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# 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.