}{\partial t} + \text{grad }\tau(\mathsf{x},t) \cdot \mathsf{v}(\mathsf{x},t).$$ The material time derivative of a material tensor field is simply its partial time derivative.

Time derivaties related to deformation

Let us now turn our attention to the time evolution of quantities associated with the deformation of a body in three dimensional space. To keep the discussion simple, only the time derivatives of a few important quantities associated with the deformation gradient $\mathsf{F}(\mathsf{X},t)$ are considered in detail. For the purpose of this discussion, we will use two sets of bases: $(\mathsf{E}_I)$ for quantities associated with the initial placement $B_0$, and $(\mathsf{e}_i)$ for the ambient three dimensional space $\mathbb{R}^3$. Note that these are chosen by a given observer. Note also that these are often identical to each other, and the distinction is retained here purely for the purpose of conceptual clarity.

Let us start with the time evolution of the deformation gradient itself. The material time derivative of the deformation gradient, which is identical to its partial derivative with respect to time, is given in coordinate form as $$\dot{\mathsf{F}}(\mathsf{X},t) = \sum \frac{\partial{F}_{iJ}(\mathsf{X},t)}{\partial t} \mathsf{e}_i \otimes \mathsf{E}_J.$$ It is worth reiterating that $\dot{\mathsf{F}}$ stands for the material time derivative of $\mathsf{F}$.

The material time derivative of the right Cauchy-Green tensor $\mathsf{C}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ is computed similarly: $$\begin{split} \dot{\mathsf{C}}(\mathsf{X},t) &= \frac{\partial}{\partial t}\left(\sum F_{kI}(\mathsf{X},t)F_{kJ}(\mathsf{X},t) \mathsf{E}_I \otimes \mathsf{E}_J\right)\\ &= \sum \left(\frac{\partial F_{kI}(\mathsf{X},t)}{\partial t}F_{kJ}(\mathsf{X},t) + F_{kI}(\mathsf{X},t)\frac{\partial F_{kJ}(\mathsf{X},t)}{\partial t}\right) \mathsf{E}_I \otimes \mathsf{E}_J. \end{split}$$ Noting that $\mathsf{F}^T(\varphi_t(\mathsf{X}),t) = \sum F_{iJ}(\mathsf{X},t) \mathsf{E}_J \otimes \mathsf{e}_i$, we see that $$\dot{\mathsf{F}}^T(\varphi_t(\mathsf{X}),t) = \sum \frac{\partial F_{iJ}(\mathsf{X},t)}{\partial t} \mathsf{E}_J \otimes \mathsf{e}_i.$$ We therefore arrive at the following expression for the material time derivative of the right Cauchy-Green tensor: $$\dot{C}(\mathsf{X},t) = \dot{\mathsf{F}}^T(\varphi_t(\mathsf{X},t)\mathsf{F}(\mathsf{X},t) + \mathsf{F}^T(\varphi_t(\mathsf{X}),t)\dot{\mathsf{F}}(\mathsf{X},t).$$ The Lagrangian rate of deformation tensor $\mathsf{D}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ is defined as $$\mathsf{D}(\mathsf{X},t) = \frac{1}{2}\frac{D\mathsf{C}(\mathsf{X},t)}{Dt}.$$ To understand the physical relevance of this tensor, consider two vectors $\Delta \mathsf{X}, \Delta \mathsf{Y} \in T_{\mathsf{X}}B_0$. The angle between these two vectors is computed using $\Delta\mathsf{X} \cdot \Delta\mathsf{Y}$. As the body deforms in space accoring to a prescribed motion, the inner product of the corresponding deformed vectors also evolves with time. Let us compute the rate at which this change: $$\begin{split} \frac{D(\mathsf{F}(\mathsf{X},t)\Delta\mathsf{X}\cdot\mathsf{F}(\mathsf{X},t)\Delta\mathsf{Y})}{Dt} &= \frac{D(\Delta\mathsf{X}\cdot\mathsf{C}(\mathsf{X},t)\Delta\mathsf{Y})}{Dt}\\ &= 2\Delta\mathsf{X}\cdot\mathsf{D}(\mathsf{X},t)\Delta\mathsf{Y}. \end{split}$$ We thus see that the Lagrangian rate of deformation tensor gives us information about how the angle between two vectors in a given tangent space evolve as a consequence of deformation.

As a final and non-trivial example of computing the material time derivatives related to the deformation, let us compute the material time derivative of the determinant of the deformation gradient, but this time around, in the Eulerian setting. Recall that the Jacobian of the deformation is defined as $J(\mathsf{X},t) = \text{det }\mathsf{F}(\mathsf{X},t)$ for any $\mathsf{X} \in B_0$ and $t \in I$. The spatial version of the Jacobian, which we will denote as $J(\mathsf{x},t)$, where $\mathsf{x} \in B_t$, is defined as $$J(\mathsf{x},t) = \text{det }\mathsf{J}(\varphi_t^{-1}(\mathsf{x}), t).$$ Note that we are using the same symbol $\mathsf{J}$ for both the material and spatial versions of the Jacobian; this is largely done because of conventional practice in the continuum mechanics literature. We will now show that the material time derivative of the Jacobian, when expressed in the Eulerian form, is given by $$\frac{DJ(\mathsf{x},t)}{Dt} = \mathsf{J}(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t).$$ Here, $\text{div }\mathsf{v}(\mathsf{x},t)$ stands for the divergence, with respect to the spatial variable $\mathsf{x}$, of the spatial velocity $\mathsf{v}_t(\mathsf{x})$. This equation will play a key role in deriving various results in the forthcoming sections; it is worthwhile working out the proof of this statement.

To keep the algebra simple, let us focus on the special case of motion in a two dimensional space $\mathbb{R}^2$. The material time derivative of the Jacobian is most easily computed in material form, and using Cartesian coordinates: if $\mathsf{X} = \varphi_t^{-1}(\mathsf{x})$, $$\begin{split} \frac{DJ(\mathsf{x},t)}{Dt} &= \frac{\partial J(\mathsf{X},t)}{\partial t}\\ &= \frac{\partial}{\partial t}\left(\text{det} \begin{bmatrix} F_{11}(\mathsf{X},t) & F_{12}(\mathsf{X},t)\\ F_{21}(\mathsf{X},t) & F_{22}(\mathsf{X},t) \end{bmatrix}\right)\\ &= \text{det}\begin{bmatrix} \dot{F}_{11}(\mathsf{X},t) & \dot{F}_{12}(\mathsf{X},t)\\ F_{21}(\mathsf{X},t) & F_{22}(\mathsf{X},t) \end{bmatrix} + \text{det}\begin{bmatrix} F_{11}(\mathsf{X},t) & F_{12}(\mathsf{X},t)\\ \dot{F}_{21}(\mathsf{X},t) & \dot{F}_{22}(\mathsf{X},t) \end{bmatrix}. \end{split}$$ To compute $\dot{F}_{iJ}(\mathsf{X},t)$ note that $$\begin{split} \dot{F}_{iJ}(\mathsf{X},t) &= \frac{\partial}{\partial t}\frac{\partial \varphi_i(\mathsf{X},t)}{\partial X_J}\\ &= \frac{\partial}{\partial X_J}\frac{\partial \varphi_i(\mathsf{X},t)}{\partial t}\\ &= \frac{\partial V_i(\mathsf{X},t)}{\partial X_J}. \end{split}$$ Using the fact that $\mathsf{V}(\mathsf{X},t) = \mathsf{v}(\varphi(\mathsf{X},t),t)$, we see that $$\begin{split} \dot{F}_{iJ}(\mathsf{X},t) &= \frac{\partial v_i(\varphi(\mathsf{X},t),t)}{\partial X_J}\\ &= \sum \frac{\partial v_i(\varphi(\mathsf{X},t),t)}{\partial x_j}\frac{\partial \varphi_j(\mathsf{X},t)}{\partial X_J}\\ &= \sum \frac{\partial v_i(\mathsf{x},t)}{\partial x_j} F_{jJ}(\mathsf{X},t). \end{split}$$ Subsituting this expression for $\dot{F}_{iJ}(\mathsf{X},t)$ in the expression for $\dot{J}(\mathsf{x},t)$ derived earlier, we get, after a tedious but straightforward expansion, $$\begin{split} \dot{J}(\mathsf{x},t) &= \text{det}\begin{bmatrix} v_{1,1}F_{11} + v_{1,2}F_{21} & v_{1,1}F_{12} + v_{1,2}F_{22}\\ F_{21} & F_{22} \end{bmatrix}\\ &+ \text{det}\begin{bmatrix} F_{11} & F_{12}\\ v_{2,1}F_{11} + v_{1,2}F_{21} & v_{2,1}F_{12} + v_{2,2}F_{22}\\ \end{bmatrix}\\ &= v_{1,1}\,\text{det}\begin{bmatrix}F_{11} & F_{12}\\ F_{21} & F_{22}\end{bmatrix} + v_{2,2}\,\text{det} \begin{bmatrix}F_{11} & F_{12}\\ F_{21} & F_{22}\end{bmatrix}\\ &= \left(\frac{\partial v_1(\mathsf{x},t)}{\partial x_1} + \frac{\partial v_2(\mathsf{x},t)}{\partial x_2}\right) J(\mathsf{x},t)\\ &= J(\mathsf{x},t) \, \text{div }\mathsf{v}(\mathsf{x},t). \end{split}$$ In the foregoing derivation, we have used the shortand notation $v_{i,j}$ to denote $\partial v_i(\mathsf{x},t)/\partial x_j$, and omitted the arguments in some of the intermediate expressions to keep the notation simple.

A similar but more tedious argument can be used to show that the same result holds in three dimensions too. This completes the proof of the important result: $\dot{J}(\mathsf{x},t) = J(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t)$.

Reynolds' transport theorem

To conclude our discussion of time derivatives of various quantities related to the motion and deformation of a body in three dimensional space, let us consider material time derivatives of the form $$\frac{d}{dt} \int_{B_t} f(\mathsf{x},t) \, dv.$$ Here, $f:\mathbb{R}^3 \times I \to \mathbb{R}$ is a scalar field. Note that both the integrand $f(\mathsf{x},t)$ and the domain of integration $B_t = \varphi_t(B_0)$ are functions of time. The reason for studying such integrals will become evident soon. In short, we will see that all the balance principles of continuum mechanics that we will study later on will be expressed primarily in an integral form; the rationale behind this will be explained when we study balance principles in continuum mechanics.

The standard procedure to evaluate the material time derivatives of time dependent integrals of the form presented above is to first pull back the integral to the reference placement using the change of variables formula for integration, and take the partial time derivative of the Lagrangian form of the integral. Let us work this out in detail: $$\begin{split} \frac{d}{dt} \int_{B_t} f(\mathsf{x},t) \, dv &= \frac{d}{dt}\int_{B_0} f(\varphi(\mathsf{X},t),t) J(\mathsf{X},t) \, dV\\ &= \int_{B_0} \frac{\partial}{\partial t}\left(f(\varphi(\mathsf{X},t),t) J(\mathsf{X},t)\right)\, dV\\ &= \int_{B_0} \left(\frac{Df(\mathsf{x},t)}{Dt} + f(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t)\right)J(\mathsf{X},t)\,dV\\ &= \int_{B_t} \left(\frac{Df(\mathsf{x},t)}{Dt} + f(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t)\right) \, dv. \end{split}$$ In deriving the foregoing relation, we have used some of the identities derived in the context of the mass continuity equation. The equation that we just derived, $$\frac{d}{dt} \int_{B_t} f(\mathsf{x},t) \, dv = \int_{B_t} \left(\frac{Df(\mathsf{x},t)}{Dt} + f(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t)\right) \, dv,$$ is called the Reynolds' transport theorem. This will play a key role in our study of continuum mechanics.

Mass continuity equation

As an application of the Reynolds' transport theorem, let us now derive an important physical principle, namely the mass continuity equation, that describes the conservation of mass during the motion of a continuum body.

Suppose that the mass density of the body at the inital instant of time $t = 0$ is given by the scalar field $\rho_0:B_0 \to \mathbb{R}$. This means that if we consider an infinitesimal volume $\Delta V$ about a point $\mathsf{X}\in B_0$, the mass contained within that volume is given by $\rho_0(\mathsf{X})\Delta V$. Let the mass density at time time over the region $B_t = \varphi_t(B_0)$ be given by the scalar field $\rho_t:B_t \to \mathbb{R}$. At any $\mathsf{x} \in B_t$, we will often write $\rho(\mathsf{x},t)$ for $\rho_t(\mathsf{x})$. Suppose that $\Delta v$ is the deformed volume at $\mathsf{x} = \varphi_t(\mathsf{X})$ corresponding to the intial infinitesimal volume $\Delta V$ at $\mathsf{X} \in B_0$. The mass inside the volume $\Delta v$ is exactly identical to the mass inside $\Delta V$ initially since both $\Delta V$ and $\Delta v$ contain exactly the same particles. We therefore have the identity $$\rho_0(\mathsf{X})\Delta V = \rho(\mathsf{x},t) \Delta v.$$ Using the fact that the $\Delta v = J(\mathsf{x},t)\Delta V$, we can write the equation above as $$\rho_0(\mathsf{X}) = J(\mathsf{x},t)\rho(\mathsf{x},t).$$ This equation is known as the mass conservation equation on account of the fact that it relates the initial and current mass densities.

The foregoing equation can be generalized to work for arbitrary volumes. Consider an initial volume $\Omega_0 \subseteq B_0$. Let us study the evolution of the mass contained within this volume, which is $$\int_{\Omega_0} \rho_0(\mathsf{X})\,dV.$$ At a later time $t$, the mass contained the in the deformed volume $\Omega_t = \varphi_t(\Omega_0)$ is given by $$\int_{\Omega_t} \rho(\mathsf{x},t)\,dv.$$ Since we expect this mass to be exactly identical to the mass contained in the initial volume $\Omega_0$, we have the identity $$\int_{\Omega_0} \rho_0(\mathsf{X})\,dV = \int_{\Omega_t} \rho(\mathsf{x},t)\,dv.$$ Using the transformation rules for multiple integrals, it follows immediately that $\rho_0(X) = J(\mathsf{x},t)\rho(\mathsf{x},t)$, which is the mass continuity equation derived earlier.

An equivalent and useful differential restatement of the mass continuity equation is obtained by noting that the integral identity just studied implies that the material time derivative of the mass contained in $\Omega_t$ has to be zero. Thus, $$0 = \frac{d}{dt}\int_{\Omega_t} \rho(\mathsf{x},t)\,dv = \int_{\Omega_t} \left(\frac{D\rho(\mathsf{x},t)}{Dt} + \rho(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t)\right) \, dv.$$ Notice how the Reynolds' transport theorem is used to differentiate under the integral sign. Since the foregoing result holds for an arbitrary volume contained in $\Omega_t$, it follows that the integrand has to be zero pointwise inside $\Omega_t$: $$\frac{D\rho(\mathsf{x},t)}{Dt} + \rho(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t) = 0.$$ This is the differential form of the mass continuity equation. An equivalent way of writing this is by expanding the material derivative: $$\frac{\partial \rho(\mathsf{x},t)}{\partial t} + \text{div}(\rho(\mathsf{x},t)\mathsf{v}(\mathsf{x},t)) = 0.$$ In deriving this equation, we used the identity $\text{div}(\rho \mathsf{v}) = \rho \, \text{div }\mathsf{v} + \text{grad }\rho \cdot \mathsf{v}$, whose validity is easily checked. Both these forms of the mass continuity equation are useful in applications.

An incorrect derivation of the mass continuity equation

An incorrect derivation of the mass continuity equation that is found in some textbooks goes as follows: since $\rho_0(\mathsf{X}) = J(\mathsf{x},t)\rho(\mathsf{x},t)$, and $\dot{\rho}_0(\mathsf{X}) = 0$, we see that $$\begin{split} 0 &= \frac{D(J(\mathsf{x},t)\rho(\mathsf{x},t))}{Dt}\\ &= \dot{J}(\mathsf{x},t) \rho(\mathsf{x},t) + J(\mathsf{x},t) \dot{\rho}(\mathsf{x},t)\\ &= J(\mathsf{x},t) \rho(\mathsf{x},t) \,\text{div }\mathsf{v}(\mathsf{x},t) + J(\mathsf{x},t) \left(\frac{\partial \rho(\mathsf{x},t)}{\partial t} + \text{grad }\rho(\mathsf{x},t) \cdot \mathsf{v}(\mathsf{x},t) \right)\\ &= J(\mathsf{x},t)\left(\frac{\partial \rho(\mathsf{x},t)}{\partial t} + \text{div}(\rho(\mathsf{x},t)\mathsf{v}(\mathsf{x},t)) \right). \end{split}$$ Since the determinant of the deformation gradient is always strictly positive, we arrive at the mass continuity equation. The problem with this derivation is the assumption that $\dot{\rho}_0(\mathsf{X}) = 0$. This is not a mathematical identity that is true because $\rho_0(\mathsf{X})$ is independent of $t$, which is the usual argument given. The derivation of the mass continuity equation via the integral formulation presented earlier is much cleaner and also illuminates the reason why this works: if we follow a small volume around $\mathsf{X}$ over time, the mass contained in this volume does not change. Thus, the mass continuity principle directly gives the identity $d(J(\mathsf{x},t)\rho(\mathsf{x},t))/dt = 0$. This is the proper way to justify the foregoing derivation.

A useful form of Reynolds' transport theorem

In the special case when the integrand is of the form $\rho f$, where $\rho$ is the mass density and $f$ is a scalar field, we can use the mass continuity equation to simplify the material time derivative of the integral as follows: $$\frac{d}{dt}\int_{B_t} \rho(\mathsf{x},t)f(\mathsf{x},t)\,dv = \int_{B_t} \rho(\mathsf{x},t)\frac{Df(\mathsf{x},t)}{Dt}\,dv.$$ It is left as a simple exercise to verify this. This form of the Reynolds' transport theorem is quite useful in applications.