14.3: Totally Inelastic Collision

In a totally inelastic collision, particles stick together. A possible example is the absorption of a photon by a massive particle, resulting in an increase in its mass, as well as possibly a change in its momentum. Let’s consider, as an example, a particle of mass \(m \) that is initially at rest, and absorbs an incoming photon with energy \(E_{\gamma} \). There are now three ways to calculate the energy and momentum of the particle after this collision.

Method 1

We have conservation of both (total) energy and momentum. Before the collision, the massive particle has energy \(E_{\mathrm{i}}=m c^{2} \) (as it is standing still), and the total energy of the system \(E_{\gamma}+m c^{2}\), which must be conserved. The total energy of the particle after the collision is \(E_{\mathrm{f}}=\gamma(v) m_{\mathrm{f}} c^{2} \), where both the velocity \(v\) and the mass \(m_{\mathrm{f}} \) are unknown. The total momentum before the collision is \(E_{\gamma} / c \), as the particle is initially standing still (and thus has momentum zero), while after the collision it is \(\gamma(v) m_{\mathrm{f}}v \). We thus have:

\[\begin{align} E_{\gamma}+m c^{2} &=\gamma(v) m_{\mathrm{f}} c^{2} \label{14.4} \\[4pt] E_{\gamma} &=\gamma(v) m_{\mathrm{f}} vc \label{14.5} \end{align}\]

We thus have two equations with two unknowns (\(v\) and \(m_{\mathrm{f}} \)). If we divide Equation \ref{14.5} by \ref{14.4}, we get an expression for the final velocity \(v\), which we can substitute back in either equation to solve for \(m_{\mathrm{f}} \) (and potentially use to calculate the momentum after the collision). This is not pretty though, as we’ll have complicated factors due to the presence of \(\gamma(v)\).

Method 2

The four-momentum of the system is conserved during the collision. We have \(\overline{\boldsymbol{p}}_{\gamma}\) for the photon, \(\overline{\boldsymbol{p}}_{1}\) for the massive particle before the collision, and \(\overline{\boldsymbol{p}}_{\mathrm{f}}\) for that particle after the collision, given by the following equations:

\[\begin{align} \overline{\boldsymbol{p}}_{\gamma} &=\left(\frac{E_{\gamma}}{c}, \frac{E_{\gamma}}{c}, 0,0\right) \\[4pt] \overline{\boldsymbol{p}}_{1} &=(m c, 0,0,0) \\[4pt] \overline{\boldsymbol{p}}_{\mathrm{f}} &=\left(\frac{E_{\mathrm{f}}}{c}, p_{\mathrm{f}}, 0,0\right) \end{align}\]

From \(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{1}=\overline{\boldsymbol{p}}_{\mathrm{f}}\) we can read off two equations:

\[E_{\gamma}+m c^{2}=E_{\mathrm{f}}\]

\[E_{\gamma} / c=p_{\mathrm{f}}\]

which immediately give us the final energy and momentum in terms of the initial ones. We can now find the final mass through Einstein’s equation (13.16):

\[ \begin{align} m_{\mathrm{f}}^{2} c^{4}=E_{\mathrm{f}}^{2}-p_{\mathrm{f}}^{2} c^{2}&=\left(E_{\gamma}+m c^{2}\right)^{2}-E_{\gamma}^{2} \\[4pt] &=\left(E_{\gamma}+m c^{2}\right) m c^{2} \end{align}\]

This approach circumvents the use of the \(\gamma(v)\) factor because we only use energy and momentum, not (classical) velocity. If we now want the velocity, we could still calculate it from the combination of \(m_{\mathrm{f}} \) and either \(E_{\mathrm{f}}\) or \(p_{\mathrm{f}} \), but since it was the mass and momentum we were after, there’s no need to do so.

Method 3

Since the total energy-momentum four-vector is conserved in the collision, so must be its length (or the square of the length), which is trivial to calculate (remember that \(\overline{\boldsymbol{p}} \cdot \overline{\boldsymbol{p}}=m^{2} c^{2} \)). We can often exploit this fact to make the maths much simpler. To see how this works, let’s consider the full four-vector equation for our example: \(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}}=\overline{\boldsymbol{p}}_{\mathrm{f}} \), so

\[ \begin{align} \left(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}}\right) \cdot\left(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}}\right) &= \overline{\boldsymbol{p}}_{\mathrm{f}} \cdot \overline{\boldsymbol{p}}_{\mathrm{f}} \label{14.9} \\[4pt] \overline{\boldsymbol{p}}_{\gamma} \cdot \overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}} \cdot \overline{\boldsymbol{p}}_{\mathrm{i}}+2 \overline{\boldsymbol{p}}_{\gamma} \cdot \overline{\boldsymbol{p}}_{\mathrm{i}} &=\overline{\boldsymbol{p}}_{\mathrm{f}} \cdot \overline{\boldsymbol{p}}_{\mathrm{f}} \label{14.10} \end{align}\]

\[0+m^{2} c^{2}+2 E_{\gamma} m=m_{\mathrm{f}}^{2} c^{2}\]

which immediately gives us \(m_{\mathrm{f}} \). If we also want \(E_{\mathrm{f}} \) or \(p_{\mathrm{f}} \), we can again use Equations \ref{14.9} and \ref{14.10} for the components, but if we only wanted the final mass, we’re done in one step.

Note that although method 3 usually is the easiest route to your answer, it is not always - and it is a good idea to at least be aware of the other options.