I have been looking for an answer to 'Why is the laboratory frame energy always greater than the center of mass frame energy during collisions?'.

A lot of resources provided mathematical explanations. I wanted to get some perspectives in terms of physics.

Could you explain this?

We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame).

If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object is not moving, so it has no kinetic energy. In the lab frame the total object is moving so it does have kinetic energy.

  • Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving? – user7852656 Oct 17 at 21:34
  • @user7852656 The "total object", in the sense of the entire system, isn't moving by definition of the center of momentum frame. If it was, you would no longer be in the COM frame. Consider say, a group of particles doing nothing but uniformly moving to the "left" in the lab frame. Clearly the group has some momentum. In the COM frame the momentum vanishes (by definition), because the particles are no longer moving relative to the frame. – mbrig Oct 18 at 5:07
  • Hi @mbrig, thanks for your comment. If the COM frame refers to the total object, could you also explain the lab frame energy as a contrast? – user7852656 Oct 18 at 5:41

I should say, in this answer I've set $c=1$.


In the centre of mass frame, the total energy is just the rest mass energy of all your particles. If you think about the system collectively (e.g. as a "ball" of your particles with momentum given by the net momentum of the system), you will find this "ball" is stationary and has no momentum, and thus does not contribute anymore to the energy.

If you are in a frame where the total momentum is not zero however, then this "ball" has some net momentum in some direction, and will therefore always add to the energy from the rest masses alone.


In response to @MaciejPiechotka's comment...

I believe my previous answer where I claimed the total energy of the centre of mass frame is $E=m_1+m_2$ (now removed) is incorrect. You do indeed need to include the momenta somehow in the centre of mass energy. After all, if two particles are flying towards each other at some velocity $v$, the centre of mass energy ought to be less than if they were flying towards each other at a larger velocity!

My equation $E^2_i=m^2_i+|\vec{p}_i|^2$ is still correct for the $i$th particle, and so the total energy in the centre of mass frame should actually be (for the two-particle collision)

$$E = E_1 + E_2 = \sqrt{m_1^2+|\vec{p}_1|^2}+\sqrt{m_2^2+|\vec{p}_2|^2}$$

The algebra gets a bit messy, so I suppose it is indeed better to view the problem in qualitatively as I have in my answer above.

I might add to this answer if I find a convincing mathematical argument...

  • Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions? – user7852656 Oct 17 at 21:26
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    My physics is quite rusty but are you sure you don't need to add momentum of individual particles into first equation? Otherwise simple electron/positron collision resulting in 2 photons of same wavelength (mass 0) breaks conservation of energy. – Maciej Piechotka Oct 18 at 6:24
  • @user7852656 It's not necessarily about momentum transfer as such - rather I am just considering the momentum of the system as a whole (doesn't matter if the collision is elastic or inelastic). If it were inelastic, the momentum would still have to go somewhere, and as long as my system is closed, I won't lose it (it'll just end up somewhere else, perhaps in a new particle etc). – Garf Oct 18 at 9:11
  • @MaciejPiechotka Please see my corrected answer, I think you are right, I made a mistake in my original answer. I've tried to do the same thing using this correct centre of mass energy, but the algebra got very messy, and I can't show nicely the result. I might return to this problem later. My wordy answer is still correct I believe. – Garf Oct 18 at 10:03

I think you can obtain physics perspective by examining a couple of simple cases. Center of mass (COM) is a mathematical construct so an exact proof does require some algebra which is out-of-scope in this question.

We can focus on kinetic energies and collisions are irrelevant to the discussion.

First, what does it mean being in the lab frame or COM frame? We are looking at all the possible inertial frames (frames moving at constant velocity, possibly zero), and asking in which of these, kinetic energy is minimal. The COM frame is well defined, it moves at the same velocity as the COM. The lab frame, can actually be a frame moving at any other velocity.

The first simple case are two objects fixed in place. Obviously, the optimal inertial frame, minimizing energy, would be the one with no relative velocity to these objects. So you prefer the COM frame to any moving frame.

Next, consider two objects moving north at equal velocities. Again, it is obvious that your optimal frame should move north and at the same velocity (that's the only frame where energy is zero). So again, you intuitively select the COM frame.

Finally, assume two objects, one standing and another moving slowly towards the first along the x axis. Obviously, you wouldn't select your optimal frame to be moving very fast towards +X. You would not select it moving very fast towards -X either. So somewhere in between, will be the optimal velocity. The rest is algebra.

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