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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} $?

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    $\begingroup$ Different authors use different notation. $\endgroup$
    – Qmechanic
    Commented Apr 19, 2019 at 12:27
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    $\begingroup$ Shouldn't the equation for the convective derivative be $\frac{Du}{Dt}=\frac{\partial{u}}{\partial t}+\vec v\cdot\vec{\nabla} u$ where $\vec v$ is the velocity of the flow and ${u}=u(x,t)$ is the material? And I believe $D/Dt$ implies it's a covariant derivative. $\endgroup$ Commented Nov 3, 2019 at 6:02
  • $\begingroup$ when the v of the frame of reference is the same as the velocity of the fluid then they are both the same. The material derivative is just a special case with a special v $\endgroup$
    – ChemEng
    Commented Nov 4, 2019 at 23:44

2 Answers 2

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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.

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  • $\begingroup$ Great answer. Would it be correct to simply say the relationship is X=x? $\endgroup$ Commented Oct 22, 2020 at 16:31
  • $\begingroup$ @electronpusher Indeed it would. $\endgroup$
    – Tofi
    Commented Oct 26, 2020 at 12:46
  • $\begingroup$ You said that $x,y,z$ are spatial coordinates independent of time but when mapping between Eulerian and Lagrangian coordinates, doesn't it make spatial coordinates the functions of time $x=X(\mathbf x_0,t)$, $y=Y(\mathbf x_0,t)$, $z=Z(\mathbf x_0,t)$? $\endgroup$
    – user366089
    Commented May 8, 2023 at 6:48
  • $\begingroup$ @user366089 The equality should not be read as an identity. $X$, $Y$ and $Z$ are functions of time, and the values of these functions at any point are equal to the spatial coordinates of that point $(x, y, z)$. The spatial coordinates, however, are independent variables and not functions of time. $\endgroup$
    – Tofi
    Commented May 8, 2023 at 7:12
  • $\begingroup$ I don't get how this equality is understood. Even though spatial coordinates are independent of time, they still vary with time? $\endgroup$
    – user366089
    Commented May 8, 2023 at 7:36
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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?

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