Elastic Collisions

Figure 10.2: One-dimensional elastic collisions of two particles in the center of momentum frame as seen in spacetime diagrams.

Suppose a particle with mass $m_1$ and initial velocity $u_1$ in the center of momentum frame collides elastically with another particle of mass $m_2$ with initial velocity $u_2$. The momenta of the two particles are

$$p_1 = \frac{m_1 u_1}{(1 - u_1^2 /c^2 )^{1/2}} ~~~ p_2 = \frac{m_2 u_2}{(1 - u_2^2 /c^2 )^{1/2}} .$$ (11.7)

In the center of momentum frame we must have

$$p_1 = - p_2 .$$ (11.8)

Figure 10.2 shows what happens when these two particles collide. The first particle acquires momentum $p_1'$ while the second acquires momentum $p_2'$. The conservation of momentum tells us that the total momentum after the collision is the same as before the collision, namely zero, so

$$p_1' = - p_2' .$$ (11.9)

In the center of momentum frame we know that $\vert p_1 \vert = \vert p_2 \vert$ and we know that the two momentum vectors point in opposite directions. Similarly, $\vert p_1' \vert = \vert p_2' \vert$. However, we as yet don't know how $p_1'$ is related to $p_1$. Conservation of energy,

$$E_1 + E_2 = E_1' + E_2' ,$$ (11.10)

gives us this information. Notice that if $p_1' = -p_1$, then $E_1'^2 = p_1'^2 c^2 + m_1^2 c^4 = p_1^2 c^2 + m_1^2 c^4 = E_1^2$. Assuming positive energies, we therefore have $E_1' = E_1$. If $p_2' = -p_2$, then we can similiarly infer that $E_2' = E_2$. If these conditions are satisfied, then so is equation (10.10). Therefore, a complete solution to the problem is

$$p_1 = -p_1' = -p_2 = p_2' \equiv p$$ (11.11)

and

$$E_1 = E_1' = (p^2 c^2 + m_1^2 c^4 )^{1/2} ~~~ E_2 = E_2' = (p^2 c^2 + m_2^2 c^4 )^{1/2} .$$ (11.12)

In other words, the particles just exchange momenta.

The left panel of figure 10.2 shows what happens in a collision when the masses of the two colliding particles are equal. If $m_1 = m_2$, then the incoming and outgoing velocities of the two particles are the same, as indicated by the inverse slopes of the world lines. On the other hand, if $m_1 > m_2$, then the velocity of particle 2 is greater than the velocity of particle 1, as is illustrated in the right panel of figure 10.2.

Figure 10.3: Elastic collisions viewed from a reference frame in which one particle is initially stationary.

Suppose we wish to view the results of an elastic collision in a reference frame in which particle 2 is initially stationary. All we have to do is to transform the velocities into a reference frame moving with the initial velocity of particle 2, as illustrated in figure 10.3. We do this by relativistically adding $U = -u_2$ to each velocity. (Note that the velocity $U$ of the moving frame is positive since $u_2$ is negative.) Using the relativistic velocity translation formula, we find that

$$v_1 = \frac{u_1 + U}{1 + u_1 U/c^2 } ~~~ v_1' = \frac{u_1'... ...{1 + u_1' U/c^2 } ~~~ v_2' = \frac{u_2' + U}{1 + u_2' U/c^2 }$$ (11.13)

where $u_1$, $u_1'$, $u_2$, and $u_2'$ indicate velocities in the original, center of momentum reference frame and $v_1$, $v_1'$, etc., indicate velocities in the transformed frame.

In the special case where the masses of the two particles are equal to each other, we have $v_1 = 2U/(1 + U^2 /c^2 )$, $v_1' = 0$, and $v_2' = 2U/(1 + U^2 /c^2 ) = v_1$. Thus, when the masses are equal, the particles simply exchange velocities.

If the velocities are nonrelativistic, then the simpler Galilean transformation law $v = u + U$ can be used in place of the relativistic equations invoked above.