We choose the Lagrange variable $$a$$ so that, in the acoustically unperturbed medium, it will coincide with the Euler coordinate x_0. We denote a certain quantity, expressed in Lagrange variables, by L(a,t) and the same quantity in Euler variables by E(x,t). The function L(a,t) refers to a fluid particle, its value at time t corresponding to the point in space a + $$\xi$$, where $$\xi$$ is the displacements of the particle. Therefore,

$$L(a,t) = E(a+\xi, t)$$

Expanding Eq. (22) in a series on the displacement and retaining only two terms of the expansion, we obtain

$$L(a,t) = E(a,t) + \xi \frac{\partial E}{\partial a}$$

We can verify that the second term of the series is much smaller than the first, or that

$$\xi \frac{\partial E}{\partial a} \ll E$$

Inasmuch as $$\frac{\partial E}{\partial a} ~ k E$$, condition (24) assumes the form $$\xi k \ll 1$$, which is true in our case, because$$\xi k ~ \frac{v}{c} \ll 1$$

$$\xi k ~ \frac{v}{c} \ll 1$$

It is apparent from Eq.(23) that the disparity between the Lagrange and Euler variables begins with second -order quantities. This permit us in (23) to replace $$\frac{\partial E}{\partial a}$$ by $$\frac{\partial L}{\partial t}$$, where upon we obtain the following equation for transforming Lagrangian variables to Euler variables, correct to and including terms in $$\xi$$.$$$$

$$E=L -\xi\frac{\partial E}{\partial x}$$

Thus, applying (25) to the expressions for the hydrodynamic velocity and density,

$$v=\frac{\partial \xi}{\partial t}$$, $$\rho = \rho_0 [1 - \frac{\partial \xi}{\partial a}+ (\frac{\partial \xi}{\partial a})^2]$$

represented in Lagrange variables, we obtain their expressions in Euler variables:

$$v=\frac{\partial \xi}{\partial t} - \xi \frac{\partial^2 \xi}{\partial t \partial x}$$, $$\rho = \rho_0 [1 - \frac{\partial \xi}{\partial x} + (\frac{\partial \xi}{\partial x})^2 + \xi \frac{\partial^2 \xi}{\partial x^2}]$$

The fundamental equation for the one-dimensional plane flow of an ideal fluid in Lagrange variables have the form (see [15 §2])

$$\rho dx = \rho_0 da$$

$$\rho_0 (\frac{\partial v}{\partial t})_a = (\frac{\partial p}{\partial t})_t$$

The equation of motion of a viscous thermally conducting fluid is obtained by the insertion of $$p^\prime$$, determined by Eq. (17), into Eq. (29), and the addition of a term to account for viscosity of the medium, where this term must have the same form as in the Euler variables. If the relation (26)* are also included, the initial equation reduced to the form $$$$

$$ \frac{\partial^2 \xi}{\partial t^2} - c^2 \frac{\partial^2 \xi}{\partial a^2} - \frac{b}{\rho_0} \frac{\partial^3 \xi}{\partial t\partial a^2} + 2 \varepsilon c^2 \frac{\partial \xi}{\partial a} \cdot \frac{\partial^2 \xi}{\partial a^2} = 0$$

Here $$\varepsilon = 1 + \frac{\rho_0 (\frac{\partial c^2}{\partial \rho})_s}{2c^2}$$

§2. The Rayleigh Radiation Pressure

We first examine a plane standing wave between two-plane-prallel rigid boundiares, one of which is stationary, the other oscillating, This problem contains as a special case the Rayleigh problem, in which both boundaries are stationary. We carry out the analysis in Lagrange variables, by means of which we readily describe the boundary condition at the oscillating surface.

Taking time average of the equation of motion written in Lagrange variables (29), and recognizing that $$\frac{\partial v}{\partial t} = 0$$, by virtue of the finteness of $$v$$, we obtain.

$$\bar{p_L^\prime} = \text{const} = A$$

in Lagrange variables $$\bar{p_L^\prime}$$ is the radiation pressure on the moving particles of the medium. On a nonmoving particle or on a stationary solid obstacle, the radiation pressure has the same value in Lagrange and Euler variables, namely $$\bar{p_L^\prime} =\bar{p_E^\prime}$$. The constancy of in Lagrange variables implies , however , that the radiation pressure on the stationary and on the oscillating solid surfaces is identical and equal to a constant $$A$$. The value of the constant is found by the equation of state (15), which we represent in the form

$$p_L^\prime = c_0^2 \rho^\prime + \frac{\gamma - 1}{2} \frac{c_0^2}{\rho} \rho^{\prime2}$$

Let the Lagrangian coordinate of the oscillating boundary $$a_0 = 0$$, and let the coordinate of the stationary boundary $$a_0 = 0$$, and let the coordinate of the stationary boundary $$a_l = l$$. Averaging Eq. (40) over the time and length l with regard to the result (39).we obtain

$$\bar{p_L^\prime} = A = \frac{c_0^2}{l} \int_0^L \bar{\rho^\prime_L} da + \frac{\gamma -1 }{2}\cdot \frac{c^2}{\rho l} \int_0^L \rho^{\prime2} d a$$

By the equation of continuity (28).

$$\rho_0 da = \rho dx$$

we transform the first integral on the right-hand side of (41) to the form

$$\int_0^l \bar{\rho_L^\prime}da=-\int_0^l\bar{\rho^\prime\frac{\partial \xi}{\partial a}}da - \rho_0 \int^l_0 d\bar{\xi} $$

Recognizing that in the linear approximation $$\rho^\prime = - \rho_0(\partial \xi /\partial a)$$, we obtain from expression (41) with regard to Eq. (42)

$$\bar{p_L^\prime} = \frac{\gamma + 1}{2l} \cdot \rho c^2 \int_0^l\bar{(\frac{\partial \xi}{\xi})^2} da - \frac{c^2 \rho}{l} [\bar{\xi}(l,t) - \bar{\xi(0,t)}]$$

For stationary of harmonically oscillating boundaries

$$\bar{\xi(l,t)} = \bar{\xi}(0,t) = 0$$

For stationary or harmonically oscillating boundaries

$$\xi(l,t) = \xi(0,t) = 0$$

When both boundaries are fixed, and the displacement field in the linear approximation has the fomr

$$\xi = \frac{v_0}{\omega} \sin ka \cos \omega t, kl = n \pi$$

Eq. (43) reduces to the Rayleigh equation (38). 八分之一

If one of the boundaries is fixed and the other oscillates harmonically, the sound field can be described in the linear approximation

by an equation of the form

$$\xi = - \xi_0 \frac{\sin k(l-a)}{\sin kl} \cos \omega t$$

when $$\xi_0$$ is the displacement amplitude of the oscillating boundary at point $$a = 0$$. Substituting expression (44) into (43), we obtain an equation for the radiation pressure in Lagrange variables:

$$\bar{p_L^\prime} = \frac{\gamma+1}{8} \rho v_0^2 (1 + \frac{\sin 2 kl} {2 kl})$$

where ,$$v_0= \xi_0 \omega /\sin k l$$ is the particle velocity amplitude of the medium. The radiation pressure on the stationary and oscillating boundaries is determined by the same expression. For $$2 k l \gg 1$$ or $$2 k l = (2n+1) \pi$$, Eq. (45) transforms to (38). For $$2k l \gg 1$$ or $$2 kl = (2n+1)\pi$$, Eq.(45) tranfroms to (38). For $$k l \ll 1$$ , as the resonance interval is approached , the velocity $$v_0$$ and, with it $$\bar{p_L^\prime}$$ increases with limit.

In Euler variables the pressure $$\bar{p_E^\prime}$$, according to Eq. (25) and the first-approximation relation $$p^\prime = -\rho c^2 (\partial \xi/\partial a)$$, has the form $$\bar{p_E^\prime} = \bar{p_L^\prime} + \rho c^2 \xi \bar{\frac{\partial ^2 \xi}{\partial x^2}}$$

Substituting $$\xi$$ as determined by Eq. (44), we obtain

(46)$$\bar{p_E^\prime} = \bar{p_L^\prime} - \rho \frac{v_0^2}{2}\sin^2 k(l-x)$$

In the absence of $$\bar{p_L^\prime}$$, the pressure $$\bar{p_E^\prime}$$ is a function of the given point in space.

We now consider a spherical standing wave between two concentric spheres, with the inner one pulsating harmonically and the outer one fixed. In they linear approximation, this standing wave is described by the formula

(47) $$\xi = \frac{R}{\omega} (\frac{k \cos k R}{R} - \frac{\sin kR} {R^2})\sin \omega t$$

where R is the Lagrangian spherical coordinate. The boundary condition on the stationary outer sphere of radius $$R_2$$, $$\xi(R_2) = 0$$ leads to the relation

$$\tan k R_2 = k R_2$$

We proceed from the equation of motion, which is written in Lagrange variables under spherical symmetry

$$\frac{\partial p}{\partial R} = - \rho_0 \frac{R^2}{r^2} \frac{\partial v}{\partial t}$$

where $$r = R + \xi (R,t)$$ is the Eulerian spherical coordinate. Integrating this equation over $$R$$ with regard to expression (47), we obtain an equation for the radiation pressure on the stationary outer sphere$$$$

$$\bar{p^\prime} = \frac{\rho B^2 k^2}{4} \frac{\sin^2 kR_2 }{R_0^2} + D$$

where D is a constant, which has to be determined by the same method as the constant $$A$$ for the plane-wave case. An analogous solution is obtained when the inner sphere is fixed and the outer one oscillates.

If $$R_1 = 0$$, i.e. if the interval sphere is omitted, then

$$\bar{p^\prime}(R_2) = \frac{\rho B^2 k^2}{8}(3\gamma - 1) \frac{\sin^2 kR_2}{R^2_2}$$

which exactly agrees with the result of Lucas [48, 49], obtained by the method of adiabatic invariants.

If both spheres are fixed $$\xi(R_1) = 0$$, then

$$\bar{p^\prime}(R_2) = \frac{\rho B^2 k^2}{4} { \frac{\sin^2 k R_2} {R^2_2} + \frac{3 (\gamma -1)}{2 (R_2^3 - R_1^3)}[R_2 \sin^2 kR_2 - R_1(1 - \frac{\sin 2kR_1}{2kR_1})]} $$

for $$kR_1 = n \pi$$, where n is an integer greater then zero.

$$\bar{p^\prime}(R_2) = \frac{\rho B^2 k^2} {4} \frac{\sin^2 kR_2}{R_2^2} + \frac{3 \rho B^2}{2 (R_2^3 - R_1^3)}[\frac{1}{R_1} + \frac{\gamma - 1}{4\pi} k^2 (R_2 \sin^2 k R_2 - R_1)]$$

For $$R_1 \approx R_2 \gg \lambda$$, where $$\lambda$$ is the wavelength, Eq. (48) goes over to Eq. (45), as it should.

Inasmuch as the radiation pressure acts on the inner sphere as well, this suggests the possibility of building rather unique "acoustic bearings", in which the central part would be supported by the radiation pressure. However, the practical realization of this motion runs up against substantial difficulties, as is particularly evident from the work of Seegal [50], who has investigated the problem in detail.

Among the problems associated with the Rayleigh radiation pressure is the one-dimensional problem treated by Airy and later analyzed in detail by Fubini-Chiron [51]. This problem concerns a traveling sound wave generated by the oscillation of a plane perpendicular to an axis a. The initial and boundary conditions of the problem are:

1) For $$t < 0$$ and any $$a > 0$$ the displacement $$\xi = 0$$.

2) For $$t≥0 $$ the displacement of the plane ($$a=0$$) is $$\xi = \xi_0 (1 - \cos\omega t) $$.

We are interested only in the radiation pressure on the radiating plane as determined by Eq. (43).

We first consider the wave in an ideal fluid. To avoid calculations of the displacement field in the second approximation, we carry out space averaging over the length $$l = ct$$, i.e., over the entire perturbed domains. Then at the leading front of the wav

e $$\xi(l,t) = 0$$, because the radiation begins with zero displacement $$\xi (0,0) = 0$$, and with zero velocity $$v(0,0) = 0$$. By making the time interval small enough, one can regard $$\xi(0,t)/l$$ as negligibly small. Consequently,

$$\bar{p^\primeL} = \frac{\gamma + 1}{2 l t} \rho c^2 \int_0^l da \int{a/c}^t (\frac{\partial \xi}{\partial a})^2 dt$$

In the time integral, the lower limit corresponds to the time of arrival of the perturbation at point $$a$$, and prior to this time the integrand is zero. Substituting into Eq. (51) the solution of the wave equation for $$\xi$$ in correspondence with the solution of our problem in the first approximation

$$\xi = \xi_0 [ 1- \cos (\omega t - ka )]$$

after integration we obtain the familiar result

$$\bar{p^\prime_L} = \frac{\gamma+1}{8} \rho \xi^2_0 \omega^2$$

Equation (43) can also be used to calculate the dissipative correction for the radiation pressure, because the dissipative terms in the equation of motion and in (17) do not affect the results (39) and (41) used for the derivation of (43). The solution of the linearized equation (30) in this case may be written in the form

$$\xi = \xi_0 1 - \cos\omega (t -\frac{a}{c})$$

The factor $$(1- \alpha a)$$ is the approximate value of the exponential $$\exp(-\alpha a) $$ at distances $$a \ll 1/{\gamma}$$ over which the solution (53) is valid.

Substituting the dissipative correction for the displacement

$$\xi^\prime = - \xi_0 \alpha a [1 - \cos (\omega t - ka)]$$

into the result (43), we find that the second-order dissipative correction to the radiation pressure is equal to

$$-\frac{c^2 \rho}{l} \bar{\xi^\prime (l,t)} = \xi_0 \rho c^2 \alpha$$

Adding the results (52) and (54), we obtain an expression for the radiation pressure on a radiating plane in a dissipative medium:

§3. Langevin Radiation Pressure

Consider a sound wave progagating in an unbounded space and decaying to zero at infinity. For the given sound field, Eq. (21) may be assumed to have the form

$$\bar{p^\prime} = -\frac{\rho v^2}{2} + \frac{c^2\bar{\rho^{\prime2}}}{2 \rho}$$

Here, first of all, the term $$b \text{div} \bar{v}$$ is omitted as a negligibly small quantity if $$\bar{v}$$ is second-order small, as mentioned already in the discussion of Eq. (37); second, the constant is chosen equal to zero, in order to meet the condition of zero perturbation at infinity. If the constant did not vanish, we would contain the parameter $$\gamma$$. Generally, in problems where the contact between the sound field and the unperturbed medium permits the constant to be set equal to zero, the parameter $$\gamma$$ must be absent in the radiation pressure equations. In such cases it is customary to speak of the Langevin radiation pressure.

We now deduce formula for the Langevin radiation pressure on an acoustically completely absorbing obstacle, and obstacle having a reflection coefficient of one, and at an interface between two fluids. We investigate these problems by the usual procedure. Then in §2 of Chap. 3, we return to the same problems in order to show that , form our point of view, certain results need to be modified in the case of plane waves.

Let the second wave be completely absorbed by the obstacle. In the filed of a plane traveling sound wave $$\rho^\prime = \rho v/c$$. and according to (57) $$\bar{p^\prime}= 0$$. Therefore, an obstacle that absorbs the plane sound wave is acted upon by the radiation pressure described by Eqs. (34) and (36) with $$p^\prime = 0$$. In vector form this radiation pressure is $$\mathbf{P}_r = \rho \mathbf{v}(\mathbf{v}\cdot \mathbf{n})$$

In particular, for n and v parallel

$$\mathbf{P}_r = \rho \bar{v^2} \mathbf{n} = 2\bar{E_k} \mathbf{n}$$

where $$\bar{E_k}$$ is the kinetic energy density in the wave.

In a spherical traveling wave the velocity potential is

$$\phi = \frac{\phi}{r} \cos (\omega t - kr)$$

By Eqs. (10) and (57), we now obtain .

Consequently, if the spherical wave is absorbed , the radiation pressure vector is described by the complete Eq. (32)