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In the particle disintegration problem in the book by Landau and Lifshit(z), it is considered a particle with velocity $\vec{V}$ in the lab frame, which disintegrates into two particles with masses $m_1$ and $m_2$. It is said that $\vec{v}$ and $\vec{v}_0$ are the velocities of one of the particles in the lab frame and the center of mass (c.m.) frame, respectively, hence being "evident" that $\vec{v}=\vec{V}+\vec{v}_0$. How is this evident? Shouldn't $\vec{V}$ be the velocity of the c.m. in the lab frame?

If the latter was already the case, from the conservation of momentum in the lab frame, we would have:

$$m_0\,\vec{V} = m_1\,\vec{v_1}+m_2\,\vec{v_2}$$ (where either $\vec{v_1}$ or $\vec{v_2}$ are equal to $\vec{v}$; each number corresponds to a particle).

This can be written as
$$\frac{m_0}{M} \vec{V}=\vec{v}_{cm}$$ with $M=m_1+m_2$ and $\vec{v}_{cm}$ being the velocity of the c.m. in the lab frame. Thus, $\vec{V}=\vec{v}_{cm}$ would require conservation of mass (which is not mentioned).

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    $\begingroup$ Looking at your link, you can tell that they are working in the non-relativistic limit almost immediately when they write the energy of particle 1 as $E_1+\frac{p_0}{m_1^2}$ in the c.m. frame. This is a classic result from special relativity for the energy in a low-speed limit. $\endgroup$
    – MonkeysUncle
    Commented Mar 20, 2015 at 18:06
  • $\begingroup$ @MonkeysUncle actually the whole book is about classical mechanics. $\endgroup$
    – Ruslan
    Commented Mar 20, 2015 at 18:22

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Center of mass before the disintegration is in the initial particle. This means that the center of mass moves with velocity $\vec V$ in lab frame. Thus, to switch from center of mass frame to lab frame you just use the Galilean transformation

$$\vec v=\vec v_0+\vec V.$$

As for conservation of mass, it is certainly implied because the system is considered to be closed, and the book concerns only non-relativistic theory, where mass doesn't depend on internal energy of the system.

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I would argue that the transformation is NOT evident, at least not to those who do not frequently perform coordinate transforms. However, it would have been evident to those who do multibody dynamics, with the help of proper notations.

Consider using Kane's notation, for example, ${}^{B}\mathbf{v}^{A}$ stands for the velocity of the object $A$ with respect to the reference frame $B$, then we can translate the notations from Landau as

$$\mathbf{v}_0={}^{C}\mathbf{v}^{P},\qquad\mathbf{v}={}^{L}\mathbf{v}^{P}, \qquad\mathbf{V}={}^{L}\mathbf{v}^{C}$$

where $C$ stands for Center of mass, $L$ stands for Lab frame, and $P$ stands for the Particle. Then just as vector addition, we have (note that ${}^{B}\mathbf{v}^{A}=-{}{}^{A}\mathbf{v}^{B}$) $${}^{C}\mathbf{v}^{P}={}^{L}\mathbf{v}^{P}-{}^{L}\mathbf{v}^{C}$$ which is just $$\mathbf{v}_0=\mathbf{v}-\mathbf{V}$$ or $$\mathbf{v}=\mathbf{V}+\mathbf{v}_0$$

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