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How cyclic coordinates affect the dimension of the cotangent manifold

  1. Dec 10, 2013 #1
    Our professor's notes say that "In general, in Hamiltonian dynamics a constant of motion will reduce the dimension of the phase space by two dimensions, not just one as it does in Lagrangian dynamics." To demonstrate this, he uses the central force Hamiltonian,

    [tex]H=\frac{P_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ \frac{p_{\phi}}{2mr^2 sin^2 \theta} + V(r).[/tex]

    Since by Hamilton's equation [itex]\dot{p_{\phi}}=0[/itex] this is a constant of the motion. So specifying $p_{\phi}=\mu$ gives us a 5 dimensional manifold. The notes go on to state that, "Furthermore, on each invariant submanifold the Hamiltonian can be written

    [tex]H=\frac{P_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ \frac{\mu}{2mr^2 sin^2 \theta} + V(r),[/tex]

    which is a Hamiltonian involving only two freedoms [itex]r[/itex] and [itex]\theta[/itex]. Therefore the motion actually occurs on a 4-dimensional submanifold of the 5-dimensional submanifold of [itex]T^*Q[/itex] . . ." However, to me it looks like we still have five degrees of freedom: [itex]p_{\theta}, p_r, r, \theta,[/itex] and [itex]\phi[/itex]. So I'm not sure what he means when he says that the presence of a constant of motion reduces the dimension of the cotangent manifold by 2. Is he saying that if w specify a numerical value for H then the dimension is reduced from 5 to 4, or does just the presence of a cyclic coordinate reduce the dimension from 6 to 4?
     
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  3. Dec 10, 2013 #2

    ChrisVer

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    but isn't you Φ fixed?
     
  4. Dec 10, 2013 #3

    dextercioby

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    Check out corollary 2, page 67 from V.I. Arnold's <Mathematical Methods of Classical Mechanics>. The phase space dimension is indeed reduced by 2xn, n=nr. of cyclic coordinates for the Hamiltonian.
     
  5. Dec 12, 2013 #4
    Thank you. I think I'm just confusing terminology.
     
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