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Volume elements of phase space

  1. May 14, 2015 #1
    First, two definitions: let ## \varrho (M)## be the probability density of macro states ##M ## (which correspond to a subgroup of the phase space) and ## \mathrm{d} \Gamma ## be the volume element of a phase space.
    In my lecture notes, the derivation for continuity equation of probability density starts with:
    $$ \frac{\mathrm{d}}{\mathrm{d}t} \int \mathrm{d} \Gamma \, \varrho= \int \frac{ \bar{n} \cdot \bar{v} \mathrm{d}A \, \mathrm{d}t}{\mathrm{d}t} \varrho + \int \mathrm{d} \Gamma \, \partial_t \varrho ,$$
    where the first term on right hand side is supposed to be the time-derivative of the volume element in phase space.
    This step bothers me. Why do we also take the derivative of ## \boldsymbol{ \mathrm{d} \Gamma} ## - are these not usually ignored (see, for instance, Leibniz integral rule for taking a derivative inside an integral)? What kind of a mathematical object is ## \boldsymbol{\mathrm{d} \Gamma}##?
     
    Last edited: May 14, 2015
  2. jcsd
  3. May 19, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. May 20, 2015 #3
    I do have some updates on this problem.
    I'm still not convinced with the above derivation. Maybe it works with some explanation, but I prefer something mathematically more straight-forward.
    However, I did find an alternative derivation, which, to my mind, goes more smoothly.
    As a reminder, we wish to get a continuity equation for ##\rho ##. Actually, we'll find Liouville's theorem at the same time.
    Firstly, we need to show that ## \mathrm{d}/\mathrm{d}t \, \mathrm{d} \Gamma = 0## i.e. that a infinitesimal volume element in phase space does not change in volume. This appears to hold approximately.
    During a small time ##\mathrm{d}t##: ## x(0) \to x(0) + \dot{x}(0) \mathrm{d}t ##, so that ## \mathrm{d}x(t) = \mathrm{d}x(0) + \partial \dot{x}(0)/\partial x(0) \mathrm{d}t##. Thus, (after a short calculation, which is easiest to perform writing out the Jacobian of the 'coordinate transformation'), ##\prod_i \mathrm{d}q_i(t) \mathrm{d}p_i(t)= \prod_i \mathrm{d}q_i(0) \mathrm{d}p_i(0) + \mathcal{O} \mathrm{d}t^2##, but as we expanded up to linear order in the first place, we drop the ## \mathrm{d}t^2##-term.
    Thus, the infinitesimal volume element stays constant in time.

    There is still something that bothers me in the next part. We want to argue that ## \mathrm{d}/\mathrm{d}t \, \mathrm{d} \Gamma \, \rho= 0 ##. We remember that the trajectories ## q_i ## and ## p_i ## follow ##i##:th particle (and so on). Let's follow the density along the trajectories of some particles. My cognitive problem is that in general, I've understood that ##\rho## is the probability to find the system from some macrostate. Say, you start from an initial state where the particles are close together and define your macro state as the volume that particles take, and set the particles free. I would definitely expect a higher volume becoming more probable as time goes on.
    Then again, in some contexts, ## \rho## is taken to be the particle density and the probability to find particles from phase space is normalised to ## N##. Then, if I follow the trajectories of some ##n## particles weighed with probability density, it is intuitive that the probability density along those paths stays constant, because we always find the same number of particles in that (deforming) volume of phase space.
    If someone is able to clarify this, I would be happy (or I'll come back to it when I understand it better). It's funny how physicists in general don't define the probability density in a clear way (read: at all).

    After asserting the above, we are practically done. We note that ## \rho= \rho(t,...,q_i,...,p_i,...)##. Thus $$ 0= \frac{\mathrm{d}\rho}{\mathrm{d}t} = \partial_t \rho + \frac{ \partial \rho }{ \partial q_i } \dot{q}_i + \frac{ \partial \rho }{ \partial p_i } \dot{p}_i $$ (sum over ##i##), and denoting ##\boldsymbol{v}= (\dot{\boldsymbol{q}},-\dot{\boldsymbol{p}}) ##
    $$ 0 = \partial_t \rho +( \nabla \rho ) \cdot \boldsymbol{v}$$
    which is the continuity equation, since in this case ## \nabla \cdot \boldsymbol{v}= 0 ## so that ## \nabla \cdot (\rho \boldsymbol{v})= (\nabla \rho) \cdot \boldsymbol{v} ##.
     
  5. May 22, 2015 #4
    Possibly a last update. I think this problem is pretty much solved (or, well, solved in an alternative way). I still couldn't see ##\mathrm{d}/\mathrm{d}t \, \mathrm{d}\Gamma \, \rho= 0## from any equations. But, as the last messages show, I was confused about the definition of the probability distribution. After some googling and thinking I realised that in any case ##\rho= \rho(t,\boldsymbol{q},\boldsymbol{p})## gives a weight to each trajectory ##(\boldsymbol{q}_i,\boldsymbol{p}_i)## in phase space. (It's only after one maximises entropy with boundary conditions that one arrives at ##\rho## as a function of some macroscopic variables.) And as such, if there are no particles created or destroyed, and we follow some trajectories (of some particles) during a time-interval, the particles should maintain the same weights.

    My only 'problem' with this is that to me, it still feels like we 'assume' the difficult part of a result, the rest is just 'trivial' algebra, and still we call it a result.
     
    Last edited: May 22, 2015
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