Phy5646/Group5RelativisticProb

Practice with relativistic 4-vectors, presented by group #5: (Anthony Kuchera, Jeff Klatsky, Chelsey Morien)

Consider the collision of an energetic positron with an electron at rest in the laboratory frame. The collision is so violent that the electron-positron pair gets converted into a pair of muons. Compute the minimum kinetic energy of the positron in the laboratory frame for the reaction to proceed.

The reaction we want to study is: $e^{+}+e^{-}\rightarrow \mu ^{+}+\mu ^{-}$

In the center of mass frame, the total linear momentum is zero, which means the two muons that are produced are at rest.

The 4-momentum of the electron and positron in the COM frame are given by:

$P_{e^{+}}^{\mu }=({\frac {E}{c}},+p)$

$P_{e^{-}}^{\mu }=({\frac {E}{c}},-p)$

By energy conservation: $E=m_{e}\gamma c^{2}=m_{\mu }c^{2}$

From this equation we find that $\gamma ={\frac {m_{\mu }}{m_{e}}}$. Recall that $\gamma$ is defined by: $\gamma ={\frac {1}{\sqrt {1-\beta ^{2}}}}$ where $\beta ={\frac {v}{c}}$, where v is the velocity if the particle we are observing.

Now that we have derived this relation, we must shift back into the laboratory frame, in which the electron is at rest. Do do this we compute the 4-momentum in a reference frame moving with the electron velocity ($\beta$ to the left). Define $k_{e^{+}}^{\mu }$ as the 4-momentum of the positron in the laboratory frame. Recall that in relativistic mechanics the following matrix (called the boost matrix) can be used to transform between frames of reference: $\left({\begin{array}{ll}\gamma &\beta \gamma \\\beta \gamma &\gamma \end{array}}\right)$

Therefore we have:

$k_{e^{+}}^{\mu }=\left({\begin{array}{ll}\gamma &\beta \gamma \\\beta \gamma &\gamma \end{array}}\right)P_{e^{+}}^{\mu }$ = $\left({\begin{array}{ll}\gamma &\beta \gamma \\\beta \gamma &\gamma \end{array}}\right)\left({\begin{array}{l}m_{e}\gamma c\\m_{e}\gamma v\end{array}}\right)$ = $m_{e}\gamma c\left({\begin{array}{ll}\gamma &\beta \gamma \\\beta \gamma &\gamma \end{array}}\right)\left({\begin{array}{l}1\\\beta \end{array}}\right)$ = $m_{e}\gamma c\left({\begin{array}{l}\gamma (1+\beta ^{2})\\2\beta \gamma \end{array}}\right)$

Since $k_{e^{+}}^{\mu }=\left({\begin{array}{l}E_{lab}^{min}\\pc\end{array}}\right)$ we can write the equation:

$E_{lab}^{min}=m_{e}c^{2}\gamma ^{2}(1+\beta ^{2})=m_{e}c^{2}\gamma ^{2}{\frac {2\gamma ^{2}-1}{\gamma ^{2}}}=m_{e}c^{2}(2\gamma ^{2}-1)$.

Therefore the minimum kinetic energy is given by:

$E_{lab}^{min}=2m_{e}c^{2}\left[\left({\frac {m_{\mu }}{m_{e}}}\right)^{2}-1\right]$