16 Disintegration of particles

In many cases the laws of conservation of momentum and energy alone can be used to obtain important results concerning the properties of various mechanical processes. It should be noted that these properties are independent of the particular type of interaction between the particles involved.

Let us consider a “spontaneous” disintegration (that is, one not due to external forces) of a particle into two “constituent parts”, i.e. into two other particles which move independently after the disintegration.

This process is most simply described in a frame of reference in which the particle is at rest before the disintegration. The law of conservation of momentum shows that the sum of the momenta of the two particles formed in the disintegration is then zero; that is, the particles move apart with equal and opposite momenta. The magnitude $p_0$ of either momentum is given by the law of conservation of energy:

\[ E_i= E_{1i}+\frac{{p_0}^2}{2m_1} +E_{2i}+\frac{{p_0}^2}{2m_2} \]

here $m_1$ and $m_2$ are the masses of the particles, $E_{1i}$ and $E_{2i}$ their internal energies, and $E_i$ the internal energy of the original particle. If $\epsilon$ is the “disintegration energy”, i.e. the difference

\[\label{equations/16.1}\tag*{(16.1)}\epsilon=E_i-E_{1i}-E_{2i}\]

which must obviously be positive, then

\[\label{equations/16.2}\tag*{(16.2)}\epsilon=\textstyle\frac{1}{2}\displaystyle{p_0}^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\frac{{p_0}^2}{2m}\]

which determines $p_0$; here $m$ is the reduced mass of the two particles. The velocities are $v_{10} = p_0/m_1, v_{20} = p_0/m_2$.

Let us now change to a frame of reference in which the primary particle moves with velocity $V$ before the break-up. This frame is usually called the laboratory system, or $L$ system, in contradistinction to the centre-of-mass system, or $C$ system, in which the total momentum is zero. Let us consider one of the resulting particles, and let $\boldsymbol{v}$ and $\boldsymbol{v}_0$ be its velocities in the $L$ and the $C$ system respectively. Evidently $ = +_0, or $\boldsymbol{v}-\boldsymbol{V} = \boldsymbol{v}_0$, and so

\[\label{equations/16.3}\tag*{(16.3)}v^2+V^2-2vV\cos\theta={v_0}^2\]

where $\theta$ is the angle at which this particle moves relative to the direction of the velocity $\boldsymbol{V}$. This equation gives the velocity of the particle as a function of its direction of motion in the $L$ system. In fig14 the velocity $\boldsymbol{v}$ is represented by a vector drawn to any point on a circle¹ of radius $v_0$ from a point $A$ at a distance $V$ from the centre. The cases $V \lt v_0$ and $V\gt v_0$ are shown in fig14a and fig14b respectively. In the former case \theta can have any value, but in the latter case the particle can move only forwards, at an angle $\theta$ which does not exceed $\theta_\text{max}$, given by

\[\label{equations/16.4}\tag*{(16.4)}\sin\theta_\text{max}=v_0/V\]

this is the direction of the tangent from the point $A$ to the circle.

The relation between the angles $\theta$ and $\theta_0$ in the $L$ and $C$ systems is evidently

\[\label{equations/16.5}\tag*{(16.5)}\tan\theta=v_0\sin\theta_0/(v_0\cos\theta_0+V)\]

If this equation is solved for $\cos\theta_0$, we obtain

\[\label{equations/16.6}\tag*{(16.6)}\cos\theta_0=-\frac{V}{v_0}\sin^2\theta\pm\cos\theta\sqrt{1-\frac{V^2}{{v_0}^2}\sin^2\theta}\]

For $v_0 > V$ the relation between $\theta_0$ and $\theta$ is one-to-one fig14a. The plus sign must be taken in (16.6) $\cos\theta_0=-\frac{V}{v_0}\sin^2\theta\pm\cos\theta\sqrt{1-\frac{V^2}{{v_0}^2}\sin^2\theta}$ , so that $\theta_0 = 0$ when $\theta = 0$. If $v_0 \lt V$, however, the relation is not one-to-one: for each value of $\theta$ there are two values of $\theta_0$, which correspond to vectors $\boldsymbol{v}_0$ drawn from the centre of the circle to the points $B$ and $C$ fig14b, and are given by the two signs in (16.6) $\cos\theta_0=-\frac{V}{v_0}\sin^2\theta\pm\cos\theta\sqrt{1-\frac{V^2}{{v_0}^2}\sin^2\theta}$ . j In physical applications we are usually concerned with the disintegration of not one but many similar particles, and this raises the problem of the distribution of the resulting particles in direction, energy, etc. We shall assume that the primary particles are randomly oriented in space, i.e. isotropically on average.

In the $C$ system, this problem is very easily solved: every resulting particle (of a given kind) has the same energy, and their directions of motion are isotropically distributed. The latter fact depends on the assumption that the primary particles are randomly oriented, and can be expressed by saying that the fraction of particles entering a solid angle element $\mathrm{d}^{}o_0$ is proportional to $\mathrm{d}^{}o_0$, i.e. equal to $\mathrm{d}^{}o_0/4\pi$. The distribution with respect to the angle $\theta_0$ is obtained by putting $\mathrm{d}^{}o_0 = 2\pi \sin\theta_0\mathrm{d}^{}o_0$, i.e. the corresponding fraction is

\[\label{equations/16.7}\tag*{(16.7)}\textstyle\frac{1}{2}\displaystyle\sin\theta_0\mathrm{d}^{}\theta_0\]

The corresponding distributions in the $L$ system are obtained by an appropriate transformation. For example, let us calculate the kinetic energy distribution in the $L$ system. Squaring the equation $\boldsymbol{v}=\boldsymbol{v}_0+\boldsymbol{V}$, we have $v^2={v_0}^2+V^2+2v_0V\cos\theta_0$, whence $\mathrm{d}^{}(\cos\theta_0) = \mathrm{d}^{}(v^2)/2v_0V$. Using the kinetic energy $T = \textstyle\frac{1}{2}\displaystyle mv^2$, where $m$ is $m_1$ or $m_2$ depending on which kind of particle is under consideration, and substituting in (16.7) $\textstyle\frac{1}{2}\displaystyle\sin\theta_0\mathrm{d}^{}\theta_0$ , we find the required distribution:

\[\label{equations/16.8}\tag*{(16.8)}(1/2mv_0V)\mathrm{d}^{}T\]

The kinetic energy can take values between $T_\text{min}= \textstyle\frac{1}{2}\displaystyle m(v_0-V)^2$ and $T_\text{max}=\textstyle\frac{1}{2}\displaystyle m(v_0+V)^2$. The particles are, according to (16.8) $(1/2mv_0V)\mathrm{d}^{}T$ , distributed uniformly over this range.

When a particle disintegrates into more than two parts, the laws of conservation of energy and momentum naturally allow considerably more freedom as regards the velocities and directions of motion of the resulting particles. In particular, the energies of these particles in the $C$ system do not have determinate values. There is, however, an upper limit to the kinetic energy of any one of the resulting particles. To determine the limit, we consider the system formed by all these particles except the one concerned (whose mass is $m_1$, say), and denote the “internal energy” of that system by $E_i'$. Then the kinetic energy of the particle $m_1$ is, by (16.1) $\epsilon=E_i-E_{1i}-E_{2i}$ and (16.2) $\epsilon=\textstyle\frac{1}{2}\displaystyle{p_0}^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\frac{{p_0}^2}{2m}$ , $T_{10} = {p_0}^2/2m_1 = (M-m_1)(E_i-E_{1i}-E_i')$ where $M$ is the mass of the primary particle. It is evident that $T_{10}$ has its greatest possible value when $E_i'$ is least. For this to be so, all the resulting particles except $m_1$ must be moving with the same velocity. Then $E_i'$ is simply the sum of their internal energies, and the difference $E_i-E_{1i}-E_i'$ is the disintegration energy $\epsilon$. Thus

\[\label{equations/16.9}\tag*{(16.9)}T_{10,\text{max}}=(M-m_1)\epsilon/M\]