a phase space density usually isn't a density of "number of points"
it's a probability density
I mean, you can think about this in terms of ensembles and numbers, but whether or not that's "correct" depends on what physical situation you're modelling
the density represents the state of your system - if it changes, the state of your system changes
what that physically means depends on what your system is
@imbAF "nearly the same energy" would only be true if distances in phase space had some sort of universal correlation with energy, but how different the energy of two points that are "close" is depends on your Hamiltonian
So then you can point out whether what I think, is correct or wrong
phase space / liouville etc etc
can we do it like this? Because I have spend the last 4 days trying to understand all this
For a system of N-3D particles, we have 6N D.O.F and therefore a 6N dimensional phase space. I know that one point in phase space represents a possible state of the system. I also understand that a tiny volume element $c\cdot dq^{3n}dp^{3n}$ contains a certain amount of points, which physically represent all the system's with energy between $E$ and $E + \Delta E$. Am I corrent until now?
Ah, see, that we're doing statistical mechanics with "big $N$" is a very relevant information!
you can also talk about phase space densities in the context of a single particle and you're just uncertain where it is, that's what I was going on about above
1. Isn't then phase space infinite ? 2. Regardless whether the system is in equilibrium or not, it's phase space is one, and should contain both cases right?
the phase space "doesn't care", in other words, if the system is in eq, we have some points or a single points (letting fluctuations aside) that doesn't move along a trajectory over time, and if it is not in eq. then this point (the system) moves along the tajectory (the different states ) until the final one (the micro state of equilibrium, when we let aside fluctuations) or the microstates, which belong to E + dE when we consider the fluctuations. Both cases are plotted in phase space
Equilibrium means that your phase space density is such that when you evolve all the points that belong to it along their trajectory at once, the overall density doesn't change
As a discrete analogy: If 1 evolves into 2 and 2 evolves into 3 and 3 evolves into 1, then a density that is one-third 1, one-third 2 and one-third 3 is constant
the idea of statistical mechanics is that you don't know in which state the system is, exactly, and so you have to track a bunch of possible states at once
so what you evolve in statstical mechanics is not a single point, but a "cloud of points" - the phase space density
the path a state traces out in phase space as it evolves in time, is this the abstract meaning or the physical one
Because you told me, that my interpretation of a point in the trajectory, as the different states a system can be in, for a period of time, is incorrect
i.e. what matters there isn't any single trajectory, it's whether or not the cloud of points, as each point follows its own trajectory, looks different from one instant to the next or not and for that it doesn't matter whether the individual points are moving around, they have no "identity", all that matters is whether the overall distribution is the same or not
"the density, the "cloud of points", that's the macrostate" . I thought until now, that when you integrate the density over the volume E and E + dE, you get the probability that the system is in this region. So I thought the region, all the points within the region, is the macro state
But if what you say it's true
and if density is the macrostate, then with this new interpretation, what do we get when we integrate over a region ?
the macrostate (i.e. phase space density) is nothing but a list of probabilities for microstates
again, the idea of statistical mechanics is that we don't know exactly which microstate the system is in. The macrostate/density quantifies that uncertainty
so the physical meaning of integrating the density over some set of states is just hte probability that the system is "really" in one of these states we integrated over
and we can integrate this over some region $R$ of phase space, that's a set of microstates
and this $\int_R \rho$ is then the probability that the system is "really" in one of these microstates
perhaps what confuses you is the probabilistic nature here: The macrostate is not something the system is "actually in", it's just an expression of our uncertainty - it exists only "in our mind"
if we were able to track the position and momentum of every particle in the world precisely and continuously, we would need no statistical mechanics and no "macrostates"
how integration over a region in phase space, using rho, which is macrostate, gives me the probability of knowing that the system is in a certain microstate
I think, this has soemthing to do with the multiplicity of the macrostate
cuz I think I got all the integrating and probability and microstate
but I need your confirmation
If we consider the density constant ( I don't know what that would mean physically, since density now is macrostate), but let;s go on and use it in the integral over some region, then since it's constant it goes out and the integral over the region, must give me idk really, macrostate over all the microstate, something like probability of each microstate for this particular macro state, I am not sure
I think density integrated over a region is the probability of the system being in this region, which physically is translated, as the probability of the system being at a certain macrostate, since I consider the volume the macrostate
and all the points inside the volume the microstates
and by knowing all the points inside this volume, which I consider the macrostate,
and also taking into consideration ALL THE POINTS (all the micro states) in the phase space (that the system can be in), their ratio
must give me the probability of the system being in this macrostate
hence the reason the macrostate with the highest mulitplicity is the one, the most probable state that we can find the system
Ok, just to backtrack a little, you said that integrating the density over a region gives me the probability of the system being in a micro state, why would we want to know that?
we usually assume that isolated systems in thermodynamic equilibrium have these equiprobable microstates/uniform densities (essentially this is another part of the definition of "equilibrium")
@imbAF I don't know, you asked me what the integral tells us ;)
equilibrium thermodynamics usually isn't very interested in that sort of thing
since the reason we introduced the macrostate in the first place is that we don't really care about the individual microstates
well, that's the thing - quantum statistical mechanics works much better because it's...just quantum mechanics, done for large systems. You don't need to invent "macrostates" or "phase space coarse graining" or any of that stuff
that $\delta$-function is just a convenient way to write "This function is zero on all microstates with energy $\neq E$ and gives 1 when integrated over all states with energy $E$"
the reason it's a $\delta$-function and not a piecewise definition with $1/V$ in one part and $0$ in the other is because the "shell" of states with energy $E$ is infinitely thin and actually has zero "volume"
so you need something weird like the $\delta$ to express that
you need the following additional assumption to arrive at the $\delta$: 1. All allowed microstates are equiprobable. 2. Only microstates with energy $E$ are allowed.
And how do the trajectories in phase space behave when the system is in equilibrium, which corresponds to the region where \rho is non-zero constant there and zero outside. I don't think they can go outside the region, cuz then that would mean that the system goes out of equilibrium
but I was making a more general statement - on trajectories (solutions to the equations of motion), energy is a constant because of energy conservation
The differential of the exponential map is known for the Riemannian case (or at least we may have some information about it).
Proposition (Gallot, Hulin, Lafontaine - Riemannian Geometry, 3.46)
Let $m$ be a point of a Riemannian manifold $(M,g)$, and $u,v$ be two tangent vectors at $m$. Let $c$ ...
I was thinking how on earth we have preservation of energy, when the system can be in a non- equilibrium state and ofc over time, by gaining or losing energy it reaches equilibrium which is to say that for the system the energy isn't conserved
By construction, the value of the Hamiltonian is a constant along solutions to Hamilton's equation of motion, and the value of the Hamiltonian is the energy (here)