# Intuition for Material Derivative

I am studying fluid mechanics and am trying to gain a finer intuition for the meaning behind the material derivative. So, firstly this is defined as $(\partial_t + \mathbf{u}\cdot\nabla )f$ where $f$ is some multivariable function and $\mathbf{u}$ is the velocity field. I know the material derivative gives the rate of change of $f$ in a particular fluid element as it flows with the fluid. The first term gives the contribution from the temporal rate of change of f at a specific point, and the second term gives the contribution from the movement of the fluid element itself, and is the directional derivative of f in the direction of the velocity. However, it is stated in my notes that the second term, $\mathbf{u}\cdot\nabla(f)$ is the rate of change of f along some streamline multiplied by the magnitude of $\mathbf{u}$, ie. $\mathbf{u}\cdot\nabla(f)= df/ds \vert{\mathbf{u}}\vert$ where s is the distance along the streamline. I dont understand this. We are following the fluid element along a particle path, why is there some random streamline important? Can someone explain?

• It is not "some" streamline but the one that goes through the point where you are evaluating the derivative. Since particles follow streamlines (for time-independent $\mathbf{u}$), $\mathbf{u} \cdot \nabla f$ is the change of $f$ the way a particle carried by $\mathbf{u}$ experiences it, if $f$ is time-independent also. I think it is good to think of the material derivative if you a looking at a property of the fluid (e.g. entropy, temperature, ..) and of the other derivative when you have an external field, like gravity – user66081 May 5 '17 at 17:35
• Thanks for the reply, you helped my understanding. It would be really helpful if you have an algebraic way of deriving the fact we are talking about: $\mathbf{u}\cdot\nabla(f)= df/ds \vert{\mathbf{u}}\vert$. I believe if we split u into its magnitude and directional unit vector, we can then obtain the magnitude of u multiplied by the dot product between this unit vector and the gradient of f, which somehow can be shown as equal to this thing. – Max Sykes May 6 '17 at 13:45
• I believe this just requires a simple understanding of directional derivative which I unfortunately lack- any explanation appreciated – Max Sykes May 6 '17 at 13:48

This is to answer the question $u \cdot \nabla f = \frac{df}{ds} |u|$, where $u$ is the velocity field. Everything is time-independent (stationary) here.
Let's hypothesize that $\frac{df}{d\tau}$ means the derivative along the stream line. A streamline is an integral line of (i.e. the solution of the ODE, traced out) $X'(\tau) = u(X(\tau))$, and let's say $X(0) = x_0$ is where we evaluate the formula in question. Here, $\tau$ is a fictitious time that parameterizes the streamline (it coincides with the real time if $u$ is stationary). Then $\frac{d f}{d \tau}(x_0)$ is the derivative of the function $h(\tau) := f(X(\tau))$ at $\tau = 0$. By the chain rule and the definitions, $$h'(\tau) = \nabla f(X(\tau)) \cdot X'(\tau) = \nabla f(X(s)) \cdot u(X(s))$$ so that $$\frac{df}{d\tau}(x_0) = h'(0) = \nabla f(x_0) \cdot u(x_0).$$
Ok, that's not what we wanted; the variable $s$ in the formula is apparently not what I called the fictitious time. Let's take $s = \tau |u|$ which has the unit of length. Now we parameterize the streamline by length rather than time. This looks uncomfortable, because $u$ depends on spatial variable; indeed, the streamline is an integral line of the unit-speed $Y'(s) = u(Y(s)) / |u(Y(s))|$ with $Y(0) = x_0$. Let us set $g(s) := f(Y(s))$, and compute the derivative $g'(0)$. We have $$g'(s) = \nabla f(Y(s)) \cdot Y'(s) = \nabla f(Y(s)) \cdot u(Y(s)) / |u(Y(s))|$$ so that $$\frac{df}{ds}(x_0) = g'(0) = \nabla f(x_0) \cdot u(x_0) / |u(x_0)| .$$ Multiply by $|u(x_0)|$ to obtain the original formula.