COM vs Lab Total energy discrepency

physmurf

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

meopemuk

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

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

where [itex] \mathbf{P} [/itex] is the total momentum and [itex] M [/itex] 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 [itex] E = Mc^2 [/itex], 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.

Doc Al

...
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.)

physmurf

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.