fluid dynamics - Conserved quantities and total derivatives?

I am having a bit of a crisis in understanding of the physical meanings of total derivatives.

When a quantity $\rho$ (be it a vector or a scalar) is said to be conserved, then (mathematically)

$$\frac{d\rho}{dt} = 0$$ (right??)

Now, if I just integrate both sides with respect to time I get $\rho$ = constant.

But the total derivative can be written as (via the chain rule)

$$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \mathbf{u}\cdot\nabla\rho$$ where I am not even sure what $\mathbf{u}$ is (the speed of a moving reference frame?).

Given these equations, if the former has $\rho$ = constant as a solution, then what is the point of partial derivative and gradients? They are all going to be zero anyway?

And what exactly is the physical meaning of the total derivative? In fluid dynamics and plasma physics I have been told that it describes how a quantity changes when observed from the frame "moving with the fluid"...

Answer

It seems to me that you are confonding a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative).

Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from $\mathbb R^3$ to $\mathbb R^3$:

$$x=x(t,y)\:, \quad x \in \mathbb R^3\:.$$ For every $t\in \mathbb R$, the map $\mathbb R^3 \ni y \mapsto x(t,y) \in \mathbb R^3$ associates the particle with initial position $y$ with the position $x$ of the same particle at time $t$.

For every fixed $t$, the map above is supposed to be inverible with differentiable (smooth) inverse, and the two maps $\mathbb R^3 \ni y \mapsto x(t,y) \in \mathbb R^3$ and $\mathbb R^3 \ni x \mapsto y(t,x) \in \mathbb R^3$ are jointly smooth in all variables simultaneously.

At time $t$, the velocity of the particle with initial position $y$ is consequently given by:

$$v_L(t,y) = \frac{\partial x}{\partial t}\:.$$ The formula above pictures the field of velocities in the so called Lagrangian description: physically relevant quantities, at a given value of time, are viewed as functions of the initial position of the particles forming the continuous body. It is however often more convenient to describe this velocity as a function of the position in space at time $t$. We obtain this way the so called Eulerian description of the field of velocities: $$v(t,x) := v_L(t,y(t,x))\:.$$ (I will not make use on an index $E$, henceforth assuming that what it is not Lagrangian is automatically Eulerian.) $x$ is a given point in space at time $t$. At that time it is crossed by the particle with initial position $y(t,x)$.

Generally speaking, if $S$ is a (Cartesian) tensor field defined on the continuous body, we have two representations. The Lagrangian one $S_L(t,y)$ and the Eulerian one $S(t,x)$, where, obviously,

$$S(t,x) = S_L(t,y(t,x))\quad \mbox{and}\quad S_L(t,y)= S(t,x(t,y))\:.$$ It is sometime convenient to compute the derivative of $S$ along the stories of the particles of continuous body while representing the obtained tensor field in Eulerian picture.

For instance, the acceleration field is

$$a(t,x)= \left.\frac{\partial }{\partial t} v_L(t,y)\right|_{y=y(t,x)}\:.$$ Notice that the time derivative is correctly computed in Lagrangian representation, since we have to follow each particle separately. The RHS of the identity above reads, only exploiting Eulerian objects: $$a(t,x) = \frac{\partial v}{\partial t} + v \cdot \nabla_x v(t,x)\:.$$ (It is an easy exercise to establish that identity from definitions.)

In general the Lagrangian derivative of a tensor field (represented in Eulerian picture) $S(t,x)$ is defined as:

$$\frac{DS}{Dt} := \frac{\partial S}{\partial t} + v(t,x)\cdot \nabla_x S(t,x)\qquad(1)$$ It turs out that: $$\left.\frac{DS}{Dt}(t,x)\right|_{x=x(t,y)}= \frac{\partial}{\partial t} S_L(t,y)\:,\qquad (2)$$ so that, the Lagrangian derivative is just a way to compute time derivatives along the stories of the particles of the continuous body remaining in Eulerian representation.

Remark. This Lagrangian derivative is the one you call total derivative, but its physical meaning is very precise as I illustrated above. In the rest of this post I keep denoting it by $D/Dt$ instead of $d/dt$.

Let us come to the conservation law of the mass. First of all, we have to endow our continuous system with a density of mass $\rho$. As before, we can adopt either an Eulerian or Lagrangian description:

$$\rho= \rho(t,x)\qquad \mbox{and}\quad \rho_L= \rho_L(t,y):= \rho(t,y(t,x))\:.$$ The most clear statement concerning conservation of mass is the following.

For every (sufficiently regular) portion $V_L$ of the initial configuration of continuous body, corresponding to a portion $V_t$ at each value $t$ of time,

$$V_t = \{x \in \mathbb R^3 \:|\: x = x(t,y)\:, y \in V_L\}\:,$$ the mass included in $V_t$ does not change in time: $$\frac{d}{dt} \int_{V_t} \rho(t,x) dx =0\:,\qquad(3)$$

Let us transform this requirement into the equivalent local statement. Passing to the Lagrangian picture, (1) reads:

$$\frac{d}{dt} \int_{V_L} \rho(t,x(t,y)) |J_t| dy =0$$ that is $$\frac{d}{dt} \int_{V_L} \rho_L(t,y) |J_t| dy =0$$ where $J_t$ is the determinant of the Jacobian matrix of elements $\frac{\partial x_i}{\partial y_j}$. Since in (2) $V_L$ does not depend on time and everything is regular (the integrand is smooth and $V_L$ can always be assumed to be bounded) we can swap the symbol of derivative and that of integral (essentially taking advantage of Lebesgue's dominated convergence theorem): $$\int_{V_L} \left(\frac{\partial}{\partial t}\rho_L(t,y)\right) |J_t| + \rho_L(t,y) \frac{\partial}{\partial t} |J_t| dy =0\:.$$ It is possible to prove (it is not so simple actually), that $\frac{\partial}{\partial t} |J_t| = |J_t|\nabla_x \cdot v(t,x)|_{x=x(t,y)}$. Using it in the LHS of the found identity and taking (1) and (2), into account, we find that (3) implies (in fact is equivalent to): $$\int_{V_L} \left( \left.\frac{D\rho}{Dt}\right|_{x=x(t,y)}+\rho_L(t,y) \left.\nabla_x \cdot v(t,x)\right|_{x=x(t,y)}\right) |J_t| dy =0\:.\qquad (4)$$ In other words, coming back to Eulerian variables: $$\int_{V_t} \frac{D\rho}{Dt}+\rho(t,x) \nabla_x \cdot v(t,x) dx =0\:.\qquad(5)$$ Form (4) or (5), using the fact that the integrand is continuous and $V_L$ or $V_t$ substantially are arbitrary, ones infers that the integral version of the law of mass conservation (3) is equivalent to the local requirement in Eulerian formulation: $$\frac{D\rho}{Dt}+\rho(t,x) \nabla_x \cdot v(t,x) = 0 \qquad (6)\:.$$ Finally, makig use of the definition of Lagrangian derivative (1), that identity can equivalently be stated as: $$\frac{\partial\rho}{\partial t}+ \nabla_x \cdot \rho(t,x) v(t,x) = 0 \qquad (7)\:.$$

Remark. A continuous body is incompressible if the volume measure $V_t$ of its portions $V_L$ remains constant along its story:

$$\frac{d}{dt} \int_{V_t} dx =0\:.\qquad(3)'$$ Dealing with as before, one immediately sees that it is completely equivalent to say that (it would be enough to everywhere replace $\rho$ for $1$) $$\nabla_x \cdot v(t,x) = 0 \qquad (6)'\:.$$ Consequently, the law of conservation of mass for an incompressible continuous body, from (6),symply reads: $$\frac{D\rho}{Dt}=0\:. \qquad (8)$$ This is the equation you pointed out in the case of the law of conservation of mass. However, with that interpretation, it holds for incompressible bodies only, the general formulation being (6)'.

For a generic quantity $\rho$ (even vectorial or tensorial), (8) simply says that the quantity is constant in time along the story of every particle of the system, though that constant may depend on the particle.

For the sake of completeness, let me say a few words about another popular formulation of the conservation mass law. Starting from (7), integrating both sides in a geometric volume $U$ (so, not a portion of continuous body, but a geometric volume at rest with the reference frame we are using), we have:

$$\int_U \frac{\partial \rho}{\partial t} dx = - \int_U \nabla_x \cdot \rho v(t,x) dx\:.$$ Thus, divergence theorem leads to the most popular version of the considered law whose meaning is illustrated in other answers to your question, so I will not spend any word on it: $$\frac{d}{dt}\int_U \rho(t,x) dx = - \int_{+\partial U} \rho v(t,x)\cdot n dS(x)\:,\qquad (9)$$ where $n$ is the outward unit vector at $x\in \partial U$.

As a final remark regarding the theory of continuous bodies, let me stress that the notion of Lagrangian derivative plays a crucial role in developing the theory of continuous bodies. For instance "$F=ma$" has to be written, for every portion $V_L$ of continuous body, exploiting the Lagrangian derivative:

$$\frac{d}{dt}\int_{V_t} \rho(t,x) v(t,x) dx = F_{V_t}$$
that is, with some elementary manipulations, taking (6) into account, $$\int_{V_t} \rho(t,x) \frac{Dv}{Dt}(t,x) dx = F_{V_t}\:.$$

Remark. Dealing with Hamiltonian mechanics, there is a similar conservation law concerning probability when one studies statistical ensembles, i.e. statistical (Hamiltonian) mechanics. In that situation, in fact, Liouville's theorem establishes that the Hamiltonian evolution preserves the canonical volume of the space of phases. In other words (6)' holds true, where $v$ is the field of Hamiltonian velocities $(dq/dt,dp/dq)$. As a consequence the law of conservation of probabilities, described by the Liouville density $\rho$, can be stated in the simpler version:

$$\frac{D\rho}{Dt}=0\:,$$ which, in turn, adopting the standard notation of statistical Hamiltonian mechanics, reads: $$\frac{d\rho}{dt}=0\:,$$ and, eventually, can be re-formulated into the celebrated Liouville equation form: $$\frac{\partial \rho}{\partial t} + \{\rho , H\}=0\:.$$