- #36
gato_
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Rasalhague said:[tex](1) \enspace m\textbf{v}.[/tex]
[tex](2) \enspace \frac{m_0 \,v}{\sqrt{1-\left ( \frac{v}{c} \right )^2}},[/tex]
[tex](3) \enspace E^2 = \left ( pc \right )^2 + \left ( mc^2 \right )^2,[/tex]
[tex](4) \enspace E = pc[/tex]
[tex](5) \enspace p = \frac{h}{\lambda},[/tex]
Actualy, all these definitions are related. All of them can be interpreted as infinitesimal generators of translations
[tex]\phi(x+\delta x)\sim \phi(x)+\delta x^{\mu}\partial_{\mu}\phi\equiv \phi(x)+\delta x\cdot p[\phi(x)] [/tex]
(A factor [tex]i\hbar[/tex] is required in the quantum definition for mathematical reasons).
So, first fact, momentum is related to space and time translations (in relativity, energy is the momentum component related to time translations). There is a similar derivation for the energy momentum tensor.
The mathematical fact is the momentum, on whatever the representation you choose to act on, is conserved whenever the underlying system is invariant under translations. And the physical principle is that the undelying space on which matter exists is itself invariant under those translations (or have other symmetries leading to other conserved quantities). All the physical interactions may be thought off as local momentum interchanges.
On a more technical side, it could be said that SO(3,1), the group representing the symmetry of flat space in general relativity has two (Cassimir) invariants: [tex]P^{2}=M^{2}[/tex] and spin. That means any irreducible representation of this group (that is, any free particle) can be labeled by those two. Eq (3) merely states that a free particle has a constant momentum. Of course, when you consider the rest of interactions, Electromagnetic weak and strong, you have to add additional labels.