Hey guys, I'm extremely confused about the distinction between equilibrium statistical mechanics and kinetics, based on literally just the first section of vol. 5 and vol. 10 of Landau.
Starting off with (classical) phase space (for simplicity), for a system of [itex]n[/itex] particles with [itex]3n[/itex] coordinate degrees of freedom and [itex]3n[/itex] momenta we have a [itex]6n[/itex] dimensional phase space [itex]\Gamma[/itex] containing points [itex](\vec{q},\vec{p})[/itex] (known as micro-states). The motion of the system over a time interval is represented as a parametrized curve [itex]\gamma[/itex], [itex]\gamma (t) = (\vec{q}(t),\vec{p}(t)) = (\vec{q},\vec{p}) [/itex]. The form of the curve [itex]\gamma[/itex] for a physical system is determined by the Hamiltonian, but what is actually going on here? Do I say that [itex]H = H(\vec{q}(t),\vec{p}(t)) = E(t)[/itex] determines the motion, or do I say that [itex]H = H(\vec{q}(t),\vec{p}(t),t) = E(t)[/itex] determines the motion? The two situations are different.
Is it that [itex]H = H(\vec{q}(t),\vec{p}(t)) = E(t)[/itex] determines the motion of the system (the curve [itex]\gamma[/itex]), which in general has no dimensional reduction or restrictions (i.e. we have [itex]6n[/itex] degrees of freedom), but that, in the time-independent case [itex]H = H(\vec{q}(t),\vec{p}(t)) = E[/itex] we have only [itex]6n-1[/itex] free coordinates (should I maybe write [itex]H = H(\vec{q},\vec{p}) = E[/itex]?), resulting in our curve living on the energy shell in the picture:
Or, is it that [itex]H = H(\vec{q}(t),\vec{p}(t),t) = E(t)[/itex] determines the motion in general, and we can draw the above picture in the general case, however the [itex]6n-1[/itex] dimensional surface changes with time? In other words, adding an extra [itex]t[/itex] coordinate does not change the dimensionality of phase space, rather it parametrizes the energy surface, allowing it to vary in time. Only in the time-independent case [itex]H = H(\vec{q}(t),\vec{p}(t)) = E[/itex] do we have a constant-energy surface, but even in the time-dependent case we can draw a [itex]6n-1[/itex] dimensional surface on which the curve lives, so long as the surface changes with time.
So far, we just have a curve in [itex]\Gamma[/itex] space, no statistical mechanics or kinetics.
Starting off with (classical) phase space (for simplicity), for a system of [itex]n[/itex] particles with [itex]3n[/itex] coordinate degrees of freedom and [itex]3n[/itex] momenta we have a [itex]6n[/itex] dimensional phase space [itex]\Gamma[/itex] containing points [itex](\vec{q},\vec{p})[/itex] (known as micro-states). The motion of the system over a time interval is represented as a parametrized curve [itex]\gamma[/itex], [itex]\gamma (t) = (\vec{q}(t),\vec{p}(t)) = (\vec{q},\vec{p}) [/itex]. The form of the curve [itex]\gamma[/itex] for a physical system is determined by the Hamiltonian, but what is actually going on here? Do I say that [itex]H = H(\vec{q}(t),\vec{p}(t)) = E(t)[/itex] determines the motion, or do I say that [itex]H = H(\vec{q}(t),\vec{p}(t),t) = E(t)[/itex] determines the motion? The two situations are different.
Is it that [itex]H = H(\vec{q}(t),\vec{p}(t)) = E(t)[/itex] determines the motion of the system (the curve [itex]\gamma[/itex]), which in general has no dimensional reduction or restrictions (i.e. we have [itex]6n[/itex] degrees of freedom), but that, in the time-independent case [itex]H = H(\vec{q}(t),\vec{p}(t)) = E[/itex] we have only [itex]6n-1[/itex] free coordinates (should I maybe write [itex]H = H(\vec{q},\vec{p}) = E[/itex]?), resulting in our curve living on the energy shell in the picture:
Or, is it that [itex]H = H(\vec{q}(t),\vec{p}(t),t) = E(t)[/itex] determines the motion in general, and we can draw the above picture in the general case, however the [itex]6n-1[/itex] dimensional surface changes with time? In other words, adding an extra [itex]t[/itex] coordinate does not change the dimensionality of phase space, rather it parametrizes the energy surface, allowing it to vary in time. Only in the time-independent case [itex]H = H(\vec{q}(t),\vec{p}(t)) = E[/itex] do we have a constant-energy surface, but even in the time-dependent case we can draw a [itex]6n-1[/itex] dimensional surface on which the curve lives, so long as the surface changes with time.
So far, we just have a curve in [itex]\Gamma[/itex] space, no statistical mechanics or kinetics.
0 commentaires:
Enregistrer un commentaire