# The center of mass & relativistic collisions

• I
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In special relativity (especially relativistic collisions), is the center of mass frame as useful as Newtonian mechanics?
In special relativity (especially relativistic collisions), is the center of mass frame as useful as Newtonian mechanics?

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Summary:: In special relativity (especially relativistic collisions), is the center of mass frame as useful as Newtonian mechanics?

In special relativity (especially relativistic collisions), is the center of mass frame as useful as Newtonian mechanics?
Even more useful! Vital, in fact.

PS Center of momentum frame, of course.

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• vanhees71
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Even more useful! Vital, in fact.
I'm going to disagree here, since I think defining the "center of mass frame" has some subtleties (system mass or component masses?). The zero momentum frame, which is the same as the center of mass frame in Newtonian physics and the center of system mass frame in relativistic physics, is most certainly as important as you say.

So I'm picking nits, but given the knock-down-drag-out fights we've seen on here over mass and its conservation, I think they're important nits to pick.

• vanhees71
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Yes, zero momentum frame was what I had in mind.

• vanhees71 and Ibix
I'm still confused. For example, in 'Introduction to Elementary Particles by Griffith', for relativistic collisions, the center of momentum frame is introduced to solve problems. But isn't the center of mass frame appropriate in relativistic collisions?

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I'm still confused. For example, in 'Introduction to Elementary Particles by Griffith', for relativistic collisions, the center of momentum frame is introduced to solve problems. But isn't the center of mass frame appropriate in relativistic collisions?
The key concept in SR is energy-momentum. In general, you should start thinking in terms of energy and momentum and not in terms of mass and velocity.

To take one example: a photon is modeled as a massless particle in SR. It has energy and momentum, hence the centre of momentum (or zero momentum) frame can be defined for collisons/decays involving photons. But, a centre of mass frame when one particle is massless is not very useful.

• The key concept in SR is energy-momentum. In general, you should start thinking in terms of energy and momentum and not in terms of mass and velocity.

To take one example: a photon is modeled as a massless particle in SR. It has energy and momentum, hence the centre of momentum (or zero momentum) frame can be defined for collisons/decays involving photons. But, a centre of mass frame when one particle is massless is not very useful.
I understand now by your good example.
Thankful.

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In relativity it's indeed the center-momentum frame, not the center of mass frame. That's because today 112 years after Minkowski's crucial article about the mathematical structure of special relativity ("Minkowski space") we express everything in covariant quantities rather than in some arbitrary confusing ones, and that's why "relativistic mass" is not used anymore anywhere in current research (though there are still some new textbooks introducing the confusion, because the authors are unwilling to learn century-old math ;-)).

The invariant mass of a system (e.g., a set of point particles or a continuum mechanics description of a fluid or some fields like the electromagnetic field) is given by the total four-momentum ##P^{\mu}## of the system by
$$M^2 c^2=P_{\mu} P^{\mu} = (E/c)^2-\vec{P}^2 \geq 0.$$
For ##M>0## you can always find an inertial frame, where ##\vec{P}=0##, and that's called the center-of-momentum frame, and it's considered as the "rest frame" of the system.

The reason is that from Noether's theorem applied to Lorentz boosts it follows that for a closed system the center energy-weighted average rather than the mass-weighted average moves with constant velocity.

E.g., take two interacting particles. Their total momentum is conserved (Noether's theorem applied to translation invariance in space and time), i.e.,
$$p_1+p_2=\text{const}.$$
Written in terms of the coordinate time that reads
$$m_1 \gamma_1 \dot{x}_1 + m_2 \gamma_2 \dot{x}_2=\text{const},$$
but
$$m_1 \gamma_1=E_1/c^2, \quad m_2 \gamma_2=E_2/c^2.$$
This implies
$$E_1 \dot{\vec{x}}_1 + E_2 \dot{\vec{x}}_2=\text{const}.$$
The temporal component means
$$E=E_1+E_2=\text{const},$$
and thus the energy-weighted average of the three-velocities (rather than the mass-averaged three-velocities) is conserved,
$$\vec{V}=\frac{E_1 \dot{\vec{x}}_1 + E_2 \dot{\vec{x}}_2}{E_1+E_2}=\text{const}.$$

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