I'm trying to solve the following homework question.

Suppose that in the laboratory frame of reference we have $2$ particles. Particle "$a$" is at rest with total energy $E_a$, while particle "$b$" is moving away with total energy $E_2$. If particle $b$ has momentum $\vec{p}$, show that the reference frame in which the center-of-mass is static moves in the direction of $\vec{p}$ with speed $$ u = \frac{c^2 p}{E_a + E_b} $$ with respect to the lab. Also, show that the total momentum of the system is $0$ in the COM reference frame.

My attempt to solve this problem was as follows. I tried to place the particle at rest on the origin, and afterward, find the speed at which the center-of-mass was moving away from the origin in the lab reference frame, which would correspond to the speed we want.

Since particle $a$ is at rest in the lab reference frame, we know that $$ E_a = m_ac^2 $$ and similarly, since partible $b$ is moving (let's say, with velocity $\vec{v}$), we know that

\begin{align*} E_b &= \gamma m_b c^2\\ p &= \gamma m_b v \end{align*} with $\gamma^{-1} = \sqrt{1 - \left(\frac{v}{c}\right)^2}$. Now, by definition we know that the center-of-mass's position (along the direction of $\vec{p}$) will be \begin{align*} R = \frac{1}{m_a + m_b}\left(m_a(0) + m_b (vt)\right) \end{align*} which would mean that \begin{align*} u = \frac{R}{t} = \frac{1}{m_a + m_b}m_bv = \frac{1}{\frac{E_a}{c^2} + \frac{E_b}{\gamma c^2}}\frac{p}{\gamma}= \frac{c^2 p}{\gamma E_a + E_b} \end{align*} And this is almost the equation I want, but not quite. I'm not sure where the mistake in my reasoning is, could anyone tell me what I did wrong? Or alternatively, is there a better way to try and solve this problem? Thank you very much!

  • $\begingroup$ Could anyone tell me what I did wrong? Check-my-work questions are off-topic on this site. $\endgroup$ – G. Smith Oct 6 '20 at 20:49

I am not exactly sure about my answer, however, I think that the relativistic form for $R$ is:

$$R = \frac{1}{\gamma_a m_a + \gamma_b m_b}\left(\gamma_a m_a(0) + \gamma_b m_b (vt)\right)\space,$$

where, according to your data, $\gamma_a=1$ and $\gamma_b=\gamma$. Therefore, we have:

\begin{align*} u = \frac{R}{t} = \frac{\gamma}{m_a + \gamma m_b}m_bv = \frac{\gamma}{\frac{E_a}{c^2} +\gamma \frac{E_b}{\gamma c^2}}\frac{p}{\gamma}= \frac{c^2 p}{E_a + E_b} \space . \end{align*}


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