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

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, $p^{(2)}_{initial}$?

## closed as unclear what you're asking by John Rennie, ACuriousMind♦, Kyle Kanos, Jim, NeuneckAug 12 '15 at 9:56

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• "The center of momentum (CoM) frame is not the same before and after the collision." Yes it is; or at least all such frames are at rest with respect to one another. I'm really very uncertain what you are trying to say in the second paragraph or why you think there is a problem. Be sure to notice that---though we us the abbreviation "CoM" for both "center of mass" and "center of momentum" at times---here we mean only center of momentum. – dmckee Aug 9 '15 at 4:37
• Do you have a reference for the frames being the same? I realize CoM here means Center of Momentum. – user1581390 Aug 9 '15 at 6:25
• In any frame, the total momentum is conserved, so if you boost to a frame where the initial total momentum is zero, the final total momentum is also zero. You don't give enough details to tell where you're going wrong in your calculation. – ragnar Aug 9 '15 at 13:27
• Please describe the setup of your problem. Momentum is conserved even if an object throws off some of its particles to have less (rest) mass. And the mass of an isolated system is also conserved when the energy and momentum are conserved (though it isn't the sum of the masses of the parts). – Timaeus Aug 9 '15 at 17:16
• I have added the calculation to show the issue. There is no third object, so I am taking account of an isolated system. This is a mass-energy conversion, and not relativistic. – user1581390 Aug 10 '15 at 14:41

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.

• udr: "$\beta = p_{Lab}c/E_{Lab}$" -- That's essentially the solution, +1; the velocities $\vec v_{Lab}[~frame~1~] := \frac{M \vec w + m \vec u}{M + m}$ and $\vec v_{Lab}[~frame~2~] := \frac{M^* \vec v + m \vec u'}{M^* + m} := \overline s^*$ from the OP are only different "classical" approximations; neither "frame 1" nor "frame 2" are therefore the COM frame, in general. p.s. "The new energies and momenta read: [...]" -- You seem to be considering only a 1-dimensional problem; $(\vec w \cdot \vec u)^2 = (\vec w \cdot \vec w) (\vec u \cdot \vec u)$, etc. Consider instead: – user12262 Aug 12 '15 at 20:53
• $$\vec p_{\mathcal K}[~\Omega~]=$$ $$\vec p_{\mathcal S}[~\Omega~]+\vec v_{\mathcal S}[~\mathcal K~]~\left(\frac{(\vec p_{\mathcal S}[~\Omega~]\cdot\vec v_{\mathcal S}[~\mathcal K~])}{|\vec v_{\mathcal S}[~\mathcal K~]|^2} \left( \frac{1}{\sqrt{1-|\vec v_{\mathcal S}[~\mathcal K~]|^2/c^2}}-1\right)-\frac{E_{\mathcal S}[~\Omega~]}{c^2~\sqrt{1-|\vec v_{\mathcal S}[~\mathcal K~]|^2/c^2}}\right),$$and$$E_{\mathcal K}[~\Omega~]=\frac{E_{\mathcal S}[~\Omega~]-(\vec p_{\mathcal S}[~\Omega~]\cdot\vec v_{\mathcal S}[~\mathcal K~])}{\sqrt{1-|\vec v_{\mathcal S}[~\mathcal K~]|^2/c^2}},$$ – user12262 Aug 12 '15 at 21:13
• where $\vec p_{\mathcal S}[~\Omega~]$ denotes the momentum of object $\Omega$ with respect to reference system (inertial frame) $\mathcal S$, $\vec p_{\mathcal K}[~\Omega~]$ denotes the momentum of object $\Omega$ with respect to reference system (inertial frame) $\mathcal K$, $E_{\mathcal S}[~\Omega~]$ denotes the energy of $\Omega$ with respect to $\mathcal S$, $E_{\mathcal K}[~\Omega~]$ denotes the energy of $\Omega$ with respect to $\mathcal K$, and $\vec v_{\mathcal S}[~\mathcal K~]$ denotes the velocity of (each member of) $\mathcal K$ with respect to $\mathcal S$. – user12262 Aug 12 '15 at 21:14
• @user12262 You are correct, provided S = Lab, K = COM, etc. But you may want to simplify your notation a bit using \beta-s, \gamma-s, & so on, to make it an easier read. I am not being mean here, just practical: I settled for the 1d case because I was too lazy to format (the compact form of) your formulas above. ;) – udrv Aug 12 '15 at 23:34
• udr: "[...] correct, provided [...]" -- Actually, as far as I understand, the two formulas of my above comment are correct in general; and therefore also in the special case $\mathcal S\mapsto Lab$, $\mathcal K\mapsto \text{ COM frame of the participants in the collision}$, and $\Omega\mapsto\text{ the participants in the collision}$; whereby $\vec p_{\text{COM}}[~\text{participants}~]\mapsto\vec 0$. "you may want to simplify your notation a bit using \beta-s, \gamma-s, & so on" -- I rather avoid introducing notions and abbreviations (trivial as they may be) which the OP didn't use already – user12262 Aug 13 '15 at 5:15