NUCLEAR PROCESSES AT HIGH ENERGIES *)

E. Fermi

Part I

1. INTRODUCTION

Meson theory has been an important factor in the development of physics during the last 15 years. One of the remarkable achievements of the theory is the prediction of the possibility of meson production in collisions of high-energy nucleons. At comparatively low energies only one meson can be formed. At higher energies multiple emission of mesons becomes possible.

In the present work an attempt is made to develop a rough theoretical approximation for calculating the outcome of nuclear collisions at very high energies. In particular, such collisions of two nucleons are considered in which several $\pi$-mesons **) and, possibly, some number of antinucleons arise as a result.

For the analysis of processes of this kind the usual theory, which treats the formation and absorption of $\pi$-mesons as a small perturbation, is wholly inapplicable. Indeed, because of the large numerical value of the interaction constant, higher approximations exceed the preceding ones and the series of successive approximations diverges. We shall attempt to investigate the possibility of another method, proceeding from entirely different considerations. The basic idea is as follows.

In the collision of two nucleons with very high energy (in the center-of-inertia system) their energy is suddenly released in

) Part I is a translation of an article by E. Fermi published in Progress of Theoretical Physics 5, No. 4 (1950), devoted to the 15th anniversary of Yukawa’s theory; Part II is a translation of an article by the same author in Physical Reviews 81*, 683 (1951).

**) The author uses the term “pions,” which, in the opinion of the editors, is unfortunate.

a small effective volume surrounding the nucleons. Visually this phenomenon may be imagined as a collision in which two nucleons, together with their surrounding pion “shell,” collide with one another in such a way that the entire effective volume of space filled with nucleons and with the pion field surrounding them suddenly becomes saturated with a very large portion of energy. Since the interaction of the nucleons with the pion field is large, we may expect that, in accordance with statistical laws, this energy will rapidly be distributed among the various degrees of freedom present in the given volume. In that case one can statistically calculate the probability of the formation, in this small volume, of a definite number of pions with a prescribed energy distribution. It is further assumed that the energy concentration rapidly decreases and that the particles into which the energy is converted fly out in all directions.

One may think that such a picture probably corresponds to reality as a limiting case opposite to the case of perturbation theory. At the same time, the study of a theory that deviates from the unknown truth in the direction opposite to that of the usual theory may prove useful, if only because in this way the actual picture of the phenomena may be enclosed between two known theoretical limits. It may be hoped that a theory of the proposed kind will be a sufficiently good approximation to reality in the region of very high energies, when the number of possible states with a given energy is large, which sharply increases the probability of the establishment of statistical equilibrium.

The state that we regard as a kind of statistical equilibrium is characterized as follows. First of all, the laws of conservation of charge and momentum must be satisfied. Further, one may assume that only those states participate in the statistical equilibrium which are most rapidly reached by the system from the initial state. For example, radiative processes with the formation of photons certainly will not have time to begin. The only transitions that can be regarded as the fastest are transitions predicted by Yukawa’s theory. A sequence of such transitions, beginning from the initial state in which two colliding nucleons are present, can lead only to the formation of a definite number of charged or neutral pions and also, presumably, to the formation of nucleon–antinucleon pairs. Below only such processes are discussed. We emphasize the existence of an additional conservation law for the difference between the number of nucleons and the number of antinucleons.

The proposed theory has some similarity with the point of view of Heisenberg[^1], who describes the collision between

nucleons of high energy, assuming that the π-meson “liquid” surrounding the nucleons is set into a kind of turbulent motion by the impact energy. He uses quantitative laws of turbulence to estimate the distribution of the energy of turbulent motion among vortices of various sizes. Turbulence represents the beginning of an approach to the thermal equilibrium of a liquid. It describes the transition of the energy of motion into many states with increasing wave numbers. From a qualitative point of view, the theory we propose is an attempt to carry Heisenberg’s point of view to its extreme consequences with respect to the possibility of the actual attainment of a state of statistical equilibrium.

Multiple production of mesons was also investigated in the interesting work of Lewis, Oppenheimer, and Wouthuysen². The authors emphasize the importance of the strong coupling expected in the pseudoscalar theory for processes of high multiplicity.

In the proposed theory there is only one arbitrary parameter—the effective volume \(\Omega\), in which the energy of the colliding nucleons is released. Since the π-meson field surrounding the nucleons has an extent of the order \(\hbar/\mu c\), where \(\mu\) is the mass of the π-meson, one may assume that \(\Omega\) also has linear dimensions of the same order. So long as the Lorentz contraction can be neglected, as \(\Omega\) one may take a sphere of radius \(\hbar/\mu c\). However, if the nucleons (in the center-of-inertia system) impinge on one another with very large energy, then the π-meson field surrounding them undergoes Lorentz contraction, and the volume \(\Omega\) correspondingly decreases.

In accordance with these considerations, the volume \(\Omega\) is taken to depend on the energy:

\[ \Omega=\Omega_0\,\frac{2Mc^2}{W}, \tag{1} \]

where \(\Omega_0\) is the effective volume without Lorentz contraction, \(W\) is the total energy of the two colliding nucleons in the center-of-inertia system, and \(M\) is the nucleon mass. The factor \(\frac{2Mc^2}{W}\) is the Lorentz-contraction coefficient. The volume \(\Omega_0\) may be taken in the form of a sphere of radius \(R\):

\[ \Omega_0=\frac{4}{3}\pi R^3. \tag{2} \]

It follows from the assumptions of the theory that, putting

\[ R=\frac{\hbar}{\mu c}=1.4\cdot10^{-13}\ \text{cm}, \tag{3} \]

one may hope to obtain satisfactory agreement with the known facts.

Such a choice of the effective volume, although satisfactory in order of magnitude, is, of course, arbitrary and may be replaced by another in order to improve agreement with experiment. Increasing \(\Omega_0\) raises the probability of processes with the formation of a larger number of particles.

According to the viewpoint being developed, the total cross section for the collision of two nucleons will be of the order of the geometrical cross section of the \(\pi\)-meson cloud. In the numerical calculations the total cross section is taken by us to be equal to the area of a circle of radius \(R\):

\[ \sigma_{\text{tot.}}=\pi R^2 . \tag{4} \]

If \(R\) is taken according to (3), then \(\sigma_{\text{tot.}}=6\cdot 10^{-26}\ \text{cm}^2\). To compute the cross section of a process in which, for example, three \(\pi\)-mesons are produced, the total cross section (4) must be multiplied by the relative probability that, as a result of the collision of two nucleons, precisely three \(\pi\)-mesons arise, and not some other number of particles of one kind or another.

The probability of a transition to a state of a given type is proportional to the square of the effective matrix element and to the density of states per unit energy interval (statistical weight). Our assumption about statistical equilibrium consists in admitting that the square of the effective matrix element is proportional to the probability that all particles corresponding to the given state are simultaneously found in the volume \(\Omega\). For example, in the case of \(n\) completely independent particles with momenta \(p_1, p_2, \ldots, p_n\), this probability is \((\Omega/V)^n\), where \(V\) is the normalization volume. The statistical weight in the present case is equal to

\[ \left(\frac{V}{8\pi^3\hbar^3}\right)^n \frac{d}{dW}Q(W), \]

where \(Q(W)\) is the volume of momentum space corresponding to the total energy \(W\). The probability of occurrence of the state under consideration is assumed by us to be proportional to the product

\[ S(n)=\left(\frac{\Omega}{8\pi^3\hbar^3}\right)^n \frac{dQ(W)}{dW}. \tag{5} \]

There are, however, complications arising from the fact that the particles are not independent.

a) In the center-of-inertia system, only the coordinates and momenta of \(n-1\) particles out of \(n\) are independent variables. For this reason, in (5) the exponent of \(\Omega\) will be \(n-1\), and not \(n\). Moreover, the momentum space \(Q(W)\) will be \(3(n-1)\)-dimensional, and not \(3n\)-dimensional.

b) Some of the particles may be identical, and this must be taken into account when computing \(Q(W)\).

c) Some of the particles may possess spin, and in that case the corresponding multiplicity of states should be taken into account.

d) Conservation of angular momentum restricts statistical equilibrium only to those states whose angular momentum is equal to the angular momentum of the two colliding nucleons. In all the cases under consideration the wavelength \(\lambda\) of the nucleons is much smaller than the radius \(\hbar/\mu c\) of the sphere of interaction. Accordingly, it makes sense to consider collisions with different values of the impact parameter \(b\) (\(b\) is equal to the distance between the two straight lines along which the nucleons move before the collision). In units of \(\hbar\), the angular momentum is \(l=b/\lambda\). The cross section for collision with impact parameter between \(b\) and \(b+db\) is \(2\pi b\,db=2\pi\lambda^2 l\,dl\). Collisions with different impact parameters must be considered separately, and for each \(l\) one must calculate the probabilities of the various possible processes. The cross section for a process of a given kind is obtained by summing over all \(l\).

It has been found that in most cases the results obtained in this way differ only by an insignificant numerical factor from the results obtained when conservation of angular momentum is neglected.

To simplify the calculations, conservation of angular momentum is, as a rule, not taken into account. In typical cases, however, corrections for conservation of angular momentum are given.

2. EXAMPLE. FORMATION OF \(\pi\)-MESONS IN COLLISIONS OF LOW-ENERGY NUCLEONS

As a first example let us consider the simplest case—the formation of \(\pi\)-mesons in the collision of two nucleons with an energy close to the threshold for \(\pi\)-meson formation. It should be noted that in this case the statistical approximation may prove erroneous because of the small magnitude of the energy and, correspondingly, the small number of possible states. To begin with, we shall simplify the problem and shall not pay attention to spin and to the various possible charges of the nucleons and \(\pi\)-mesons. Let us consider the process in the center-of-mass system. Let \(T/2\) be the kinetic energy of each of the colliding nucleons. A \(\pi\)-meson can be emitted for \(T>\mu c^2\). We shall assume that this inequality is satisfied, but shall suppose that the kinetic energy is so close to the threshold that, as a result of the reaction, only nonrelativistic nucleons and a \(\pi\)-meson can arise.

Conservation of energy leads only to two kinds of possible states: states (a), in which elastic scattering of two nucleons occurs without formation of a \(\pi\)-meson, and states (b),

in which a \(\pi\)-meson is formed and after the reaction three particles are present.

The statistical weight of the states \((a)\) is obtained as follows. Since the momenta of the nucleons are equal in magnitude and antiparallel, momentum space will be three-dimensional, and in formula (5) one must put \(n=1\). The reduced mass in our case is \(M/2\), the momentum is \(p=\sqrt{MT}\), and the volume of the phase sphere is \(Q(T)=4\pi p^3/3\); correspondingly, according to (5), we arrive at the usual expression for the statistical weight:

\[ S_2=\frac{\Omega\cdot M^{3/2}}{4\pi^2\hbar^3}\sqrt{T}. \tag{6} \]

\(S_2\) should be compared with the statistical weight \(S_3\) of the states \((b)\), in which three particles are present: two nucleons and a \(\pi\)-meson. Since in this case of the three particles only two are independent (conservation of momentum), in (5) one must put \(n=2\). Conservation of momentum must also be taken into account in calculating the volume of momentum space. Let \(p\) be the momentum of the \(\pi\)-meson and let \(-\frac{1}{2}p\pm q\) be the momenta of the nucleons. In that case the kinetic energy is

\[ T_1=\left(\frac{1}{2\mu}+\frac{1}{4M}\right)p^2+\frac{1}{M}q^2, \tag{7} \]

where \(T_1=T-\mu c^2\) is the kinetic energy remaining after formation of the \(\pi\)-meson. Formula (7) describes a six-dimensional ellipsoid in the momentum space of the vectors \(p\) and \(q\). The volume of this ellipsoid is

\[ Q_3=\frac{\pi^3}{3!}\left(\frac{4M^2\mu}{2M+\mu}\right)^{3/2}T_1^3; \tag{8} \]

the factor \(\pi^3/3!\) for the six-dimensional sphere is the analogue of the factor \(4\pi/3\) for the three-dimensional sphere. Substituting \(Q_3\) into (5), we have:

\[ S_3=\frac{\Omega^2}{16\pi^3\hbar^6}\left(\frac{M^2\mu}{2M+\mu}\right)^{3/2}T_1^3\simeq \frac{\Omega^2 M^3\mu^{3/2}T_1^3}{32\sqrt{2}\,\pi^3\hbar^6}, \tag{9} \]

where we have made a simplification, taking into account that \(\mu\ll M\). The probabilities of the events \((a)\) and \((b)\) are proportional to \(S_2\) and \(S_3\). Since \(S_3\) is very small, as the probability of \(\pi\)-meson formation one may take \(S_3/S_2\):

\[ \frac{S_3}{S_2}= \frac{\Omega\mu^{3/2}}{8\sqrt{2}\pi\hbar^3}\, \frac{(T-\mu c^2)^2}{\sqrt{T}} \simeq \frac{\Omega\mu\,(T-\mu c^2)^2}{8\sqrt{2}\pi\hbar^3 c}; \tag{10} \]

we have replaced \(T\) in the denominator by \(\mu c^2\), since in our case the kinetic energy only slightly exceeds the threshold of the reaction. The Lorentz contraction of the volume in this case may be neglected, and for \(\Omega\) one may take \(\Omega_0\) from formulas (2) and (3). Finally,

\[ \frac{S_3}{S_2}=\frac{1}{6\sqrt{2}}\left(\frac{T}{\mu c^2}-1\right)^2. \tag{11} \]

[The cross section for formation of a $\pi$-meson is the product of the total cross section (4) and the probability (11). For example, in the bombardment of nucleons at rest by nucleons with energy $345\ \mathrm{Mev}$ (the highest proton energy at Berkeley), $T=165\ \mathrm{Mev}$ in the center-of-inertia system. On the other hand, $\mu c^2=140\ \mathrm{Mev}$, and the probability of formation of a $\pi$-meson is $S_3/S_2=0.0038$. Thus, at this bombardment energy, $0.4$ percent of nucleon collisions should lead to the formation of a $\pi$-meson.

A more detailed analysis shows that in collisions of two protons the probability of formation of a positive $\pi$-meson is twice as large as (11) and is equal to $0.0076$. Indeed, when a positive $\pi$-meson is formed, a proton and a neutron remain in the final state, and the statistical weight of these particles, owing to their nonidentity, is twice as large as the statistical weight of two protons. Similarly, the probability of formation of a positive $\pi$-meson in a collision of a proton with a neutron is twice as small as (11) and is equal to $0.0019$, as is the probability of formation of a negative $\pi$-meson under the same conditions.

As an example let us consider the bombardment of carbon by protons with energy $345\ \mathrm{Mev}$. In this case collisions of the incident proton with nuclear protons and neutrons are equally probable. Therefore, the probability of formation of a positive $\pi$-meson is $0.0076/2+0.0019/2=0.0048$, while the probability of formation of a negative $\pi$-meson is only $0.0019/2=0.001$. The nuclear cross section of carbon is approximately $3\cdot 10^{-25}\ \mathrm{cm}^2$, and, multiplying it by the probabilities of the processes, we obtain a cross section of $1.4\cdot 10^{-27}\ \mathrm{cm}^2$ for the formation of positive $\pi$-mesons and $3\cdot 10^{-28}\ \mathrm{cm}^2$ for negative ones. Taking into account the extreme crudeness of the calculations, it must be acknowledged that the obtained quantities are in unexpectedly good agreement with the experimental data. Up to this point we have not taken into account conservation of angular momentum. After formation of the $\pi$-meson the energy of the three particles (the $\pi$-meson and two nucleons) is small, and they will be in an $s$-state.

Consequently, for the process the essential role will be played by initial states only with zero angular momentum, for which the largest cross section is $\pi\lambda^2$, which is much smaller than (4). However, in the present case the relative weight of elastic scattering compared with $\pi$-meson generation is also smaller, since only particles with zero angular momentum are scattered.

Calculations show that these two effects almost cancel each other, and taking angular-momentum conservation into account changes the previously obtained cross sections for $\pi$-meson formation by a factor of $2/3$. The angular distribution of the nucleons changes more substantially. If angular-momentum conservation is neglected, one should expect that in the center-of-inertia system the scattering of nucleons will be spherically symme-

tric. In the case where momentum is conserved, however, the cross section for elastic scattering per unit solid angle is no longer constant, but varies approximately as \(1/\sin \theta\), where \(\theta\) is the scattering angle in the center-of-inertia system.

3. FORMULAS FOR STATISTICAL WEIGHTS

We shall give the standard formulas for the statistical weights \(S\) for a number of simple cases.

Let us first consider the case of the production of \(n\) particles with masses \(m_1, m_2, \ldots, m_n\). Neglecting spin and conservation of momentum, and regarding the particles as statistically independent, we obtain for \(S\) two formulas, corresponding to the classical and the extremely relativistic cases:

\[ S_n= \frac{(m_1m_2\ldots m_n)^{3/2}\Omega^n} {2^{3n/2}\pi^{3n/2}\hbar^{3n}} \frac{T^{\frac{3n}{2}-1}} {\left(\frac{3n}{2}-1\right)!} \quad \text{(class.)}, \tag{12} \]

\[ S_n= \frac{\Omega^n}{\pi^{2n}\hbar^{3n}c^{3n}} \frac{W^{3n-1}}{(3n-1)!} \quad \text{(ext. rel.)}. \tag{13} \]

In (12) \(T\) is the total kinetic energy of the \(n\) particles, while in (13) \(W\) is their total kinetic energy plus the rest energy. It is easy to obtain \(S\) also for the case in which \(s\) particles (usually nucleons) are classical, and \(n\) particles (usually \(\pi\)-mesons) are extremely relativistic. Again neglecting spin, statistical dependence, and conservation of momentum, and assuming that all classical particles have the nucleon mass \(M\), we find:

\[ S(s,n)= \frac{M^{\frac{3s}{2}}\Omega^{n+s}} {2^{\frac{3s}{2}}\pi^{2n+\frac{3s}{2}}\hbar^{3s+3n}c^{3n}} \frac{(W-sMc^2)^{3n+\frac{3s}{2}-1}} {\left(3n+\frac{3s}{2}-1\right)!}. \tag{14} \]

Or, equivalently,

\[ S(s,n)= \frac{M^{\frac{3s}{2}}\Omega^{\frac{s}{2}+\frac{1}{3}}} {2^{\frac{3s}{2}}\pi^{\frac{s}{2}+\frac{2}{3}}\hbar^{\frac{3s}{2}+1}c^{\,1-\frac{3s}{2}}} \frac{ \left( \dfrac{\Omega^{1/3}(W-sMc^2)} {\pi^{2/3}\hbar c} \right)^{3n+\frac{3s}{2}-1} } {\left(3n+\frac{3s}{2}-1\right)!}. \tag{15} \]

Using (15), one can obtain the sum \(S(s,n)\) over all possible \(n\). If the mean value \(\bar n \gg 1\), then

\[ \sum_{n=0}^{\infty} S(s,n) \simeq \frac{M^{\frac{3s}{2}}\Omega^{\frac{s}{2}+\frac{1}{3}}} {3\cdot 2^{\frac{3s}{2}}\pi^{\frac{s}{2}+\frac{2}{3}}\hbar^{\frac{3s}{2}+1}c^{\,1-\frac{3s}{2}}} \exp\left( \frac{\Omega^{1/3}(W-sMc^2)} {\pi^{2/3}\hbar c} \right). \tag{16} \]

The numerical values of (15) and (16), taking for \(\Omega\) (1), (2), and (3), are:

\[ Mc^2 S(s,n)= \frac{6.31}{w^{1/3}} \left(\frac{98.8}{w}\right)^{s/2} \frac{\left(6.31\frac{w-s}{w^{1/3}}\right)^{3n+\frac{3s}{2}-1}} {\left(3n+\frac{3s}{2}-1\right)!}, \tag{17} \]

\[ Mc^2 \sum_{n=0}^{\infty} S(s,n) \simeq \frac{2.10}{w^{1/3}} \left(\frac{98.8}{w}\right)^{s/2} \exp\left(6.31\frac{w-s}{w^{1/3}}\right), \tag{18} \]

where

\[ w=W/Mc^2, \tag{19} \]

and it is assumed that \(\mu/M=0.15\).

In the formulas given, we have neglected conservation of momentum. The condition that the total momentum be zero is easy to take into account approximately. The approximation consists in regarding the mass of the \(\pi\)-meson as negligibly small in comparison with the mass of the nucleon. In that case the momentum of the nucleons will be much greater than the momentum of the \(\pi\)-mesons, since the kinetic energy is distributed approximately uniformly among all particles, and for conservation of momentum it is sufficient to require that the sum of the momenta of the nucleons alone be zero. Formula (15) is then changed as follows: a) instead of \(s\) everywhere (except in the expression \(W-sMc^2\)) one should write \(s-1\), since now we have \(s-1\) statistically independent momenta of heavy particles; b) the factor \(M^{3s/2}\) must be replaced by

\[ \frac{M^{3(s-1)/2}}{s^{3/2}}, \]

since instead of \(M\) one must take a quantity analogous to the reduced mass. With this account of momentum conservation, formulas (17) and (18) take the form:

\[ Mc^2 S(s,n)= \frac{6.31}{s^{3/2}w^{1/3}} \left(\frac{98.8}{w}\right)^{\frac{s-1}{2}} \frac{\left[6.31\,(w-s)w^{-1/3}\right]^{3n+\frac{3s}{2}-1}} {\left(3n+\frac{3s}{2}-1\right)!}, \tag{20} \]

\[ Mc^2 \sum_{n=0}^{\infty} S(s,n) \simeq \frac{2.10}{s^{3/2}w^{1/3}} \left(\frac{98.8}{w}\right)^{\frac{s-1}{2}} \exp\left[6.31\,(w-s)/w^{1/3}\right]. \tag{21} \]

In all the formulas given, the particles are regarded as statistically independent. So long as a small number of nucleons and \(\pi\)-mesons participate in the process, the error will be small, but for processes of high multiplicity the neglect of the statistical dependence of the particles may lead to substantial errors. The exact formulas are very complicated, since there are, at least,

measure, three kinds of \(\pi\)-mesons and four kinds of nucleons and antinucleons. We did not attempt to introduce the corresponding complications in the region of comparatively low energies. In this case the statistical dependence was taken into account approximately, assuming that among statistically independent particles there is only one kind of \(\pi\)-meson, one kind of nucleon, and one kind of antinucleon. Such an assumption is in general not valid, and in the region of high energies it leads to an overestimate of the multiplicity of the processes.

For the case of very high energies, statistical correlations can be taken into account in a simple way by replacing the statistical model by a thermodynamic one, as will be done in Section 6.

All the formulas given above do not take account of the conservation of momentum. The errors arising from this will be discussed in Section 6, where the corresponding corrections will be given.

4. MULTIPLE PRODUCTION OF \(\pi\)-MESONS

In Section 2 we considered the production of one \(\pi\)-meson at comparatively low energy. We shall now consider collisions at higher energies, as a result of which several \(\pi\)-mesons may arise. A rough picture of these processes can be obtained by computing, according to (20), the relative probabilities for the production of \(0, 1, 2, \ldots, n\ldots\) \(\pi\)-mesons. In this formula one should put \(s=2\). We shall neglect statistical correlations and the conservation of momentum. The probabilities of various values of \(n\) are proportional to the quantity

\[ \left\{\frac{251}{w}(w-2)^3\right\}^{n}\Big/\left(\frac{3}{2}\cdot\frac{5}{2},\ldots,\frac{6n+1}{2}\right). \tag{22} \]

Table I gives the probabilities, calculated by this formula, of the generation of \(\pi\)-mesons for various multiplicities of the process. In the first column is given the energy \(w\) of the nucleons in units of \(Mc^2\) in the center-of-mass system.

Table I

In the second column is given the energy \(w'\) of the primary particle (including also its rest energy) in the laboratory

system of coordinates. In the following eight columns are given, expressed in percent, the probabilities of processes of different multiplicity \(n\). The last column shows the mean number of \(\pi\)-mesons formed.

We see that for a kinetic energy of the bombarding particle of the order of \(1000\) MeV, corresponding to the first line of the table, the probability of elastic scattering of two nucleons is \(50\%\). This probability rapidly decreases and becomes less than one percent for a bombardment energy of the order of \(5000\) MeV. With increasing bombardment energy the probability of multiple processes increases, as follows from the table. According to (22), the most probable value of \(n\) is approximately equal to

\[ 2.1(w-2)w^{\frac{1}{3}}. \]

In Section 6 it will be shown that in the region of high energies neglect of conservation of angular momentum and of statistical correlations leads to very substantial errors. Table I gives only qualitative indications of the transition from elastic scattering to the production of single \(\pi\)-mesons and then to multiple production of these particles. The quantitative regularities of multiple generation of \(\pi\)-mesons are more reliably expressed by formula (32).

5. FORMATION OF ANTINUCLEONS

If the total energy of two colliding particles in the center-of-inertia system is \(>4Mc^2\), then processes leading to the formation of nucleon–antinucleon pairs become possible. If the energy only slightly exceeds the threshold \(4Mc^2\), pair formation cannot be accompanied by the formation of additional \(\pi\)-mesons. With increasing energy, the pairs, as a rule, will be accompanied by a certain number of \(\pi\)-mesons. For intermediate energies \(w<10\) one may use formula (20). Substituting in it \(s=4\), we obtain the statistical weight for the formation of a nucleon pair accompanied by the emission of \(n\) \(\pi\)-mesons. For \(s=2\) we obtain a quantity proportional to the probability that a pair is not formed and that at the end of the reaction there are two initial nucleons and \(n\) \(\pi\)-mesons.

Omitting in (20) the common factor \(Mc^2\), we have:

\[ S(4,n)=\frac{775}{w^{\frac{11}{6}}}\, \frac{\left[6.31(w-4)w^{-\frac{1}{3}}\right]^{3n+\frac{7}{2}}} {\left(3n+\frac{7}{2}\right)!}. \tag{23} \]

When normalizing these probabilities to unity, it should be taken into account that in the energy region under consideration the probability of pair formation does not exceed \(1\%\), in consequence of which the probability of pair formation may be neglected in the normalization denominator, and then the normalization denominator becomes \(\sum S(2,n)\). For the calculation

of this sum one can apply (21), as a result of which we arrive at the following expression for the probability of formation of a nucleon pair and \(n\) \(\pi\)-mesons:

\[ P(4,n)=\frac{105}{w} \left(6.31\,\frac{w-4}{w^{1/3}}\right)^{3n+\frac{7}{2}} \frac{\exp\left[-6.31\frac{w-2}{w^{1/3}}\right]} {\left(3n+\frac{7}{2}\right)!}. \tag{24} \]

The numbers in the columns from \(n=0\) to \(n=4\), inclusive, should be multiplied by \(10^{-4}\).

Table II has been compiled according to formula (24). \(w\) and \(w'\) have the same meaning as in Table I. The following five columns give the probability of formation of a pair and simultaneous formation of \(n\) \(\pi\)-mesons. These probabilities are multiplied by \(10^{4}\). The eighth column gives the total probability \(P\) of formation of a pair.

As above, we have neglected the statistical correlations mentioned in Section 3, and conservation of momentum. For this reason the quantities in Table II give only

Table II

a qualitative indication of the results that will be obtained below by means of more correct calculations.

The greatest probability of formation of an antinucleon in the range of energies considered by us is 0.005. In collisions with such energies, in all probability, two or three \(\pi\)-mesons are formed on the average, whence it follows that the average number of antinucleons per \(\pi\)-meson is of the order of 0.002. We see that detecting antinucleons is very difficult even in the region of rather high energies.

6. COLLISIONS AT ULTRAHIGH ENERGIES

The study of collisions of superenergetic nucleons can be simplified by assuming that all particles formed in this process are extremely relativistic, and by allowing that the detailed statistical calculation of the probabilities of various events in this case can be replaced by a thermodynamic one.

At first we shall neglect the conservation of momentum. The effect of conservation will be taken into account at the end of this section.

The superhigh energy density, suddenly arising in the volume \(\Omega\), leads to multiple formation of \(\pi\)-mesons and nucleon–antinucleon pairs. Since both kinds of particles are extremely relativistic, the energy density will, according to Stefan’s law, be proportional to the fourth power of the temperature \(T\).

\(\pi\)-mesons, like photons, obey Bose–Einstein statistics. We assume that the temperature is so high that the rest mass may be neglected. In this case the relation between the energy and the momentum of \(\pi\)-mesons will be the same as for photons, and Stefan’s law for \(\pi\)-mesons will be the same as in the usual case of black radiation. The only difference lies in the statistical weight. For a photon the statistical factor is 2, in accordance with the two possible directions of polarization. If one assumes that \(\pi\)-mesons have spin zero and differ only by the magnitude of the charge \(\pm e\) or 0, then for them the statistical factor will be 3. Thus, multiplying the Stefan density by \(3/2\), we obtain the energy density of \(\pi\)-mesons:

\[ \frac{3\cdot 6{,}494}{2\pi^{3}\hbar^{3}c^{3}}(kT)^{4}, \tag{25} \]

where the numerical factor

\[ 6{,}494=\frac{\pi^{4}}{15}=6\sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^{4}. \]

The energy density of nucleons and antinucleons is expressed in an analogous way. It is only necessary to take into account that the statistical factor in this case is 8 (since we have four sorts of nucleons and antinucleons with two possible spin directions) and that these particles obey the Pauli principle. In the extremely relativistic case the energy density of nucleons and antinucleons is

\[ \frac{4\cdot 5{,}682}{\pi^{2}\hbar^{3}c^{3}}(kT)^{4}, \tag{26} \]

where the numerical factor

\[ 5{,}682=6\sum_{n=1}^{\infty}(-)^{n+1}\left(\frac{1}{n}\right)^{4}. \]

To obtain the temperature, the total energy of the system should be equated to the product of the volume \(\Omega\) and the total density (25) and (26). Taking (1) into account, we have:

\[ (kT)^{4}=0{,}152\,\frac{\hbar^{3}c^{3}W^{3}}{Mc^{2}\Omega_{0}}. \tag{27} \]

To calculate the number of \(\pi\)-mesons, nucleons, and antinucleons it is necessary to know the density of these particles, which can be obtained by the standard methods of statistical mechanics. In the extremely relativistic case the particle density is proportional to the third power of the temperature. For the total density of \(\pi\)-mesons and nucleons we have:

\[ n_{\pi}=0.367\,\frac{(kT)^3}{\hbar^3c^3},\qquad n_N=0.855\,\frac{(kT)^3}{\hbar^3c^3}. \tag{28} \]

The total number of \(\pi\)-mesons and nucleons is obtained by multiplying \(\Omega\) by (28), with the subsequent substitution of the temperature according to (27). The result, however, must be changed in accordance with conservation of momentum. We shall give the results with the correction for conservation. Conservation of momentum lowers the total number of particles and introduces into the formulae (28) an additional numerical factor 0.51. In addition, conservation of momentum leads to an anisotropy of the angular distribution, in the sense of a predominance of particles arising with momentum parallel to the momentum of the initial nucleons. Taking all these corrections into account, one can show that

\[ \text{number of }\pi\text{-mesons} =0.091\left(\frac{\Omega_0MW^2}{c\hbar^3}\right)^{\frac14} =0.54\sqrt{\frac{W}{Mc^2}}, \tag{29} \]

\[ \text{number of nucleons and antinucleons}= \]

\[ =0.21\left(\frac{\Omega_0MW^2}{c\hbar^3}\right)^{\frac14} =1.3\sqrt{\frac{W}{Mc^2}}. \tag{30} \]

In accordance with these results, the total number of charged particles arising in superenergetic collisions is equal to

\[ 1.2\left(\frac{W'}{mc^3}\right)^{\frac14}, \]

where \(W'\) is the energy of the incident primary particle in the laboratory coordinate system. In these formulae the values (2) and (3) were taken for \(\Omega_0\).

The formulae derived are valid only for ultrahigh energies, whose order is easy to estimate. Substituting into (27) \(\Omega_0\), according to (2) and (3), we obtain the relation between temperature and energy:

\[ \frac{kT}{Mc^2}=0.105\sqrt{\frac{W}{Mc^2}}, \tag{31} \]

whence it follows that the relativistic condition for nucleons

\[ \left(\frac{kT}{Mc^2}>1\right) \]

is fulfilled only for \(W>100\,Mc^2\). In the laboratory coordinate system this corresponds to energies of the bombarding particles \(\gtrsim 5\cdot10^{12}\) eV. At lower energies the number

formed antinucleon pairs will decrease rapidly, since the formation of each pair requires a minimum of \(2Mc^2\) energy. In this energy interval, pair formation is, in all probability, better described by the formulas of Section 5.

Comparison of (29) with (30) shows that the number of nucleons and antinucleons arising in super-energetic collisions exceeds the number of \(\pi\)-mesons. The reason for this lies in the difference of the statistical weights (8 for nucleons and 3 for \(\pi\)-mesons). Antiprotons, which are the most interesting from the experimental point of view, evidently constitute only one quarter of the total number of particles (30). Therefore, even in the region of superhigh energies, antiprotons are formed in smaller numbers than \(\pi\)-mesons.

In the region of intermediate energies, where multiple formation of \(\pi\)-mesons is still possible, one may also use the method of thermodynamics, assuming, however, that only the \(\pi\)-meson gas is in thermodynamic equilibrium and that the activation energy of the pairs is too high for these particles to be formed in appreciable quantity. The energy density in this case will be determined by formula (25). The numerical coefficient in formula (27) under these conditions will be 0.046 instead of 0.152, and the quantity \(W\) in this formula must be replaced by \(W-2Mc^2\), since the rest energy of the nucleons does not enter into the energy of the \(\pi\)-meson gas. Introducing also a factor of 0.51 to take account of conservation of momentum, we obtain the following approximate expression for intermediate energies:

\[ \text{Number of } \pi\text{-mesons} = 0.323\, \frac{M^{\frac14} R^{\frac34}(W-2Mc^2)^{\frac32}} {\hbar^{\frac34}c^{\frac14}W} = 1.34\,\frac{(w-2)^{\frac32}}{w}, \tag{32} \]

where \(w=\dfrac{W}{Mc^2}\). For intermediate energies from \(10^4\) to \(10^5\) MeV this formula, in all probability, gives better quantitative results for multiple processes than Table I. In particular, it should be noted that the multiplicities given in the last two rows of Table I are greatly overestimated: according to (32), at these energies the average number of \(\pi\)-mesons should be only of the order of 2. The discrepancy is due to two effects that were not taken into account in compiling Table I: statistical correlations between the different kinds of \(\pi\)-mesons and conservation of momentum. Formula (32) approximately takes both these circumstances into account.

At present, multiple formation of nucleons is not yet accessible to experiment, which makes it difficult to compare our results with experience. Contemporary theory apparently leads to too low a multiplicity, except in the energy region of order

\(10^{12}—10^{13}\) ev. As experimental data accumulate, it will be possible to improve the agreement of our theory with experiment by changing the value (3) for \(R\). If the observed multiplicity of the processes proves to be greater than the theoretical one, then the quantity \(R\) will have to be increased, and in the opposite case decreased.

In the present theory we have considered only one type of meson, namely \(\pi\)-mesons. If there exist heavier particles closely associated with nucleons (as, apparently, follows from Anderson’s recent experiments\(^3\)), then they too can attain statistical equilibrium. However, the large magnitude of their rest masses places them in unfavorable conditions as regards competition with the formation of \(\pi\)-mesons (except in the region of superhigh energies), and one may expect that in the majority of cases the number of \(\pi\)-mesons will exceed the number of heavier mesons.

Part II

ANGULAR DISTRIBUTION OF \(\pi\)-MESONS ARISING IN COLLISIONS OF HIGH-ENERGY NUCLEONS

1. INTRODUCTION

Above (Part I) a method was discussed for calculating the probability of formation of \(\pi\)-mesons in collisions of high-energy nucleons. The method is based on the assumption that the interaction between \(\pi\)-mesons and nucleons is so strong that statistical equilibrium has time to be established among all states compatible with the conservation laws of energy, charge, momentum, etc. More precisely, it is assumed that, in the collision of two energetic nucleons, the entire store of energy present in the center-of-mass system is released in a small effective volume, whose dimensions are of the order of the dimensions of the \(\pi\)-meson cloud surrounding the nucleons. By a succession of Yukawa processes this energy may produce states in which, along with the initial nucleons, some number of \(\pi\)-mesons is present. The states into which the system can pass from the initial state of two nucleons are restricted by a number of conservation laws. Next the basic assumption is made that the probability of occurrence of one or another state is proportional to the probability that all particles corresponding to that state will simultaneously be located in the effective volume.

At comparatively small energies, when the Lorentz contraction may be neglected, the effective volume \(\Omega_0\) may be taken equal to the volume of a sphere of radius \(R\)

\[ \Omega_0 = \frac{4\pi R^3}{3}, \tag{33} \]

where \(R\) is of order \(\dfrac{\hbar}{\mu c}\). The numerical value

\[ R=\frac{\hbar}{\mu c}=1.4\cdot 10^{-13}\ \text{cm}, \tag{34} \]

apparently gives results that agree well with experiment, and this value was adopted for the numerical calculations.

In studying collisions in the high-energy region, one must take Lorentz contraction into account. This was done in Part I by replacing the effective volume \(\Omega_0\) by the quantity

\[ \Omega=\frac{(2Mc^2)}{W}\Omega_0, \tag{35} \]

where \(W\) is the total energy (including the rest energy) of the two colliding nucleons in the center-of-inertia system, and the quantity \(\dfrac{2Mc^2}{W}\) is the familiar Lorentz contraction factor. This point may be justified as follows.

At all the energies considered, the de Broglie wavelength of the nucleons is much smaller than \(R\), so that the motion of the nucleons may be treated as quasiclassical. The \(\pi\)-meson field surrounding the nucleons will be Lorentz-contracted and, as a result of the collision, all the energy will first be released in the contracted volume (35). Soon after this, various reactions will begin, and at the same time the effective volume will begin to expand, which will lead to a gradual decrease in the energy concentration. Assumption (35) will be valid if it is assumed that the effective time of the reaction is determined by the maximum energy concentration and that the equilibrium reached at this early stage of the process is “frozen in” before the expansion of the effective volume becomes appreciable.

On these grounds, expression (35) was used in Part I to calculate the probable number of charged particles produced in nuclear collisions in the very-high-energy region. Two different formulas were obtained for this number, namely,

\[ 1.2\left(\frac{W'}{Mc^2}\right)^{\frac14} \tag{36} \]

and

\[ 1.06\left(\frac{W'}{Mc^2}\right)^{\frac14}, \tag{37} \]

the first corresponding to the case of nucleon–anti-

nucleon pairs, and the second—their absence. \(W'\) is the energy of the primary nucleon in the laboratory coordinate system.

For checking these formulas, only a small amount of experimental material is available at present. The largest number of stars in cosmic rays, caused by energetic protons, are explosions of heavy nuclei occurring under conditions in which processes of multiple particle production must play a very large role. Cases in which it may be assumed that the collision occurred between a cosmic-ray proton and a single nucleon are very rare. Recently Schein\(^4\) and collaborators discovered a very interesting case of this kind, apparently caused by a proton with an energy of the order of \(3\cdot 10^{13}\) ev. Fifteen tracks of minimum ionization were observed, most of them lying in a cone with an aperture of order \(0.003\) radian. The presence of only two tracks of lower energy indicates that the collision occurred at the surface of the nucleus and that only one nucleon could have participated in the process. Another photograph with similar characteristics was published by the Bristol group\(^5\). In this case the energy of the primary proton was considerably smaller, and it may be estimated as \(3\cdot 10^{12}\) ev. For \(W' = 3\cdot 10^{13}\) ev formulas (36) and (37) give, for the probable number of charged particles, the values 16 and 14 (15 observed). For the second star, assuming \(W' = 3\cdot 10^{12}\) ev, these formulas give the most probable values 9 and 8 (7 observed). Agreement with experiment shows that the assumption of Lorentz contraction of the effective volume should not contain a large error. It should be remembered, however, that the mean number of particles arising in processes of the given type depends only on the fourth root of \(\Omega\), so that a change of this volume by a factor of 2 or 3 will cause only small changes in the mean number of particles.

2. ANGULAR DISTRIBUTION

The main purpose of the present work is to study the angular distribution on the basis of a statistical model. In the laboratory coordinate system, the angular distribution of particles arising in high-energy nuclear collisions exhibits a striking tendency to split into two distributions: one of them, very narrow, contains about 50% of the particles, while the second, containing the remaining particles, has the form of a cone with a much wider aperture. Schein\(^4\) and collaborators firmly adhere to this point of view and conclude that, upon transformation to the center-of-mass system, the angular distribution reduces to two bundles of particles in two polar directions. Although such a concentration seems strange, it nevertheless does not reduce, in the syste-

in the center-of-mass system to very small angles, for a noticeable number of particles are deflected from the polar axis by angles up to 40 or 50 degrees. In the case of Schein’s star the total energy in the center-of-mass system is of the order of \(250 Mc^2\). Including also the unobserved neutral particles, one may assume that the star contains approximately 25 particles, each of which has on the average an energy of the order of \(10^{10}\) eV—an estimate quite compatible with direct measurements of energies in the laboratory coordinate system. A particle with energy \(10^{10}\) eV has a wavelength \(\lambda \simeq 2\cdot 10^{-15}\) cm. The radius of action of nuclear forces is \(R = 1.4\cdot 10^{-13}\) cm, from which one may think that the greatest deflection from the polar axis should be of the order

\[ \frac{\lambda}{R}\simeq 0.015 \simeq 1^\circ . \]

Such an assumption is obviously in sharp contradiction with the observations, which indicate a much broader angular distribution.

Let us proceed to discuss the angular distribution following from the statistical theory. At first glance it may seem that this theory will lead to an isotropic distribution in the center-of-mass system, since one might think that the evaporation of particles from the effective volume \(\Omega\) will obey an analogue of the Maxwell distribution, leading to spherical symmetry. A more detailed analysis shows, however, that such a conclusion is erroneous, and that the theory leads to a distribution very similar to the observed one. Quantitatively this follows from the following considerations.

It is extremely improbable that the collision of two nucleons should prove to be strictly central. Far more probable are such collisions in which the particles pass at some distance from one another and in which the system possesses a considerable angular momentum. In the case of Schein’s star the angular momentum may reach many hundreds or thousands of units \(\hbar\). Since angular momentum must be conserved between the initial and final states, the emitted particles must carry off the entire initial store of angular momentum. As a consequence, the angular distribution of the emitted particles will no longer be spherically symmetric, even if one follows literally the statistical or thermodynamic method of calculation developed in Part I. The law of statistical distribution is usually written on the assumption that the only conserved quantity is the energy, and the form of this law changes if the system is such that other quantities are also conserved in it, for example, angular momentum. Changes in the law of statistical equilibrium disturb the isotropic distribution of particle velocities.

Before proceeding to quantitative conclusions, let us consider qualitatively the role of the Lorentz contraction, which flattens

effective volume \(\Omega\). In Fig. 1 the straight lines \(a\) and \(b\) are the trajectories of two nucleons in the center-of-inertia system, and the curve \(c\) represents a section of the flattened effective volume \(\Omega\), in which the energy is released at the initial instant of time. The angular momentum in this case is directed upward, perpendicular to the plane of the figure.

Let us introduce a coordinate system \(x, y, z\), whose origin coincides with the center of the volume \(\Omega\); the \(y\)-axis is directed along the trajectories \(a\) and \(b\) in the two nucleons, and the \(z\)-axis (not shown in Fig. 1) is perpendicular to the plane \(ab\) and coincides with the direction of the angular momentum. After the collision all the particles formed fly out of the volume \(\Omega\). Since the effective volume is strongly flattened and the \(y\)-dimensions may be neglected, the \(z\)-component of the angular momentum of a particle flying out from the point \(x, z\) (\(y\) is very small) will be

Fig. 1. Trajectories of nucleons in the center-of-mass system.

\[ Z = xp\cos\vartheta, \tag{38} \]

where \(p\) is the momentum of the particle, and \(\vartheta\) is the angle between the momentum and the \(y\)-axis. The emitted particles must carry away a large amount of angular momentum; hence it follows that particles emitted from points with \(x>0\) must have \(\cos\vartheta>0\), and \(\cos\vartheta\) must be maximal, i.e. the angle \(\vartheta\) for such particles will be very small. Particles emitted from points with \(x<0\) also carry a large amount of positive angular momentum; for them \(\cos\vartheta<0\) and is close to \(-1\), i.e. the angle \(\vartheta\) is close to \(180^\circ\). From these qualitative considerations it follows that the angular distribution of the particles will not be isotropic and that the particles will be emitted predominantly in two polar directions, \(\vartheta=0^\circ\) and \(\vartheta=180^\circ\). For a quantitative estimate of the effect it is necessary to know the true shape of the effective volume \(\Omega\). The assumption that this volume is a strongly flattened ellipsoid obtained by Lorentz contraction of the sphere (33) is too simplified, especially in the case depicted in Fig. 1, when the nucleons \(a\) and \(b\) collide at large impact distances. On the other hand, without considerable arbitrariness it is impossible to specify the actual shape of the effective volume. Therefore, in the subsequent calculations we shall regard the volume \(\Omega\) as a flattened ellipsoid with transverse axis \(R\) and with a very small axis of symmetry, fully recognizing the crudeness of such a model.

Let us carry out the calculations in the thermodynamic approximation, which is applicable in the region of very high energies. If

if in the collision process only one quantity, the energy \(W\), were conserved, then the mean number of particles in a state with energy \(w\) in the volume \(\Omega\) would be equal to

\[ \frac{1}{e^{\beta w}-1}, \tag{39} \]

where \(\beta=\frac{1}{kT}\). The formula is valid for particles obeying Bose–Einstein statistics. In the opposite case the \(-1\) in the denominator should be replaced by \(+1\). In the approximation under consideration the angular distribution proves to be independent of the type of particle statistics, on the basis of which in what follows only the case of Bose–Einstein statistics will be considered in detail. Formula (39) is inapplicable in the case when, along with the energy, the \(z\)-component of the momentum is also conserved. In such a case, instead of (39) one should write

\[ \frac{1}{e^{\beta w-\lambda z}-1}, \tag{40} \]

where the constants \(\beta\) and \(\lambda\) are chosen in such a way that the total energy and the \(z\)-component of the momentum have the correct values\(^*\).

We shall restrict ourselves only to the case when all particles are ultrarelativistic in the center-of-inertia system, in accordance with which we put:

\[ w=cp,\qquad z=xp\eta, \tag{41} \]

where \(\eta=\cos\vartheta\). Introducing the notation

\[ \gamma=c\beta,\qquad \rho=\frac{\lambda R}{c\beta}, \tag{42} \]

we write the distribution law (40) in the form:

\[ \frac{1}{\exp\left[\gamma p\left(1-\frac{\rho\eta x}{R}\right)\right]-1}. \tag{43} \]

The number of particles in an element of phase-space volume is equal to expression (43), multiplied by the element of phase volume and divided by \((2\pi\hbar)^3\). Let us consider the element of phase volume corresponding to the spatial volume element

\[ \left(\frac{2Mc^2}{W}\right)\pi(R^2-x^2)\,dx, \tag{44} \]

\(^*\) Let us note that not only the \(z\)-component of the momentum is conserved, but also its \(x\)- and \(y\)-components. In (40) only the \(z\)-component appears because coefficients of the type \(\lambda\) for the other components become zero, if the momentum components in the two remaining directions vanish. For this same reason, in (40) there are no terms corresponding to the momentum of the particles formed, although the momentum is also conserved (in the center-of-inertia system the total momentum vanishes).

enclosed between the abscissae \(x\) and \(x+dx\), and to the element of momentum volume

\[ 2\pi p^{2}\,dp\,d\eta, \tag{45} \]

corresponding to particles with momentum magnitude between \(p\) and \(p+dp\), for which \(\cos\vartheta\) lies between \(\eta\) and \(\eta+d\eta\). The number of particles in such an element is

\[ dn=\frac{Mc^{3}}{2\pi\hbar^{3}W}\, \frac{(R^{2}-x^{2})\,dx\,p^{2}dp\,d\eta} {\exp\left[\gamma p\left(1-\frac{\rho\eta x}{R}\right)\right]-1}. \tag{46} \]

Integration of this expression (with respect to \(p\) from 0 to \(\infty\), with respect to \(x\) from \(-R\) to \(+R\), and with respect to \(\eta\) from \(-1\) to \(+1\)) leads to the following value of the total number \(N\) of particles of the given kind:

\[ N=\frac{a}{2\pi}\,\frac{Mc^{3}R^{3}}{W\hbar^{3}\gamma^{3}} \left(\frac{1+\rho^{2}}{\rho^{3}}\ln\frac{1+\rho}{1-\rho}-\frac{2}{\rho^{2}}\right), \tag{47} \]

where the numerical factor

\[ a=2\sum_{n=1}^{\infty}\frac{1}{n^{3}}=2.413 \tag{48} \]

arises from integration with respect to \(p\). Multiplying (46) by the energy \(cp\) of the particles and integrating with respect to \(p\), \(x\), and \(\eta\), we obtain the total energy of the particles:

\[ W'=\frac{b}{3\pi}\,\frac{Mc^{3}R^{3}}{W\hbar^{3}\gamma^{4}} \left(\frac{1}{\rho}\ln\frac{1+\rho}{1-\rho}+\frac{2}{1-\rho^{2}}\right), \tag{49} \]

where

\[ b=6\sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{\pi^{4}}{15}=6.494. \tag{50} \]

Finally, multiplying (46) by \(xp\eta\) and integrating, we obtain the total \(z\)-component of the angular momentum of the particles:

\[ M_{z}=\frac{b}{2\pi}\,\frac{Mc^{2}R^{4}}{W\hbar^{3}\gamma^{4}} \left( \frac{2}{\rho^{3}} +\frac{\left(\frac{4}{3\rho}\right)}{1-\rho^{2}} -\frac{1+\frac{1}{3}\rho^{2}}{\rho^{4}}\ln\frac{1+\rho}{1-\rho} \right). \tag{51} \]

Formulas (49) and (51) express the energy and angular momentum of particles of one definite kind, for example, neutral \(\pi\)-mesons. If particles of different kinds are present in the system—for example, neutral and charged \(\pi\)-mesons or various kinds of nucleons and antinucleons—then (49) and (51) retain their form, but the numerical value of the constant \(b\) becomes different. From formulas (49) and (51) one can determine the parameters \(\gamma\) and \(\rho\). In the case of the collision shown in Fig. 1, each of the initial nucleons has

energy \(\dfrac{W}{2}\) and momentum \(\dfrac{W}{2c}\). Consequently, the total angular momentum of this system is equal to \(M_z=W\dfrac{r}{c}\), where \(r\) is the impact distance of the nucleons from the center of the effective volume \(\Omega\). Starting from this expression for \(M_z\) and dividing (51) by (49), we have:

\[ \frac{r}{R}=\frac{3 f_1(\rho)}{2 f_2(\rho)}, \tag{52} \]

where \(f_1\) and \(f_2\) are the functions standing in brackets in (51) and (49), respectively. The right-hand side of (52), as a function of \(\rho\), is given in the second column of Table III. Relation (52) can be used to calculate the parameter \(\rho\) appearing in the distribution law (43). The parameter \(\rho=0\) when \(r=0\), i.e., for a strictly central collision. In this case \(\eta\) drops out of (43) and the angular distribution proves to be spherically symmetric. For larger values of \(r\), \(\rho\) approaches unity, and in this case the distribution function (43) depends sharply on the angle \(\vartheta\). To obtain the explicit form of the angular distribution, we integrate (46) with respect to \(p\) and \(x\), as a result of which we obtain the number of particles for which \(\eta\) lies between \(\eta\) and \(\eta+d\eta\). We have:

Table III

Numerical data for (52) and (54)

\[ \frac{dn}{d\eta}=\frac{\alpha M c^2 R^3 f_4(\rho \eta)}{2\pi W \hbar^3 \eta^3}, \tag{53} \]

where

\[ f_4(\alpha)=\frac{2}{\alpha^2(1-\alpha^2)}-\frac{1}{\alpha^3}\ln\left(\frac{1+\alpha}{1-\alpha}\right). \tag{54} \]

The numerical values of this function are given in Table III.

3. DISCUSSION OF RESULTS

The angular distribution of particles is given by formula (53). Since \(d\eta=-\sin\vartheta\,d\vartheta\) is proportional to the element of solid angle, \(\dfrac{dn}{d\eta}\) will be constant for an isotropic distribution. This will occur, for example, for a strictly central collision, when \(r=\rho=0\). If the impact parameter \(r\) has become

be, and \(\rho\) is different from zero, then in this case the function \(f_4(\rho\eta)\) is not constant and attains its largest values for \(\eta=\pm 1\). The maxima of this function are the more pronounced, the closer \(\rho\) is to unity.

It should be expected that \(r\) varies from collision to collision within the limits from \(0\) to \(R\). The mean value of \(r\) is equal to \(\dfrac{R}{\sqrt{2}}\). Collisions for which \(r\) is greater (or less) than this value will lead to an angular distribution more

A — Schein with co-workers
B — Camerini with co-workers

Fig. 2. Dependence of \(f_4(0.959\,\eta)\) on \(\eta\).

(or less) sharply expressed than in the case of an average collision with \(r=\dfrac{R}{\sqrt{2}}\).

Fig. 2 represents the function \(f_4(\rho\eta)\) as a function of \(\eta\) for \(\rho=0.959\), which corresponds to \(r=\dfrac{R}{\sqrt{2}}\). The curve represents an angular distribution for the average collision; for an isotropic distribution we would obtain a straight line parallel to the \(\eta\)-axis. It is easy to see that the particles are sharply grouped in two directions, \(\eta=+1\) and \(\eta=-1\), and the concentration of particles still has an appreciable magnitude for \(\eta=\pm 0.8\), corresponding to angles \(37^\circ\) and \(143^\circ\). For comparison with experiment we have shown in Fig. 2 the assumed form of the angular distribution for the two above-mentioned stars \(^{4,5}\). In processing the experimental data, a certain arbitrariness was allowed in passing from the laboratory system to the system of the center of inertia, so that the results obtained are only a general indication of the character of the experimental

data. It is easy to see that the theoretical angular distribution is by no means incompatible with experiment.

As was already noted above, it is possible that different collisions will be accompanied by different angular distributions of the particles produced. Central collisions should lead to a more isotropic distribution than in Fig. 2, while in the case of very eccentric collisions the angular distribution will be sharper.

The angular distribution calculated by us does not depend on the collision energy. For small energies, however, the simple theory set forth here is inapplicable for two reasons: first, in this case the Lorentz contraction of the effective volume is small and, second, at small energies the particles cannot be regarded as extremely relativistic in the center-of-inertia system.

References

1. W. Heisenberg, Nature 164, 65 (1949); Zeits. f. Physik 126, 569 (1949).
2. H. W. Lewis, J. R. Oppenheimer and S. A. Wouthuysen, Phys. Rev. 73, 127 (1948).
3. A. J. Seriff, R. B. Leighton, C. Hsiao, E. W. Cowan and C. D. Anderson, Phys. Rev. 78, 290 (1950).
4. Lord, Fainberg and Schein, Phys. Rev. 80, 970 (1950).
5. Camerini, Fowler, Lock and Muirhead, Phil. Mag. 41, 413 (1950).

From the Translator

I. Ya. Pomeranchuk [DAN 78, 889 (1951)] has pointed out that the conclusions drawn by Fermi may prove to be quantitatively incorrect, owing to the illegitimacy of applying the statistics of a relativistic ideal gas to a strongly interacting system. Since the dimensions of the region in which the particles are produced are comparable with the radius of their interaction, they (the particles) will interact strongly with one another, which in turn will lead to the production of new particles.