# Calculate number microstates? Why not include acceleration?

1. Apr 28, 2014

### llisuhrtgslir

I want to calculate the number of microstates in a system for, say, n indistinguishable particles in a 3D box. Some examples I see just represent one microstate as a list of positions. Other examples use a list of positions and a list of velocities (translational and rotational). And if you're supposed to use position and velocity, why not acceleration too? I can maybe see why you wouldn't use "jerk" and fifth derivatives of position and so on.

2. Apr 28, 2014

### Staff: Mentor

Because the state of such is system is completely described by the canonical coordinates of the particles. Specifying the position and momentum of all the particles uniquely locates it in phase space.

3. May 1, 2014

### Jano L.

Those examples work within Newtonian statistical physics, where the state of many-particle system is specified by stating coordinates and their first derivatives. Higher derivatives are then determined by the equations of motion.

This is not valid in relativistic theory, where the interaction of the particles is no longer describable by such ordinary differential equations.

4. May 1, 2014

### Khashishi

The acceleration on each particle is calculated from the force on the particles, which you can determine from a snapshot of the system, whereas the velocity is just a free parameter.

5. May 1, 2014

### WannabeNewton

States of individual particles of a system in classical relativistic stat mech are given (in flat space-time) by the phase space of 4-positions and 4-momenta with the latter constrained to lie on the mass hyperboloid so it isn't much different from the Newtonian case.

6. May 2, 2014

### Jano L.

Only if the particles are non-interacting. State of a system of interacting particles in relativistic theory is not specified by their positions and momenta only, because forces acting on them cannot be functions of the positions and momenta only.

7. May 2, 2014

### micromass

Let's wait for the OP to come back before going into the intricacies of relativity theory. As the question was posed, no relativity theory was needed, so let us keep it nice and elementary.