In the phase space of a simple harmonic oscillator, the trajectory is a ellipse. If we assume that the system is initially in a certain microstate, which in the phase space has a point assigned to it , what does it mean when the point moves in this trajectory?

If I have to give my personal opinion, since this trajectory is made out of many points, and each point represents a microsystem, basically we are jumping from a micro state to another, but that is not possible, since (for an isolated system)

a trajectory represents a constant energy

and different micro states = different energies

am I missing something here?

It seems questionable to identify Microsystems with phase space points in this way. After all, different points on a given trajectory just represent the same motion at different times

What do you mean in this way?

Aren't microstates represented with points in the phase space?

@imbAF I have no idea what you mean by "different micro states = different energies"

@ACuriousMind thx for coming back first of all

Let me try and clarify 1 thing

A microstate is just a point in phase space, a macrostate is a probability distribution over microstates.

yes

@ACuriousMind I remember and I understood all what you taught me

but there are some ambiguities that I have

Give me a moment

Let's say a system is in a particular microstate, which ofc we cannot measure, but the system is in that. In phase space this micro system has a point assigned to it. And a bunch of points represent the density function which is a macrostate

am I corrent until now?

correct*

you seem to be using "system" instead of "state" at random points, but otherwise, yes

no

the system

is in a state,

and the state is represented with a point in the phase space

"In phase space this micro system has a point assigned to it." if you don't mean microstate in this sentence I have no idea what you're trying to say

microstate*

sorry

typo

Ok

I will try to explain this as good as I can

becuase all this is related to a question I have regarding Liouville

Let's say that a simple system in an initial moment t_0 is at a particular micro state, which is an arrangement of positions and velocities, so it has an energy value assigned to it

Then we observe the point moving in a trajectory

which translates in phase space, different coordinates of position and momentum

shouldn't that imply that our initial microstate of time t_0 , over time changes it's energy to another microstate, which has another energy value?

If I can give my answer to this

why would it have another energy value?

because the values of position and momentum change as we move in the trajectory?

But If I can give my answer

sure, but all points along a trajectory have the same energy

yes but why?

that's what we mean by "energy conservation" - the total energy does not change

two points

I understand that

I don't understand the WHY. If you move along a trajectory, you have a change of coordinates for x and p correct?

well, trajectories are solutions to Hamilton's equations of motion, and these equations are precisely the equations for curves along which the Hamiltonian is constant (they are integral curves of the vector field associated with the Hamiltonian)

I understand that, mathematically

But I am trying to have an intuitive understanding

which for me is as follows, and I need your confirmation

Actually I have no idea what Landau is on about

Assuming that the system has energy conserved

a microstate is an eigenstate

eigenstate : \psi_n

time evolution of eigenstate \psi_n(t)= exp[iE_n t /\hbar]psi_n

this is classical mechanics, there are no eigenstates

Oh yeah

ok backtracking

In a trajectory (leaving Hamiltonian explanation aside), the reason why I ALSO think that the energy is conserved, is because you can have the same microstate, but in an initial time t_0 some particles have a certain energy value greater then some others. On a later time, this changes, but the total energy that characterises the micro state, doesn't change, hence even though we move along the line, we still have the same energy

Does the micro vs macrostate distinction even quite make sense in QM? You do have separable states but those are the exception

picture the ideal gas

the energy for a microstate is dependent from the momentums, of each particle. so if you have different values of momentum, then you have different energy

that's why I am questioning the conservation of energy

@Semiclassical it's easier there - a macrostate is just a density matrix :P

Hmm. Yes, fair enough

when you move along a trajectory, as you change the coordinates of momentum of each from the N particles

I hope I somehow made sense

I can't explain it better then this

you seem to think that because momenta and positions change, the overall energy should change, too

YES

isn't that the case for an ideal gas?

And I guess the closest analog for microstate would be a reduced density matrix

@Semiclassical it's just a pure state

Collisions typically just move energy around

@imbAF but that's the whole point - because the trajectories are not random lines but solutions to the equations of motion, their positions and momenta change exactly such that the overall value of energy doesn't change

oooo thank you man

finally

that has nothing to do with statistical mechanics, that's just a fact of classical mechanics (without friction) in general

a proper answer

I need to study more the hamiltonian

But we did it one semester only, some years ago and I don't remember much of it

Maybe this is the reason why I was searching for an intuitive understanding, rather then getting it straight from the Hamiltonian

Cuz I forgot most stuff about it

Ok then two the Liouville

I was rereading it

And one other def. of it was that : " It asserts that the phase-space distribution function is constant along the trajectories of the system—that is that the density of system points in the vicinity of a given system point traveling through phase-space is constant with time."

Can we unpack this a bit?

Saw something today which I probably knew how to show at one point

phase space distribution = the pdf ?

which as you said multiple times is the macrostate

it says that for a phase space density $\rho(q,p,t)$ that's a solution of the equation of motion for $\rho$ and a trajectory $q(t),p(t)$ that's a solution of Hamilton's equations, $\frac{\mathrm{d}}{\mathrm{d}t}\rho(q(t),p(t),t) = 0$

yes

I remember and I have your notes

mathematical interpretation that is

Suppose you have a cart moving along a horizontal aluminum track. If we ignore the wheels ie rolling without slipping then there’s ofc no horizontal acceleration and everything is boring

Now suppose we attach a small magnet to the bottom

what this means is that if you take a bunch of points in phase space at time $t_0$ and note down how probable it is the system is in one of them, then jump to $t_1$ and look at the points the original ones have evolved into (by moving along their trajectories), the probabilities for these are still the same as for the ones you started with

The claim is that there will now be a velocity-dependent damping force, due to magnetic induction / eddy currents in the aluminum

That sounds plausible, but I don’t remember how to fill in the details

Give me a moment

@ACuriousMind i think I am finally understanding, just processing it currently

Because I was always, in my mind, picturing 1 point 1 trajectory, and speaking about distribution function for 1 trajectory doesnt make sense, when we already said that a bunch, a cluster of points is the distribution function

one sec

and

I assume the volume doesn't change

because the probability should be the same

before and after

that's one implication - because the density is constant along all the trajectories, the volume on which it is non-zero cannot change, only deform

yes

But this time evolution of the points, does physically represent a system in non equilibrium going in equilibrium or just fluctuations of the system in equilibrium?

doesn't matter

it doesn't ?

Liouville's theorem is a general theorem about phase space dynamics

it doesn't care whether you're in equilibrium or not

it doesn't even care if you're doing thermodynamics or have some other reason to consider phase space densities

and it also doesn't care whether energy or other quantities are converved ?

it's just math

conserved

@imbAF oh, it does!

so if it does

if energy is not conserved, then it doesn't work

does it mean that the phase space for micro cannonical ensemble is different from cannonical or grand cannonical ?

but if energy is not conserved then you basically don't have a Hamiltonian system

so liouville is relevant only for micro canonical ?

no

the energy conservation is about energy conservation along the individual trajectories, not about whether or not the macrostate has a definite constant energy

ahaa

then what about canonical and grand canonical ensembles

do we have energy conservation along the trajectories?

I mean, do they have the same phase space as the micro canonical ensemble ?

those are two very different questions

you can briefly touch upon both of them

if it ain't to taxing for you

you have energy conservation along the trajectories simply because you're doing Hamiltonian mechanics - the Hamiltonian is the energy, after all, and the way Hamiltonian mechanics is constructed, it is always conserved along trajectories

as for the phase space, no: The grand canonical ensemble has varying particle numbers, so while the other two ensembles can have $\mathbb{R}^{6N}$ for some fixed $N$ as their phase space, the grand canonical ensemble need to live on a space where different microstates can have a different number of particles

but don't think too hard about what these spaces are, they're just some gigantic $\mathbb{R}^{\text{very large number}}$ and nothing else matters for thermodynamics

I see

I have a better understanding

One thing more, in the simple case of micro canonical ensemble

for different variables in \rho (\q^{3n}, p^{3n},t)

we are observing different macro states

I hope that is correct

I'm not sure what you mean

the distribution fucntion

what does "for different variables in ..." mean?

or probability density function

i mean

If you move around phase space

observing the volume around random points (microstates)

the density function has different values

which correspond, physically, to different macro states that the system can be in

right?

no, the density function is the macrostate

the whole function

yes

but a system can have many macro states

and in equilibrium

it ends up in one of them

sure, as many as there are (normalized) functions on phase space

what do you mean?

any function whose integral over phase space is 1 is a possible macrostate (a priori, without any assumptions about equilibrium or whatever)

yes

any function = different values of \rho

that's how I understand it at least

you probably mean the right thing but that's not how one would say it :P

and if you integrate \rho, you must always get 1, when integrating over phase space

why, did I say it in a bad way?