Collisional Conservation Laws

Consider

$$\int C_{12}\,d^3{\bf v}_1 = \int\!\!\int\!\!\int\!\!\int u\,\sigm... ...f_1'\,f_2'-f_1\,f_2)\,d^3{\bf v}_1\,d^3{\bf v}_2\,d^3{\bf v}_1'\,d^3{\bf v}_2',$$ (3.25)

which follows from Equation (3.20). Interchanging primed and unprimed dummy variables of integration on the right-hand side, we obtain

$$\int C_{12}\,d^3{\bf v}_1 = \int\!\!\int\!\!\int\!\!\int u'\,\sig... ...f_1\,f_2-f_1'\,f_2')\,d^3{\bf v}_1'\,d^3{\bf v}_2'\,d^3{\bf v}_1\,d^3{\bf v}_2.$$ (3.26)

Hence, making use of Equation (3.16), as well as the fact that $u'=u$ , we deduce that

$$\int C_{12}\,d^3{\bf v}_1 =- \int\!\!\int\!\!\int\!\!\int u\,\sigma({\bf v}_1, {\bf v}_2; {... ...(f_1'\,f_2'-f_1\,f_2)\,d^3{\bf v}_1\,d^3{\bf v}_2\,d^3{\bf v}_1'\,d^3{\bf v}_2'$$

$$= - \int C_{12}\,d^3{\bf v}_1,$$ (3.27)

which implies that

$$\int C_{12}\,d^3{\bf v}_1=0.$$ (3.28)

The previous expression states that collisions with particles of type $2$ give rise to zero net rate of change of the number density of particles of type $1$ at position ${\bf r}$ and time $t$ . In other words, the collisions conserve the number of particles of type $1$ . Now, it is easily seen from Equations (3.20) and (3.21) that

$$C_{12}\,d^3{\bf v}_1=C_{21}\,d^3{\bf v}_2.$$ (3.29)

Hence, Equation (3.28) also implies that

$$\int C_{21}\,d^3{\bf v}_2=0.$$ (3.30)

In other words, collisions also conserve the number of particles of type $2$ .

Consider

$$(m_1+m_2) \int {\bf U}\,C_{12}\,d^3{\bf v}_1 = {\bf0}.$$ (3.31)

This integral is obviously zero, as indicated, as a consequence of the conservation law (3.28), as well as the fact that the center of mass velocity, ${\bf U}$ , is a constant of the motion. However, making use of Equations (3.10) and (3.29), the previous expression can be rewritten in the form

$$\int m_1\,{\bf v}_1\,C_{12}\,d^3{\bf v}_1 = - \int m_2\,{\bf v}_2\,C_{21}\,d^3{\bf v}_2.$$ (3.32)

This equation states that the rate at which particles of type $1$ gain momentum due to collisions with particles of type $2$ is equal to the rate at which particles of type $2$ lose momentum due to collisions with particles of type $1$ . In other words, the collisions conserve momentum.

Finally, consider

$$\int K\,C_{12}\,d^3{\bf v}_1 = 0.$$ (3.33)

This integral is obviously zero, as indicated, as a consequence of the conservation law (3.28), as well as the fact that the kinetic energy, $K$ , is the same before and after an elastic collision. It follows from Equations (3.15) and (3.29) that

$$\int \frac{1}{2}\,m_1\,v_1^{\,2}\,C_{12}\,d^3{\bf v}_1 = - \int\frac{1}{2}\,m_2\,v_2^{\,2}\,C_{21}\,d^3{\bf v}_2.$$ (3.34)

This equation states that the rate at which particles of type $1$ gain kinetic energy due to collisions with particles of type $2$ is equal to the rate at which particles of type $2$ lose kinetic energy due to collisions with particles of type $1$ . In other words, the collisions conserve energy.