Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to count the degree of freedom?

  1. Mar 29, 2006 #1
    Please tell me how to count the degree of freedom in ADM gravity.

    6 independent components of [tex]q_{ab}[/tex],
    and 4 constraints
    = 2 degrees of freedom

    shouldn't it be
    6 independent components of [tex]q_{ab}[/tex]
    implies 12 phase space variables
    4 constraints
    =8 phase space variables
    =4 degrees of freedom?
  2. jcsd
  3. Mar 29, 2006 #2


    User Avatar

    Convention. How many degrees of freedom does a particle on a line have? one, but two phase space variables.
  4. Mar 30, 2006 #3
    I asked why 4 constraints mean 4 degrees of freedom are deducted? (that's 8 phase space variables!) It seems only 4 phase space variables become dependent on each other by 4 constraints.
  5. Mar 31, 2006 #4
    Hi, in the Hamiltonian formulation one has the canonical variables q_ab and p_ab which determine 12 local degrees of freedom and have to satisfy 4 constraints, so 8 remain. However, one has to take into account the lapse and shift variables (originating from the choice of foliation). Those - contracted with the constraints - generate 4 gauge transformations which correspond with the lie derivative of those phase space variables which satisfy the constraints in the direction of the associated vectorfield. Since the dirac algebra closes, no further constraints arise and therefore 8 - 4 = 4 (vectorfields determine 4 local degrees of freedom) local phase space variables are relevant. Note that in three spacetime dimensions, the counting would be 3+3 = 6 phase space variables satisfying 3 constraints and 3 gauge transformations -> gravity is topological. In general, for a D+1 dimensional spacetime : D(D+1) phase space variables - 2(D+1) coming from constraints and gauge trans = (D-2)(D+1) phase space variables.

    At the level of the spacetime metric (Lagrangian viewpoint) : you have (D+1)(D+2)/2 local degrees of freedom. Now, (D+1) degrees of freedom are pure tensorial, that means dependent upon choice of a local coordinate system. Once you have chosen one particular system, you still have the active diffeomorphism freedom: that is you can consider mappings shifting up spacetime points, these represent again D+1 local degrees of freedom. In total (D+1)(D+2)/2 - 2(D+1) = (D+1)(D-2)/2. In four d, these are the so called 2 graviton degrees of freedom. Since these obey hyperbolic (second order) partial differential equations, this results in 4 phase space variables.


    Last edited: Mar 31, 2006
  6. Mar 31, 2006 #5


    User Avatar

    A bit more abstractly then Careful's answer:

    Correct, but first class constraint means two things:
    1: C = 0
    2: {C, O} = 0

    That's basically two unknowns, on the constraintsurface where condition one holds you ALSO have to compute gaugeorbits which imply condition 2.
    The space of physical states is characterized by equivalence classes of points on the contstraint surface.

    Quantum mechanically your wavefunction depends only on x not on p, it's on configuration space not phase space, and therefore implementing the first condition already implies the second one in some precisely definable sense.
  7. Apr 1, 2006 #6
    thank you.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?