Convective derivative vs total derivative

Question

I was wondering what is the difference between the convective/material derivative and the total derivative. We were introduced to the notion of material derivative

$$ \frac{D\vec{u}}{Dt}=\frac{\partial \vec{u}}{\partial t}+(\vec u\cdot\vec{\nabla})\vec u $$

in fluid mechanics, but I cannot see how this is any different to the time derivative of a vector field $\vec{u}({\vec{x(t)}},t)$ and just applying the chain rule?

If there is no difference, why introduce new notation of $ \frac{ D\vec{u}}{D t} $ instead of just using the standard notation $\frac{d\vec{u}}{dt} $?

Answer 1

I think you have to be careful when calling the material derivative a total time derivative calculated by applying the chain rule, because it is not just that.

For starters, you need to distinguish between the Lagrangian and Eulerian formalisms in continuum mechanics. The Eulerian formalism is expressed in terms of spacial coordinates $(x,y,z)$ and time. Spatial coordinates here are independent variables that describe fixed positions in space, and they're not a function of time. On the contrary, the Lagrangian formalism is based on following a certain specific fluid element in the flow, whose position is given as $\mathbf X(\mathbf x_0,t)$, where $\mathbf x_0=(x_0,y_0,z_0)$ is the initial position of the fluid element.

Consider a certain flow quantity $q$ expressed in the Lagrangian formalism: $$q=q(\mathbf X(\mathbf x_0,t),t).$$ The total time derivative of $q$, calculated by applying the chain rule is: $$\frac{dq}{dt}=\left(\frac{\partial q}{\partial t}\right)_{\mathbf X=cst}+(\mathbf u\cdot\nabla_\mathbf X)q.$$ Note that the partial derivative with respect to time is calculated at constant $\mathbf X$, and the gradient in the second term at the right hand side is calculated with respect to $\mathbf X$, whereas the material derivative is actually expressed in the Eulerian formalism: $$\frac{Dq}{Dt}=\left(\frac{\partial q}{\partial t}\right)_{\mathbf x=cst}+(\mathbf u\cdot\nabla_\mathbf x)q.$$ where $\mathbf x=(x,y,z)$ is the independent position vector.

So how do we pass from the total time derivative to the material derivative? Well lucky we are; the mapping between $\mathbf X$ and $\mathbf x$ is a very simple one: $$X(\mathbf x_0,t)=x$$ $$Y(\mathbf x_0,t)=y$$ $$Z(\mathbf x_0,t)=z$$ which means that the partial time derivative of $q$ at constant $\mathbf X$ is the same as the partial time derivative at constant $\mathbf x$, and the gradient with respect to $\mathbf X$ is the same as the gradient with respect to $\mathbf x$. Thus, the two expressions become equivalent.

Answer 2

Take a look at the definitions below

Total derivative: $$\frac{d}{dt} = \frac{\partial}{\partial t} + \frac{dx_i}{dt}\frac{\partial}{\partial x_i}$$

Material derivative: $$\frac{D}{Dt} = \frac{\partial}{\partial t} + u_i\frac{\partial}{\partial x_i}$$

Would you agree they are the same if $u_i=\frac{dx_i}{dt}$?

Why people use inconsistent notation for both? I do not know.

In my opinion the total derivative is used most often in mathematics whereas the material derivative is used most often in physics. In physics, $\frac{dx_i}{dt}$ has a clear physical interpretation as the instantaneous velocity. In mathematics, there is not necessarily a physical interpretation and the variable notation may be arbitrary.

The next question is, do you understand what the physical interpretation of the material derivative is?