Do the degrees of freedom change for particles in a relativistic system?

[LarryS]

Consider one particle traveling at a relativistic velocity in 3-space. Then the configuration space of the system consisting of that one particle would have 3 degrees of freedom – 1 particle times 3 dimensions.

Because of its high energy, the particle decays into, say, 2 particles. Now the configuration space of the system has 6 degrees of freedom.

Is the above correct?

But, the 2 particles are now entangled in position (as well as momentum). So does the configuration space truly have 6 degrees of freedom, or something less than that due to position correlation between the 2 particles?

Thanks in advance.

 

[strangerep]

Consider one particle traveling at a relativistic velocity in 3-space. Then the configuration space of the system consisting of that one particle would have 3 degrees of freedom – 1 particle times 3 dimensions.

For a relativistic particle, there are two invariants: the mass and the (total) spin. For simplicity, let's suppose we're talking about a spinless particle. Then, although it might seem like there's 4 degrees of freedom (1 energy and 3 momenta), these are constrained by the relativistic mass relation:
<br /> M^2 ~=~ E^2 - |{\mathbf P}|^2<br />
so there's really only 3 degrees of freedom. The equation above defines the "mass hyperboloid", or "mass shell" for the particle.

Because of its high energy, the particle decays into, say, 2 particles. [...]

Total energy and total momentum are both conserved. This places further constraints on any particles that might emerge from a decay, thus reducing the degrees of freedom applicable to such situations.