Can we say that the center-of-momentum frame is the frame in which the center of mass is at rest? Isn't the center-of-momentum frame is same as the frame in which the center-of-mass is at rest? Since the position of the center-of-mass of a system of particles is defined as $\vec R=\sum_i m_i \vec r_i/M\Rightarrow\frac{d\vec R}{dt}=\sum_i m_i\vec v_i/M=\sum_i\vec p_i/M$. Therefore, the frame in which the center of mass is at rest, $d\vec R/dt=0\Rightarrow \sum_i \vec p_i=0$. Now, by definition, the center-of-momentum frame is one in which the total linear momentum vanishes. So am I right in saying that the "center-of-momentum frame is the same as the frame in which the center-of-mass is at rest."?
 A: In newtonian mechanics, center-of-momentum frame is the same as the frame in which the center-of-mass is at rest as showed in your derivation.
A: Let's check a simple 1D situation: 2 equal mass particles on a collision course. One (1) at rest at the origin, and another (2) moving in the $+$ direction towards it:
$$ x_1m = 0$$
$$ x_2(t)m = (x_2(0)+\beta t)m$$
Center of mass is clearly:
$$ X_{cm}(t) = \frac 1 2 (x_2(0)+\beta t) $$
Wait. We could do this problem in the center of mass frame where (maybe) both particles have equal and opposite velocities, so let's boost everything by $\beta/2$:
$$ v_1 = \frac{0-\beta/2}{1+\frac 1 2 0\beta}= -\beta/2 $$
$$ v_{cm} = \frac{\beta/2 - \beta/2}{1+\frac 1 4 \beta^2} = 0 $$
Now here comes the rub:
$$ v_2 = \frac{\beta + \beta/2}{1+\frac 1 2\beta^2} = \frac 3 2 \frac{\beta}{1 + \frac 1 2 \beta^2} \ne + \beta/2 = -v_1$$
So there is clear asymmetry.
Another way of looking at this is that the center of mass (in 1D) is the mean of $x$ weighted by $m$, while the center of momentum is weighted by $\gamma\beta m$.
