About the material derivative of a fluid particle's velocity

Question

Given that the acceleration of a fluid particle in a velocity field is the substantial or material derivative of the velocity of that field. And this derivative includes the derivative with respect to space and that with respect to time.So the acceleration of a fluid particle is due to two reasons:

• *it may be convected into a region of higher or lower velocity

• It may have moved into a region of unsteady flow (where the velocity field depends on time)*

My question is that I'm not able to see the difference between the two causes.. Moving into a region of higher or lower velocity, means that the particle has moved in the fluid but that also means that it has moved in time since time is passing by any way. So they don't seem independent to me. And if we supposed that it weren't changing its velocity with respect to its spatial coordinates then this only means that its not moving so the velocity is zero. So it doesn't seem logical to me to separate the partial derivative w.r.t time and that w.r.t space..

Answer 1

To understand how the two terms in the usual expression for the material derivative arise, let's recall what the material derivative is first.

To calculate the material derivative of a quantity $Q$, we imagine to be riding an element (a very small volume) of fluid, and we are equipped with a clock and some device that shows us the current value of $Q$. Let's denote by $\boldsymbol{X}$ the fluid element we're riding on.

Let's emphasize that the velocity of the element of fluid is exactly zero in our reference frame (we're sitting on it!). And we might see nearby elements of fluid getting closer to us (compression), or moving away (expansion), or some of them getting closer while others moving away (shear).

To measure the material derivative of $Q$ at time $t$ – imagine for example to be measuring the temperature – we record the value of $Q$, wait a very short time interval $\Delta t$, and record again the value, and then take the difference $$\frac{Q(\boldsymbol{X},t+ \Delta t) - Q(\boldsymbol{X},t)}{\Delta t}.$$ The material derivative of $Q$ measured at our fluid element $\boldsymbol{X}$ is the limit of the above fraction: $$\dot{Q}(\boldsymbol{X},t) := \lim_{\Delta t \to 0}\frac{Q(\boldsymbol{X},t+ \Delta t) - Q(\boldsymbol{X},t)}{\Delta t} \equiv \frac{\partial Q(\boldsymbol{X},t)}{\partial t}.$$

The last expression shows that the material derivative is just a time derivative taken keeping the position fixed on a particular fluid element. We can also say that it's the time derivative taken in a frame of reference instantaneously at rest with respect to a particular fluid element.

If the quantity $Q$ is a vector, the difference above will be a vector difference, but the definition is unchanged.

It's important to note that if the measurement of the quantity $Q$ does not depend on any frame of reference, then its material derivative doesn't depend on any frame of reference either. This is the case for the temperature or the internal energy, for example. If the measurement of $Q$ does depend on the choice of frame of reference instead, then its material derivative depends on that choice too. This is the case for velocity, for example: velocity with respect to what?

Now let's see how an observer in a different frame, for example a frame fixed with respect to a laboratory's walls, can calculate the material derivative. This new frame of reference doesn't need to be inertial, that is, fixed with respect to the distant stars.

Our new observer must do exactly what the observer riding on the fluid element $\boldsymbol{X}$ does: measure $Q$ at $\boldsymbol{X}$ at time $t$, then again at $\boldsymbol{X}$ at time $t+ \Delta t$, and then take the limit of the ratio above. But $\boldsymbol{X}$ does not stay in a fixed place for our new observer: at time $t$ it will be at position $\boldsymbol{x}$ with respect to his or her frame, and at time $t+\Delta t$ it will be at a new position $\boldsymbol{x}'$. If the time interval is small, this position will be given by $$\boldsymbol{x}' \approx \boldsymbol{x} + \boldsymbol{v}\,\Delta t,$$ where $\boldsymbol{v}$ is the instantaneous velocity the fluid element $\boldsymbol{X}$ has at position $\boldsymbol{x}$ at time $t$ with respect to our new observer.

So our observer must take the limit of this ratio: $$\frac{Q(\boldsymbol{x} + \boldsymbol{v}\,\Delta t,t+ \Delta t) - Q(\boldsymbol{x},t)}{\Delta t}.\tag{*}\label{ratio}$$

Use a Taylor expansion and the derivative of a composite function, $f[g(t+\Delta t),h(t+\Delta t)] \approx f[g(t),h(t)]+\left(\frac{\partial f}{\partial g}\frac{\partial g}{\partial t}+\frac{\partial f}{\partial h}\frac{\partial h}{\partial t}\right)\,\Delta t$, for the first term in the numerator: $$Q(\boldsymbol{x} + \boldsymbol{v}\,\Delta t,t+ \Delta t) \approx Q(\boldsymbol{x},t) + \left[\boldsymbol{v} \cdot\nabla Q(\boldsymbol{x},t)+\frac{\partial Q(\boldsymbol{x},t)}{\partial t}\right]\,\Delta t.$$ Substituting this expansion in the ratio $\eqref{ratio}$, simplifying, and taking the limit, we finally obtain $$\dot{Q}(\boldsymbol{x},t) = \boldsymbol{v} \cdot\nabla Q(\boldsymbol{x},t) + \frac{\partial Q(\boldsymbol{x},t)}{\partial t}.$$

The two terms appear because our observer must keep track of the motion of the fluid element. You can see the two terms also if you adopt a space-time perspective: imagine to trace the trajectory of the fluid element $\boldsymbol{X}$ on a $t$-$\boldsymbol{x}$ diagram. You can calculate the difference between $Q(\boldsymbol{x}',t+\Delta t)$ and $Q(\boldsymbol{x},t)$ in two steps: first move in the $t$ direction keeping $\boldsymbol{x}'$ fixed; this gives you the $\partial/\partial t$ term. Then move in the $\boldsymbol{x}$ direction keeping $t$ fixed. This gives you the $\boldsymbol{v}\cdot\nabla$ term. This decomposion depends on the frame we're using to describe the motion.

So it is as you say:

the two terms are not independent

and their separation is arbitrary as much as a choice of frame is arbitrary. This is why I personally don't like the explanation of acceleration in terms of this or that reason: I can always produce an arbitrary acceleration by arbitrarily changing my frame of reference. And if we're speaking of the acceleration with respect to the fixed stars, Newton's second law simply says that it's due to the total forces acting on the fluid element, be they pressure-like, viscous, or external (like the force of the wind on the ocean's surface). Convection, unsteady flow, etc. are also effects due to such forces. I don't find it useful to use effects to explain other effects, better to use the causes.

The above explanation is very pictorial. If you want to give it some more rigour, consider the description of the motion of bodies commonly done in continuum mechanics, rather than the one specialized to fluid mechanics:

Given a body, which could be a mass of fluid or a piece of material, we imagine to label each of its elements with a coordinate $\boldsymbol{X}$. The body is thus represented by an abstract differential manifold: it has topological and differential properties, but no metric ones, like a sort of abstract blob. Each point of this blob occupies, at each time $t$, a point in space, which we identify by a coordinate system in some particular frame. In other words we are considering the map $$(\boldsymbol{X},t) \mapsto \boldsymbol{x}(\boldsymbol{X},t).$$ For each $t$, this map is one-to-one between $\boldsymbol{X}$ and $\boldsymbol{x}$, so we have also $$(\boldsymbol{x},t) \mapsto \boldsymbol{X}(\boldsymbol{x},t).$$

The instantaneous velocity of the element $\boldsymbol{X}$ at time $t$ in that frame and coordinate system is defined as $$\boldsymbol{v}(\boldsymbol{X},t) := \frac{\partial \boldsymbol{x}(\boldsymbol{X},t)}{\partial t},$$ which we can also refer to $\boldsymbol{x}$ by a change of variables in the argument: $\boldsymbol{v}(\boldsymbol{x},t) = \boldsymbol{v}[\boldsymbol{X}(\boldsymbol{x},t),t].$

The material derivative is defined as the time derivative of the velocity with respect to the manifold of the body: $$\dot{\boldsymbol{v}}(\boldsymbol{X},t) := \frac{\partial \boldsymbol{v}(\boldsymbol{X},t)}{\partial t},$$ and when we express it in terms of the coordinate and frame $\boldsymbol{x}$ we obtain the two usual terms because of the derivative of a composite function, as explained above. See the references below for this point of view.

Three final remarks:

The first is important when $Q$ is the velocity $\boldsymbol{V}$ of the fluid element with respect to some reference frame. You can calculate the material derivative of this velocity in the same frame this velocity refers to, as is usually done; this will be $$\frac{\partial\boldsymbol{V}}{\partial t} + \boldsymbol{V}\cdot\nabla\boldsymbol{V}.$$ But you could very well calculate the material derivative of this velocity $\boldsymbol{V}$ in a different frame than that this velocity refers to – even this is exceedingly rare (more common in general relativity). In this other frame the fluid element will have velocity $\boldsymbol{v}$, and the material derivative will be $$\frac{\partial\boldsymbol{V}}{\partial t} + \boldsymbol{v}\cdot\nabla\boldsymbol{V};$$ note the two different velocities in the second term. I hope this doesn't confuse you, but it reminds us that the two velocities that appear in the $\nabla$ term have slightly different origins and meaning.

The second remark is that an expression like "$\partial/\partial t$" doesn't mean anything if we don't specify which frame of reference we're using: in a partial time derivative we keep the position fixed; but fixed with respect to which frame?

Final remark: the formula of the material derivative is also connected to time variation of the integral of an extensive quantity over a region of the fluid; it's called the Reynolds transport theorem. For this and a mathematically more rigorous analysis of the material derivative see for example

• I. Samohýl, M. Pekař (2014): The Thermodynamics of Linear Fluids and Fluid Mixtures (Springer), § 3.1,

• C. A. Truesdell (1991): A First Course in Rational Continuum Mechanics. Vol. 1: General Concepts (2nd ed., Academic Press), § II.6.

Answer 2

My question is that I'm not able to see the difference between the two causes.

The material derivative combines temporal and spatial changes in a quantity. If the quantity in question is velocity, then the first term, $\partial \mathbf{V} / \partial t$, defines temporal changes in velocity but not spatial.

A specific example that nicely illustrates the difference would be for shallow water waves in the runup region (i.e., where there depth decreases near shore). In this case, the second term, $\mathbf{V} \cdot \nabla \mathbf{V}$, describes two things: the spatial dependence of the wave speed and the steepening of the wave front as it propagates (e.g., see discussion at https://physics.stackexchange.com/a/139436/59023).

Side Note: Although water waves are probably the most well recognized wave in nature, they can be horribly complicated as they can exhibit things like shoaling, dispersion, breaking, etc. In short, water waves can exhibit nearly all the possible complications one could imagine for oscillations in a neutral, non-relativistic fluid which makes them both fascinating and messy.

So they don't seem independent to me. And if we supposed that it weren't changing its velocity with respect to its spatial coordinates then this only means that its not moving so the velocity is zero. So it doesn't seem logical to me to separate the partial derivative w.r.t time and that w.r.t space.

Let's take a step back and take the linearized limit of the velocity, i.e., let $\mathbf{V} \rightarrow \mathbf{V}_{o} + \delta \mathbf{V}$, where $Q_{o}$ is a quasi-static quantity and $\delta Q$ is a fluctuating quantity such that $\langle Q_{o} \rangle = 0$. Then we assume that $\delta \mathbf{V} \propto e^{i \left( \mathbf{k} \cdot \mathbf{x} - \omega \ t \right)}$ which allows us to rewrite the derivatives in terms of $\mathbf{k}$ and $\omega$. That is, we can show that $\partial / \partial t \rightarrow -i \ \omega$ and $\nabla \rightarrow i \ \mathbf{k}$. I discuss this in more detail at https://physics.stackexchange.com/a/225067/59023 .

Now we see that the material derivative can be rewritten as $\left( -i \ \omega + i \mathbf{V} \cdot \mathbf{k} \right)$. The point is that some modes grow in space but not time, some grow in time but not space, some do both, and some do neither. Yes, it is a little confusing since if the wave propagates and only grows spatially, then it seems arbitrary that one says it does not grow in time given that at a later time it may have a larger amplitude. So perhaps the fault is in the phrasing. It's not really that it "grows in time/space" so much as the growth depends upon time or space.

Side Note: If the imaginary part of $\omega$ is finite, then there will be a temporal growth in the quantity. If the imaginary part of $\mathbf{k}$ is finite, then there will be a spatial growth in the quantity (e.g., similar to the runup region for water waves propagating toward a shoreline).

Answer 3

Take a fixed point in space where you measure the velocity of whatever particle happens to be under that spot. If the velocity measured does not change, then it is said we have steady flow, and the time based acceleration is zero.

The material acceleration may not be zero because on a neighboring point the velocity vector has changed in magnitude or direction. This is the first case in your question.

On the contrary, with unsteady flow the measured velocity at a fixed location changes over time. This is the second case in your question.