Problem conserving 4-momentum at CoM frame in an inelastic collision [closed]

Question

I am confused about the case where mass is not conserved in a collision (not due to relativistic factors). The center of momentum (CoM) frame is not the same before and after the collision.

Let's call frame 1 the initial CoM frame, and frame 2 the final CoM frame. So on the one end in frame 2: $p^{(2)}_{final}=0$ because it is the CoM frame. But on the other end in frame 1: $p^{(1)}_{initial}=0$, because it is also the CoM frame. As I understand it, there cannot be another frame where $p_{initial}=0$ apart from frame 1, therefore $p^{(2)}_{initial}\neq0$. So 4 momentum is not conserved.

Actually the Wiki page on CoM frame suggests there are several CoM frames, but I am assuming they do not differ by velocity but rather by point of origin?

In the simpler, non relativistic case, I found that by calculating $p_{initial}$ in the 2nd frame:

$$p^{(2)}_{initial} = p^{(Lab)}_{initial} \cdot \text{const} \neq 0$$

EDIT:

Here is the non-relativistic calculation (relativistic does not help). The process (in lab frame) is: $(M,\bar{\omega}) + (m,\bar{u}) \rightarrow (M+\delta,\bar{v}) + (m,\bar{u'})$. The velocity of the center of momentum in frame 2 is $\bar{s}^*$.

$$M^*=M+\delta \qquad \mu = \frac{M}{m} \qquad \mu^*=\frac{M^*}{m} \qquad \bar{s}^* = \frac{\mu^*\bar{v}+\bar{u'}}{1+\mu^*}$$

$$p^{(2)}_{initial}=M (\bar{w}-\bar{s^*}) + m (\bar{u}-\bar{s^*})=\left\{ M \bar{w} + m \bar{u} \right\} - \left\{ M \bar{s^*} + m \bar{s^*} \right\} $$ Using conservation of momentum in lab frame: $$= \left\{ M^* \bar{v} + m \bar{u}' \right\} - \left\{ M \bar{s^*} + m \bar{s^*} \right\}$$ $$= m \left( \left\{ \mu^* \bar{v} + \bar{u}' \right\} - \bar{s^*} \left\{ \mu + 1 \right\} \right)$$ $$= m \left( \left\{ \mu^* \bar{v} + \bar{u}' \right\} - \frac{\mu^*\bar{v}+\bar{u'}}{1+\mu^*} \left\{ \mu + 1 \right\} \right)$$ $$= m \left\{ \mu^* \bar{v} + \bar{u}' \right\} \left( 1 - \frac{1}{1+\mu^*} \left\{ \mu + 1 \right\} \right)$$ $$= \left\{ M^* \bar{v} + m \bar{u}' \right\} \frac{\delta }{\delta +m+M}$$ $$= p^{(2)}_{initial} = p^{(lab)}_{initial} \frac{\delta }{\delta +m+M} \neq 0$$

But hey, you might think $\bar{s^*}$ isn't the right velocity, well lets see; Here is the (final) momentum in the center of momentum frame:

$$p^{(2)}_{final} = M^* (\bar{v}-\bar{s^*}) + m (\bar{u'}-\bar{s^*}) $$ $$= M^* \left(\bar{v}-\frac{\mu^*\bar{v}+\bar{u'}}{1+\mu^*}\right) + m \left(\bar{u'}-\frac{\mu^*\bar{v}+\bar{u'}}{1+\mu^*}\right)$$ $$= \left(M^* \bar{v}+m \bar{u'}\right) - (m + M^*) \left(\frac{\mu^*\bar{v}+\bar{u'}}{1+\mu^*}\right)$$ $$= \left(M^* \bar{v}+m \bar{u'}\right) - (m + M^*) \left(\frac{M^*\bar{v}+m \bar{u'}}{m + M^*}\right)$$ $$= \left(M^* \bar{v}+m \bar{u'}\right) - \left(M^*\bar{v}+m \bar{u'}\right) = 0$$

What am I not taking into consideration? What is the initial state of frame 2, $p^{(2)}_{initial}$?

Answer 1

What is missing: this must be recognized as a relativistic process and be treated accordingly.

Why is it a relativistic process: energy transforms into mass or conversely. This is something that cannot be accounted for using only classical mechanics. In older times I would've added "because it needs to account also for relativistic mass", but since using the term "relativistic mass" seems to be discouraged as of lately, I will "refrain".

What to do: Write energy-momentum conservation properly in the Lab frame:

$$ ( E_\omega/c, p_\omega ) + ( E_u/c, p_u ) = ( E_\nu/c, p_\nu ) + ( E_{u'}/c, p_{u'} ) = ( E_{Lab}/c, p_{Lab} ) $$

Here energies and momenta relate to velocities and rest masses as $$ E_\omega/c = \gamma_\omega M_0c, \;\; p_\omega = \beta_\omega(E_\omega/c) = (\beta_\omega c)(\gamma_\omega M_0) = (\gamma_\omega M_0)\omega $$ $$ E_u/c = \gamma_u m_0c, \;\; p_u = \beta_u(E_u/c) = (\beta_u c)(\gamma_u m_0) = (\gamma_u m_0)u $$ etc, with the usual relativistic notation for adimensional velocities $\beta_\omega$, $\beta_u$, and $M_0$, $m_0$, etc, the rest masses for the corresponding particles. Note that you can write the momentum conservation law in the quasi-classical form $$ M\omega + mu = (M + \delta)\nu + mu' $$ only provided the masses $M$, $m$ are understood to be the "relativistic" masses $\gamma_\omega M_0$, $\gamma_u m_0$ in the Lab frame, not the rest masses $M_0$, $m_0$. This is what caused you trouble.

If we want to transform to the COM frame, we need only identify the correct boost velocity from the total energy and momentum, $\beta = p_{Lab}c/E_{Lab}$, and apply the corresponding Lorentz transform. The new energies and momenta read:

$$ p'_\omega = \gamma [ p_\omega - \beta E_\omega/c ], \;\; E'_\omega/c = \gamma [E_\omega/c - \beta p_\omega ] $$

$$ p'_u = \gamma [ p_u - \beta E_u/c ], \;\; E'_u/c = \gamma [E_u/c - \beta p_u ] $$

$$ p'_\nu = \gamma [ p_\nu - \beta E_\nu/c ], \;\; E'_\nu/c = \gamma [E_\nu/c - \beta p_\nu ] $$

$$ p'_{u'} = \gamma [ p_{u'} - \beta E_{u'}/c ], \;\; E'_{u'}/c = \gamma [E_{u'}/c - \beta p_{u'} ] $$

Now we can check that, given energy-momentum conservation in the Lab frame, in the COM we find

$$p'_\omega + p'_u = \gamma [ (p_\omega + p_u) - \beta (E_\omega + E_u)/c ] = \gamma ( p_{Lab} - \beta E_{Lab}/c ) = \gamma ( \beta E_{Lab}/c - \beta E_{Lab}/c ) = 0 $$

and similarly,

$$ p'_\nu + p'_{u'} = \gamma [ (p_\nu + p_{u'}) - \beta (E_\nu + E_{u'})/c ] = \gamma ( p_{Lab} - \beta E_{Lab}/c ) = \gamma ( \beta E_{Lab}/c - \beta E_{Lab}/c ) = 0 $$

So there is a single COM after all, as expected.