If I understand you correctly, you're basically trying to get a physical understanding of the meaning of the total derivative.
In continuum mechanics, of which both solid and fluid mechanics are subsets, we can think of material as an infinite collection of infinitely small particles. In applying conservation laws, we use either a Lagrangian reference frame, in which we follow and track the material particles throughout their evolution, or we can use an Eulerian reference frame, in which we watch a fixed location in space through which material particles move. To use a (somewhat oversimplified) analogy, if police are trying to catch people speeding on a highway, a Lagrangian reference frame would be akin to following the cars, while an Eulerian reference frame would be akin to standing on the side of the road with a radar gun.
The important thing is that the conservation concepts must apply to each material particle whether or not we are following them directly. That is to say, mass, momentum, and energy must be conserved for each particle, i.e., F=ma, etc., applies at the particle level.
In solid mechanics, a Lagrangian reference frame is typically used, and the use of conservation laws is fairly transparent, if not misleadingly so. In fluid mechanics, however, we typically do not track material particles; instead, we use an Eulerian reference frame, which presents a problem -- how do we apply, among others, F=ma to material particles if we can't track them?
This is where the concept of the total derivative (or rather, the material derivative, as "total derivative" has specific meaning outside that of mechanics) comes in. The material derivative of a quantity represents the change in that quantity that a specific material particle experiences as it moves through space. This is the term that we need to use for conservation laws -- i.e., a = Dv/Dt.
I should note that you have an error in your understanding of how a material derivative is applied -- to avoid confusion, I will use a quantity 'q'. The correct expansion is as follows (I'm just going to type the LaTeX code here and forgo the actual compiling of it -- it should be fairly apparent what I'm doing):
\frac{Dq}{Dt} = \frac{\partial q}{\partial t} + (\mathbf{u} \cdot \nabla) qNote that the (\mathbf{u} \cdot \nabla) term persists, regardless of what you are taking the material derivative of. The reason why will become apparent shortly.
To highlight the differences between Lagrangian and Eulerian reference frames, we need to look at exactly how we track the particles -- this means asking what *coordinate system* we are using to define our equations. Remember that all conservation laws are partial differential equations in space and time coordinates. We have a choice, however, in how we define our space coordinates.
In Lagrangian reference, we track material particles throughout an evolution by looking at everything in relation to their original location, so our spatial coordinates are defined in terms of X_0, Y_0, and Z_0. A material particle at specific X_0, Y_0, and Z_0 coordinates always stays at those coordinates -- the values of those coordinates never change. In Eulerian reference, however, we look at everything in terms of where it is "now." This means that we have x, y, and z coordinates that are independent of the material particles. A particle at a particular (x,y,z) location at one point in time will have different coordinates at another point in time.
Why is this distinction important?
Let's go back to the concept of the total derivative. Looking at this from a partial differential equations standpoint, if we want to know how a quantity associated with a moving particle changes with time, we need to apply the chain rule. If we admit that a particle's spatial coordinates may change with time, then we have:
\frac{D}{Dt} q(x,y,z,t) = \frac{D}{Dt} q(x(t), y(t), z(t), t)
= \frac{\partial q}{\partial x} \frac{\partial x}{\partial t}
+ \frac{\partial q}{\partial y} \frac{\partial y}{\partial t}
+ \frac{\partial q}{\partial z} \frac{\partial z}{\partial t}
+ \frac{\partial q}{\partial t} \frac{\partial t}{\partial t}(The partial derivative of t with respect to itself in the last term is just me being pedantic about the chain rule expansion).
If we look at the first three terms, we see that they contain the partial derivatives of spatial coordinates with respect to time. Those are, by definition, the velocity components with respect to the coordinate system. Simplifying the equation, we can write (using u,v,w, for velocity components -- not my favorite thing to do, but in the absence of actually using LaTeX to make it look pretty, this will have to do):
\frac{Dq}{Dt} = u \frac{\partial q}{\partial x}
+ v \frac{\partial q}{\partial y}
+ w \frac{\partial q}{\partial z}
+ \frac{\partial q}{\partial t}Keep in mind that this is only valid if we are examining the material particle using a coordinate system in which its spatial coordinates can change, i.e., the Eulerian reference frame. If, instead, we use the Lagrangian system, we expand it using the chain rule in the same way:
\frac{D}{Dt} q(X_0,Y_0,Z_0,t) = \frac{D}{Dt} q(X_0(t), Y_0(t), Z_0(t), t)
= \frac{\partial q}{\partial X_0} \frac{\partial X_0}{\partial t}
+ \frac{\partial q}{\partial Y_0} \frac{\partial Y_0}{\partial t}
+ \frac{\partial q}{\partial Z_0} \frac{\partial Z_0}{\partial t}
+ \frac{\partial q}{\partial t} \frac{\partial t}{\partial t}But in this case, X_0, Y_0, and Z_0 are not actually functions of time -- they are constant, so their time derivatives are all *zero*. That gives the (almost deceptively) trivial relationship:
\frac{D}{Dt} q(X_0,Y_0,Z_0,t) = \frac{\partial q}{\partial t}Ok, tangent over -- back to your actual question: the physical interpretation of the terms in the Eulerian material derivative.
The first term as you have written, i.e., the partial derivative with respect to time, refers to how a particular quantity changes at a fixed location in space. Whether or not this term is zero doesn't mean anything in particular for how that quantity varies for any particular material particle. It is saying, say, what is the air temperature at 29 degrees north, 95 degrees west, 3 meters altitude? It is not saying, what is the temperature of the air molecule (ignoring for a moment that the temperature of a single molecule is nonsensical) that exists at that location? If the temperature is changing at a location in space, any material particles that exist at that location will experience that change as long as they are in that location.
The remaining term, (with the nabla), is how a material particle changes *as a result of moving through space*. If the temperature here is steady at 30 C, and the temperature over there is steady at 40 C, then I will experience a temperature change by walking from here to there. Even though the temperature at any given point in space never changes, the material particles experience a change by virtue of the fact that they are moving through space.
The combination of those two changes is the total change in quantity that a material particle experiences, which is what is used in conservation law equations.
Remember, in applying the material derivative expansion to a field quantity, the "u" that is inside the parentheses stays there. The other is what changes. When you apply the expansion, remember that the quantity in the derivative fractions is the same in *every* term.
Hope this helps!
