In the Eulerian description of velocity field, suppose $x,y,z$ are fixed coordinates, the velocity at that point at time $t$ is $\mathbf{u}(x,y,z,t)$.

I am confused whether $x,y,z$ depend on time or not. Does the word "fixed" indicate time independence? It seems like $x,y,z$ is time-dependent due to its material derivative:

$$\frac{D\mathbf{u}}{Dt}=\frac{\partial \mathbf{u}}{\partial t}+\frac{\partial \mathbf{u}}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial \mathbf{u}}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial \mathbf{u}}{\partial z}\frac{\partial z}{\partial t}$$

How come $x,y,z$ depends on time?

How do I interpret the term "fixed" in this context? Does it mean



$$\frac{\partial x}{\partial t}\ne0 \quad \frac{\partial y}{\partial t}\ne0 \quad \frac{\partial z}{\partial t}\ne0$$

Also I am not sure how to interpret $$\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}, \frac{\partial z}{\partial t}$$

  • $\begingroup$ Seems related: physics.stackexchange.com/questions/594895/… Although I'm not sure it contains the answer to your question.. you may want to give this thread a look. $\endgroup$
    – Amit
    May 6 at 21:30
  • 3
    $\begingroup$ Very good question! If you were my student, I would give you A just for this question! Every time I wait students ask me this, and nobody asks. It is very difficult to give an intelligible explanation to this question. I believe, you're using "cubes" (elementary control volumes) to derive the Navier-Stoke's (and, probably, using White's book). That way of derivation is simplified and leads to the confusion you brought up. The real way of the Navier-Stoke's derivation is different and doesn't lead to this confusion. $\endgroup$ May 6 at 23:38
  • $\begingroup$ In this context, "fixed" means "specified." $\endgroup$ May 7 at 10:37
  • $\begingroup$ My answer to a similar question: physics.stackexchange.com/a/511999/111055 $\endgroup$
    – Tofi
    May 8 at 5:40

1 Answer 1


In the Eulerian specification we consider a field $\vec u(\vec r, t)$ of the velocity of the fluid at certain spatial and temporal coordinates. That is, $\vec r$ and $t$ are independent variables, of which $\vec u$ depends, so by definition of the partial derivative, $\partial_t \vec r = 0$ (and also $\nabla t = \vec 0$).

The (non-partial) derivatives like $\frac{dx}{dt}$ are the same as the partial derivatives, since $x$ is just a coordinate (and does not have any additional "hidden" dependency on $t$).

In general, the total derivatives only differ from the partial ones, if we take them of some function that has hidden dependencies on the coordinate with respect to which we differentiate. There are no such dependencies in the case of the Eulerian view where the coordinates are just coordinates instead positions of fluid parcels in time. (In contrast, remember how in, e.g., Hamiltonian mechanics, we have a function where one of the arguments has a dependency on time: $f(x(t), t)$, then $d_t f = (\partial_t x) \partial_x f + \partial_t f$.)

Finally, your material derviate looks wrong, or like it somehow mixes up the Lagrangian and Eulerian view (perhaps your $x, y, z$ in the expression are not the coordinates of the Eulerian view, but rather the components of the Lagrangian description?).

The material derivative in the Eulerian view is given by $$ \frac{D f}{Dt} = \partial_t f + (\vec u \cdot \nabla) f. $$ As you can see, there are no time derivatives in the second term.

The idea of the material derivative, is that it gives the change along the flow, so you have a term from the intrinsic change of the quantity with time plus a term due to the change of the position of the fluid parcel (the convective term).

Do understand this, we look at the positions $\vec R(\vec r_0, t)$ of the fluid parcels that are at $\vec r_0$ at $t = t_0$ and at a function $f(\vec R(t), t)$ – in other words we change to the Lagrangian view. The very definition of the velocity field is, that $\vec u(\vec r_0, t_0) = \partial_t \vec R(\vec r_0, t_0)$.

The derivative $$ \frac{d}{dt} f(\vec R(t), t) $$ will be the change of the quantity along the flow.

We can now express this in terms of $\vec r_0$ and $\vec u$ at $t = t_0$ via the chain rule: $$ \left. \frac{d}{dt} f\bigl(\vec R(\vec r_0, t), t\bigr) \right|_{t=t_0} = \bigl(\partial_t \vec R(\vec r_0, t_0)\bigr) \cdot \nabla f(\vec r_0, t_0) + \partial_t f(\vec r_0, t_0) = \big(\vec u(\vec r_0, t_0) \cdot \nabla\big) f(\vec r_0, t_0) + \partial_t f(\vec r_0, t_0) := \frac{Df(\vec r_0, t_0)}{dt} $$

In other words, the material derivative, is just the expression for the derivative of a quantity along the flow – which is a natural derivative in the Lagrangian view – as expressed in the Eulerian view.

  • $\begingroup$ Thanks but shouldn't $\vec u(\vec r_0, t) = \partial_t \vec R(\vec r_0, t)$ be $\vec u(R(\vec r_0, t), t) = \partial_t \vec R(\vec r_0, t)$ according to Wikipedia? $\endgroup$
    – user366089
    May 6 at 23:16
  • $\begingroup$ Yes, that's a typo by me – it should have been $\vec u(\vec r_0, t_0) = \partial_t \vec R(\vec r_0, t_0)$ (which is your equation at $t = t_0$ – and I only need it at that point in time later on). $\endgroup$ May 6 at 23:27
  • $\begingroup$ (In the sense, that I use the Lagrangian view with $t = t_0$ as reference point – but since $t_0$ is arbitrary, it works for all times.) $\endgroup$ May 6 at 23:29
  • $\begingroup$ I mean are you sure that it is $\vec u(\vec r_0, t_0) = \partial_t \vec R(\vec r_0, t_0)$, not $\vec u(R(\vec r_0, t_0), t_0) = \partial_t \vec R(\vec r_0, t_0)$? Wikipedia page I linked wrote $\mathbf{u}\left(\mathbf{X}(\mathbf{x}_0,t), t \right) = \frac{\partial \mathbf{X}}{\partial t} \left(\mathbf{x}_0,t \right)$. $\endgroup$
    – user366089
    May 6 at 23:59
  • $\begingroup$ Both expressions are the same, because I choose the label cooridinates $\vec r_0$ by the initial condition $\vec R(\vec r_0, t_0) = \vec r_0$, then $\vec R(\vec r_0, t)$ is the path of a parcel that's at $\vec r_0$ at the reference time $t_0$. $\endgroup$ May 7 at 10:48

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