COM vs Lab Total energy discrepency

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SUMMARY

The discussion centers on the discrepancy between total energy in the Center of Mass (COM) frame and the lab frame during elastic neutron scattering with a nucleus. It is established that the energy in the COM frame is always lower than in the lab frame due to the non-zero momentum of the system in the lab frame. The equation E = √(M²c⁴ + P²c²) illustrates that while the rest mass remains constant, the total energy varies with momentum, leading to this difference. This concept is rooted in classical physics and is crucial for understanding relativistic energy transformations.

PREREQUISITES
  • Understanding of elastic scattering in nuclear physics
  • Familiarity with relativistic energy equations, specifically E = √(M²c⁴ + P²c²)
  • Knowledge of reference frames in physics, particularly Center of Mass (COM) and lab frames
  • Basic grasp of momentum and its role in energy calculations
NEXT STEPS
  • Study the implications of relativistic energy transformations in different reference frames
  • Explore the concept of momentum conservation in elastic collisions
  • Learn about the differences between classical and relativistic mechanics
  • Investigate practical applications of COM frame analysis in nuclear physics
USEFUL FOR

Students and professionals in nuclear physics, physicists interested in relativistic mechanics, and anyone seeking to deepen their understanding of energy transformations in different reference frames.

physmurf
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I am reading from the book, "Nuclear Reactor Theory", by Lamarsh. I have run across an idea that I am struggling to understand:

It states that for a neutron that scatters elastically with a nucleus, the Energy in the Center of Mass (COM) frame of reference will always be slightly less than the total energy in the lab frame of reference.

For some reason I don't like this. Sure I can follow the mathematics and the explanation, but it doesn't seem right that the total energy would be the same. When I contacted my professor about this, he indicated that this is merely a classical idea. However, for some reason, this just doesn't jive with what I think should be going on.

Anybody have any conceptual explanations on why this is true?

Thanks
 
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Relativistic energy of any system in any reference frame is given by

E = \sqrt{M^2c^4 + \mathbf{P}^2c^2}...(1)

where \mathbf{P} is the total momentum and M is the rest mass, which is independent on the reference frame. In the center-of-mass frame the total momentum is zero, so the energy is E = Mc^2, which is lower than in any other reference frame (where the total momentum is non-zero).

So, the reason for the excess energy in the lab frame is the fact that the system (neutron + nucleus) has a non-zero momentum.

Eugene.
 
physmurf said:
It states that for a neutron that scatters elastically with a nucleus, the Energy in the Center of Mass (COM) frame of reference will always be slightly less than the total energy in the lab frame of reference.
...
For some reason I don't like this. Sure I can follow the mathematics and the explanation, but it doesn't seem right that the total energy would be the same.
I guess I don't get what's bugging you. The total energy will always be less in the COM frame.

(Looks like Eugene already answered.)
 
Doc Al said:
...

I guess I don't get what's bugging you. The total energy will always be less in the COM frame.

(Looks like Eugene already answered.)

Intuitively I would think that the total energy of the system should be the same no matter what frame of reference you are in... However, that is where my misconception lies...Energy is not always going to be the same in every reference frame!

I apologize for the stupid question. I appreciate your responses.
 

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