Centre of mass speed and composite mass in relativistic collisions

[mike217]

A particle of rest mass m is accelerated to a kinetic energy K in a nuclear reactor. This particle is incident on a stationary target particle, also of rest mass m.

a) Show that the speed of the centre of mass (that is the speed of the frame in which the total momentum is zero) is $$\gamma v/(\gamma +1)$$ where v is the speed of the incident particle and $$\gamma =(1-v^2/c^2)^-1/2$$. Verify that this expression reduces to the usual one in the non relativistic case when v<<c.

b) If the collision is perfectly inelastic-show by using the conservation of energy and momentum that the mass M of the resulting composite is $$M=m\sqrt{2(\gamma +1)}$$ Verify that this reduces to the usual value in the non relativistic case v<<c

I have the solution to this problem, but there are a couple of things I don't understand. How do you get the following expressions

$$(Ptot-Vcm*Etot/c^2)/\sqrt(1-(Vcm)^2/c^2)=0$$

and
$$\gamma*m*v=MVcm/\sqrt(1-(Vcm)^2/c^2)=\gamma*M*v/((\gamma+1)*\sqrt((1-\gamma^2*v^2/c^2)/(\gamma+1)^2)$$

 

[member 553083]

I understand the conservation of momentum and energy and the definition of gamma, but I don't see how to get the expressions above.

 

[thepatientmental]

To understand these expressions, we need to first understand the concept of the center of mass (CM) frame in relativistic collisions. The CM frame is the frame of reference in which the total momentum of the system is zero. In this frame, the particles move with a common velocity called the center of mass velocity (Vcm). This is the velocity that we are trying to find in part (a) of the problem.

Now, let's break down the first expression: (Ptot-Vcm*Etot/c^2)/\sqrt(1-(Vcm)^2/c^2)=0. Here, Ptot is the total momentum of the system, and Etot is the total energy of the system. In the CM frame, the total momentum is zero, so Ptot=0. This means that we can simplify the expression to -Vcm*Etot/c^2=0. Now, we know that Etot is given by the sum of the kinetic energies of the two particles, which can be written as Etot=K+mc^2. Substituting this into the equation, we get -Vcm*(K+mc^2)/c^2=0. Rearranging, we get Vcm=K/(K+mc^2). Now, we also know that the relativistic factor \gamma is given by \gamma=(1-v^2/c^2)^-1/2. Substituting this into the expression for Vcm, we get Vcm=\gamma v/(\gamma+1). This is the expression that we were asked to verify in part (a) of the problem.

Moving on to the second expression: \gamma*m*v=MVcm/\sqrt(1-(Vcm)^2/c^2)=\gamma*M*v/((\gamma+1)*\sqrt((1-\gamma^2*v^2/c^2)/(\gamma+1)^2). This expression is derived from the conservation of energy and momentum in a perfectly inelastic collision. In this type of collision, the two particles stick together and move with a common velocity after the collision. This common velocity is the center of mass velocity (Vcm). Now, the total momentum before the collision is given by Ptot=m*v, where m is the mass of the incident particle and v is its velocity. After the collision, the total momentum is given by Ptot=M*Vcm, where M is the