The h Bar

12:10 AM

@imbAF I seem to have been begging the question. I think of a "macrostate" as a region of phase space whose shape and probability density don't evolve over time. Your linked video shows that, if you were to identify the microstate of such a system as a small pointlike region of the phase space, later on that probability would have smeared out over the region corresponding to the macrostate.

Liouville's theorem basically says that the total probability is conserved during this process.

So I guess I'm still a little confused about your initial question:

3 hours ago, by imbAF

How do two different micro states of the same macrostate, differ from each other?

The answer is that they're in different parts of the phase space because they have different positions and momenta.

Leaky Nun

2:18 AM

idk if this is a stupid question, but let's say an electron goes from state A to state B and releases a photon of energy E; and then another electron absorbs the photon and goes from state B to state A

my question is, during the path of the photon, since the universe is expanding, the energy of the photon should decrease, so it wouldn't be possible anymore for the same photon to cause the second electron to go from state B to state A?

so lasers shouldn't be possible?

antimony

2:31 AM

what is the energy loss experienced by a photon from universe expansion per unit time & distance?

@LeakyNun

Leaky Nun

however small it is, it is still relevant, because you need the exact amount of energy right

i don't know if i'm misunderstanding photon absorptions

but i'm quite sure you need at least the energy, if not exactly, right

antimony

i don't know enough to know either, i was just curious if you had a number

i don't know how its quantized if you need a certain amount or not

hopefully someone smart here can tell us

lasers have finite linewidth, which suggests they can lase over a bandwidth?

again i don't know enough about that either

Leaky Nun

> [...] revealed a combined transverse Doppler and gravitational redshift up to 200 km/s/c, in agreement with general relativity predictions.

from wiki

antimony

wow that article is incredible

but i'm not equipped to understand it

Debanjan Biswas

2:52 AM

@JohnRennie Thank you so much sir for your cooperation. Now I know about what is included in a PhD thesis in Physics since many years ago. Please regret me if I wasted your time.

antimony

definitely did not waste time because i enjoyed reading it alot ^

John Rennie

5:30 AM

@LeakyNun Spacetime isn't expanding inside a physics lab!

The expansion isn't uniform at small scales. At large scales, i.e. larger than galaxy clusters, spacetime is expanding uniformly, but at smaller scales it isn't and specifically on the scale of the Solar System it isn't.

So a photon travelling along a laser tub is not red shifted by the expansion of spacetime.

Leaky Nun

5:42 AM

@JohnRennie but isn't the "large scale" made up of trillions of "small scale"?

John Rennie

@LeakyNun No. The spacetime curvature and the distribution of matter are linked together. On a large scale the matter in the universe is on average moving apart, and spacetime is expanding.

In the Solar System matter is not moving apart because it is gravitationally bound, and likewise locally spacetime is not expanding.

Have a read through:

186

Q: Why does space expansion not expand matter?

I have looked at other questions on this site (e.g. "why does space expansion affect matter") but can't find the answer I am looking for. So here is my question: One often hears talk of space expanding when we talk about the speed of galaxies relative to ours. Why, if space is expanding, does ma...

Leaky Nun

@JohnRennie is the expansion rate completely 0, or just very very very small?

John Rennie

Completely zero.

Leaky Nun

so how large does something need to be before spacetime starts expanding around it?

John Rennie

The spacetime geometry in the Solar system is approximately a geometry called the Schwarzschild metric, and this is completely different to expanding spacetime, which has a geometry called the FLRW metric.

@LeakyNun Somewhere in the millions to billions to light years.

Leaky Nun

5:51 AM

but that's just how you model the two different scales right

like on earth you can use newtonian physics, but that doesn't mean the relativistic effects are completely 0 right

@JohnRennie so if an object is 1,000,000 light years wide then it causes spacetime expansion, but not if it is only 999,999 light years wide?

John Rennie

That isn't how it works. An object doesn't cause the spacetime to expand around it. Many objects all moving away from each other cause the spacetime to expand around the whole group of objects.

The expansion is a collective property of all the objects in the universe.

Leaky Nun

ok so if two electrons move away from each other, the spacetime between them do not expand; but if two galaxies move away from each other, then the spacetime between them expands

John Rennie

Suppose your two galaxies moving away from each other are surrounded by many other galaxies that are not moving away from each other. This could be the case in a galaxy cluster where many galaxies are bound together by their gravitational fields.

The spacetime geometry is a collective property of all the galaxies, not just the two moving away from each other.

Leaky Nun

actually even in the first case wouldn't the doppler effect cause the energy of the photon to be different anyway?

John Rennie

So the spacetime in between the two galaxies isn't expanding.

@LeakyNun well yes, electrons could be moving away from each other, towards each other or parallel and it would Doppler shift light from one electron to the other. We wouldn't regard this as spacetime changing.

Leaky Nun

5:59 AM

would this make the laser not work then?

John Rennie

The transition energy isn't an exact value. There are actually a range of values. There are various reasons for this, but it means that even the light from a perfect laser consists of a small range of wavelengths not a single wavelength. We call this the line width.

If the electrons in the laser are also moving randomly the associated Doppler shift causes an additional broadening of the wavelength range called Doppler broadening:

Doppler broadening

In atomic physics, Doppler broadening is broadening of spectral lines due to the Doppler effect caused by a distribution of velocities of atoms or molecules. Different velocities of the emitting (or absorbing) particles result in different Doppler shifts, the cumulative effect of which is the emission (absorption) line broadening. This resulting line profile is known as a Doppler profile. A particular case is the thermal Doppler broadening due to the thermal motion of the particles. Then, the broadening depends only on the frequency of the spectral line, the mass of the emitting particles, and...

Leaky Nun

@JohnRennie are the energy levels exact?

@JohnRennie oh so it actually happens :o that's interesting

John Rennie

@LeakyNun No, the energy levels are slightly broadened due to the Heisenberg uncertainty principle. The excited state has a short lifetime and this causes an uncertainty in its energy. This is called lifetime broadening.

Leaky Nun

oh wow

ok this explains the emission part; i'm not quite sure if i understand the absorption part; let me formulate my question as follows

let's say i have an electron at (x,y)=(0,0), and then I aim a photon at (1,1) from (2,0); would the electron be able to absorb the photon?

John Rennie

An isolated electron cannot absorb a photon. Do you mean a hydrogen atom at (0, 0) absorbing (for example) a 10.2 eV photon to transition to the 2p state?

Leaky Nun

6:08 AM

sure

John Rennie

I'm not sure what you mean by:

> I aim a photon at (1,1) from (2,0)

Leaky Nun

my photon emitter is located at (2,0) and it is directed towards (1,1)

i.e. i'm missing the atom by 45 degrees

John Rennie

If the photon doesn't hit the atom the atom can't absorb it.

Leaky Nun

is this a 0% chance or "very very low" chance?

John Rennie

Well, quantum particles like photons are not like little balls that either collide or miss each other. They are more like fuzzy clouds spread out over some region of space and with fuzzy edges.

And in general there is a probability for the particles to interact the depends on how much the clouds overlap.

Leaky Nun

6:13 AM

i thought the clouds are actually infinitely big, just that the amplitude gets really really really small after a while

John Rennie

There is no preset cloud shape. It depends on the history of the particle i.e. how it has interacted with things in the past. Though as a first approximation your statement is correct.

Leaky Nun

at least for 1s the amplitude never becomes 0

John Rennie

But in most cases the cloud density falls fast enough fare from the cloud that any interaction is so unlikely that it makes sense to say it's zero.

Leaky Nun

is there a classical analogue for "absorption"? like if i imagine a photon as a water wave... not sure what an electron should be

John Rennie

@LeakyNun Well yes, but if the interaction probability is so low that we'd need to do the experiment for many times the age of the universe to see an interaction then it seems silly to claim the probability is non-zero.

Yes, technically it's non-zero, but in practice it's zero!

Leaky Nun

6:18 AM

also I think I'm still confused about the implication of the uncertainty principle; it only means that we don't know exactly how much energy the atom needs to be excited from 1s to 2p, but it is still a definite amount, right; like even if the amount is from 10.1 to 10.3 eV, and the photon's energy is also in this range, that doesn't guarantee an absorption right

@JohnRennie something something quantum tunnelling maybe

i don't understand QM

John Rennie

The states 1s, 2s, etc are solutions to the time independent Schrodinger equation. They are the states we get for an atom that has existed unchanged forever and will exist unchanged for an infinite time into the future.

Leaky Nun

ok

John Rennie

When we say a real hydrogen atom is in the 2p state what we actually mean is that it's in a time dependent state that is almost the same as the time independent 2p state but not quite.

So the excited states we see in the lab are actually not quite the same as the states computed from the Schrodinger equation.

And that's why their energies are not precisely defined.

Leaky Nun

but it's defined for a specific hydrogen atom in a lab right

it just isn't exactly the same as the theoretical value

John Rennie

No, because quantum objects can exist in superposition of many different states.

A real hydrogen atom in the (alleged) 2p state is actually in a mixture of all states so it is a mixture of all the energies.

So the excited state exists as a range of energies not a single precisely defined energy.

Leaky Nun

6:25 AM

ok so for example let's suppose the energy required is uniformly distributed between 10.1 and 10.3; and the photon's energy follows the same distribution; does it follow that the atom can absorb the photon?

or does this depend on the joint distribution as well

John Rennie

Yes, as long as the photon energy range overlaps the atom energy range there is a non-zero probability that the photon will be absorbed.

Leaky Nun

I think this is what I don't understand, because the probability that the two energies are the same is 0 right; because the subset {x=y} has measure 0 in R^2

John Rennie

You're still thinking that the photon and atom have some precisely defined energy, it's just that the energy can lie in some range.

We already agreed that particles are like clouds. Yes?

Leaky Nun

aha, what you mean is that the resulting "thing" is in a superposition between absorbed and not absorbed?

Silly Goose

squeesh

John Rennie

6:29 AM

No, I'm going to use the fact that particles are like clouds to try and explain what it means to be in a superposition.

Leaky Nun

ok, please go on

John Rennie

Well, consider this question: what is the position of a cloud?

It has an average position, e.g. a centre of mass, but that's just an average position. Yes?

Leaky Nun

it is not defined until you measure it

honestly the concept of measurement causes me headache

John Rennie

The cloud exists over a range of positions. Yes?

Leaky Nun

yes

John Rennie

6:31 AM

That is, the cloud simultaneously has many positions.

Leaky Nun

sure

John Rennie

Well the energy of a state works in the same way. A state has a range of energies in the same way (mathematically!) that a cloud has a range of positions.

A state may have a well defined average energy, but this is like saying the cloud has a well defined average position.

It's just an average.

Leaky Nun

should i be thinking about the mathematical wavefunction for this case?

John Rennie

Let's avoid any maths for now.

Leaky Nun

like Ψ(e1,e2) where e1 is the energy required for transition and e2 is the photon's energy

@JohnRennie ok

John Rennie

6:33 AM

My point is that when a photon and H atom interact both objects exist as a range of energies. They don't have a precise energy any more than a cloud has a precise position.

What matters is how much the two ranges of energies overlap.

Leaky Nun

is the formula very complicated?

John Rennie

The maths can get involved, but I think the hard bit is grasping the basic ideas because they are so different from the classical mechanics that we are used to.

In Newtonian mechanics an object has some kinetic energy. We may not be able to measure it exactly, but we know it has an exact value.

But in QM this is not longer true.

Leaky Nun

@JohnRennie even if they overlap completely, the problem is that this still doesn't guarantee an absorption, and I don't know how to interpret this using clouds

John Rennie

@LeakyNun Like all analogies the cloud analogy works up to a point but can be misleading if we push it too far. There isn't a way to use the analogy of two clouds to understand why the interaction only has a probability rather than being guaranteed.

Leaky Nun

what do you think about measurement?

Silly Goose

6:41 AM

you'd need qft to model photon - electron interaction right?

John Rennie

@LeakyNun In my experience most physicists just ignore questions about what the measurement process actually means in QM. When we're doing calculations we rarely need to consider it.

Many physicists (and philosophers) have wasted many lifetimes arguing about what exactly the measurement process means.

I suggest you ignore it too :-)

Leaky Nun

@JohnRennie so if I shoot a photon at an electron does it become absorbed and not absorbed "at the same time"?

or does that already constitute a measurement

John Rennie

@SillyGoose It depends. You need QFT to explain the creation and annihilation of particles, but for the interaction probability you can use Fermi's golden rule and that's regular QM not QFT.

@LeakyNun What happens is the interaction between the photon and the atom entangles them and you get a new single state that is a mixture of a photon ad the atom.

You can see this as a superposition of the original photon and ground state atom with no photon and the excited state atom.

So this entangled state is indeed a superposition of absorbed and not absorbed.

Silly Goose

@LeakyNun if you ever wish to take a peak I've heard "Decoherence" by Schlosshauer, any papers on "einselection", "quantum darwinism", or related terms by Zurek are some places to start. the textbook on decoherence is more of a textbook, but the program of Zurek is an attempt to explain how the classical emerges from the quantum and in the process to eventually answer the measurement problem

Leaky Nun

is this how beam splitters work? :D

John Rennie

6:48 AM

But note that the maths involved in decoherence is pretty scary!

Silly Goose

yeah def not a place to start with QM in general :P it seems

John Rennie

@LeakyNun Yes, with a beam splitter the photon becomes a superposition of the reflected and transmitted states.

i.e. it's both at the same time.

Leaky Nun

I see

Silly Goose

@JohnRennie is the interaction between the photon and electron embedded in an additional term of the usual, say, hydrogen hamiltonian?

or i guess how do you introduce something encoding the photon electron interaction in nonrela qm

John Rennie

Have a read through:

Fermi's golden rule

In quantum physics, Fermi's golden rule is a formula that describes the transition rate (the probability of a transition per unit time) from one energy eigenstate of a quantum system to a group of energy eigenstates in a continuum, as a result of a weak perturbation. This transition rate is effectively independent of time (so long as the strength of the perturbation is independent of time) and is proportional to the strength of the coupling between the initial and final states of the system (described by the square of the matrix element of the perturbation) as well as the density of states. It...

Basically we have an interaction operator and we calculate the matrix element of the initial and final states with this operator.

Silly Goose

6:51 AM

oh interesting

John Rennie

Conceptually it's pretty straightforward though as usual the calculation can be hard.

imbAF

10:03 AM

Is there a difference between the phase space density and the density of states?

ACuriousMind

4:11 PM

@SillyGoose The number of people who would agree that decoherence is an important tool in modelling quantum measurements is far greater than the number of people who would agree it "answer[s] the measurement problem"

several different interpretations might use decoherence in their explanation of the measurement problem (e.g. MWI-like interpretations will use decoherence as an argument why the branched universes can't interact with each other), but I think anyone who suggests decoherence solves the measurement problem has either misunderstood what decoherence does or what the measurement problem is about :P

ACuriousMind

4:33 PM

@JohnRennie It's not the Heisenberg uncertainty principle, this is either the lifetime formula for resonances or a heuristic from the golden rule (see e.g. physics.stackexchange.com/a/222385/50583 for a simple heuristic derivation). People often call this the HUP so they don't have to derive it, but the HUP does not apply to time and energy as time is not an operator.

I understand the urge to simplify but this particular simplification has provoked so many confused questions on physics.SE that I don't think it's good to propagate it

Silly Goose

5:11 PM

it seems like zurek in particular thinks strongly of QD's ability to answer the measurement problem but i am definitely not certain; from one of his abstracts: "hows how effective ‘wave-packet collapse’ arises as a result of the proliferation throughout the environment of imprints of the state of the system...these advances mark considerable progress towards settling the quantum measurement problem." @ACuriousMind

it seems to me that decoherence itself just pushes the measurement problem back because from what I've heard "wave function collapse" is still an integral part of decoherence theory

ACuriousMind

of course Zurek thinks highly of it, he's the one inventing it ;)

Silly Goose

Hehe

schn

7:33 PM

A very basic question, but when we talk about quantization of energy in an atom, which energy are we referring to? Is it the total, potential or kinetic energy of electrons that is quantized?

bolbteppa

7:48 PM

Total

schn

Okay, thank you. Another basic question; what gives rise to the subshells in an atom?

Mr. Feynman

damn, L&L III (QM) is a wonderful book

I've been looking for something about WKB that I couldn't find anywhere else and L&L saved me again

bolbteppa

Compare even just the first two chapters to Sakurai etc it's a different ballgame

Subshells of an atom is not a basic question, it comes from the quantization of the hydrogen atom and properties of the resulting spectrum

user 726941

And hydrogen is as simple as it gets :p

Mr. Feynman

@bolbteppa I love and hate Sakurai's book at the same time

user 726941

8:02 PM

Sounds... bipolar

Mr. Feynman

I think it could be the best introductory book for QM, it has great content but sometimes I think there are pointless things (or, well, I would say those things differently :P)

user 726941

"the best introductory book" is a coveted title to hold

Mr. Feynman

I'm not sure whether you're being sarcastic but I don't think it's advanced so what should I say?

user 726941

I'm not being sarcastic at all, given most institutions leave introductory material to lecturers.

My initial query was concerning the "love-hate" relationship.

Mr. Feynman

I see

ACuriousMind

8:17 PM

@user726941 that's...not what bipolar means

user 726941

Love and Hate are not polar opposites.

ACuriousMind

"Bipolar" denotes a mental health condition where someone alternates between states of mania and depression, it's not having two contradictory feelings about something at once at all.

user 726941

I see

imbAF

8:34 PM

For an isolated system, the microstates, are eigenstates of the hamiltonian, is that the case?

ACuriousMind

@imbAF no

A classical microstate is just any point in phase space. A quantum microstate is just any vector/ray in Hilbert space.

imbAF

I don't understand. The case made in my notebook is that a microstate, for an isolated system, is an eigenstate of the system

in equilibrium ofcourse

ACuriousMind

do we have to have the conversation again that "eigenstate of a system" is a meaningless phrase?

imbAF

I wasn't sure xD

whether you told me to say eigenstaes of H or of the system

so I said it wrong again

ACuriousMind

and I also have the feeling we've already had multiple conversations about why real-world systems tend to be found in energy eigenstates but that doesn't mean the non-energy eigenstates aren't possible in principle

imbAF

8:43 PM

Understandably that is what I was going to ask

But I do remember, in previous discussion we had

that we agreed upon the fact that a microstate, in the MCE, is an eigenstate of the Hamiltonian of the system

Now I am confused as why it's not the case

Because, I wanted to ask smth related to that

ACuriousMind

@imbAF I'm pretty sure we did not agree upon that

but you're using words weirdly again

there is no such thing as "a microstate in the MCE"

what you are probably trying to say is that the MCE assigns a uniform non-zero probability to the microstates with a specific energy

imbAF

That happens for when the system is in equilibrium right?

that the phase space probability density is non zero for a certain region, which corresponds to a certain energy, and zero elsewhere

ACuriousMind

no, that happens when it is in equilibrium and in the MCE

imbAF

yeah

ACuriousMind

the CE is also an equilibrium distribution

imbAF

8:51 PM

I didn't mention MCE

because we were talking about it

but I probably should have

ACuriousMind

but the CE doesn't assign a uniform probability; this distinction matters: "equilibrium" isn't enough to know what the density matrix/probability density of the system is, you also need to think about which ensemble is appropriate for the situation you're trying to model

imbAF

if energy is conserved, and particle number too

than MCE is the one ,right?

ACuriousMind

yes

imbAF

Yesterday i tried to ask a question, about something that I thought, and confused me a lot.

Without extra details, just that MCE+ eq.

Let's consider two different micro states. In phase space, these are represented with two points. Also let's assume that both belong to the same macro state. So, they have roughly the same energy, or exactly the same. A difference between the two, can be the energy and the position at an arbitrary moment in time. Am I ok with what I said, or soemthing needs to change?

Silly Goose

Townsend is basically sakurai but less details (also not as much content but both books probably cover the same usual ug material) :P and townsend is an ug book so sakurai prob could be used

imbAF

9:03 PM

@ACuriousMind here, when I made the claim that micro states are eigenstates of the Hamiltonian, no one from those who replied , pointed it out as a mistake

0

Q: Microstates/ Eigenstates of Hamiltonian and multiplicity

I have 2 specific questions regarding micro-states, entropy and the relation with the hamiltonian. If we observe a closed system, then the microstates of it are eigenstates of the hamiltonian. Is there a way to prove this? We took it as a fact in our class, but it wasn't proven. When the closed...

quantum-mechanics statistical-mechanics

ACuriousMind

@imbAF I'm not sure why you say the difference might be "the energy and the position"

the difference can be either in the generalizedpositions or the generalized momenta, as long as the two points in phase space have the same energy

imbAF

so there is a difference between two microstates, of the same macrostate

ACuriousMind

uh, sure? if there wasn't a difference, in what sense could they be two different microstates?

imbAF

exactly

so now, when we consider the time evolution of a point in phase space, which is a trajectory

this means, something changes for the microstate, NOT energy, since it satisfices the Hamilton eq.

The one thing I can think of that changes, are the generalized coordinates

But if that's the case

ACuriousMind

...why can you not think of the generalized momenta changing?

imbAF

9:07 PM

or those

either way

isn't our initial microstate, changing to another microstate after some time t?

ACuriousMind

the generalized positions and momenta are literally the full list of things that define a point in phase space, so I don't know why you're saying "only" :P

@imbAF sure

imbAF

if that's the case

ACuriousMind

equilibrium means that the overall probability distribution - the macrostate - doesn't change, not that the individual states are stationary in time

imbAF

probability distribution-macrostate= phase space probability density ?

ACuriousMind

yes, distribution here is just another word for density

(strictly speaking a distribution is slightly more general than a density mathematically but let's really not get into that, it's irrelevant :P)

imbAF

9:10 PM

I mean it threw me off

probability distribution should be kind of different then the pdf, but I can;t put it into words

and the pdf in phase space, includes a bunch of points, right?

ACuriousMind

what exactly do you mean by "including" points?

imbAF

if a microstate is a point in phase space

the macrostate is represented with what there?

ACuriousMind

a probability density

imbAF

and probability density, is represented as what in phase space,in case where we deal with just mass (of a substance), mass density, is the amount of substance per unit volume

so the phase space pdf, must be interpreted in a similar way, in phase space

phase space is "filled" with points

and in the unit volume in phase space, a certain amount of points are included

bolbteppa

A microstate is a point of phase space, a macrostate is a region of phase space, similar to the probability theory distinction between 'outcomes' (points) vs 'events' (subsets) in a probability space. Knowing the microstates means you know the macrostate, but knowing the macrostate doesn't mean you know the microstates.

A macrostate might be fixed by saying the total energy of all the particles is equal to some value, so regions of the phase space (where the particles have certain energies) may be less or more likely to contribute to the total energy which is what the probability density on phase space tells you.

ACuriousMind

9:21 PM

@imbAF the probability density for the MCE with energy $E$ is $\rho(q,p) = V^{-1}\delta(H(q,p) - E)$ with $V$ the volume of all phase space point with energy $E$, or you can choose any nascent $\delta$ instead of $\delta$, see en.wikipedia.org/wiki/…

imbAF

but is it really a volume? Isn't in the case of MCE, a surface of the hypersphere, the region with the micro states, which have energy within the desired interval?

ACuriousMind

yes, if you use the $\delta$ it's a hypersphere

imbAF

Btw integrating this over the entire phase space (I don't know the boundaries of integration): $\rho(q,p) = V^{-1}\delta(H(q,p) - E)$, what do we get?

ACuriousMind

@imbAF of course 1 because that's the definition of a probability distribution

that's why people use the nascent $\delta$s so that the sphere has a small thickness so it has an actual volume and they don't need to discuss $\delta$ functions

imbAF

and small delta

implies energy in the interval E - E+dE right?

ACuriousMind

9:28 PM

yes

imbAF

but what you are saying is that

$\int\delta(H(q,p) - E)d\vec x$ gives you V?

ACuriousMind

yes, but it more or less defined to yield that

I don't really want to get into the theory of how to define multi-dimensional $\delta$s correctly

imbAF

I find it weird how to integral gives me V

ACuriousMind

forget about it

just choose some $\mathrm{e}^{-\pi x^2}$ instead of the $\delta(x)$ like the Wiki article does

imbAF

Ok

Might do it like that

This is one of your answers, when we previously discussed about quantum statistical mechanics

"
also, note that it's not "every microstate in this volume has probability 1/V", it's "the probability density over this volume is 1/v""

What did you mean with the last part?

ACuriousMind

9:39 PM

I was talking about the difference between a probability and a probability density

imbAF

probability is what you get once you integrate over a region, yes

ACuriousMind

if you have a probability density that is uniform over a region $S$ with volume $V$, then the density is $\rho(x) = \begin{cases}1/V & x\in S \\ 0 & x\notin S \end{cases}$

this does not mean that "the probability of a point in $S$ is $1/V$"

I wasn't saying anything more than that

imbAF

I see

In our class, we defined $\rho$ (for a system with constant energy(this was how it was specified)) $\rho(x) = \begin{cases}1/{\Sigma(E)dE} & x\in S \\ 0 & x\notin S \end{cases}$ where $\Sigma(E)$ is the nr. of microstates in the surface of the hypersphere with radius proportinal to E

ACuriousMind

yes, that's another way to express what I meant by the $\delta$ above

imbAF

And one can continuous by saying, that the probability of the system being in any of the microstates (that are included in this volume) is the same, 1/V right?

ACuriousMind

9:48 PM

6 mins ago, by ACuriousMind

this does not mean that "the probability of a point in $S$ is $1/V$"

imbAF

then what do we mean when we say, the probability of the system being in any of the micro states, is constant

for the MCE of course

ACuriousMind

The probability of a point is zero. That's how probability densities work

imbAF

yes

that's why I find it weird that we make the above mentioned claim

ACuriousMind

that means the probability density is uniform on $S$

which is the correct continuous generalization of assigning a constant non-zero probability $1/N$ to $N$ finitely many states

imbAF

@ACuriousMind I don't understand how this statement statement, doesn't contradict your last one. As said above, a microstate(which I assume is what you meant with:"$N$ finitely many states")=point in phase space. Maybe perhaps $1/V$ and $1/N$ are two different things, but ultimately, N is the nr. of states in V. Or perhaps, it doesn't have to do anything with physics, and it's pure math complexity

I am not sure

ACuriousMind

9:58 PM

@imbAF the number of states in the surface is infinitely many

that's why you can't assign a finite 1/N probability to each individual state

imbAF

because x and p are continuous ?

ACuriousMind

yes

imbAF

but you assign this 1/N to finite amoun N of them?

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