# Feynman's argument that at thermal equilibrium average kinetic energy of two gases are same

I had a hard time following Feynman's argument that when two different gases are present in the same box their average kinetic energy is the same at equilibrium. To quote from his book:

....Next, we have to worry about the actual case, where we do not have the collision in the CM system, but we have two atoms which are coming together with vector velocities $$v_1$$ and $$v_2$$. What happens now? We can analyze this collision with the vector velocities $$v_1$$ and $$v_2$$ in the following way:

We first say that there is a certain CM; the velocity of the CM is given by the “average” velocity, with weights proportional to the masses, so the velocity of the CM is $$v_{CM}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$. If we watch this collision in the CM system, then we see a collision just like that in Fig. 39–3, with a certain relative velocity $$w$$ coming in. The relative velocity is just $$v_1−v_2$$. Now the idea is that, first, the whole CM is moving, and in the CM there is a relative velocity $$w$$, and the molecules collide and come off in some new direction. All this happens while the CM keeps right on moving, without any change.

Now then, what is the distribution resulting from this? From our previous argument we conclude this: that at equilibrium, all directions for w are equally likely, relative to the direction of the motion of the CM.1 There will be no particular correlation, in the end, between the direction of the motion of the relative velocity and that of the motion of the CM. Of course, if there were, the collisions would spray it about, so it is all sprayed around. So the cosine of the angle between $$w$$ and $$v_{CM}$$ is zero on the average. That is,

$$⟨w⋅v_{CM}⟩=0.$$ But $$w⋅v_{CM}$$ can be expressed in terms of $$v_1$$ and $$v_2$$ as well:

$$w⋅v_{CM}=\frac{(V_1−V_2)⋅(m_1V_1+m_2V_2)}{m_1+m_2} =\frac{(m_1v_1^2−m_2v_2^2)+(m_2−m_1)(V_1⋅V_2)}{m_1+m_2}$$

First, let us look at the $$v_1⋅v_2$$; what is the average of $$v_1⋅v_2$$? That is, what is the average of the component of velocity of one molecule in the direction of another? Surely there is just as much likelihood of finding any given molecule moving one way as another. The average of the velocity $$v_2$$ in any direction is zero. Certainly, then, in the direction of $$v_1$$, $$v_2$$ has zero average. So, the average of $$v_1⋅v_2$$ is zero! Therefore, we conclude that the average of $$m_1 v_1^2$$ must be equal to the average of $$m_2 v_2^2$$ That is, the average kinetic energy of the two must be equal: $$⟨\frac{1}{2}m_1 v_1^2⟩=⟨\frac{1}{2}m_2 v_2^2⟩$$.....

The part I don't get is where he he says COM is moving with Vcm and in the system there is relative velocity of $$|V_1 - V_2|$$. Why if we "watch in CM" the molecules have relative velocity? And how is that $$⟨w⋅v_{CM}⟩=0$$? Can someone explain what he is trying to say here.