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The concept of relativistic energy comes from it's conservation in relativistic mechanics for an elastic collision.

It seems to me that another possible derivation could equate the energy of a single particle before, with the kinetic energy after of N particles after the energy is distributed amongst them in the form of the Boltzman distribution, as well as individually being of the form $mv^2$.

Are there any problems in doing this?

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Uh, Boltzmann distribution? At what temperature? –  Marek Apr 11 '11 at 17:59
    
@Marek, yes, the problem is determining the equivalent temperature for some known volume $V$ containing the particles. –  Larry Harson Apr 12 '11 at 14:41
    
Why the down vote? –  Larry Harson Apr 12 '11 at 14:42
    
@user2146: you didn't understand. The problem here is that your system isn't statistical nor thermodynamical at all. If it were, then all temperatures would be fine and system would behave differently with variation of temperature. There is obviously no such behavior in your model. Besides this, I honestly don't understand where do those $N$ particles come from. Are they supposed to be internal degrees of freedom? Or are they products of collision? In any case, your question doesn't make much sense as it stands now. –  Marek Apr 13 '11 at 6:30
    
@marek suppose you have an eleastic container of a large number N of particles with the 1st one having an inital velocity V, the rest at rest. After a while, the energy of the system will be distributed amongst the particles statistically. –  Larry Harson Apr 13 '11 at 15:17
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1 Answer

There are problems, I shall list some of them:

1) In special relativity, relativistic energy concept is not coming from consideration of elastic collisions. In fact, relativistic energy is a covariant generalisation of non-relativistic energy. As a viable approach to do this one may generalise the action for a free particle first, and then derive relativistic 3-momenta from lagrangian and energy from hamiltonian. The point I want to stress is that no collisions are needed for derivation. In contrast, covariant generalisation procedure is crucial.

2) You can link the energy of the particle to the energy of the whole ensemble (even a non-relativistic one). However, as you cannot derive the relativistic energy of a particle from a single collision, you cannot do that for a collision with an ensemble.

P.S. On the contrary, if you think that you can 'derive' relativistic energy from 1 collision, you shall then be able to do it for collision with an ensemble.

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The derivation from covariance of collisions is well known, and repeated in many places. The "Lagrangian" derivation assumes that the Lagrangian formalism is not modified in relativity, and this requires physical reasoning--- why should the derivatives in the Lagrangian approach still be time derivatives, and not proper time derivatives, for example. –  Ron Maimon Apr 15 '12 at 22:50
    
@Ron: Covariance of collisions is one thing, and just conservation of energy is another. As for lagrangian formalism, start from least action principle. Action is invariant, and is produced from scalar lagrange function integrated over proper time. Energy conservation arises from symmetry reasons. It is genuinely foundamental and doesn't involve any artificial reasoning. –  Alexey Bobrick Apr 16 '12 at 20:14
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The problem is that you have to justify why least action still works, physically. It's not good enough to show that everything works formally, you have to convince yourself it is really true, and for this, there is no substitute for a dozen good thought experiments on collisions, constant forces, etc. Then you can be confident the formalism is ok even in the relativistic setting. Of course, we nowadays justify Hamilton's mechanics from QM, so you can try to take that route, but relativistic QM is much much harder than getting the kinematics right. –  Ron Maimon Apr 17 '12 at 8:30
    
@Ron: Intuitively you would go as follows: Motion in Newtonian mechanics is all about functions (positions, forces, etc.). Action is a functional. As functionals are a much broader set then functions, you may expect at least to find one which satisfies a finite of conditions. In other words, the assumption behind the theory is general enough for it to hold in any case, you just select the real form of the action by consecutively applying physical statements, such as symmetries, invariance, existence of space-time etc. –  Alexey Bobrick Apr 17 '12 at 19:05
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@Alexie Bobrick: Where do you get your certainty that when you rethink all the foundations of physics a closed system will still conserve energy? You have to check if the assumptions are unchanged. Bohr believed that Quantum effects lead energy conservation to get violated, but he was wrong. It is no good to make arguments based on foramlism--- the formalism follows the physics not the other way around. You must consider physical situations, and enough of them to be sure that the formalism corresponds to reality. Friction example was about hypothetical fundamental friction, not open systems. –  Ron Maimon Apr 18 '12 at 3:10
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