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Question

Suppose a relativistic particle of mass $M$ and initial momentum $p_i$ is fired towards a stationary electron of mass $m_e$. Following the collision, the two particles move in the same direction with the electron having momentum $p_e$ and the particle now having momentum $p_f$. Show that the kinetic energy of the electron is given by $$ T = \dfrac{2 m_e c^6 p_i^2}{m_e^2 c^4 + M^2 c^4 + 2m_e c^2 \sqrt{p_i^2 c^2 + M^2 c^4}} $$

What I've tried before is writing down the conservation of momentum and energy, namely $p_i = p_e + p_f $ and $E_i = E_e + E_f$ along with the mass energy relation $ E^2 = p^2 c^2 + m^2 c^4 $ for each of the particles. In addition to this, I noted that the kinetic energy is given by $ E = T + mc^2 $ allowing us to derive that $ E^2 = T^2 + 2Tmc^2 + m^2 c^4 = p^2 c^2 + m^2 c^4 $, hence $ T^2 + 2Tmc^2 = p^2 c^2$ with appropriate subscripts for each particle. I'm not sure how to eliminate both $p_e$ and $p_f$ to get the final result into the desired form without getting lost in a sea of algebra. I've heard about four-vectors and their applications to these types of problems, although we haven't directly covered them in class so I don't think this is the desired approach.

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Here's a possible route. Let's start with the identities $$ \begin{align} E_i^2 &= p_i^2c^2 + M^2c^4, \tag{1}\\ E_f^2 &= p_f^2c^2 + M^2c^4, \tag{2}\\ E_e^2 &= p_e^2c^2 + m_e^2c^4 = (T + m_ec^2)^2.\tag{3} \end{align} $$ Conservation of momentum implies $$ p_i = p_e + p_f.\tag{4} $$ We can write this as $$ p_f = \frac{1-k}{2}p_i,\qquad p_e = \frac{1+k}{2}p_i,\tag{5} $$ for some unknown value $k$. Next, conservation of energy gives $$ E_i + m_ec^2 = E_f + E_e,\tag{6} $$ or $$ E_f = E_i - T.\tag{7} $$ Combining Eqs. (3) and (5), we get $$ 4p_e^2c^2 = p_i^2c^2(1+k)^2 = 4(T^2 + 2m_ec^2T),\tag{8} $$ and combining Eqs. (2) and (5) leads to $$ 4E_f^2 = p_i^2c^2(1-k)^2 + 4M^2c^4 = 4(E_i - T)^2. \tag{9} $$ I will leave the rest to you. All that's left is to eliminate $1+k$ from Eq. (8). You can do this by first subtracting (9) from (8), and write this as an equation for $(1+k)$ in terms of $T$. You can then plug this back into (8), and you should get your result.

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  • $\begingroup$ Thanks for detailing this approach, I've tried and ended up getting the correct answer but it just doesn't seem like something I would think of myself. What made you think to introduce the k-factor and is it possible to tackle this question without it? $\endgroup$ – backstrapp Apr 26 '18 at 23:54
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    $\begingroup$ @backstrapp Actually, you don't have to use this $k$ term; you can keep $p_e^2$ in Eq. (8) and use (p_i-p_e)^2 in Eq. (9), then eliminate $p_e^2$ from both and put the expression for $p_e$ back into Eq. (8). The result will be the same. $\endgroup$ – Pulsar Apr 27 '18 at 0:36
  • $\begingroup$ Thanks for the advice, managed to complete the question in this manner too but still took a while to muddle through all the algebra. I imagine four vectors would give an easier $\endgroup$ – backstrapp Apr 27 '18 at 23:10

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