HYDRODYNAMIC THEORY OF MULTIPLE PARTICLE FORMATION

S. Z. Belenkii and L. D. Landau

§ 1. INTRODUCTION

As experience shows, in collisions of ultrafast particles a large number of new particles are formed (multiple showers).

The energy of the particles producing such showers is of the order of \(10^{12}\) ev and higher. It is characteristic that such collisions occur not only between a nucleon and a nucleus, but also between two nucleons. Thus, the formation of two mesons in collisions of neutrons with protons was observed already at comparatively small energies of the order of \(10^9\) ev in experiments carried out at the Cosmotron\(^1\).

Fermi\({}^{2,3}\) is responsible for the ingenious idea of considering the collision process at very high energies by means of thermodynamic methods.

The basic premises of Fermi’s theory reduce to the following:

1. It is assumed that, in the collision of two very energetic nucleons, the energy in the system of their center of gravity is released in a very small volume \(V\). Since the nuclear interaction is very strong, and the dimensions of the volume are small, the distribution of energy will be determined by statistical laws. This makes it possible to consider the collision of particles of high energy without making use of any particular theories of nuclear interaction.

2. The volume \(V\) in which the energy is released is determined by the size of the meson cloud around the nucleons, whose radius is of the order \(\hbar/\mu c\), where \(\mu\) is the mass of the \(\pi\)-meson. However, since the nucleons move with high velocity, the meson cloud surrounding them undergoes Lorentz contraction along the direction of motion of the nucleons. Thus, the volume \(V\), in order of magnitude, will be equal to:

\[ V=\frac{4\pi}{3}\left(\frac{\hbar}{\mu c}\right)^3\cdot\frac{2Mc^2}{E'}, \tag{1.1} \]

where \(M\) is the mass of the nucleon, and \(E'\) is the energy of the nucleons in the center-of-gravity system.

3. Fermi assumes that the particles are formed, in accordance with the laws of statistical equilibrium, in the volume \(V\) at the very moment of the collision. The particles thus formed, no longer interacting with one another, fly out of the volume in a “frozen” state.

4. Fermi considered both central collisions, which according to his calculations lead to an isotropic angular distribution, and noncentral (peripheral) collisions. In this case, in the statistical calculation, in addition to conservation of energy, conservation of angular momentum was also taken into account. In noncentral collisions one obtains an angular distribution that is no longer isotropic in the center-of-mass system.

Fermi’s basic idea of applying statistical methods to the study of collision processes is undoubtedly very fruitful; however, the individual assumptions and quantitative calculations are unconvincing (see \(^{4,5}\)).

The assertion that the number of particles in a multipion star is determined by the number of particles arising in the volume \(V\) at the very moment of the collision is unfounded. At that moment, because of the large density of the particles and the strong interaction between them, it is in general meaningless to speak of the number of particles. Even if one assumes that the particles are formed at the first moment, the assumption of strong interaction cannot be reconciled with the supposition that this interaction immediately ceases after the particles fly out of the volume.

In reality, the system expands, and the number of particles becomes definite only when the interaction between them becomes small. It is precisely then that the free flight of the particles takes place. This circumstance was pointed out by I. Pomeranchuk \(^{4}\). Furthermore, Fermi \(^{3}\) incorrectly calculated peripheral collisions and, consequently, the angular and energy distributions of the particles. Fermi’s calculation is difficult to reconcile with the theory of relativity. According to Fermi, in a substantially noncentral collision, during a collision time of order

\[ \left(\frac{\hbar}{\mu c}\right)\left(\frac{Mc^2}{E'}\right)\frac{1}{c} \]

the interaction propagates over the entire volume of the meson cloud, i.e. over a distance of order \(\hbar/\mu c\). This means that the disturbance would have to propagate with a velocity considerably exceeding the velocity of light.

The shortcomings of Fermi’s theory are mainly connected with the fact that the expansion of the composite system is incorrectly taken into account. As Landau showed \(^{6}\), the expansion of the system can be considered on the basis of relativistic hydrodynamics. The use of hydrodynamics in the present case is just as consistent as the use of thermodynamics, since the domains of applicability of the one and the other coincide*).

*) Indeed, the conditions for the applicability of thermodynamics and hydrodynamics consist in the requirement

\[ \frac{l}{L}\ll 1, \]

where \(l\) is the “mean free path,” and \(L\) is the smallest dimension of the system.

From the qualitative point of view, the collision process appears as follows:^6

1. In the collision of two nucleons, a composite system arises, with energy being released in a small volume \(V\), Lorentz-contracted in the transverse direction.

At the instant of collision a large number of “particles” arises; the “mean free path” in the system that has arisen is small in comparison with its dimensions, and statistical equilibrium is established in the system.

1. The second stage of the collision consists in the expansion of the system. A hydrodynamic approach should be applied to this stage, and the expansion may be regarded as the motion of an ideal (nonviscous and non-heat-conducting) fluid*). In the course of expansion the “mean free path” continues to remain small in comparison with the dimensions of the system, which also justifies the use of hydrodynamics.

Since the velocities in the system are comparable with the velocity of light, it is necessary to use not ordinary but relativistic hydrodynamics. During the first and second stages of the collision, formation and absorption of particles occur in the system all the time. Here the large energy density in the system is essential. In this case the system is not characterized at all by the number of particles, owing to the strong interaction between the individual parts of the system.

1. As the system expands, the interaction weakens and the mean free path increases. The number of particles as a physical characteristic appears when the interaction is sufficiently small. When the mean free path becomes comparable with the linear dimensions of the system, the system breaks up into separate particles. We shall call this stage the “rarefaction.” Rarefaction takes place at a system temperature of the order of \(T \simeq \mu c^2\), where \(\mu\) is the mass of the \(\pi\)-meson. (Temperature is everywhere measured in energy units.)

§ 2. THERMODYNAMIC RELATIONS

IN THE RAREFACTION OF THE SYSTEM

In the rarefaction of the system the interaction between particles may be neglected, and therefore a number of simple relations is obtained.

Let us consider some region of the system that is at the rarefaction temperature \(T_k\). We shall use the following expression for the rela-

*) Let us explain this by the following qualitative considerations. In order that viscosity and heat conduction could be neglected, the Reynolds number

\[ \frac{LV}{l v} \]

must be greater than 1. Here \(L\) is the smallest dimension of the system, \(V\) the “macroscopic” velocity, \(v\) the “molecular” velocity, and \(l\) the mean free path. Since \(V\) and \(v\) are of order \(c\), the condition \(R \gg 1\) coincides with the condition

\[ \frac{l}{L} \ll 1 . \]

tivistic density of \(\pi\)-mesons in this region:

\[ n_\pi=\frac{g_\pi}{2\pi^2}\left(\frac{T_k}{\hbar c}\right)^3\cdot F(z_\pi), \tag{2.1} \]

where

\[ z_\pi=\frac{\mu c^2}{T_k}. \]

Here \(g_\pi\) is the number of possible states of the particle; in our case (respectively, the presence of \(\pi^+\)-, \(\pi^-\)- and \(\pi^0\)-mesons), \(g_\pi=3\);

\[ F(z_\pi)=z_\pi^3\int_0^\infty \frac{x^2\,dx}{e^{z_\pi\sqrt{1+x^2}}-1}. \tag{2.2} \]

The function \(F(z_\pi)\) may be represented in the following form:

\[ F(z_\pi)=z_\pi^2\sum_{m=0}^{\infty} \frac{K_2\!\left[z_\pi(1+m)\right]}{1+m}, \tag{2.3} \]

where \(K_2(z)\) is the modified Bessel function of the second kind (see \(^{7,8}\)). The series (2.3) converges rapidly. We give the asymptotic expressions for \(K_2(z)\) for large and small values of \(z\):

\[ \begin{aligned} K_2(z)&=\left(\frac{\pi}{2z}\right)^{1/2}e^{-z} \left[1+\frac{15}{8}\frac{1}{z}\right] && \text{for } z\gg 1,\\ K_2(z)&=\frac{2}{z^2} && \text{for } z\ll 1. \end{aligned} \tag{2.4} \]

Using (2.4), it is not difficult to verify that at high temperatures

\[ n_\pi=0.365\left(\frac{T}{\hbar c}\right)^3, \]

and at low temperatures

\[ n_\pi=g_\pi\left(\frac{T\mu}{2\pi\hbar^3}\right)^{3/2}e^{-\mu c^2/T}. \]

For the energy density of the \(\pi\)-mesons, \(\varepsilon_\pi\), we have the following expression:

\[ \varepsilon_\pi=T_k\left(\frac{g_\pi}{2\pi^2}\right) \left(\frac{T_k}{\hbar c}\right)^3\Phi(z_\pi), \tag{2.5} \]

where

\[ \Phi(z_\pi)=z_\pi^2\sum_{m=0}^{\infty} \frac{3K_2\!\left[z_\pi(1+m)\right]+z_\pi(1+m)K_1\!\left[z_\pi(1+m)\right]} {(1+m)^2}. \tag{2.6} \]

Here \(K_1(z)\) is the modified Bessel function of the first kind. Let us recall the asymptotic expressions for \(K_1(z)\) for large and small values of \(z\):

\[ \begin{aligned} K_1(z)&=\left(\frac{\pi}{2z}\right)^{1/2}e^{-z} \left[1+\frac{3}{8}\frac{1}{z}\right] && \text{for } z\gg 1,\\ K_1(z)&=\frac{1}{z} && \text{for } z\ll 1. \end{aligned} \tag{2.7} \]

It is not difficult to verify that for \(z_\pi \to 0\) (very high temperatures) \(\Phi(0)=6.49\), while for \(z\gg 1\) we obtain that \(\dfrac{\varepsilon_\pi}{n_\pi}=\mu c^2+\dfrac{3}{2}T\). Finally, let us give the expression for the entropy density of \(\pi\)-mesons (in absolute units):

\[ s_\pi=\left(\frac{T}{\hbar c}\right)^3 \frac{g_\pi}{2\pi^2}\,G(z_\pi), \tag{2.8} \]

where

\[ G(z_\pi)=z_\pi^2\sum_{m=0}^{\infty} \frac{4K_2\!\left[z_\pi(1+m)\right]+z_\pi(1+m)K_1\!\left[z_\pi(1+m)\right]}{(1+m)^2}. \tag{2.9} \]

The values of the functions \(F(z)\), \(\Phi(z)\), and \(G(z)\) are given in Table 1.

Table 1

\(z\)	\(F(z)\)	\(\Phi(z)\)	\(G(z)\)	\(F^*(z)\)	\(\Phi^*(z)\)	\(G^*(z)\)
0	2.40	6.49	8.65	1.80	5.68	7.57
0.5	2.17	6.30	8.31	1.72	5.58	7.37
0.7	2.02	6.12	8.02	1.65	5.47	7.19
0.9	1.86	5.90	7.67	1.56	5.33	6.95
1	1.78	5.78	7.48	1.52	5.24	6.81
1.2	1.62	5.51	7.07	1.41	5.05	6.51
1.5	1.39	5.06	6.42	1.25	4.72	6.00
2	1.05	4.27	5.31	0.982	4.07	5.07
3	0.561	2.78	3.33	0.546	2.72	3.27
6	0.0599	0.471	0.531	0.0599	0.471	0.531
7	0.0268	0.237	0.263	0.0268	0.237	0.263
8	0.0117	0.115	0.127	0.0117	0.115	0.127

In Table 1 the values of the functions \(F^*(z)\), \(\Phi^*(z)\), \(G^*(z)\) refer to the case of a Fermi gas, whereas the functions \(F(z)\), \(\Phi(z)\), \(G(z)\) refer to a Bose gas. For the Fermi gas, instead of formula (2.2), we shall have:

\[ F(z)=z^3\int_0^\infty \frac{x^2\,dx}{e^{z\sqrt{1+x^2}}+1}, \]

and the series (2.3), (2.6), (2.8) will be alternating in sign.

For \(z\gg 1\)

\[ F(z)\simeq F^*(z)\simeq z^2\left(\frac{\pi}{2z}\right)^{1/2}e^{-z} \left(1+\frac{15}{8}\frac{1}{z}\right), \]

\[ \Phi(z)\simeq \Phi^*(z)\simeq z^3\left(\frac{\pi}{2z}\right)^{1/2}e^{-z} \left(1+\frac{27}{8}\frac{1}{z}\right), \]

\[ G(z)\simeq G^*(z)\simeq z^3\left(\frac{\pi}{2z}\right)^{1/2}e^{-z} \left(1+\frac{35}{8}\frac{1}{z}\right), \]

In addition to $\pi$-mesons, generally speaking, other particles may also be formed in the system. As shown in [7], the equilibrium number of mesons with masses exceeding the $\pi$-meson mass may, generally speaking, be significant when $T_k \approx \mu c^2$. However, some of the heavy mesons apparently interact only weakly with nucleons and therefore are not produced in stars. Furthermore, other heavy particles apparently are formed in stars in pairs (for example, $\Lambda$-particles are formed together with $K$-mesons)*), but the fraction of these particles is relatively small, and the experimental data are insufficient to draw definite conclusions. Therefore we shall not consider them here.

It is necessary, moreover, to take into account the presence of nucleons in the system. Since several nucleons (not fewer than two) take part in high-energy collisions, conservation of nuclear charge should also be taken into account (see [9]).

The expressions for the densities of nucleons $n_{\mathrm{нн}}$ and antinucleons $n_{\mathrm{ан}}$ have the following form:

\[ \left. \begin{aligned} n_{\mathrm{нн}} &= \frac{g_{\mathrm{н}}}{2\pi^2}\left(\frac{T_k}{\hbar c}\right)^3 F_1(z_{\mathrm{н}},y_{\mathrm{нн}}),\\ n_{\mathrm{ан}} &= \frac{g_{\mathrm{н}}}{2\pi^2}\left(\frac{T_k}{\hbar c}\right)^3 F_1(z_{\mathrm{н}},y_{\mathrm{ан}}). \end{aligned} \right\} \tag{2.10} \]

Here $g_{\mathrm{н}}$ is the number of possible states of a particle at a given momentum; for nucleons $g_{\mathrm{н}}=4$ (two charge states and two spin directions)

\[ F_1(z_{\mathrm{н}},y)=z_{\mathrm{н}}^3\int_0^\infty \frac{x^2\,dx}{e^{-y+z_{\mathrm{н}}\sqrt{1+x^2}}+1}, \tag{2.11} \]

where $z_{\mathrm{н}}=\dfrac{Mc^2}{T}$, $M$ is the nucleon mass, $y=\dfrac{\mu}{T}$, and $\mu$ is the chemical potential.

The equilibrium condition with respect to the formation and annihilation of pairs is $y_{\mathrm{нн}}+y_{\mathrm{ан}}=0$. Denoting $y_{\mathrm{нн}}$ by $y$, we obtain that $y_{\mathrm{ан}}=-y$.

In the cases of interest to us $y \ll z$ and $z>1$. Therefore in formula (2.11) the denominator may be expanded in a series in powers of the quantity $\bigl(y-z\sqrt{1+x^2}\bigr)$. Restricting ourselves to the first term in the expansion, we obtain:

\[ n_{\mathrm{нн}}=\frac{g_{\mathrm{н}}}{2\pi^2}\left(\frac{T_k}{\hbar c}\right)^3 F_\nu(z_{\mathrm{н}})e^y;\quad n_{\mathrm{ан}}=\frac{g_{\mathrm{н}}}{2\pi^2}\left(\frac{T_k}{\hbar c}\right)^3 F_\nu(z_{\mathrm{н}})e^{-y}. \tag{2.10'} \]

* The relatively long lifetimes of all heavy particles are difficult to reconcile with the assumption of their strong interaction with nuclei, unless one admits that these particles are formed in pairs.

For the energy densities \((\varepsilon_{\mathrm{nn}}\) and \(\varepsilon_{\mathrm{an}})\) and the entropy densities \((s_{\mathrm{nn}}\) and \(s_{\mathrm{an}})\) (2.10) of nucleons and antinucleons we have:

\[ \varepsilon_{\mathrm{nn}}= T_k\left(\frac{g_{\mathrm{n}}}{2\pi^2}\right) \left(\frac{T_k}{\hbar c}\right)^3 \Phi_\nu(z_{\mathrm{n}})\,e^y; \]

\[ \varepsilon_{\mathrm{an}}= T_k\left(\frac{g_{\mathrm{n}}}{2\pi^2}\right) \left(\frac{T_k}{\hbar c}\right)^3 \Phi_\nu(z_{\mathrm{n}})\,e^{-y}; \tag{2.12} \]

\[ s_{\mathrm{nn}}= \left(\frac{T_k}{\hbar c}\right)^3 \left(\frac{g_{\mathrm{n}}}{2\pi^2}\right) \left[G_\nu(z_{\mathrm{n}})-yF_\nu(z_{\mathrm{n}})\right]e^y; \]

\[ s_{\mathrm{an}}= \left(\frac{g_{\mathrm{n}}}{2\pi^2}\right) \left(\frac{T_k}{\hbar c}\right)^3 \left[G_\nu(z_{\mathrm{n}})+yF_\nu(z_{\mathrm{n}})\right]e^{-y}. \tag{2.13} \]

The quantity \(y\) (the chemical potential) is determined by the condition that the difference between the number of nucleons and antinucleons in the entire system must remain constant and equal to the number of initial nucleons \(N_0\).

Let us use the formulas given above, (2.1)—(2.13), to calculate several quantities.

First of all, let us estimate the breakup temperature of the system \(T_k\). Since the density of \(\pi\)-mesons is considerably greater than the density of nucleons, this quantity is determined mainly by the density of \(\pi\)-mesons. In order for breakup of the system to be possible at the temperature \(T_k\), the mean free path of the particles \(l\) must, at \(T_k\), be of the order of the characteristic dimensions of the system \(L\). As can be shown on the basis of the more detailed hydrodynamic calculation given in § 4, \(L\) may approximately be written as

\[ L \simeq \left(\frac{\hbar}{\mu c}\right) \left(\frac{E}{Mc^2}\right)^{1/12}, \]

where \(E\) is the energy of the primary nucleon and \(M\) is its mass. The mean free path

\[ l \simeq \frac{1}{n\sigma}. \]

Let us assume, for an estimate, that in order of magnitude

\[ \sigma=\pi\left(\frac{\hbar}{\mu c}\right)^2. \]

If for \(n\) we take the expression determined by formula (2.1), we obtain:

\[ \frac{l}{L}= \frac{2\pi}{3}\, \frac{z_\pi^3}{F(z_\pi)} \cdot \left(\frac{Mc^2}{E}\right)^{1/12}. \tag{2.14} \]

The results of the calculation by formula (2.14) are given in Table II.

It is seen from this that the breakup temperature \(T_k\) lies within the limits \(0.7—1.5\mu c^2\). Indeed, for the value \(\frac{l}{L}\ll 1\) breakup cannot occur, since the mean free path is too small, while for \(\frac{l}{L}\gg 1\) the system cannot exist because its dimensions are too small. It follows

It should be noted that the uncertainty of our calculation, connected with the fact that both \(\sigma\) and \(L\) are known only to order of magnitude, does not allow one to determine theoretically the decay temperature \(T_k\).

Table II

\(T_k/\mu c^2\)	\(l/L\)*)
0.5	9
0.7	3
1.0	0.7
1.5	0.21
2	0.07

*) Here \(l/L\) is taken at \(E=10^{13}\) eV.

The quantity \(T_k\) depends rather weakly on the quantity \(l/L\) (when \(l/L\) changes by almost a factor of 100, \(T_k\) changes by a factor of four). On the other hand, since \(l/L\) depends extremely weakly on the initial energy, \(T_k\) depends on it even more weakly. Thus, the introduction of the quantity \(T_k\), practically independent of the properties of the system (i.e., of the initial energy), has physical meaning.

Let us now turn to the calculation of the number of antinucleons. We shall assume that each of the parts of the system decays at the same temperature of the system \(T_k\), although the decay moments for different parts may not coincide. Then the ratio of the total number of nucleons in the system to the total number of \(\pi\)-mesons will be equal to the ratio of their densities. From formulas (2.1) and (2.10′) it is not difficult to obtain:

\[ \operatorname{sh} y=\frac{N_\pi^0}{N_{\mathrm{H}}^0}\cdot\frac{N_0}{N_\pi}; \qquad \operatorname{ch} y=\frac{N_\pi^0}{N_{\mathrm{H}}^0}\cdot\frac{N_{\mathrm{H}}}{N_\pi}. \tag{2.15} \]

Here \(N_\pi^0\) and \(N_{\mathrm{H}}^0\) are the total numbers of \(\pi\)-mesons and nucleons (nucleons and antinucleons) in the system, under the condition that initial nucleons are absent. \(N_0\) is the number of initial nucleons, and \(N_\pi\) and \(N_{\mathrm{H}}\) are the total numbers of \(\pi\)-mesons and nucleons (nucleons and antinucleons) in the presence of initial nucleons. From formulas (2.15) we obtain:

\[ \frac{N_{\mathrm{H}}}{N_\pi} = \sqrt{ \left(\frac{N_{\mathrm{H}}^0}{N_\pi^0}\right)^2 + \left(\frac{N_0}{N_\pi}\right)^2 }. \tag{2.16} \]

In an analogous manner we obtain the ratio of the total energy density of nucleons and antinucleons \(\varepsilon_{\mathrm{H}}\) to the energy density \(\varepsilon_\pi\) of \(\pi\)-mesons:

\[ \frac{\varepsilon_{\mathrm{H}}}{\varepsilon_\pi} = \sqrt{ 1+ \left(\frac{N_\pi^0}{N_{\mathrm{H}}^0}\right)^2 \left(\frac{N_0}{N_\pi}\right)^2 \cdot \left(\frac{\varepsilon_{\mathrm{H}}^0}{\varepsilon_\pi^0}\right) }. \tag{2.17} \]

Here \(\varepsilon_{\mathrm{H}}^0\) and \(\varepsilon_\pi^0\) are the energy densities of nucleons and \(\pi\)-mesons when the number of initial nucleons \(N_0=0\). The ratio \(\dfrac{\varepsilon_{\mathrm{H}}}{\varepsilon_\pi}\) does not depend on

of the coordinate system in which the collision process is considered.

Indeed, let a particle of mass \(M\) have, in the coordinate system where the given element of matter is at rest, energy \(E\). In another coordinate system, moving relative to the given one with velocity \(v\), this particle will have energy

\[ E'=\frac{E+p_x v}{\sqrt{1-\frac{v^2}{c^2}}}, \]

where \(p_x\) is the projection of the particle momentum on the direction of the velocity \(v\). Since in the given coordinate system the distribution is isotropic, averaging over all directions, we obtain

\[ E'=\frac{E}{\sqrt{1-\frac{v^2}{c^2}}}. \]

The energy of a particle of another mass \(\mu\) will transform according to the same law. Thus, the ratio of the energies will not depend on the velocity \(v\).

The ratios \(\dfrac{N_{\mathrm{n}}^0}{N_\pi^0}\) and \(\dfrac{\varepsilon_{\mathrm{n}}^0}{\varepsilon_\pi^0}\) are easily found from formulas (2.1), (2.5), (2.10), and (2.12), if in formulas (2.10) and (2.12) one sets \(y=0\). The corresponding values are given in Table III.

If the expansion temperature is of order \(1.2\text{--}1.5\,\mu c^2\), then the excitation of “isobaric” states of the system and the formation of \(\Lambda\)-particles may play a known role. Excitation of “isobaric” states will lead to some increase in the number of produced \(\pi\)-mesons, which is insignificant for \(N_\pi>N_0\). As for \(\Lambda\)-particles, the role of their admixture has been estimated in works \(^{7,9}\). However, since the spin of the \(\Lambda\)-particle is unknown and they are formed in pairs with \(K\)-particles (this is not taken into account in the cited works), this estimate is very preliminary.

Table III

\(\dfrac{T_k}{\mu c^2}\)	\(\dfrac{N_{\mathrm{n}}^0}{N_\pi^0}\)	\(\dfrac{\varepsilon_{\mathrm{n}}^0}{\varepsilon_\pi^0}\)
1.0	0.06	0.14
1.2	0.126	0.30
1.5	0.27	0.57
2.0	0.56	1.0

Let us consider the following example. Let \(T_k=1.5\,\mu c^2\); then

\[ \frac{N_{\mathrm{n}}^0}{N_\pi^0}=0.27, \]

and the number of antinucleons is equal to \(0.135\). Suppose that

\[ \frac{N_0}{N_\pi}=0.15 \]

(which corresponds, for example, to three initial nucleons for 20 mesons formed). Then from formula (2.16) we obtain that

\[ \frac{N_{\mathrm{n}}}{N_\pi}=0.3. \]

The number of all nucleons formed is \(0.15\), and that of antinucleons is \(0.075\). Thus, in this case the number of antinucleons, taking into account the initial nucleons, will decrease by almost a factor of two.

Formula (2.8) gives the ratio of the energy carried away by nucleons to the energy carried away by \(\pi\)-mesons. Let \(T_k=1.2\,\mu c^2\). Then

\[ \frac{\varepsilon_{\mathrm{H}}^0}{\varepsilon_\pi^0}=0.3,\qquad \frac{N_{\mathrm{H}}^0}{N_\pi^0}=0.13. \]

For \(\frac{N_0}{N_\pi}=0.15\), \(\frac{\varepsilon_{\mathrm{H}}}{\varepsilon_\pi}=0.42\). If, however, \(\frac{N_0}{N_\pi}\simeq 1\), then \(\frac{\varepsilon_{\mathrm{H}}}{\varepsilon_\pi}\simeq 2.3\), i.e. the nucleons carry away about \(70\%\) of the total energy.

It should be emphasized, however, that for \(N_0\simeq N_\pi\) the number of particles formed is small and the theory developed is a rather rough approximation.

Let us note that the number of antinucleons may, generally speaking, be determined by a higher temperature than the disintegration temperature; in this sense formula (2.16) gives a lower bound. On the other hand, formulas (2.16) and (2.17) contain two parameters—the disintegration temperature of the system \(T_k\) and the quantity \(\frac{N_0}{N_\pi}\). It is possible, however, to form a quantity which does not depend on \(\frac{N_0}{N_\pi}\) and which is determined only by the disintegration temperature of the system \(T_k\), even if the number of nucleons is fixed also at a higher temperature. It is not difficult to see that the ratio of the energy per nucleon to the energy per \(\pi\)-meson depends only on the disintegration temperature \(T_k\), and an experimental determination of this ratio would apparently be the best way of determining the disintegration temperature. Let us note that this ratio does not depend on the reference frame. The values of this ratio are given in Table IV.

Table IV

\(\dfrac{T_k}{\mu c^2}\)	\(a=\dfrac{\varepsilon_{\mathrm{H}}}{\varepsilon_\pi}\,\dfrac{n_\pi}{n_{\mathrm{H}}}\)
\(T_k\gg \mu c^2\)	6.8
0.5	3.76
0.7	3.28
0.8	2.92
1	2.65
1.5	2.16
2	1.83

It is very important to find the relation between entropy and the number of particles. Let us recall that the expansion of the system (the second stage) is regarded as the motion of an ideal liquid and therefore is adiabatic. Shock waves could violate this adiabaticity, but they are not formed in the process of expansion. Therefore the entropy of the system and of its separate elements changes in the first stage of the collision and remains unchanged throughout the entire hydrodynamic stage of expansion up to disintegration into individual particles.

From equations (2.1), (2.8), (2.10), (2.13) we obtain the following expression for the entropy of the system:

\[ S=\left\{\frac{G_0(z_{\mathrm{H}})}{F_0(z_{\mathrm{H}})}-y\cdot \operatorname{th} y\right\}N_{\mathrm{H}}+\frac{G_0(z_\pi)}{F(z_\pi)}\,N_\pi . \tag{2.18} \]

If \(N_0=0\), then \(y=0\) as well, and formula (2.18) reduces to the following:

\[ N^*=\alpha S, \tag{2.19} \]

where \(N^*=N+N_\pi\), i.e. \(N^*\) is equal to the sum of the particles formed (nucleons and \(\pi\)-mesons), and the coefficient \(\alpha\) is equal to

\[ \alpha=\frac{8F_0(z_{\mathrm H})+3F(z_\pi)} {8G_0(z_{\mathrm H})+3G(z_\pi)}. \tag{2.20} \]

The function \(\alpha\) is given in Table V.

As follows from this table, the function \(\alpha\) depends only weakly on the breakup temperature \(T_k\). It is not difficult to see that, in the interval of variation of \(T_k\) from \(0.7\) to \(1.5\,\mu c^2\), the coefficient \(\alpha\) practically does not change.

Passing to the case when the number of initial nucleons is not equal to zero, we turn to the equation of conservation of nuclear charge:

\[ N_{\mathrm{нн}}-N_{\mathrm{ан}}= \]

\[ =\left(\frac{T_k}{\hbar c}\right)^3 \frac{g_{\mathrm H}}{2\pi^2}F(z_{\mathrm H}) \left[e^y-e^{-y}\right]V_{\mathrm{ср}}=N_0. \tag{2.21} \]

Here \(V_{\mathrm{ср}}\) is the total volume of the system.

On the other hand, the total number of nucleons is equal to

\[ N_{\mathrm H}=N_{\mathrm{нн}}+N_{\mathrm{ан}}= \]

\[ =\left(\frac{T_k}{\hbar c}\right)^3 \frac{g_{\mathrm H}}{2\pi^2}F(z_{\mathrm H}) \left[e^y+e^{-y}\right]V_{\mathrm{ср}}. \tag{2.22} \]

Table V

\(\dfrac{T_k}{\mu c^2}\)	\(\alpha\)
0.17	0.113
0.5	0.198
0.67	0.215
0.83	0.222
1	0.223
1.43	0.216
2	0.213
\(T_k\gg \mu c^2\)	0.25

Let us now return to equation (2.18). Instead of \(S\) we take the ratio of the number of particles \(N^*\), formed when \(N_0=0\), to the initial number of nucleons \(N_0\). To do this, we multiply the right- and left-hand sides of equation (2.12) by \(a\) and divide by \(N_0\) (see (2.21)).

As a result of the transformations we obtain:

\[ \frac{N^*}{N_0} = \left[ \frac{G_0(z_{\mathrm H})}{F_0(z_{\mathrm H})}\operatorname{cth} y -y+ \frac{g_\pi G(z_\pi)}{2g_{\mathrm H}F_0(z_{\mathrm H})\operatorname{sh} y} \right]\alpha. \tag{2.23} \]

With the aid of equations (2.1), (2.10), (2.22) we obtain:

\[ \frac{N}{N_0} = \frac{g_\pi F_0(z_\pi)}{2g_{\mathrm H}F_0(z_{\mathrm H})} \frac{1}{\operatorname{sh} y} +\operatorname{cth} y. \tag{2.24} \]

From these equations one can find the ratio \(\dfrac{N}{N_0}\) as a function of \(\dfrac{N^*}{N_0}\) (the quantity \(y\) is a parameter). Calculations show that, over a wide temperature interval (from \(T_k=0.5\) to \(T_k=2\,\mu c^2\)) and up to values \(\dfrac{N^*}{N_0}\simeq 2\), \(\dfrac{N}{N_0}\) will always be less than \(\dfrac{N^*}{N_0}\). In this case the difference of \(\dfrac{N}{N_0}\) from \(\dfrac{N^*}{N_0}\) will be the greater, the lower the temperature \(T_k\). However,

already for \(\frac{N^*}{N_0}=3\) and \(T_k=0.5\,\mu c^2\), \(\frac{N}{N_0}=2.54\), i.e. the ratio \(\frac{N}{N^*}=0.85\). For \(\frac{N^*}{N_0}<2\) the number of newly formed particles rapidly decreases and \(\frac{N}{N_0}\) tends to 1. Thus the relation

\[ N=aS \tag{2.25} \]

is valid also in the presence of initial nucleons, up to values of \(\frac{N^*}{N_0}\) of order 2, where \(N\) should be understood as the sum of those formed in collisions of particles and the initial nucleons.

§ 3. TOTAL NUMBER OF PARTICLES

For the hydrodynamic treatment of the system, which is used in the first two stages of the collision, it is necessary to know the equation of state of the substance.

As the equation of state of strongly compressed matter at temperatures \(T \gg \mu c^2\), we shall adopt the following (see\(^6\)):

\[ p=\frac{\varepsilon}{3}, \tag{3.1} \]

where \(p\) is the pressure, and \(\varepsilon\) is the energy density. As is known, for the pressure and density of macroscopic bodies the inequality \(p \leqslant \varepsilon/3\) holds, with the equality sign valid in the extremely relativistic case. However, this inequality was derived under the assumption of electromagnetic interaction between particles, and at present there is no proof that it must be valid in the case of an arbitrary interaction. Nevertheless, the choice of the equation of state (3.1) seems to us very plausible. Since the number of particles in the system is not specified, but is itself determined from statistical equilibrium, the chemical potential is equal to zero. Hence \(\varepsilon-Ts+p=0\), where \(s\) is the entropy per unit volume. Using the equation of state, we have \(Ts=\varepsilon+p=4/3\,\varepsilon\). Since at a given volume \(d\varepsilon=T\,ds\), it is not difficult to see that

\[ s \simeq \varepsilon^{3/4}; \qquad T \simeq \varepsilon^{1/4}. \tag{3.2} \]

The relations (3.2) coincide with the relations for “black radiation,” which, of course, was to be expected.

We have already indicated that the entropy of the system remains constant during the hydrodynamic stage of expansion and changes only at the first stage, at the initial moment of the collision. On the other hand, the number of particles in the star is connected with the entropy by relation (2.25). It follows from this that, in order to determine the total number of particles, one should calculate the change of entropy at the initial moment of the collision. This change of entropy is simplest to calculate in the case of the collision

two identical particles, for example two nucleons. Let \(E'\) be the energy of the nucleons in the center-of-inertia system. In the center-of-inertia system, by virtue of the symmetry of the system, the matter at the instant immediately following the collision is at rest. The total entropy of the system is proportional to \(\varepsilon^{3/4}V_1\), where \(\varepsilon\) is the energy density, \(V_1\) is the volume in which the energy is concentrated. Therefore \(\varepsilon = E'/V_1\); the entropy, and consequently also the number of particles, are proportional to \(E'^{3/4}V_1\). Since, owing to the Lorentz contraction, the volume \(V_1\) transforms inversely proportional to \(E'\), while the energy \(E\) in the laboratory system is proportional to \(E'^2\), we finally obtain:

\[ N \simeq E^{1/4}, \tag{3.3} \]

where \(E\) is the energy in the laboratory system. From dimensional considerations formula (3.3) can be written in the form

\[ N = k \left(\frac{E}{2Mc^2}\right)^{1/4}, \tag{3.3'} \]

where \(k\) is a constant of order 1. Experiment shows that \(k \simeq 2\). Relation (3.3) coincides with that obtained by Fermi, which is natural, since in deriving it we proceeded from the equation of state for black radiation, which Fermi also adopts, and from the relation \(N \simeq S\), which is valid at any “temperature” of expansion of the system. According to Fermi, the expansion temperature is determined at the first instant after the collision. This assumption, which, as was shown\(^4\), is internally contradictory, nevertheless does not affect the validity of relation (3.3).

We have considered head-on collisions, when the particles pass at distances comparable with their radius of action. We now turn to peripheral collisions, when the distance of passage is large compared with the radius of action of the forces of both particles. At first glance it may seem that the average number of particles produced should rapidly decrease with increasing impact parameter. This is connected with the fact that the energy of the meson field of the colliding nucleons, concentrated in each region, would seem (in the system in which this region is at rest) to fall rapidly with increasing distance from their centers. Such a point of view is adopted in the works of Heisenberg\(^ {10}\) and Bhabha\(^ {11}\), devoted to multiple production of particles at high energies. Meanwhile, this point of view is in contradiction with the quantum uncertainty relation and is therefore erroneous\(^6\). This question is examined in greater detail in the work of E. L. Feinberg and D. S. Chernavskii\(^ {12}\). Indeed, if the energy of a region of the system were to decrease, then it would very soon become small compared with the uncertainty \(\Delta E \simeq \dfrac{\hbar c}{l}\), where \(l\) is the thickness

region, reduced as a result of Lorentz contraction*). It follows from this that it is not the true total energy of the system that is small, but its mathematical expectation. In other words, what is decreased is not the total energy of the system and the total number of particles, but only the probability that such a collision will occur at all. In a detailed analysis of peripheral collisions a quantum treatment is necessary, since the classical one is invalid because of the violation of the uncertainty relation.

A consistent quantum treatment of the problem of peripheral collisions presents great difficulties. Such a treatment requires the use of the meson theory of nuclear forces, which has not yet been developed. It is not excluded that in a peripheral collision (which in itself is unlikely) only some fraction of the energy of the primary nucleon is transferred in the collision process. At present, however, the theory is not in a position to settle this question. It seems to us that, in general, at the present level of the theory there is no particular sense in distinguishing central and peripheral collisions in collisions of two nucleons. The effective cross sections for collisions with the formation of a many-ray star are determined, in order of magnitude, by the “radius” of the nucleon \(\hbar/\mu c\).

Heisenberg\({}^{10}\), because of an incorrect account of peripheral collisions, arrives at the erroneous conclusion that the collision cross section grows logarithmically with energy. Let us dwell on this question in more detail (see\({}^{12}\)). Heisenberg considers the collision of superfast nucleons in the system of their common center of gravity. The meson fields of the nucleons decrease in the perpendicular direction \(y\) according to the law \(e^{-y/r_0}\), where \(r_0=\dfrac{\hbar}{\mu c}\). To estimate the fraction \(\gamma\) of the initial energy of the nucleon \(E\) that goes into the particles produced, the region is considered in which the meson clouds of both particles overlap at the collision parameter \(b\). In this case one obtains \(\gamma \simeq e^{-b/r_0}\), and the number of particles produced is, in order of magnitude,

\[ N \simeq \frac{\gamma E}{\mu c^2} \simeq \frac{E}{\mu c^2} e^{-b/r_0}. \tag{3.4} \]

The cross section for the formation of some number \(N\) of mesons is equal to \(\pi b_0^2\), where \(b_0\) corresponds to \(N\). For example, for \(N\simeq 2\) we have:

\[ \sigma_{N\geq 2}\simeq \pi r_0^2\left(\ln \frac{E}{\mu c^2}\right)^2 . \tag{3.5} \]

It follows from this that there should arise chiefly showers with a small number of particles, while the cross section grows logarithmically with ener-

*) Since \(l \simeq \dfrac{\hbar}{\mu c}\dfrac{Mc^2}{E}\), then \(\dfrac{\Delta E}{E}\simeq \dfrac{\mu}{M}\).

gies. But, as was indicated above, Heisenberg’s approach to noncentral collisions is incorrect, and therefore formula (3.5) is erroneous.

Indeed, for \(\gamma E \lesssim \Delta E\) its derivation comes into contradiction with the uncertainty relation. This corresponds to collision parameters \(b \simeq r_0\).

In Baba’s work\(^{11}\) an attempt is made, on the basis of the study of multiple stars, to draw certain conclusions about the structure of the nucleon. In doing so the author, from the classical point of view, considers distant collisions and, consequently, also comes into contradiction with the uncertainty relation. A quantum treatment, however, as has already been indicated, cannot at present be carried out. Therefore no conclusions about the structure of the nucleon can be drawn from the analysis of multiple stars.

Collisions of nucleons with nuclei, as well as of nuclei with nuclei, will be considered in § 5.

§ 4. ENERGY AND ANGULAR DISTRIBUTION OF PARTICLES

In this section we shall consider the expansion stage in the collision of two nucleons. To investigate the expansion of the system one should use the equations of relativistic hydrodynamics:

\[ \frac{\partial T_{ik}}{\partial x_k}=0, \tag{4.1} \]

\[ T_{ik}=\omega u_i u_k+p g_{ik}, \tag{4.2} \]

where \(\omega=\varepsilon+p\) is the heat function per unit volume; \(u_i\) is the 4-velocity; \(g_{11}=g_{22}=g_{33}=1;\ g_{44}=-1;\) below we put \(c=1\), \(x_i=x_1,x_2,x_3,it\).

Projection of equation (4.1) onto the direction \(u_i\) leads to the following equation:

\[ \frac{\partial s u_k}{\partial x_k}=0, \tag{4.3} \]

where \(s\) is the entropy density. Equation (4.3) expresses the adiabatic character of the motion. In deriving equation (4.3), the thermodynamic relations \(d\varepsilon=Tds\) and \(\omega=Ts\) were taken into account. (Let us recall that, since the number of particles in the system is not specified, the chemical potential \(\mu\) is equal to zero.)

Let us now “project” equation (4.1) onto a direction perpendicular to \(u_i\). This projection is, obviously,

\[ \frac{\partial T_{ik}}{\partial x_k}+u_i u_k\frac{\partial T_{kl}}{\partial x_l}, \]

since, when scalar-multiplied by \(u_i\), it gives zero. Calculation\(^{13}\) leads to the following equation:

\[ \omega u_k \frac{\partial u_i}{\partial x_k}+u_i u_k \frac{\partial p}{\partial x_k}+\frac{\partial p}{\partial x_i}=0. \tag{4.4} \]

Taking into account that in our case \(\omega=Ts\) and \(dp=s\,dT\) (since \(d\omega=dp+Tds\)), from equation (4.4) we obtain:

\[ u_k\frac{\partial T u_i}{\partial x_k}+\frac{\partial T}{\partial x_i}=0. \tag{4.5} \]

In the center-of-mass system the system has, at the instant of collision, the form of a strongly flattened disk (at first the transverse dimensions of this disk \(a\) are \(E'/2M\) times greater than its thickness \(\Delta\)). This character of the motion is preserved during a considerable part of the expansion stage. The motion of the matter may then be regarded as one-dimensional. The solution both for the one-dimensional and for the subsequent three-dimensional stages of expansion was first obtained by Landau\({}^{6}\). Later I. M. Khalatnikov\({}^{14}\) obtained a more exact solution for the one-dimensional stage, proceeding from the study of the general one-dimensional problem in relativistic hydrodynamics. In the present paragraph, in considering the one-dimensional stage, we shall first follow paper\({}^{14}\) and then paper\({}^{6}\).

In the one-dimensional case we shall have only the coordinates \(x_1\) and \(x_4\). It is not difficult to see that equation (4.5) can then be written in the following form (see\({}^{14}\)):

\[ \frac{\partial T u_1}{\partial x_4} = \frac{\partial T u_4}{\partial x_1} \tag{4.6} \]

and, consequently, there exists a function \(\varphi\) such that

\[ T u_1=\frac{\partial\varphi}{\partial x_1},\qquad T u_4=\frac{\partial\varphi}{\partial x_4}. \tag{4.7} \]

The function \(\varphi\) is the potential of one-dimensional motion in relativistic hydrodynamics. For the potential \(\varphi\) we have the following differential relation:

\[ d\varphi=T u_4\,dx_4+T u_1\,dx_1. \tag{4.8} \]

In what follows, instead of \(x_4\) we shall use the quantity \(t\) \((x_4=it)\), instead of \(u_4\) the quantity \(u_0=\dfrac{1}{\sqrt{1-v^2}}\) \((u_4=iu_0)\), and instead of \(x_1\) the quantity \(x\). Equation (4.8) will then take the following form:

\[ d\varphi=-T u_0\,dt+T u_1\,dx. \tag{4.8'} \]

Let us introduce a variable \(\alpha\), connected with the velocities \(u_0\) and \(u_1\) by the relations

\[ u_1=\operatorname{sh}\alpha,\qquad u_0=\operatorname{ch}\alpha. \tag{4.9} \]

Now we perform a Legendre transformation with respect to the variables \(\omega\) and \(\alpha\). For the potential \(\chi\) we obtain:

\[ d\chi=d(\varphi+T u_0 t-T u_1 x)= \]

\[ =(t\,\operatorname{ch}\alpha-x\,\operatorname{sh}\alpha)\,dT +(t\,\operatorname{sh}\alpha-x\,\operatorname{ch}\alpha)\,T\,d\alpha. \tag{4.10} \]

From relation (4.10) it follows:

\[ \frac{\partial \chi}{\partial T}=t\,\operatorname{ch}\alpha-x\,\operatorname{sh}\alpha;\qquad \frac{\partial x}{\partial \alpha}=T\,(t\,\operatorname{sh}\alpha-x\,\operatorname{ch}\alpha), \tag{4.11} \]

\[ t=\frac{\partial \chi}{\partial T}\operatorname{ch}\alpha-\frac{1}{T}\frac{\partial \chi}{\partial \alpha}\operatorname{sh}\alpha;\qquad x=\frac{\partial \chi}{\partial T}\operatorname{sh}\alpha-\frac{1}{T}\frac{\partial \chi}{\partial \alpha}\operatorname{ch}\alpha . \tag{4.12} \]

Now let us turn to equation (4.3), which in the case of one-dimensional motion is written in the following form:

\[ \frac{\partial s u_0}{\partial t}+\frac{\partial s u_1}{\partial x}=0. \]

In this equation we pass from the variables \(t\) and \(x\) to the variables \(T\) and \(\alpha\):

\[ \frac{\partial(t,x)}{\partial(T,\alpha)} \left\{ \frac{\partial(s\,\operatorname{ch}\alpha,x)}{\partial(t,x)} - \frac{\partial(s\,\operatorname{sh}\alpha,t)}{\partial(t,x)} \right\} = \frac{\partial(s\,\operatorname{sh}\alpha,x)}{\partial(T,\alpha)} - \frac{\partial(s\,\operatorname{ch}\alpha,t)}{\partial(T,\alpha)} . \]

Since in our case \(s\) is completely determined by the temperature \(T\), we obtain:

\[ \frac{1}{s}\frac{ds}{dT} \left\{ \frac{\partial}{\partial \alpha}(x\,\operatorname{ch}\alpha-t\,\operatorname{sh}\alpha) - (x\,\operatorname{sh}\alpha-t\,\operatorname{ch}\alpha) \right\} + \frac{\partial}{\partial T}(t\,\operatorname{ch}\alpha-x\,\operatorname{sh}\alpha)=0. \tag{4.13} \]

Using relations (4.11), (4.12), we transform equation (4.13) to the following form:

\[ \frac{1}{s}\frac{ds}{dT} \left\{ \frac{\partial \chi}{\partial T} - \frac{1}{T}\frac{\partial^2\chi}{\partial \alpha^2} \right\} + \frac{\partial^2\chi}{\partial T^2}=0. \tag{4.14} \]

But

\[ \frac{s}{T}\frac{dT}{ds} = \frac{1}{T}\frac{dp}{d\varepsilon}\frac{d\varepsilon}{ds} = \frac{dp}{d\varepsilon} = c_0^2, \]

where \(c_0\) is the speed of sound in the medium.

Introducing, instead of the variable \(T\), the variable \(y=\ln T\), we obtain:

\[ \frac{\partial^2\chi}{\partial \alpha^2} - c_0^2\frac{\partial^2\chi}{\partial y^2} + (c_0^2-1)\frac{\partial\chi}{\partial y}=0. \tag{4.15} \]

Since the equation of state of the medium is \(p=\frac{\varepsilon}{3}\), we have \(c_0^2=\frac{1}{3}\). Substituting this value into equation (4.15), we have:

\[ 3\frac{\partial^2\chi}{\partial \alpha^2} - \frac{\partial^2\chi}{\partial y^2} - 2\frac{\partial\chi}{\partial y} =0^*). \tag{4.15′} \]

Thus, the solution of the one-dimensional problem of relativistic hydrodynamics is reduced to the solution of a linear differential equation with constant coefficients. The transition from the potential \(\chi\) to the variables \(x\) and \(t\) is made by formulas (4.12).

*) As a result of misprints in the paper by I. M. Khalatnikov\(^{14}\), the coefficients in equations (4.15) and (4.15′) differ from the corresponding coefficients given in paper\(^{14}\).

Let us now return to the problem of the expansion of a plane disk of thickness \(\Delta\) that interests us. In the center-of-gravity system the disk expands symmetrically in both directions. Therefore we shall consider the expansion only in one direction. Let us choose \(x=0\) so that the expansion is symmetric with respect to the plane \(x_1=-l\), where \(l=\dfrac{\Delta}{2}\). First of all it is necessary to clarify the boundary conditions for our problem. It is obvious that in the plane \(x_1=-l\) (which, by symmetry, may be regarded as a stationary wall) the medium must be at rest, and this means that \(\alpha=0\). Using equation (4.12), we write this condition in the following form:

\[ \left(\frac{\partial \chi}{\partial \alpha}\right)_{\alpha=0}=l e^{-y}. \tag{4.16} \]

On the side of the vacuum the required solution must border on a simple (Riemann) wave, for which the self-similar solution holds:

\[ x=t\,\frac{\vartheta-c_0}{1-\vartheta c_0}, \tag{4.17} \]

where \(\vartheta\) is the (ordinary) velocity of motion. It can be shown (see \({}^{13}\)) that in a simple wave the quantities \(\alpha\) and \(y\) are related by

\[ \alpha=-\frac{y}{c_0}. \tag{4.18} \]

In order to find the matching condition of the simple wave with the general solution, one must substitute the expressions (4.12) for \(x\) and \(t\) into equation (4.17). Carrying out this substitution and using condition (4.18), we obtain that \(\chi=0\).

Thus, the second boundary condition is written as follows:

\[ \chi=0 \quad \text{for} \quad \alpha=-\frac{y}{c_0}. \tag{4.19} \]

Instead of the potential \(\chi\), introduce the potential \(\chi_1\):

\[ \chi=\chi_1 e^{-y}, \tag{4.20} \]

and take as variables the quantities \(\alpha\) and \(z\) (instead of \(y\)):

\[ z=-\alpha-\frac{y}{c_0}. \tag{4.21} \]

Equation (4.15) and the boundary conditions (4.16) and (4.19) take, in the new variables, the following form:

\[ \left(6\,\frac{\partial^3}{\partial \alpha\,\partial z^2} -3\,\frac{\partial^2}{\partial \alpha^2} -1\right)\chi_1=0, \tag{4.22} \]

\[ \chi_1=0 \quad \text{for} \quad z=0, \tag{4.23} \]

\[ \frac{\partial \chi_1}{\partial \alpha} -\frac{\partial \chi_1}{\partial z} = -l e^{-z/\sqrt{3}} \quad \text{for} \quad \alpha=0. \tag{4.24} \]

Next we perform the Laplace transform with respect to the variable \(z\):

\[ \psi=\int \chi_1(z)e^{-zq}\,dz . \tag{4.25} \]

Then equation (4.22) becomes

\[ 6q\frac{\partial\psi}{\partial a}-3\frac{\partial^2\psi}{\partial a^2}-\psi=0, \tag{4.26} \]

and the boundary conditions reduce to the condition

\[ \frac{\partial\psi}{\partial a}-q\psi=l-\frac{1}{q+\dfrac{2}{\sqrt{3}}} \quad \text{for } a=0. \tag{4.27} \]

We seek a solution in the form \(\psi=a(q)e^{p(q)a}\). For \(p(q)\) we obtain from equation (4.26) a quadratic equation. We choose the solution corresponding to the damping of \(\psi\) as \(a\to\infty\). The function \(a(q)\) is determined from equation (4.27). As a result we obtain:

\[ \psi = l\, \frac{ e^{\left(q-\sqrt{q^2-\frac{1}{3}}\right)a} }{ \sqrt{q^2-\frac{1}{3}}\left(q+\dfrac{2}{\sqrt{3}}\right) }. \tag{4.28} \]

Next, according to the inversion formula, we find the function \(\chi_1\):

\[ \chi_1= \frac{1}{2\pi i}\,l \int_{\delta-i\infty}^{\delta+i\infty} \frac{ e^{-\sqrt{q^2-\frac{1}{3}}\,a-yq\sqrt{3}}\,dq }{ \sqrt{q^2-\frac{1}{3}}\left(q+\dfrac{2}{\sqrt{3}}\right) }, \quad \operatorname{Re} q<\delta . \tag{4.29} \]

The integral in (4.29) is taken so that the poles of the integrand lie to the left of the straight line along which the integration is performed.

Finally, the potential \(\chi\), satisfying the boundary conditions, has the following form:

\[ \chi= l\sqrt{3}\,e^y \int_{\frac{a}{\sqrt{3}}}^{-y} e^{2y'} I_0\left(\sqrt{y'^2-\frac{a^2}{3}}\right)\,dy', \tag{4.30} \]

where \(I_0\) is the Bessel function of imaginary argument *).

Using the relations (4.12), which are conveniently rewritten in the form

\[ \begin{aligned} t&=e^{-y}\left(\frac{\partial\chi}{\partial y}\operatorname{ch}a-\frac{\partial\chi}{\partial a}\operatorname{sh}a\right),\\ x&=e^{-y}\left(\frac{\partial\chi}{\partial y}\operatorname{sh}a-\frac{\partial\chi}{\partial a}\operatorname{ch}a\right), \end{aligned} \qquad \tag{4.12'} \]

*) See Appendix I.

one can now find the dependence of \(y\) (i.e. \(T\)) and \(\alpha\) (i.e. \(v\)) on \(x\) and \(t\), and thus obtain the complete solution of the problem.

Let us consider the late stage of a one-dimensional expansion, when \(y\) and \(\alpha\) are large and \(y>\alpha\). These conditions mean a relatively low temperature (\(y\) negative) and ultrarelativistic particle velocities. For this stage we shall use the asymptotic expression for \(I_0\):

\[ I_0\left(\sqrt{y^2-\frac{\alpha^2}{3}}\right)\simeq e^{\sqrt{y^2-\frac{\alpha^2}{3}}}. \]

The potential \(\chi\) will then be of the order

\[ \chi \simeq \Delta \cdot e^{-y+\sqrt{y^2-\frac{\alpha^2}{3}}}. \]

If we confine ourselves to logarithmic accuracy, then, using (4.12), it is not difficult to obtain:

\[ \frac{t+x}{\Delta}\simeq e^{\alpha-2y+\sqrt{y^2-\frac{\alpha^2}{3}}}, \]

\[ \frac{t-x}{\Delta}\simeq e^{-\alpha-2y+\sqrt{y^2-\frac{\alpha^2}{3}}}.^{*} \tag{4.31} \]

Hence we have:

\[ \alpha=\frac{1}{2}\ln\frac{t+x}{t-x}; \]

\[ y=-\frac{1}{3}\left[ \ln\frac{t+x}{\Delta}+\ln\frac{t-x}{\Delta} -\sqrt{\ln\frac{t+x}{\Delta}\cdot\ln\frac{t-x}{\Delta}} \right]. \tag{4.32} \]

Let us introduce the following notation:

\[ \ln\frac{t+x}{\Delta}=\tau;\qquad \ln\frac{t-x}{\Delta}=\eta. \]

Since \(y=\ln T\), and the energy density \(\varepsilon\simeq T^4\), the second of equations (4.32) can be written as

\[ \varepsilon=\varepsilon_0 e^{-\frac{4}{3}\left(\eta+\tau-\sqrt{\tau\eta}\right)}. \tag{4.33} \]

Further, for \(\alpha\gg 1\), \(u_1\simeq u_0\simeq \dfrac{1}{\sqrt{1-v^2}}\) (see (4.9)); the quantity

\[ \frac{1}{\sqrt{1-v^2}} \]

\[ ^{*}\text{ It is easy to see that }\quad \frac{t-x}{t+x}\simeq e^{-2\alpha}\ll 1. \]

denote by \(u\). From equation (4.9) it follows that \(a=\operatorname{Arsh} u \simeq \ln 2u\); the first of equations (4.32) then gives:

\[ u^2 \simeq \frac{t+x}{t-x}. \tag{4.34} \]

Equations (4.33) and (4.34) coincide with the equations (16) and (24) obtained earlier by Landau by another method.^6

For the simultaneous stage of expansion it is possible (starting from formula (4.30)) to obtain a more accurate solution than the solution (4.33)—(4.34), which is valid only to logarithmic accuracy. However, since the next three-dimensional stage is solved only very approximately, there is at present no need for further refinement.

Let us consider the distribution of energy and entropy along the thickness of the disk. The energy density, as is easy to see, is of order \(\varepsilon u^2\)*). For the energy \(dE\) falling on a layer of thickness \(d\delta\), where \(\delta=t-x\), we have:

\[ dE \simeq \varepsilon a^2 u^2 d\delta, \]

where \(a\) is the radius of the disk.

Using formulas (4.33), (4.34), we obtain:

\[ dE \simeq e^{-\frac{1}{3}\left(\sqrt{\tau}-2\sqrt{\eta}\right)^2}\,d\eta . \tag{4.35} \]

It follows from formula (4.35) that the energy distribution has a maximum at \(\eta=\frac{\tau}{4}\). This means that the energy is concentrated,

mainly, in the region \(\delta \simeq \sqrt[4]{t\Delta^3}\). For \(t \gg \Delta\), \(\delta \ll t\). Thus, most of the energy is concentrated in a narrow layer \(\delta \ll t\), close to the boundary of the region. Let us now find the entropy distribution. The entropy density is equal to \(su\)**). Since \(s\simeq \varepsilon^{3/4}\), for the entropy falling on a layer of thickness \(d\delta\), we find:

\[ dS \simeq sua^2 d\delta \]

or, using (4.33) and (4.34),

\[ dS \simeq e^{-\frac{1}{2}\left(\sqrt{\tau}-\sqrt{\eta}\right)^2}\,d\eta . \tag{4.36} \]

This distribution has a maximum at \(\eta=\tau\). This means that the entropy is concentrated in the region \(L\simeq t\). Hence it is seen (see (4.34)) that in the region of maximum particles \(u^2\simeq 1\) and, consequently, the condition \(a>1\) is satisfied approximately. The matter in this region moves with velocities of the order of the speed of light, but not ultrarelativistic ones.

*) Indeed, the energy density is given by the component \(T_{44}\) of the energy-momentum tensor: \(T_{44}\simeq \varepsilon u^2\).

**) The entropy density is given by the component \(s^0\) of the four-vector of entropy-flow density: \(s^0\simeq u\varepsilon^{3/4}\).

Let us now turn to the spatial stage of the expansion. The solution obtained is valid as long as the motion may be regarded as one-dimensional. For this it is necessary that the angle of expansion \(\vartheta\), i.e., the angle formed by the trajectory of a given element of matter with the \(x\)-axis, be small. More precisely, it is necessary that the distance traversed by the given element in the transverse direction be smaller than the dimensions of the system \(a\). This condition may be written in the form

\[ t\vartheta \ll a . \tag{4.37} \]

To estimate the angle \(\vartheta\), let us turn to the unused transverse components of equation (4.1). We have:

\[ \frac{\partial T_{42}}{\partial t} \simeq \frac{\partial T_{22}}{\partial y} \]

or, in order of magnitude,

\[ \frac{T_{42}}{t} \simeq \frac{T_{22}}{y}. \]

But \(T_{42} \simeq \varepsilon u^2 \vartheta\) and \(T_{22} \simeq \varepsilon\), whence \(u^2 \vartheta \simeq \dfrac{t}{a}\). Since, according to (4.34), \(u^2 \simeq \dfrac{t}{\delta}\), we finally obtain:

\[ \vartheta \simeq \frac{\delta}{a}. \tag{4.38} \]

This means that the farther a particle is from the front and, consequently, the smaller its energy, the more strongly it is deflected. Formulas (4.37) and (4.38) give:

\[ t \ll \frac{a^2}{\delta}. \tag{4.39} \]

Inequality (4.39) determines the condition of applicability of the one-dimensional solution. The boundary of applicability of the one-dimensional solution lies the farther away, the closer the particle is to the front. Starting from the moment \(t_1 = \dfrac{a^2}{\delta}\), the lateral deviation becomes substantial. The resulting motion can be considered only approximately. Qualitatively, it may be characterized as conical expansion. Let us turn to the derivative \(\dfrac{d\delta}{dt}\); \(\dfrac{d\delta}{dt} = 1 - \dfrac{dx}{dt} = 1 - v \simeq \dfrac{1}{2u^2}\). In the second (three-dimensional) stage of expansion, \(u\) is much larger than in the first. Therefore the quantity \(\delta\) remains practically constant for each element of matter. In addition, it can be shown that all derivatives of hydrodynamic quantities, both in the direction \(\delta\) and in the transverse direction, may be neglected. This means that the lateral forces are small and we have, as it were, an expansion by inertia (radial expansion). Let us summarize the qualitative description of the expansion: the angular deviation of the trajectory of each element, attained in the first, essentially one-dimensional stage of expansion, remains unchanged in the subsequent, spatial stage.

From this it follows that both the energy flux and the entropy flux carried inside any cone with a given opening angle \(\vartheta\) must remain constant. Since \(\vartheta=t-x\ll t\), the cross-sectional area of this cone is \(\sim t^2\), and the constancy of the energy and entropy fluxes means:

\[ \varepsilon u^2 t^2=\text{const};\quad su t^2 \simeq \varepsilon^{3/4} u t^2=\text{const}. \tag{4.40} \]

Hence we find:

\[ u\simeq t,\quad \varepsilon\simeq \frac{1}{t^4}. \tag{4.41} \]

These relations determine the change with time of the quantities \(u\) and \(t\) during conical expansion. In the first stage, as follows from (4.34), \(u\simeq t^{1/2}\); in the second stage \(u\simeq t\), and consequently the velocity in this case differs less from the velocity of light than in the first stage. The transition region from the first to the second stage is very complicated and will not be considered in detail. Instead, the problem is posed of “matching” the solution obtained for the second stage with the one-dimensional solution at the point where the latter becomes invalid, i.e. at the point

\[ t_1=\frac{a^2}{\delta}. \]

Introduce the following notation:

\[ \frac{\delta}{a}=e^{-\lambda},\quad \frac{\Delta}{a}=e^{-L}. \tag{4.42} \]

Recall that

\[ \frac{\Delta}{a}\simeq \frac{M}{E'}=\sqrt{\frac{2M}{E}}, \]

whence

\[ L=-\frac{1}{2}\ln\frac{E}{2M}, \]

where \(E'\) is the energy in the center-of-gravity system, and \(E\) is the energy in the laboratory system. Then

\[ \eta=\ln\frac{\delta}{\Delta}=L-\lambda;\quad \tau_1=\ln\frac{t_1}{\Delta}=\ln\frac{a^2}{\delta\Delta}=L+\lambda. \tag{4.43} \]

Substituting these expressions into formula (4.36), we obtain:

\[ dS=e^{\sqrt{L^2-\lambda^2}}\,d\lambda. \]

In the spatial stage, as has already been clarified, \(\delta=\text{const}\), and consequently \(\lambda=\text{const}\) for each element of matter. Therefore the entropy of an element of matter is determined by the same formula up to the breakup of the system into separate parts. But we know that the number of particles is proportional to the entropy. Consequently, for the particle distribution we have:

\[ dN\simeq e^{\sqrt{L^2-\lambda^2}}\,d\lambda, \tag{4.44} \]

\[ \vartheta=\frac{\delta}{a}=e^{-\lambda}. \tag{4.45} \]

The opening angle \(\vartheta\) remains constant, together with \(\delta\), for each element of matter. Formulas (4.44), (4.45) determine, in parametric form, the angular distribution of particles in the center-of-inertia system. Most particles are formed at \(\delta \simeq t\), since in this region the entropy maximum is located. For these particles the end of the one-dimensional stage occurs at \(t_k \simeq a\). Therefore the opening angle of such particles is \(\simeq 1\). At small angles fewer particles are formed, but they have greater energy. It follows from formulas (4.44)—(4.45) that in any collision the angular distribution in the center-of-inertia system is by no means isotropic, as Fermi assumed\(^2\). Let us note that the largest value of the quantity \(\lambda\) for which it still makes sense to take it into account is determined from the condition

\[ \int_{\lambda_{\max}}^{L} dN \simeq 1 . \]

From this one can derive that

\[ \lambda_{\max}=\frac{\sqrt{3}}{2}\,L. ^*) \tag{4.46} \]

Let us determine the distribution of particles with respect to energy. In the stage of one-dimensional motion \(u \simeq \sqrt{\dfrac{t}{\delta}}\), while in the stage of conical expansion \(u \simeq t\). From the condition of matching these quantities at \(t=t_1\), we obtain:

\[ u \simeq \frac{t}{a}. \tag{4.47} \]

Next we match the expressions for the energy density \(\varepsilon\). At \(t=t_1\), from formula (4.33), we obtain:

\[ \varepsilon=\varepsilon_0 e^{-\frac{4}{3}\left(2L-\sqrt{L^2-\lambda^2}\right)} . \]

On the other hand, in the stage of conical expansion \(\varepsilon \simeq \dfrac{1}{t^4}\). From the matching condition we obtain:

\[ \varepsilon=\varepsilon_0\left(\frac{t_1}{t}\right)^4 e^{-\frac{4}{3}\left(2L-\sqrt{L^2-\lambda^2}\right)} . \tag{4.48} \]

\(^*)\) Since \(dN \approx e^{\sqrt{L^2-\lambda^2}}\,d\lambda\), from the condition

\[ \int_{\lambda_{\max}}^{L} dN \simeq 1 \]

we have:

\[ C e^{\sqrt{L^2-\lambda_{\max}^2}} \simeq 1 \]

(to logarithmic accuracy). But \(N \simeq e^{\frac{L}{2}}\). Therefore

\[ \int_{\lambda=0}^{\lambda=L} dN \simeq c e^L \simeq e^{\frac{L}{2}} . \]

This means that \(C \simeq e^{-\frac{L}{2}}\) and \(\lambda_{\max}=\dfrac{\sqrt{3}}{2}\,L\).

Free dispersion of particles begins at the moment \(t_k\), when the temperature of the region of the system becomes equal to the “critical” temperature \(T_k\) (see § 2). This moment corresponds to a definite energy density \(\varepsilon_k\). From formula (4.48) we find an expression for the moment \(t_k\):

\[ t_k \simeq t_1\left(\frac{\varepsilon_0}{\varepsilon_k}\right)^{1/4} e^{-\frac{1}{3}\left(2L-\sqrt{L^2-\lambda^2}\right)} . \]

Hence, according to (4.47), we find the value of the quantity \(u\) at the moment of dispersion:

\[ u_k \simeq \frac{t_k}{a}. \tag{4.49} \]

The quantity \(u_k\) is simply related to the energy of the dispersing particles. If the energy of a particle in the frame of reference in which the given element of matter is at rest is equal to \(E_0\), then in the system of the center of inertia it is equal to \(E_0u_k\). From formula (4.49) it is clear that the later the given element of the system decays, the greater the energy possessed by the emitted particles:

\[ u_k=\mathrm{const}\, e^{\lambda+\frac{1}{3}\sqrt{L^2-\lambda^2}} . \]

The constant in this relation is determined from the equality \(\int E_0 u_k\, dN = E'\). Hence for the energy of a particle \(\mathcal{E}=E_0u_k\) we obtain the expression

\[ \mathcal{E}\simeq M e^{-\frac{L}{6}+\lambda+\frac{1}{3}\sqrt{L^2-\lambda^2}} . \tag{4.50} \]

Equations (4.44) and (4.50) give the distribution over energies in parametric form. If particles with different masses are formed, or if the energy carried away by nucleons plays an essential role, then this circumstance may be taken into account on the basis of the considerations developed in § 2.

From formulas (4.44), (4.45), (4.49) it is clear that the energy of the particles is uniquely related to the angle at which they fly out. The majority of particles (\(\lambda=0\)) have comparatively small energy. Those particles which go at small angles possess, however, considerable energy. Thus there is a concentrated flux of energy going at small angles.

It remains to pass from the system of the center of inertia to the laboratory frame of reference. Without dwelling on the details of the recalculation (see Appendix II), we give the final formulas:

\[ dN=\frac{k}{\sqrt{2\pi L}}\, e^{-\frac{L}{2}+\sqrt{L^2-\lambda^2}}\, d\lambda, \tag{4.51} \]

\[ \mathcal{E}=\frac{5\sqrt{5}}{2\sqrt{3}}\, M\, e^{-\frac{5L}{6}+\lambda+\frac{1}{3}\sqrt{L^2-\lambda^2}} . \tag{4.52} \]

The coefficients in formulas (4.51) and (4.52) are obtained from the relations

\[ \int dN=N \quad \text{and} \quad \int \mathcal{E}\,dN=E. \]

The emission angle in the laboratory system is equal to:

\[ \chi=e^{-L-\lambda}. \tag{4.53} \]

Let us note that the following useful formula also holds:

\[ \overline{\ln \chi}=-L, \tag{4.54} \]

valid for the collision of two identical particles. This means that the geometric mean of all scattering angles determines the value of the energy of the incident particle (in the collision of identical particles).

Figure 1 gives the particle distributions calculated from formulas (4.42)—(4.54).

Fig. 1. Differential energy spectra of secondary particles formed in collisions of nucleons; \(E_0\) is the energy of the primary particle. (Calculations performed by L. I. Sarycheva.)

The particle distributions both in angles and in energies turn out to be close to Gaussian when the logarithms of these quantities are used as variables. The distributions have long “tails” on both sides of the maximum. In Appendix II it is indicated that it is convenient to define \(\lambda\) in the following way: \(\lambda=-\ln \operatorname{tg}\frac{\vartheta}{2}\). When

in this case formula (4.44) becomes the following:

\[ dN \simeq e^{\sqrt{L^2-\ln^2 \tan \frac{\vartheta}{2}}}\, \frac{d\vartheta}{\sin^2 \vartheta}. \tag{4.55} \]

This formula gives the explicit distribution of particles over angles. Using formula (4.50), we obtain directly the relation of the particle energy to the angle \(\vartheta\) (in the center-of-inertia system):

\[ \mathcal{E} \simeq M e^{-\frac{L}{6}-\ln \tan \frac{\vartheta}{2} +\frac{1}{3}\sqrt{L^2-\ln^2 \tan \frac{\vartheta}{2}}}. \tag{4.56} \]

An important characteristic of the collision event is the angle within which half of the energy is emitted. We give the values of this quantity, calculated both according to Landau’s theory (formulas (4.51)—(4.53)) and according to Fermi’s theory, which takes noncentral collisions into account (Table VI).

Table VI*)

Energy in eV	\(\vartheta_{1/2}\) according to Fermi	\(\chi_{1/2}\) according to Fermi	\(\vartheta_{1/2}\) according to Landau	\(\chi_{1/2}\) according to Landau
\(10^{14}\)	0.31	\(0.7\cdot 10^{-4}\)	0.10	\(1.3\cdot 10^{-4}\)
\(10^{15}\)	—	\(2.1\cdot 10^{-4}\)	0.0154	\(2.2\cdot 10^{-5}\)
\(10^{16}\)	—	\(6.7\cdot 10^{-5}\)	0.031	\(3.8\cdot 10^{-6}\)

The values of the collision parameter \(\dfrac{r}{R}\), entering into Fermi’s theory, are equal to \(\dfrac{r}{R}=0.712\). \(\vartheta_{1/2}\) is the angle in the center-of-inertia system, \(\chi_{1/2}\) is the angle in the laboratory system.

Despite the fact that the calculation of the angular distribution carried out by Fermi seems to us incorrect, we give its results here for comparison, since they are often used in comparisons with experiment.

It is evident from Table VI that the distribution of energy fluxes is considerably narrower according to the theory being developed than according to Fermi’s theory.

§ 5. COLLISIONS OF PARTICLES WITH DIFFERENT MASSES

Until now only collisions of nucleons with nucleons have been considered. Meanwhile, in experiment, at high energies, collisions of nucleons with nuclei and of nuclei with nuclei are observed, and without a theoretical interpretation of them, strictly speaking, it is altogether impossible to compare the results of the theory with experiment.

*) Borrowed from the article by I. L. Rozental and D. S. Chernavskii\(^{15}\).

It would be wrong to regard the collision of nucleons with nuclei, or of nuclei with nuclei, as a series of collisions of nuclear protons and neutrons. Since the distance between particles in the nucleus is of the order of the radius of nuclear forces, and since in each act of collision several particles are produced, the collision must lead to the process of particle production at once throughout the entire region through which the nucleon passes in the nucleus, or the smaller nucleus in the larger one. The nucleon will interact not with the whole nucleus, but only with a part of the nucleus; that is, it will cut a “tube” out of the nucleus (see ^15)). The number of particles arising in the collision will be connected with the entropy by relation (2.25), where by \(N\) one should understand the sum of the newly formed particles and the nucleons participating in the collision.

It is simplest of all to consider a head-on collision of two identical nuclei of atomic weight \(A\). In this case the considerations that led us to formula (2.28) are fully applicable. It is not difficult to see how formula (2.28) changes in this case (see ^6)). Let the velocity of the incident nucleus be equal to the velocity of the nucleon in a nucleon–nucleon collision. For this its energy must be \(A\) times greater. Since the mass density in the nucleus is approximately equal to the mass density of the proton referred to its sphere of action, the energy density immediately after the collision remains the same as in a nucleon collision. Since, evidently, the Lorentz contraction also remains unchanged, the change in entropy will simply be proportional to the volume of the nuclei, i.e. to \(A\). Thus we finally obtain

\[ N = kA\left(\frac{E}{2AMc^2}\right)^{\frac14} = kA^{3/4}\left(\frac{E}{2Mc^2}\right)^{\frac14}. \tag{5.1} \]

The calculation of the collision of a nucleon with a nucleus, or the collision of nuclei of different weights, is considerably more complicated. We shall now dwell on the collision of a nucleon with a nucleus. The calculation of the number of particles formed in a star reduces to the calculation of the entropy at the first stage of the collision, since in the process of hydrodynamic expansion the entropy does not change. In the collision of a nucleon with a nucleon, for the calculation of the entropy there was no need to consider the mechanism of compression, since the result followed at once from symmetry considerations. The situation is different in the collision of a nucleon with a nucleus. (This problem was considered in the work ^16).)

Let us extend the idea of nuclear matter as a continuous medium also to the consideration of the first stage of the collision—the compression of nuclear matter. Such a consideration is, of course, very approximate and is used only for an estimate. One may speak of the “propagation of a shock wave or traveling wave along a nucleon” only in a conditional sense.

Let us choose a reference frame in which the nucleon and the nucleus have equal but oppositely directed velocities. Owing to Lorentz contraction, in this frame the nucleon and the nucleus are two very thin disks. As a result, at the beginning of the collision the problem may be regarded as one-dimensional. The collision of a nucleon with a nucleus is thus a collision of a nucleon with a tube cut out of the nucleus, whose cross section is equal to the cross section of the nucleon and whose length varies from the diameter of the nucleus to the “diameter” of the nucleon. As the nucleon approaches the tube, shock waves are “propagated” through the substance of the tube and through the substance of the nucleon on both sides (Fig. 2).

Fig. 2.                Fig. 3.

In the reference frame we have chosen, the substance between the shock waves will be at rest. The shock wave traveling through the lighter particle will reach the edge earlier than the shock wave traveling through the tube. When the wave reaches the edge, outflow of matter will begin from the edge (Fig. 3). A rarefaction wave will propagate through the nuclear matter with a velocity equal to the speed of sound in the medium. At the same time, and in the same direction, the shock wave which has not yet reached the edge will be moving. The calculation of the entropy will differ depending on whether the running wave manages to overtake the shock wave before the shock wave reaches the edge and the compression process ends. Since we start from the equation of state \(p=\varepsilon/3\), the speed of sound is

\[ c_0=\sqrt{\left(\frac{\partial p}{\partial \varepsilon}\right)_S}=\frac{1}{\sqrt{3}}. \]

We shall now calculate the velocity of the shock wave \(D\). Let us pass to the coordinate system in which the shock wave is at rest. Then the velocity of the matter behind the shock wave (see Fig. 2) will be equal to \(D\), while in front of the wave

\[ v_2'=\frac{v_2+D}{1+v_2D}, \]

where \(v_2\) is the velocity of the particles in the “equal-velocities” frame. By virtue of the continuity of the energy and momentum fluxes through the discontinuity surface, we have (see \(^{13}\)):

\[ \frac{p_1+D^2\varepsilon_1}{1-D^2} = \frac{p_2+v_2'^2\varepsilon_2}{1-v_2'^2}, \tag{5.2} \]

\[ \frac{D(p_1+\varepsilon_1)}{1-D^2} = \frac{v_2'(p_2+\varepsilon_2)}{1-v_2'^2}. \tag{5.3} \]

Here \(p_1\) and \(\varepsilon_1\) are the pressure and energy density behind the shock wave, and \(p_2\) and \(\varepsilon_2\) are the pressure and energy density ahead of the shock wave. Dividing the first equation by the second, we obtain:

\[ \frac{p_1 + D^2 \varepsilon_1}{D(p_1+\varepsilon_1)} = \frac{p_2 + v_2^{\prime 2}\varepsilon_2}{v'_2(p_2+\varepsilon_2)}. \tag{5.4} \]

Since the velocity \(v_2\) of the incident particles is very close to the speed of light, the velocity \(v'_2\) will also be close to the speed of light. Putting \(v'_2=1\), we see that the right-hand side of equation (5.4) is equal to 1 and, consequently, our results do not depend on the equation of state ahead of the shock wave. Using behind the shock wave, for the substance, the equation of state

\[ p_2=\frac{\varepsilon_2}{3}, \]

we obtain for \(D\) the following relation:

\[ \frac{1}{3}+D^2=\frac{4}{3}D, \]

whence

\[ D=1/3. \tag{5.5} \]

Now one can find the minimum length of the tube \(l_k\) at which the running rarefaction wave catches up with the shock wave. It is determined from the relation

\[ \frac{l_k}{d}=\frac{D+c_0}{c_0-D}, \]

where \(d\) is the “diameter” of the nucleon. Substituting the corresponding values of \(D\) and \(c_0\), we obtain

\[ \frac{l_k}{d}=3.7. \tag{5.6} \]

For tube lengths \(l<l_k\), the change in entropy is calculated simply. One should compute the entropy of the separate portions of the system immediately after the shock wave has passed through them and when they are therefore at rest in our reference frame. It is not difficult to see that the change in entropy of the whole system will then be equal to:

\[ \frac{S}{S_0}=\frac{1}{2}\left(\frac{l}{d}+1\right) \quad \text{for } \quad \frac{l}{d}\leqslant 3.7, \tag{5.7} \]

where \(S_0\) is the change in entropy in a collision of a nucleon with a nucleon.

Formula (5.7) is inaccurate to a significant degree because of the uncertainty of the nucleon “diameter.” If this formula is applied to the collision of a light nucleus with a heavy one, it becomes more accurate. In this case, \(d\) should be understood as the size of the light nucleus.

For tube lengths exceeding \(l_k=3.7d\), the solution becomes more complicated. In this case the rarefaction wave will overtake the shock wave, but will not be able to pass through the discontinuity, since the shock wave moves in the material of the incident core with a velocity exceeding the speed of sound in the material. The rarefaction wave will be reflected from the shock wave. A region will arise, bounded by the shock wave on the right and by the rarefaction wave on the left (Fig. 4). In order to describe the motion of the medium in this region, one should turn to equations (4.15)—(4.15′) for arbitrary one-dimensional motion of a relativistic gas:

\[ 3\frac{\partial^2 \chi}{\partial a^2} -\frac{\partial^2 \chi}{\partial y^2} -2\frac{\partial \chi}{\partial y}=0. \tag{4.15′} \]

Fig. 4.

The independent variables here are the quantities \(a=\operatorname{Arcth} v\), where \(v\) is the velocity of the medium, and \(y=\ln \dfrac{T}{T_0}\), where \(T_0\) is the temperature at \(v=0\). The coordinate \(x\) and the time \(t\) are expressed through the function \(\chi\) by formulas (4.12′). The region of the solution that interests us is bounded by a rarefaction wave on one side and by a shock wave on the other. Let us determine the boundary conditions for the function \(\chi\).

On the boundary with the rarefaction wave we shall have a condition analogous to (4.19):

\[ \chi=0 \quad \text{for } a=\frac{y}{c_0},\ \text{i.e. } a=\sqrt{3}\,y. \tag{5.8} \]

Let us now turn to the condition at the shock wave. Since in this case the material behind the shock wave is not at rest, upon passing to a coordinate system in which the velocity of the discontinuity is zero, the velocity behind the discontinuity will be:

\[ v'_1=\frac{v_1+D}{1+v_1D}, \tag{5.9} \]

where \(v_1\) is the velocity behind the shock wave in the system in which the velocities of the incident particles are equal. Formulas (5.2)—(5.4) remain applicable in the case under consideration if in them \(v'_1\) is substituted in place of \(D\). Hence it follows, in particular, that \(v'_1=\dfrac{1}{3}\). Equation (5.9) gives a relation between \(v_1\) and \(D\) (for \(v'_1=1/3\)), which can be written in the following form:

\[ \frac{dx}{dt}=\frac{1+3\,\operatorname{th} a}{3+\operatorname{th} a}, \tag{5.10} \]

since \(v_1=-\operatorname{th} c\). From relations (5.2)—(5.4) (in which instead of \(D\)

one should substitute \(v_1'\), and the following relation on the shock wave can be obtained:

\[ \left(\frac{\varepsilon}{\varepsilon_0}\right) = \left(\frac{T}{T_0}\right)^4 = \frac{1-v_1}{1+v_1}\,{}^{*}). \tag{5.11} \]

In the variables \(\alpha\) and \(y\) \(\left(v_1=-\operatorname{th}\alpha,\ y=\ln \frac{T}{T_0}\right)\), this means:

\[ \alpha=2y. \tag{5.12} \]

Substituting into equation (5.10) the value of \(\dfrac{dx}{dt}\), calculated by formulas (4.12), and taking into account relation (5.12), we finally obtain the following condition on the shock wave:

\[ \left(3\frac{\partial}{\partial y}+5\frac{\partial}{\partial \alpha}\right) \left(1-\frac{\partial}{\partial \alpha}\right)\chi=0 \quad \text{for } \alpha=2y. \tag{5.13} \]

We are interested in the change of entropy during the passage of the shock wave. The change of entropy before the running wave has caught up with the shock wave is given by (5.7). The further change of entropy is written in the following form:

\[ S_1=\sigma_0\int_{t_1'}^{t_2'} s u_1\,dt'. \tag{5.14} \]

Here \(\sigma_0\) is the cross-section of the tube, \(S\) is the entropy density behind the shock wave, \(u_1=\dfrac{v_1'}{\sqrt{1-v_1'^2}}\), where \(v_1'\) is the velocity of the matter behind the shock wave, \(t_1'\) is the instant of time when the running wave catches up with the shock wave, and \(t_2'\) is the instant of time when the shock wave reaches the right edge and, thus, the compression process ends. \(dt'\) is the element of time in the system in which the discontinuity is at rest:

\[ dt'=dt\sqrt{1-D^2}. \]

Using formulas (4.12′), (5.12), we obtain:

\[ S_1=\sigma_0\,\frac{1}{9}\int_0^{y_k} e^{2y} \left[ \frac{\partial}{\partial\alpha} \left(\frac{\partial}{\partial y}-1\right)\chi \right]_{\alpha=2y} \,dy. \tag{5.15} \]

Here \(y_k\) is the value of \(y\) at the instant when the shock wave reaches the edge of the system.

\[ {}^{*})\ \text{See Appendix III.} \]

Carrying out the corresponding rather complicated calculations, the details of which are given in work \({}^{16}\), we obtain:

\[ \frac{S}{S_0}=0.92\left(\frac{l}{d}-\frac14\right)^{\frac34} \quad \text{for } \frac{l}{d}>3.7, \tag{5.16} \]

where \(S_0\) is the change in entropy in a collision of a nucleon with a nucleon, and \(S\) is the total change in entropy in a collision of a nucleon with a tube of length \(l\).

If formulas (5.7) and (5.16) are averaged over all possible collisions in the nucleus, from any collision up to the collision of the incident nucleon with a peripheral one, then we obtain the following expressions:

\[ \frac{S}{S_0} = \frac13\, \frac{A-(2A^{1/3}-1)^{3/2}}{(A^{1/3}-1)^2} +0.5 \quad \text{for } A<51, \tag{5.17} \]

\[ \frac{S}{S_0} = \frac{4}{(A^{1/3}-1)^2} \left[ 0.167\left(A^{11/12}-A_0^{11/12}\right) +\frac{1}{12}\left\{A_0-(2A^{1/3}-1)^{3/2}\right\} \right] - 0.6\,\frac{A^{2/3}-A_0^{2/3}}{(A^{1/3}-1)^2} +0.5 \quad \text{for } A>51,\ A_0=51. \tag{5.18} \]

The dependence obtained can, with an accuracy of up to \(5\%\), be approximated by the following formula:

\[ \frac{S}{S_0}=A^{0.19}. \tag{5.19} \]

Here \(A\) is the atomic weight. Hence for the number of particles we have:

\[ \frac{N}{N^{\mathrm H}}=A^{0.19}, \tag{5.20} \]

where \(N^{\mathrm H}\) is the number of particles formed in a collision of a nucleon with a nucleon. The dependence of the multiplicity on the atomic weight proves, in agreement with experiment (see \({}^{15}\)), to be weak. Thus, the multiplicity in lead \((A=207)\) differs from the multiplicity in nitrogen \((A=14)\) by only a factor of \(1.7\).

The angular and energy distribution in collisions of a nucleon with a nucleus apparently differs little, within the accuracy of our calculations, from the corresponding distributions in a collision of a nucleon with a nucleon.

APPENDIX I

Let us dwell in more detail on the derivation of formula (4.30) from (4.29). Formula (4.29) can be written in the following form:

\[ h(-y)=-\frac{\chi_1}{l\sqrt{3}} = \frac{1}{2\pi i} \int_{\delta-i\infty}^{\delta+i\infty} \frac{ e^{-\sqrt{p^2-1}\,\frac{\alpha}{\sqrt{3}}-yp} }{ \sqrt{p^2-1}\,(p+2) }\,dp. \tag{1.1} \]

According to the basic relations of operational calculus (see [17], p. 29):

\[ \left. \begin{aligned} f(p)&=p\int_{-\infty}^{+\infty} e^{-pt}h(t)\,dt,\\[4pt] h(t)&=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty} e^{pt}\frac{f(p)}{p}\,dp. \end{aligned} \right\} \tag{1.2} \]

Thus, the function

\[ f(p)=\frac{e^{-\sqrt{p^{2}-1}\,\frac{\alpha}{\sqrt{3}}\frac{1}{p}}}{\sqrt{p^{2}-1}(p+2)} \]

is the “image” of the “original” \(h(-y)\).

We shall next use a rule from operational calculus known as the rule of “multiplication of images,” or “convolution” (see [17], p. 55). According to this rule, if the “image” \(f_1(p)\) corresponds to the “original” \(h_1(t)\), and the “image” \(f_2(p)\) to the “original” \(h_2(t)\), then the “image” \(f(p)=\dfrac{f_1(p)f_2(p)}{p}\) corresponds to the “original”

\[ h(t)=\int_{-\infty}^{+\infty} h_2(\xi)h_1(t-\xi)\,d\xi . \tag{1.3} \]

Thus, if the “originals” \(h_1(p)\) and \(h_2(t)\) are known, then, using relation (1.3), one can also find the “original” for the “image” \(f(p)\). We decompose the image \(f(p)\) into the following factors:

\[ f_1(p)=\frac{p}{p+2},\qquad f_2(p)=\frac{e^{-\sqrt{p^{2}-1}\,\frac{\alpha}{\sqrt{3}}\frac{1}{p}}}{\sqrt{p^{2}-1}} . \tag{1.4} \]

To the “image” \(f_1(p)\) there corresponds the “original” \(h_1(-y)\):

\[ h_1(-y)=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty} \frac{1}{p+2}e^{-yp}\,dp =e^{2y}U(-y). \tag{1.5} \]

To the “image” \(f_2(p)\) there corresponds the “original” \(h_2(-y)\) (see [17], formula XV, 15):

\[ h_2(-y)=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty} e^{-yp}\frac{1}{\sqrt{p^{2}-1}} e^{-\frac{\alpha}{\sqrt{3}}\sqrt{p^{2}-1}}\,dp = \]

\[ =I_0\left(\sqrt{y^{2}-\frac{\alpha^{2}}{3}}\right) U\left(-y-\frac{\alpha}{\sqrt{3}}\right). \tag{1.6} \]

Here \(I_0\) is the Bessel function of imaginary argument,

\[ U(t)= \begin{cases} 1, & \text{for } t>0,\\ 1/2, & \text{for } t=0,\\ 0, & \text{for } t<0. \end{cases} \]

Substitution of expressions (I.5) and (I.6) into formula (I.3) gives:

\[ h(-y)=\int_{a/\sqrt{3}}^{-y} e^{z(y+y_1)} I_0\left(\sqrt{y_1^2-\frac{a^2}{3}}\right)\,dy_1. \tag{I.7} \]

From this we immediately obtain formula (4.30)

\[ \chi=i\sqrt{3}\,e^y\int_{a/\sqrt{3}}^{-y} e^{2y_1} I_0\left(\sqrt{y_1^2-\frac{a^2}{3}}\right)\,dy_1. \]

In Khalatnikov’s paper\(^{14}\), in formula (4.36) for \(\chi\), the factor \(i\sqrt{3}\) is omitted.

APPENDIX II

In order to obtain formulas (4.51), (4.52), we shall transform from distributions in the center-of-gravity system to distributions in the laboratory coordinate system.

Let us first consider the transformations of angles.

The emission angle \(\chi\) in the laboratory system is related to the angle \(\vartheta\) in the center-of-inertia system by the following formula:

\[ \operatorname{tg}\chi \simeq \chi = \frac{v'\sqrt{1-V^2}\sin\vartheta}{v'\cos\vartheta+V}, \tag{II.1} \]

where \(v'\) is the particle velocity in the center-of-inertia system, and \(V\) is the velocity of the center of inertia with respect to the laboratory system. Let us write the denominator of this expression in the form:

\[ v'\cos\vartheta+V \simeq v'\cos\vartheta+v' + \frac{1}{2}(V^2-v'^2), \]

since

\[ V^2-v'^2=(V+v')(V-v')\simeq 2(V-v'). \]

Taking into account that \(V\) is closer to 1 than \(v'\), we have:

\[ v'\cos\vartheta+V \simeq (1+\cos\vartheta)+\frac{1}{2}(1-v'^2). \]

Using the relation \(u'=\dfrac{1}{\sqrt{1-v'^2}}\), we obtain:

\[ v'\cos\vartheta+V = 1+\cos\vartheta+\frac{1}{2u'^2}. \tag{II.2} \]

The last term can play an essential role only for angles \(\vartheta\) close to \(\pi\). In order that in this case as well the last term be small, the inequality

\[ \vartheta \gg \frac{1}{u'} \tag{II.3} \]

must be satisfied.

According to (4.45) \(\vartheta \approx e^{-\lambda}\), and from (4.50) it follows that

\[ u' \approx e^{-\frac{L}{6}+\lambda+\frac{1}{3}\sqrt{L^{2}-\lambda^{2}}}. \]

Thus inequality (II.3) means that

\[ e^{\frac{L}{6}-\frac{1}{3}\sqrt{L^{2}-\lambda^{2}}} \ll 1 . \tag{II.4} \]

This inequality is always satisfied, since for \(\lambda=0\) the right-hand side is equal to \(e^{-L/6}\), and only for \(\lambda=\lambda_{\max}\) does it become equal to 1. Putting \(v'=1\) in the numerator of expression (II.1), and replacing \(v'\cos\vartheta+V\) by \(1+\cos\vartheta\), we obtain

\[ \chi=\sqrt{1-V^{2}}\,\operatorname{tg}\frac{\vartheta}{2}. \tag{II.5} \]

Independently of the concrete calculations, the scattering in the center-of-inertia system in any collision of two identical particles is symmetric. This means that the angles \(\vartheta\) occur as often as the angles \(\pi-\vartheta\). But

\[ \operatorname{tg}\frac{\pi-\vartheta}{2}=\frac{1}{\operatorname{tg}\frac{\vartheta}{2}} . \]

Let us average \(\ln\chi\) over all particles. Then we obtain:

\[ \overline{\ln\chi}=\ln\sqrt{1-V^{2}}=-L, \tag{II.6} \]

i.e. formula (4.54).

In order to take into account angles of order unity, let us define \(\lambda\) as follows:

\[ \lambda=-\ln\operatorname{tg}\frac{\vartheta}{2}. \tag{II.7} \]

If in formula (II.5) one substitutes \(\operatorname{tg}\frac{\vartheta}{2}=e^{-\lambda}\) and \(\operatorname{tg}\frac{\pi-\vartheta}{2}=e^\lambda\) for particles flying in the center-of-inertia system in opposite directions, then it is not difficult to see that the transition from particles going to the right to particles going to the left (in the center of inertia) is effected in it by changing the sign of \(\lambda\). Therefore one may write:

\[ \chi=e^{-L-\lambda}. \tag{II.8} \]

Formulas (II.8) and (4.44) thus give the angular distribution of particles in the laboratory system, with \(\lambda\) taking positive and negative values. These formulas essentially coincide with formulas (4.51) and (4.53).

Let us now turn to the transformation of the energy from the center-of-gravity system to the laboratory system.

Let \(\mathcal E\) be the energy of a particle in the center-of-inertia system. If the particle flies to the right, then its energy in the laboratory system is equal to:

\[ \mathcal E'=\frac{\mathcal E+P_xV}{\sqrt{1-V^2}}; \tag{II.9} \]

where \(P_x\) is the projection of the momentum onto the direction of the velocity of the center of inertia,

\[ P_x=\sqrt{\mathcal E^2-(\mu c^2)^2-P_y},\qquad P_y=P\sin\vartheta, \]

where \(P\) is the absolute value of the momentum, \(\mu\) is the mass of the particle. If the velocity of the particles in the center-of-inertia system approaches the speed of light, then

\[ \mathcal E'=\frac{\mathcal E(1+\cos\vartheta)}{\sqrt{1-V^2}}. \]

Thus,

\[ \mathcal E'=\mathcal E e^L\frac{2}{1+e^{-2\lambda}}\simeq \mathcal E e^L. \tag{II.10} \]

If the particles move to the left, then

\[ \mathcal E'=\frac{\mathcal E-P_xV}{\sqrt{1-V^2}}=\frac{1-\cos\vartheta}{\sqrt{1-V^2}}\,\mathcal E, \]

since

\[ \vartheta \gg \frac{1}{u'}\simeq \frac{\mu c^2}{\mathcal E}. \]

Thus,

\[ \mathcal E'=\frac{\mathcal E e^{L-2\lambda}\cdot 2}{1+e^{-2\lambda}}\simeq \mathcal E e^{L-2\lambda}. \tag{II.11} \]

Substituting these expressions into equation (4.50), we obtain:

\[ \mathcal E'\simeq M e^{\frac{5L}{6}+\lambda+\frac{1}{3}\sqrt{L^2-\lambda^2}}. \tag{II.12} \]

This formula has the property that it describes, in the laboratory system, particles which move in the center-of-gravity system both to the right and to the left, if \(\lambda\) is assigned both signs.

APPENDIX III

Let us derive relation (5.11).

Let, in the coordinate system in which the shock wave is at rest, the velocity behind the shock wave be \(v'_1\), and the velocity ahead of the wave be \(v'_2\).

Equations (5.2) and (5.3) will then be written in the following form:

\[ \frac{p_1+v_1^{\prime 2}\varepsilon_1}{1-v_1^{\prime 2}} = \frac{p_2+v_2^{\prime 2}\varepsilon_2}{1-v_2^{\prime 2}}, \tag{III.1} \]

\[ \frac{v_1'(p_1+\varepsilon_1)}{1-v_1^{\prime 2}} = \frac{v_2'(p_2+\varepsilon_2)}{1-v_2^{\prime 2}}. \tag{III.2} \]

The quantities \(v_1'\) and \(v_2'\) are related to the velocities \(v_1\) and \(v_2\) in the coordinate system in which the velocities of the incident particles are equal, by the relations:

\[ v_1'=\frac{v_1+D}{1+v_1D}, \tag{III.3} \]

\[ v_2'=\frac{v_2+D}{1+v_2D}. \tag{III.4} \]

Eliminating \(v_1'\) or \(v_2'\), respectively, from equations (III.1) and (III.2), we obtain the following expressions:

\[ v_1^{\prime 2}=\frac{(p_1-p_2)(p_1+\varepsilon_2)}{(\varepsilon_1-\varepsilon_2)(p_2+\varepsilon_1)}, \tag{III.5} \]

\[ v_2^{\prime 2}=\frac{(p_2-p_1)(p_2+\varepsilon_1)}{(\varepsilon_2-\varepsilon_1)(p_1+\varepsilon_2)}. \tag{III.6} \]

From the last two formulas we obtain:

\[ v_1'v_2'=\frac{p_1-p_2}{\varepsilon_1-\varepsilon_2}. \tag{III.7} \]

To simplify the derivation, let us assume that on both sides of the shock wave the equation of state is the same \(\left(p=\frac{\varepsilon}{3}\right)\). We note, however, that at large velocities \(v_2\) it is sufficient to assume the validity of this equation only behind the shock wave. From (III.7) it follows:

\[ v_1'v_2'=\frac{1}{3}. \tag{III.7'} \]

Equation (III.5) takes the following form:

\[ v_1^{\prime 2}=\frac{1}{3}\cdot\frac{p_1+\varepsilon_2}{\frac{\varepsilon_2}{3}+3p_1} \tag{III.5'} \]

and hence:

\[ p_1=\varepsilon_2\,\frac{1-v_1^{\prime 2}}{9v_1^{\prime 2}-1}. \tag{III.8} \]

Substituting formulas (III.3) and (III.4) into equation (III.7), by means of simple transformations we obtain the expression for \(D\):

\[ D=\frac{-(v_1+v_2)+\sqrt{(v_1+v_2)^2-(3v_1v_2-1)(3-v_1v_2)}}{3-v_1v_2}. \tag{III.9} \]

We now find the ratio \(\dfrac{p_{1v_1}}{p_{10}}\), where \(p_{1v_1}\) is the value of the pressure at the velocity behind the shock wave equal to \(v_1\), and \(p_{10}\) is the value of the pressure at the velocity behind the shock wave equal to zero. This ratio, according to (III.8), is equal to:

\[ \frac{p_{1v_1}}{p_{10}} = \left[\frac{(1-v_1^{\prime 2})}{(9v_1^{\prime 2}-1)}\right] \cdot \left[\frac{(9v_1^{\prime 2}-1)}{(1-v_1^{\prime 2})}\right]_0 . \tag{III.10} \]

Since, at large velocities \(v_2\) tending to 1, also \(v_2' \to 1\) (see III.4), while, according to relation (III.7), \(v_1' \to \dfrac{1}{3}\), the ratio (III.10) becomes indeterminate. Therefore one must take the limiting transition:

\[ \left[\frac{p_{1v_1}}{p_{10}}\right]_{v_2\to 1} = \frac{[9v_1^{\prime 2}-1]_{v_1=0}}{[9v_1^{\prime 2}-1]_{v_1}} = \left\{ \frac{ \left[\dfrac{\partial v_1'}{\partial v_2}\right]_{v_1=0} }{ \left[\dfrac{\partial v_1'}{\partial v_2}\right]_{v_1} } \right\}_{v_2=1}. \tag{III.11} \]

But

\[ \frac{\partial v_1'}{\partial v_2} = \left[ \frac{\partial v_1'}{\partial D} \frac{\partial D}{\partial v_2} \right]_{v_2=1}. \tag{III.12} \]

The derivative \(\dfrac{\partial v_1'}{\partial D}\) is determined from equation (III.3), and the derivative \(\dfrac{\partial D}{\partial v_2}\) from equation (III.9):

\[ \left(\frac{\partial v_1'}{\partial D}\right)_{v_2=1} = \frac{(3-v_1)^2}{9(1-v_1)}; \qquad \left(\frac{\partial D}{\partial v_2}\right)_{v_2=1} = -\frac{3(v_1+1)^2}{2(3-v_1)^2}. \tag{III.13} \]

Hence it is not difficult to obtain the expression for \(\dfrac{p_{1v}}{p_{10}}\):

\[ \frac{p_{1v}}{p_{10}} = \frac{1-v_1}{1+v_1}. \tag{III.14} \]

By virtue of the equation of state, formula (5.11) follows directly from expression (III.14).

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