@imbAF Macrostate: you have ten balls distributed across one hundred buckets. Microstates: … the first ten buckets. … the last ten buckets. … every tenth bucket. … in MY BUCKET SUCK IT 99-PERCENTERS. … in his bucket, let’s get him.

Length of the previous comment to be determined by combinatorics.

ok then

when we see the evolution of two different micro states in phase space, the evolution is represented with a line in phase space

now what does change for the micro state, over time?

and the system is that of N particles

so if we consider the probability density $\rho$, which is the macro-state, and it includes a bunch of points (microstates), the time evolution of all these microstates, is translated in the phase space as a bunch of lines. So what changes for a microstate over time.

We know for certain, that the energy of the micro state, doesn't

so what does?

you have two microstates, belonging to the same macrostate, and both evolve with time. What do they change?

A nice example is the “Einstein solid,” in which each particle in the solid is a harmonic oscillator. This is “atoms jiggling about equilibrium,” but quantum-mechanical, so that the oscillations are quantized in units of hf.

Suppose I start with an Einstein solid at zero temperature, and then heat it up by some energy E=Nhf. That energy gets distributed among all of the oscillators.

The distribution mechanism is a different question than which states can occur.

You can handwave “this atom jiggles its neighbors,” so maybe the excitation energy is mechanically unlikely to stay put.

Maybe you have a anisotropic crystal, so the heat spreads out more rapidly along some directions than along others.

The “fundamental assumption of thermodynamics” is that, in equilibrium, all of the indistinguishable microstates are equally likely.

true

but, if you would consider two microstates

there must be some quality that is different

because then, you;d have the same state two times

the way I understand this

for a system of N particles, two different microstates, can differ by the position of the particles, or a bunch of particles from the entire group. While they differ in this

they do have same energy

is that a wrong way of thinking about microstates?

Sure. Consider that the balls-in-buckets model is isomorphic to the Einstein solid. If all of the balls are in the first ten buckets, then that side of the solid contains more heat than the other side. Probabilistically, you expect the random fluctuations to end up in one of the innumerable more states where the heat is evenly distributed.

Oh, you asked a question after I wrote “sure”

Ok so we just established that

Now, in phase space a microstate is represented with a point, is this a correct statement ?

Maybe? “Phase space” means different things under different circumstances.

In statistical mechanics at least

I am speaking in these terms

at the same time $\rho$ which, if I am correct, is the probability density in phase space right?

What are the dimensions in your phase space?

for a system of N 3D particles

it would be a 6N dimensional space

of the 3N positions and 3N momentums

so position and momentum

How is an oscillator state a “point” in that phase space? You have to transform to some (amplitude, phase offset) coordinates for an oscillation to be a “point”?

Well at least for a gas, a point, which is a microstate, it's a point with 6N dimensions, all the position vectors of all the particles and all the momentums

so bunch of these points, give us, the probability density

Let’s pick one model. Gases need a different toolset from solids. Should we switch?

yes

In my course, it wasn't specified the system, simply that we have a system comprised of N particles in 3D

and that in phase space, a point, is a microstate

Okay. Poof! In a gas, every particle has a position and a momentum, which we can label as a point in phase space. But the positions evolve continuously, while the momenta only change when two gas atoms collide.

ideal

gas

for simplicity

That’s what I meant, too.

So, we agree that a microstate is represented with a point, and the phase space probability $\rho$, (that includes a bunch of these points), is a macrostate

Maybe I also need to mention

that these view

or interpretation

is done for the micro canonical ensemble

I don’t understand “the phase space probability is a macro state.” Let me think out loud a moment.

I think those are different things.

Suppose the macrostate is "this gas has total energy E"

I heard in a youtube video, by a guy who derived the liouville theorem, and a member here

both said the same thnig

phase space probability density, can be considered as a macrostate

hold on

I would say this macrostate corresponds to a surface in momentum space, and the entire volume of position space.

The momentum surface is p_1^2 + p_2^2 + p_3^2 + ... = E

if you make "nice" assumptions about units so I don't have to write p^2/2m

The thing that I am trying to get on is the following

A microstate in phase space is a point.

A point, follows a trajectory, when we consider the time evolution of the state of the system

And most importantly

two trajectories, of two points (two microstates) never cross each other

My question is, what changes with time for a microstate

the position and momentum of the particles?

Like I said a few minutes ago: for an ideal gas, the particles' positions are changing continuously. The momenta in an ideal gas only change when two particles' positions get too close to each other — a collision.

@imbAF You mentioned Liouville's theorem. That Wikipedia page has some animations of how regions of phase space evolve in some single-particle systems.

But in those animations, the initial regions are all square blobs which don't correspond to thermodynamical macrostates.

why not?

A thermodynamical macrostate has a well-defined thermodynamical observable, like the total energy. But if you look at the bottom of the first animation, you see that the potential energy surface is for a single particle, and the different parts of the phase space map to different total energies.

potential energy surface? what do you mean with this?

For a one-dimensional simple harmonic oscillator, the macrostate "has energy E" corresponds to a circle in (position, momentum) space. Macroscopically, all of the oscillators with energy E are indistinguishable, whether they happen to instantaneously be at their left turning point, their right turning point, or somewhere in the middle..

@imbAF Perhaps you should send a link to the video you watched — we seem to be missing some context between us.

@imbAF I'll leave it on while I'm making dinner, and come back later.

ok