The acceleration of a particle of a continuum; $\frac{\partial v_i}{\partial t}$ vs $\frac{\mathrm{d}v_i}{\mathrm{d}t}$ Assume that the particles of a continuum body $\mathscr{B}$ with mass $\mathscr{M}$ are labeled by vectors $X_i$; then the momentum balance equation reads
$$
F_j = \int_{\mathscr{M}} a_j \mathrm{d}m
$$
where $F_j$ is the applied force, and $a_j$ is acceleration. If $v_j=f(X_i,t)$, where $v_j$ is velocity, and $t$ is time, the question is
$$
\mathbf{Why}  
\ \ a_j=\frac{\mathrm{d}f}{\mathrm{d}{t}}\ \
\mathbf{and\ not}\ \
a_j=\frac{\partial f}{\partial {t}}?
$$
 A: Coordinates: physical and material coordinates
Assume we introduce 2 sets of coordinates:

*

*a set of physical coordinates $\mathbf{x}$

*set of material coordinates $\mathbf{X}$, that can be interpreted as a set of labels associated with each material point.

and assume that these two sets of labels are related by $\mathbf{x}(\mathbf{X},t)$.
Fields
Continuum mechanics usually deals with fields, i.e. functions of space and time representing physical quantities. You can represent the space using either physical or material coordinates,
$f(\mathbf{x},t) = f(\mathbf{x}(\mathbf{X},t),t) = f_0(\mathbf{X},t)$,
using the index $_0$ for functions whose independent variables are material coordinates.
Motion of a material particle
Now, let's focus on the $k$-th material particle labelled with $\mathbf{X}_k$, and evaluate its position, velocity and acceleration as a a function of time.

*

*the position is readily $\mathbf{x}_k(t) = \mathbf{x}(\mathbf{X}_k,t)$;


*the velocity is the time derivative of the position, i.e.
$\mathbf{v}_k(t) = \dfrac{d \mathbf{x}_k}{dt} = \dfrac{\partial \mathbf{x}}{\partial t} \Bigg|_{\mathbf{X}} (\mathbf{X}_k,t) = \mathbf{v}_0(\mathbf{X}_k,t) = \mathbf{v}(\mathbf{x}_k,t)$


*the acceleration is the time derivative of the velocity, i.e. $\mathbf{a}_k(t) = \dfrac{d \mathbf{v}_k}{dt} = \dfrac{\partial \mathbf{v}}{\partial t} \Bigg|_{\mathbf{X}} (\mathbf{X}_k,t)$.
Now, writing the velocity field as a function of the physical coordinates $\mathbf{x}$, $\mathbf{v}(\mathbf{x}(\mathbf{X},t),t) $, and using the law for derivatives of composite functions you get
$\mathbf{a}_k(t) = 
\dfrac{d \mathbf{v}_k}{dt} = 
\dfrac{\partial \mathbf{v}_0}{\partial t} \Bigg|_{\mathbf{X}} (\mathbf{X}_k,t) = 
\dfrac{\partial \mathbf{v}}{\partial t} \Bigg|_{\mathbf{X}} (\mathbf{x}(\mathbf{X}_k,t),t) = 
\left\{ \dfrac{\partial \mathbf{v}}{\partial t} \Bigg|_{\mathbf{x}} (\mathbf{x}_k, t) + \dfrac{\partial \mathbf{x}}{\partial t}\Bigg|_{\mathbf{X}}(\mathbf{X}_k, t) \cdot \dfrac{\partial \mathbf{v}}{\partial \mathbf{x}}(\mathbf{x}_k, t) \right\} =
 \left\{ \dfrac{\partial \mathbf{v}}{\partial t}\Bigg|_{\mathbf{x}} (\mathbf{x}_k, t) + \mathbf{v}_0(\mathbf{X}_k, t) \cdot \nabla \mathbf{v}(\mathbf{x}_k, t) \right\} =
 \left\{ \dfrac{\partial \mathbf{v}}{\partial t} \Bigg|_{\mathbf{x}} (\mathbf{x}_k, t) + \mathbf{v}(\mathbf{x}_k, t) \cdot \nabla \mathbf{v}(\mathbf{x}_k, t) \right\} = \dfrac{D \mathbf{v}}{Dt} (\mathbf{x}_k,t)$,
where I introduced the definition of the material derivative, to write the acceleration of the material $k$-th particle as the material derivative of the velocity field, evaluate in the position of the particle itself, $\mathbf{x}_k$, i.e. $\mathbf{a} = \frac{D\mathbf{v}}{Dt} (\mathbf{x}_k, t)$.
Meaning of the material derivative
The material derivative is the mathematical operator that describe the time evolution of a physical quantity, as perceived by a material particle moving with the medium. The time derivative of physical quantity $f$ (whatever it is, from scalar to tensor fields of any order) perceived by material particles, reads
$\dfrac{D f}{D t} = \left\{  \dfrac{\partial }{\partial t} \Bigg|_{\mathbf{x}} + \mathbf{v}(\mathbf{x}, t) \cdot \nabla  \right\}  f (\mathbf{x},t)$
A: The function $f(\vec{x}, t)$ tells you the velocity of whatever particle happens to be at the position $\vec{x}$ at time $t$. Therefore, $\frac{\partial f(\vec{x},t)}{\partial t}$ calculates the difference in velocities of different particles. This is obviously not physically meaningful.
To calculate the physically meaningful acceleration, you have to calculate the difference in velocities of the same particle. If a particle is initially at spacetime position $(\vec{x},t)$, then its initial velocity is $f(\vec{x},t)$. After a time $dt$, the change in the particle's spacetime position is $(dt, f(\vec{x},t)dt)$. The change in velocity is the dot product of the four-gradient of $f$ and the spacetime displacement vector:
$$\frac{\partial f(\vec{x},t)}{\partial t} dt+ f(\vec{x},t)dt \cdot \nabla f(\vec{x},t)$$
Divide out $dt$ to get the acceleration.
