• Viscous scale of turbulence (misprint on p. 63)
    The viscous scale is defined by the requirement that the Reynolds number is of order unity:

    The kinematic viscosity is the mean free path times the thermal molecular velocity (comparable to the speed of sound). Therefore, the viscous scale is much larger than the mean free path as long as the velocity difference on the viscous scale is much smaller than the speed of sound. This is practically always the case for turbulence in liquids and gases. For example, in atmosheric turbulence, the external scale L can be in hundreds of meteres with velocity v(L) in tens of meters per second, but the viscous scale l is usually less than a millimiter so that 

  • Solution of the Problem 3.7

    In the middle of p. 151 it must be (3.29) instead of (3.5). Note also that not all features of the phase portrait are shown in the left panel of Fig. 4.3.

  • Solution of the problem 1.11

    Since Omega was given as 10 revolutions per second then the angular frequency is 2 pi times 10. To get the estimates for the force, accelerations etc, one needs extra factor 2 pi ≈ 6.3.
     

  • Solution 1.14

    In the Basic Solution the formula for the explosive velocity growth is missing:v(t)=v(0)/(1-r0Agt/B)2.
     

  • Section 1.3.2 Moving sphere

    There is a sign "-" missing in the formula for the potential in the figure on p. 23 and in the formula for the force (1.25). On p. 24, the last term in pressure must have 1/2 factor and the next sentence must be "the force is minus the pressure integral over surface".
     

  • Problem 1.15 in the Russian edition

    The frame speed must be V/\cos\theta. The velocity at infinity is V rather than u.

  • Potential vorticity on p. 14

    In the last derivation on p. 14 the last term in the first and second lines must be multiplied by omega:


     

  • Pint and point

    On page 11 must be "point" instead of "pint".
     

  • Momentum and mass flux of a propagating sound wave

    Nonzero momentum does not necessarily mean net flow. The statement to the contrary originates from Sect 65 of Landau&Lifshits, it incorrect and contradicts, in particular,  the explicit solution of Problem 4 in Sect.~101 there. Fluid particles exchange momentum, which thus can be transported without mass transfer. One can generate a wave (say, by moving piston back and forth) without producing any net flow. In this case, every fluid particle oscillate in the wave and returns to its original position after the wave train has passed by it. The time average of the mass flux must be zero at any point: . That produces the constant counterflow,, which exactly cancels the Lagrangian drift (2.20),  derived under the assumption of no mean Eulerian velocity. One lesson is that the drift is a quadratic quantity and must be determined by using the Eulerian velocity valid up to quadratic terms as well. Another lesson is that the average over space at a given time differs from the average over time at given point   for a wave train, which changes as it propagates. Note that (65,8) in Landau&Lifshits is the total momentum of the fluid, while (3.54) in Faber's "Fluid Dynamics for Physicists" is the momentum of the wave.

  • misprint on p.88

    In the paragraph after (3.4) the equation is for horizontal acceleration due to gravity acting on an inclined surface: dvx /dt=-...