How does the chain rule work in sound wave analysis using fluid mechanics? $\tfrac{d x}{dt}\neq v$?

Question

Context: I am reading Landau & Lifshitz's book on Fluid mechanics. Specifically its section on Sound waves.

In section 101, the book's authors discuss about nonlinear traveling waves in one dimension. In this section they are assuming that $v=v(\varrho)$ and $p=p(\varrho)$ where $v$ is the velocity of the fluid particles and $p, \varrho$ are perturbations due to the sound wave. They begin with the usual Euler's system of equations

$$ \frac{\partial\varrho}{\partial t} + \frac{\partial (\varrho v)}{\partial x}=0,\qquad \frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} + \frac{1}{\varrho} \frac{\partial p}{\partial x}=0. $$

Using the chain rule I arrived at $$\tfrac{\partial p}{\partial x}=\frac{d p}{d\varrho} \frac{\partial \varrho}{\partial x}= \frac{d p}{d v}\frac{d v}{d\varrho} \frac{\partial v}{\partial x}= \frac{d p}{d v} \frac{\partial v}{\partial x}$$ where the last equality comes from the assumption that the velocity depends on the density. Using this, I arrived at another representation of the momentum equation

$$ \frac{\partial v}{\partial t} + \left(v+ \frac{1}{\varrho}\frac{d p}{d v}\right)\frac{\partial v}{\partial x}=0 $$

The author's then used this result to arrive to:

$$ \left(\frac{\partial x}{\partial t} \right)_v = v + \frac{1}{\varrho} \frac{d v}{d\varrho} $$

I arrived to this result using

$$ \frac{d v}{dt} = \frac{\partial v}{\partial x} \frac{d x}{dt} + \frac{\partial v}{\partial t}=0 $$ Comparing this equation with the alternative version of the momentum's equation give the result we want. The authors are using partial derivative but I suppose that it is "the same" as long as we understand that this expression for $\tfrac{dx}{dt}$ is valid for constant v.

MY QUESTION: Why is $\frac{dx}{dt}\neq v$?. If I just closed my eyes to this I think that the equation is saying that all points with the same velocity are having their local velocity affected by the presence of the acoustic wave. That makes sense to me, but I cannot attain a good intuition on what $\frac{dx}{dt}$ means in this context and why is it different from the "particle velocity".

I read In what frame of reference are the Euler and Lagrange time derivatives taken in? but I don't fully grasp what they say.

Answer 1

L&L are deriving the simple wave solution here. If you have a function, say, $f(x,t)$ and then define a curve $x=\varphi(t)$ in the $x$-$t$ plane, then the derivative of that function along this particular curve is given by

$$\left(\frac{df}{dt}\right)_\varphi = \frac{\partial f}{\partial t} + \left(\frac{\partial x}{\partial t}\right)_\varphi \frac{\partial f}{\partial x}$$

where $(\partial x/\partial t)_\varphi$ just quantifies the slope of the chosen curve in the $x$-$t$ particle.

By analogy, the quantity $(\partial x/\partial t)_v$ that L&L uses is the derivative of the curve $x=v(t)$ or more appropriately on a curve on which $v$ is a constant. The fact that

$$\left(\frac{dv}{dt}\right)_v=0$$

is then obvious; you are taking derivate of a quantity on a curve upon which it is not changing. The $(\partial x/\partial t)_\rho$ can also be interpreted in the same way.

These are not Lagrangian derivatives of the function $x=x(t)$ for a given fluid particle, say labelled by its initial location $x=x_0$, that experiences a variable velocity field $v=v(x,t)$ as it evolves in time by the equation

$$\frac{dx}{dt}=v(x,t), \quad x_0=x(0).$$