1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculate number microstates? Why not include acceleration?

  1. Apr 28, 2014 #1
    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. jcsd
  3. Apr 28, 2014 #2

    DrClaude

    User Avatar

    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.
     
  4. May 1, 2014 #3

    Jano L.

    User Avatar
    Gold Member

    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.
     
  5. May 1, 2014 #4
    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.
     
  6. May 1, 2014 #5

    WannabeNewton

    User Avatar
    Science Advisor

    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.
     
  7. May 2, 2014 #6

    Jano L.

    User Avatar
    Gold Member

    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.
     
  8. May 2, 2014 #7

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook