Scattering in the Laboratory Frame

We have seen that two-particle scattering looks fairly simple when viewed in the center of mass frame. Unfortunately, we are not usually in a position to do this. In the laboratory, the most common scattering scenario is one in which the second particle is initially at rest. Let us now investigate this situation.

Suppose that, in the center of mass frame, the first particle has velocity ${\bf v}_1$ before the collision, and velocity ${\bf v}_1'$ after the collision. Likewise, the second particle has velocity ${\bf v}_2$ before the collision, and ${\bf v}_2'$ after the collision. We know that

$$m_1\,{\bf v}_1 + m_2\,{\bf v}_2 = m_1\,{\bf v}_1'+m_2\,{\bf v}_2' = {\bf0}$$ (368)

in the center of mass frame. Moreover, since the collision is assumed to be elastic (i.e., energy conserving),

$$v_1' = v_1,$$ (369)

$$v_2' = v_2.$$ (370)

Let us transform to a new inertial frame of reference--which we shall call the laboratory frame--which is moving with the uniform velocity $-{\bf v}_2$ with respect to the center of mass frame. In the new reference frame, the first particle has initial velocity ${\bf V}_1= {\bf v}_1-{\bf v}_2$, and final velocity ${\bf V}_1' = {\bf v}_1'-{\bf v}_2$. Furthermore, the second particle is initially at rest, and has the final velocity ${\bf V}_2' = {\bf v}_2'-{\bf v}_2$. The relationship between scattering in the center of mass frame and scattering in the laboratory frame is illustrated in Figure 23.

Figure 23: Scattering in the center of mass and laboratory frames.

In the center of mass frame, both particles are scattered through the same angle $\theta$. However, in the laboratory frame, the first and second particles are scattered by the (generally different) angles $\psi$ and $\zeta$, respectively.

Defining $x$- and $y$-axes, as indicated in Figure 23, it is easily seen that the Cartesian components of the various velocity vectors in the two frames of reference are:

$${\bf v}_1 = v_1\,(1,\, 0),$$ (371)

$${\bf v}_2 = (m_1/m_2)\,v_1\,(-1,\,0),$$ (372)

$${\bf v}_1' = v_1\,(\cos\theta,\,\sin\theta),$$ (373)

$${\bf v}_2' = (m_1/m_2)\,v_1(-\cos\theta,\,-\sin\theta),$$ (374)

$${\bf V}_1 = (1+m_1/m_2)\,v_1\,(1,\,0),$$ (375)

$${\bf V}_1' = v_1\,(\cos\theta+m_1/m_2,\,\sin\theta),$$ (376)

$${\bf V}_2' = (m_1/m_2)\,v_1\,(1-\cos\theta,\,-\sin\theta).$$ (377)

In the center of mass frame, let $E$ be the total energy, let $E_1=(1/2)\,m_1\,v_1^{\,2}$ and $E_2=(1/2)\,m_2\,v_2^{\,2}$ be the kinetic energies of the first and second particles, respectively, before the collision, and let $E_1'=(1/2)\,m_1\,v_1'^{\,2}$ and $E_2'=(1/2)\,m_2\,v_2'^{\,2}$ be the kinetic energies of the first and second particles, respectively, after the collision. Of course, $E=E_1+E_2=E_1'+E_2'$. In the laboratory frame, let ${\cal E}$ be the total energy. This is, of course, equal to the kinetic energy of the first particle before the collision. Likewise, let ${\cal E}_1'=(1/2)\,m_1\,V_1'^{\,2}$ and ${\cal E}_2'=(1/2)\,m_2\,V_2'^{\,2}$ be the kinetic energies of the first and second particles, respectively, after the collision. Of course, ${\cal E} = {\cal E}_1' + {\cal E}_2'$.

The following results can easily be obtained from the above definitions and Equations (371)-(377). First,

$${\cal E} = \left(\frac{m_1+m_2}{m_2}\right)E.$$ (378)

Hence, the total energy in the laboratory frame is always greater than that in the center of mass frame. In fact, it can be demonstrated that the total energy in the center of mass frame is less than the total energy in any other inertial frame. Second,

$$E_1 = E_1'= \left(\frac{m_2}{m_1+m_2}\right) E,$$ (379)

$$E_2 = E_2'=\left(\frac{m_1}{m_1+m_2}\right) E.$$ (380)

These equations specify how the total energy in the center of mass frame is distributed between the two particles. Note that this distribution is unchanged by the collision. Finally,

$${\cal E}_1' = \left[\frac{m_1^{\,2}+2\,m_1\,m_2\,\cos\theta+m_2^{\,2}}{(m_1+m_2)^2}\right]{\cal E},$$ (381)

$${\cal E}_2' = \left[\frac{2\,m_1\,m_2\,(1-\cos\theta)}{(m_1+m_2)^2}\right] {\cal E}.$$ (382)

These equations specify how the total energy in the laboratory frame is distributed between the two particles after the collision. Note that the energy distribution in the laboratory frame is different before and after the collision.

Equations (371)-(377), and some simple trigonometry, yield

$$\tan\psi = \frac{\sin\theta}{\cos\theta+m_1/m_2},$$ (383)

and

$$\tan\zeta = \frac{\sin\theta}{1-\cos\theta} = \tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right).$$ (384)

The last equation implies that

$$\zeta = \frac{\pi}{2}- \frac{\theta}{2}.$$ (385)

Differentiating Equation (383) with respect to $\theta$, we obtain

$$\frac{d\tan\psi}{d\theta} = \frac{1+(m_1/m_2)\,\cos\theta}{(\cos\theta+m_1/m_2)^2}.$$ (386)

Thus, $\tan\psi$ attains an extreme value, which can be shown to correspond to a maximum possible value of $\psi$, when the numerator of the above expression is zero: i.e., when

$$\cos\theta = -\frac{m_2}{m_1}.$$ (387)

Note that it is only possible to solve the above equation when $m_1>m_2$. If this is the case then Equation (383) yields

$$\tan\psi_{max} = \frac{m_2/m_1}{\sqrt{1-(m_2/m_1)^2}},$$ (388)

which reduces to

$$\psi_{max} = \sin^{-1}\left(\frac{m_2}{m_1}\right).$$ (389)

Hence, we conclude that when $m_1>m_2$ there is a maximum possible value of the scattering angle, $\psi$, in the laboratory frame. This maximum value is always less than $\pi/2$, which implies that there is no backward scattering (i.e., $\psi>\pi/2$) at all when $m_1>m_2$. For the special case when $m_1=m_2$, the maximum scattering angle is $\pi/2$. However, for $m_1<m_2$ there is no maximum value, and the scattering angle in the laboratory frame can thus range all the way to $\pi$.

Equations (378)-(385) enable us to relate the particle energies and scattering angles in the laboratory frame to those in the center of mass frame. In general, these relationships are fairly complicated. However, there are two special cases in which the relationships become much simpler.

The first special case is when $m_2\gg m_1$. In this limit, it is easily seen from Equations (378)-(385) that the second mass is stationary both before and after the collision, and that the center of mass frame coincides with the laboratory frame (since the energies and scattering angles in the two frames are the same). Hence, the simple analysis outlined in Section 6.4 is applicable in this case.

The second special case is when $m_1=m_2$. In this case, Equation (383) yields

$$\tan\psi = \frac{\sin\theta}{\cos\theta+1} =\tan(\theta/2).$$ (390)

Hence,

$$\psi = \frac{\theta}{2}.$$ (391)

In other words, the scattering angle of the first particle in the laboratory frame is half of the scattering angle in the center of mass frame. The above equation can be combined with Equation (385) to give

$$\psi + \zeta =\frac{\pi}{2}.$$ (392)

Thus, in the laboratory frame, the two particles move off at right-angles to one another after the collision. Equation (378) yields

$${\cal E} = 2\,E.$$ (393)

In other words, the total energy in the laboratory frame is twice that in the center of mass frame. According to Equations (379) and (380),

$$E_1 = E_1' = E_2=E_2' = \frac{E}{2}.$$ (394)

Hence, the total energy in the center of mass frame is divided equally between the two particles. Finally, Equations (381) and (382) give

$${\cal E}_1' = \left(\frac{1+\cos\theta}{2}\right){\cal E} = \cos^2(\theta/2)\,\,{\cal E} = \cos^2\psi\,\,{\cal E},$$ (395)

$${\cal E}_2' = \left(\frac{1-\cos\theta}{2}\right){\cal E} = \sin^2(\theta/2)\,\,{\cal E} = \sin^2\psi\,\,{\cal E}.$$ (396)

Thus, in the laboratory frame, the unequal energy distribution between the two particles after the collision is simply related to the scattering angle $\psi$.

What is the angular distribution of scattered particles when a beam of particles of the first type scatter off stationary particles of the second type? Well, we can define a differential scattering cross-section, $d\sigma(\psi)/d{\mit\Omega}'$, in the laboratory frame, where $d{\mit\Omega}'=2\pi\,\sin\psi\,d\psi$ is an element of solid angle in this frame. Thus, $(d\sigma(\psi)/d{\mit\Omega}')\,d{\mit\Omega}'$ is the effective cross-sectional area in the laboratory frame for scattering into the range of scattering angles $\psi$ to $\psi+d\psi$. Likewise, $(d\sigma(\theta)/d{\mit\Omega})\,d{\mit\Omega}$ is the effective cross-sectional area in the center of mass frame for scattering into the range of scattering angles $\theta$ to $\theta+d\theta$. Note that $d{\mit\Omega} =2\pi\,\sin\theta\,d\theta$. However, a cross-sectional area is not changed when we transform between different inertial frames. Hence, we can write

$$\frac{d\sigma(\psi)}{d{\mit\Omega}'} \,d{\mit\Omega}' = \frac{d\sigma(\theta)}{d{\mit\Omega}}\,d{\mit\Omega},$$ (397)

provided that $\psi$ and $\theta$ are related via Equation (383). This equation can be rearranged to give

$$\frac{d\sigma(\psi)}{d{\mit\Omega}'} = \frac{d{\mit\Omega}}{d{\mit\Omega}'}\,\frac{d\sigma(\theta)}{d{\mit\Omega}},$$ (398)

or

$$\frac{d\sigma(\psi)}{d{\mit\Omega}'} = \frac{\sin\theta}{\si... ...,\frac{d\theta}{d\psi}\,\frac{d\sigma(\theta)}{d{\mit\Omega}}.$$ (399)

The above equation allows us to relate the differential scattering cross-section in the laboratory frame to that in the center of mass frame. In general, this relationship is extremely complicated. However, for the special case where the masses of the two types of particles are equal, we have seen that $\psi=\theta/2$ [see Equation (391)]. Hence, it follows from Equation (399) that

$$\frac{d\sigma(\psi)}{d{\mit\Omega}'} = 4\,\cos\psi\,\,\frac{d\sigma(\theta=2\,\psi)}{d{\mit\Omega}}.$$ (400)

Let us now consider some specific examples. We saw earlier that, in the center of mass frame, the differential scattering cross-section for impenetrable spheres is [see Equation (356)]

$$\frac{d\sigma(\theta)}{d{\mit\Omega}} = \frac{a^2}{4},$$ (401)

where $a$ is the sum of the radii. According to Equation (400), the differential scattering cross-section (for equal mass spheres) in the laboratory frame is

$$\frac{d\sigma(\psi)}{d{\mit\Omega}'} = a^2\,\cos\psi.$$ (402)

Note that this cross-section is negative for $\psi>\pi/2$. This just tells us that there is no scattering with scattering angles greater than $\pi/2$ (i.e., there is no backward scattering). Comparing Equations (401) and (402), we can see that the scattering is isotropic in the center of mass frame, but appears concentrated in the forward direction in the laboratory frame. We can integrate Equation (402) over all solid angles to obtain the total scattering cross-section in the laboratory frame. Note that we only integrate over angular regions where the differential scattering cross-section is positive. Doing this, we get

$$\sigma = \pi\,a^2,$$ (403)

which is the same as the total scattering cross-section in the center of mass frame [see Equation (357)]. This is a general result. The total scattering cross-section is frame independent, since a cross-sectional area is not modified by switching between different frames of reference.

As we have seen, the Rutherford scattering cross-section takes the form [see Equation (367)]

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{q^4}{16\,(4\,\pi\,\epsilon_0)^2\,E^{\,2}}\frac{1}{\sin^4(\theta/2)}$$ (404)

in the center of mass frame. It follows, from Equation (400), that the Rutherford scattering cross-section (for equal mass particles) in the laboratory frame is written

$$\frac{d\sigma}{d{\mit\Omega}'} = \frac{q^4}{(4\,\pi\,\epsilon_0)^2\,{\cal E}^{\,2}}\frac{\cos\psi}{\sin^4\psi}.$$ (405)

Here, we have made use of the fact that ${\cal E} = 2\,E$ for equal mass particles [see Equation (393)]. Note, again, that this cross-section is negative for $\psi>\pi/2$, indicating the absence of backward scattering.