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Problem 1.11 - free kick

The question seems to address passing the ball around the wall of defenders and so the horizontal deflection of interest seems to be different than the one calculated in the solution - it is the difference between the radius of the ball's trajectory and the distance of the straight trajectory from the center of that circle (the straight trajectory is a chord on that circle of some 1km radius).

Guy Gaziv , Tue, 02/10/2015

Answer: This is true but the difference is small because the curvature radius is much larger than all the other distances.

Problem 1.8

In the solution to the Problem 1.8, it is said that the air density must be small to consider the oscillations quasi-static. That statement follows by the adiabatic law. Adiabaticity presumes fast changes. Is there a contradiction?

Anna Frishman , Tue, 12/02/2014

No, there is no contradiction, but extra explanation is needed indeed (will be added to the second edition). The period of oscillations is the radius of the bubble divided by the speed of sound in the water, as follows from (4.12). That period is supposed to be much larger than the radius of the bubble divided by the speed of sound in the air (which is assured by much lower air density), so that the pressure can be considered uniform inside the bubble. On the other hand, the period is supposed to be much smaller than the heat exchange time,  which is the radius squared divided by the thermal diffusivity. In distinction from the viscosity (see Section 1.4.3), the diffusivity in a liquid requires the transfer of molecules and can be estimated as the speed of sound times the mean free path. Therefore, the heat exchange is negligible when the radius is much larger than the mean free path. That allows us to use the adiabatic law.

Quasi-momentum and velocity potential

Is the relation between the quasi-momentum and the velocity potential on page 30 is valid for compressible flows?

AF, UL , Thu, 12/04/2014

No, it is valid only for incompressible flows. More accurately, the density is supposed to be non-changing during the pressure pulse which brings the body into motion.

Flow past a wing

Why it is claimed in sect 1.5.4 that fluid elements, separated by a wing, do not meet at the trailing edge? Why the upper one travels faster?

AB , Sun, 07/06/2014

Pressure difference P2-P1 between upper and lower sides of a wing is proportional to the wing thickness h, or in other words, to the small parameter h/L, where L is the width. The same is true for the velocity difference since by the Bernoulli theorem

P2-P =ρ(v12-v22 )/2=ρ(v1-v2 )(v1+v2 )/ 22 2 2≈ρ (v1-v2 )u.

Incidentally, this formula gives straightforward derivation of the lift via circulation: ∫(P2-P1 )dx ≈ρu (v1-v2 )dx.

Now, velocity difference is proportional to h/L ,,, ,, while the difference in the path lengths is proportional to h2/L. Therefore, for a slender wing, the upper fluid element reaches the trailing edge before the lower one. That can be see in the photograph in the Additional Material "Flow past a wing".Flow past a wing Flow . for a slender wing  

Zero resistance force

In Sect 11, Landau & Lifshits relate the absense of frag force to the fact that velocity of the potential flow decays too fast at infinity so that no energy flux goes there. Do you agree?

GK , Tue, 04/23/2013

Answer.  It is a little bit more complicated. First, it is not the flux of energy but that of quasi-momentum which determines the force. Second, the flow pattern by itself is not that relevant to the existence of the flux at infinity and force acting on the body. What is important is left-right symmetry of the potential flow and the breakdown of this symmetry by the wake, which provides for a nonzero drag. Of course, wake has a slower decay of velocity at large distances than potential dipole flow, so symmetry breaking and law of decay are related. 

Convection threshold

Is there a simpler way to derive the convection criterium (1.9)?

during Berkeley colloquium , Sun, 03/31/2013

Answer. Yes. It is simply the statement that the gain in potential energy gdz must exceed the decrease in thermal energy  cpdT.

 

Induced mass

Is the induced mass always smaller than the displaced mass (like half for a sphere)?

Weizmann student , Mon, 07/23/2012

Answer.  Not at all. Induced mass can be much larger (for a thin disc moving perpendicular to its plane) and much smaller (for a needle moving "end on") than the displaced mass.

 

Kelvin theorem and angular momentum

Angular momentum is proportional to the velocity circulation for a rotating cylinder. Does it mean that the conservation of the velocity circulation, i.e. Kelvin theorem, is equivalent to the conservation of angular momentum (see Feynman Lectures)?

KR , Wed, 07/18/2012

With all due respect to Feynman, the answer is no, angular momentum has a local density while Kelving theorem is a nonlocal conservation law.

 

Hamiltonian in Lagrangian coordinates

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.

KR , Wed, 05/02/2012

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.

 

Phase and group velocity of surface waves

Is the symbol 

means the same in the figures on pp 94 and 145? 

Anonymous , Mon, 01/23/2012

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.

Pascal Law

Is not this law just a simple consequence of randomness and isotropy of molecular motion?

HM Sher , Tue, 11/01/2011

 

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. 

 

Pitot tube

 

What if the velocity is not directed parallel to the tube and actually the B point is on some other streamline.

 

HM Sher , Tue, 11/01/2011

Answer: Pitot tube measures the velocity on the streamline which comes to the point B.

 

Flow in a curved space

How fluid flows in a curved space?

, Mon, 10/03/2011

Answer: Probably the main difference with the flat space is that Galilean invariance no longer works. See e.g. "Turbulence on hyperbolic plane..". 

 

Fluctuations of the wake boundary

On p. 67, you mention the fluctuations of the wake boundary. What is known about the statistics of these fluctuations?

, Sat, 10/01/2011

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.

 

Turbulence generation by shear

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?

, Thu, 08/18/2011

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.

 

Acoustic pulse in 2d

Section 2.3.1 describes acoustic pulses which propagate withour changing shape in 1d and 3d. What about 2d? 

KR , Sat, 07/16/2011
Answer: Apparently, two-dimensional axially symmetric wave equation, 

does not have a solution of the form
Indeed, one can consider a short pulse of a localized force, which adds δ(t)δ(r) to the rhs of the equation. By the treatment analogous to (4.41)-(4.42) one can obtain that h=0 for r>ct and 

 for r<ct. In other words, the shape changes. See Fluid Mechanics by Landau & Lifshits, Sect. 71 for a general solution.

Generally, any localized perturbation produces large time asymptotics of the form 

 

About drag crisis

What is the explanation for the drag crisis?

, Thu, 02/03/2011

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.

 

 

About cumulative jet, Exercise 1.15

Detonation of explosives actually determines 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?

Anonymous , Sun, 12/19/2010

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. Apparently, the smaller the angle, the larger is the jet speed. In most shells U is of order 2 km/sec while v can be tens of km/sec.