Questions and Answers
If Lagrangian description of an ideal fluid is a canonical Hamiltonian formalism (sect 1.3.4), why it cannot be used for statistical theory? Introducing the distribution in the phase space etc.
Answer. Using the canonical structure in Lagrangian variables is not going to be easy. It is quite an unusual formalism with a Hamiltonian that changes its form with time - indeed, the form of how pressure (and enthalpy) depends on the coordinates changes with time. For quasi-momentum in Sect 1.3 it did not matter since it exploits only spatial invariance. For other purposes, it is quite a headache. I am also not sure how to introduce an ensemble to make sense of a phase-space distribution. Which does not mean that using this formalism for statistics is impossible, just nobody did it before to the best of my knowledge.
Is the symbol
means the same in the figures on pp 94 and 145?
Answer: No. On p.95, it is the wavelength that corresponds to the minimum of the phase velocity, which takes place exactly when the phase and group velocities are equal:
On p. 145, it is the wavelength that corresponds to the minimum of the group velocity which is the speed of the caustics.
I apologize for denoting different quantities by the same symbol.
Is not this law just a simple consequence of randomness and isotropy of molecular motion?
Answer: Pascal Law in hydrostatics does not appeal to the molecular motion. In hydrostatics this is just the requirement of the absense of net momentum flux (i.e. mechanical equilibrium). Note that we do not assume isotropy; for instance, fluid can be under gravity and have its density height-dependent. The fact that we can use Pascal law not only in hydrostatics by also in motion is based on the separation of scales between molecular motion and fluid flows. For this law to be true, we need to assume that deviations from thermal equilibrium (which any inhomogeneous flow creates) are small.
What if the velocity is not directed parallel to the tube and actually the B point is on some other streamline.
Answer: Pitot tube measures the velocity on the streamline which comes to the point B.
How fluid flows in a curved space?
Answer: Probably the main difference with the flat space is that Galilean invariance no longer works.
On p. 67, you mention the fluctuations of the wake boundary. What is known about the statistics of these fluctuations?
Answer: Not much, I'm afraid. There are some old experimental data, mentioned at the end of Sect 5.8 of Statistical Hydrodynamics I by Monin & Yaglom. No theory.
Formula (2.4) and Figure 2.5 show that shear dV0x/dz>0 increases the energy of the secondary flow v1 if v1xv1z<0. How turbulence can be sustained by shear if in a random flow streamlines with opposite signs of v1xv1z are equally probably, so on average there is no energy flux from shear to turbulence?
Answer: Turbulence is not isotropic when the mean flow has a shear. It is clear that the mean value of <v1xv1z> must be negative since it is the mean vertical flux of horizontal momentum, which must flow from high- to low-V0 region. It means that in a shear-generated turbulence, more streamlines are oriented as shown in Figure 2.5 than at a right angle to it.
Section 2.3.1 describes acoustic pulses which propagate withour changing shape in 1d and 3d. What about 2d?
for r<ct. In other words, the shape changes. Generally, any localized perturbation produces large time asymptotics of the form
What is the explanation for the drag crisis?
Answer: Drag crisis is a sudden drop of the drag coefficient from 0.5 to 0.15 when Re increases from 15000 to 25000. It was discovered by Eiffel (in 1912) and explained by Prandtl as due to an onset of turbulence inside the boundary layer. Turbulent boundary layer entrains more fluid from outside, has thus more momentum and separates later downstream, which decreases the wake area and the drag.
Detonation of explosives actually determine the compression speed of the cone, that is the velocity U normal to the cone in its reference frame. How to relate this speed to the speed of the cumulative jet?
Answer: All jets have velocity V on their surface. In the reference frame moving with the velocity Vcosα the cone moves normal to itself with the speed Vtgα=U. In this frame the cumulative jet moves with the speed v=V(1+1/cosα)=U(1+cosα)/sinα.