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1:16 AM
This is a very basic question, but when raising and lowering indices, does the order of the indices mater?
For example, is this correct $\epsilon_{ij}\psi^j = \psi_i \neq \epsilon_{ji} \psi^j$
So can I only contract indices when the index I want to change to is first in the metric tensor, and the one I am summing over is to the right?

2:10 AM
Also does $G^{\alpha \beta \dots}{}_{\alpha \gamma \dots} = -G_{\alpha}{}^{ \beta \dots}{}^{\alpha}{}_{ \gamma \dots}$?

6 hours later…
8:00 AM
@DIRAC1930 that...depends on what your objects are
$\epsilon$ is antisymmetric, so $\epsilon_{ij} = -\epsilon_{ji}$
If $G$ is a generic tensor, then there is no general relation between one order of indices and another

9:01 AM
Morning

9:29 AM
Hm
I wonder if there's a higher category version of the worldfunction
I guess something like a map from two branes to the Polyakov action of their unique cobordism
Do branes have a Riemann normal neighbourhood
question is, is there always an on-shell cobordism between two branes?
in a small neighbourhood at least

9:58 AM
I'm not even sure that's true in flat space for "straight" strings, unless they are coplanar
though I guess strings can probably reasonably have some torque
moving along mysterious helicoidal Dbranes

10:21 AM
@ACuriousMind Do on-shell branes have a name in differential geometry

what's an on-shell brane

Submanifolds that minimize the volume functional

as opposed to just a brane

with some boundary conditions
Like say lines for $\mathbb{R}^2$

I think that's a "minimal submanifold"

10:27 AM
could be
Let's see

What does it mean if the density in phase space changes over time at a certain region?

10:48 AM
What sort of "meaning" are you looking for?

well, in general or in a simple case
the trajectories do not intersect, correct?
So now if at a certain region in space, we have a change of the nr. of points in this region
that means that more points are coming in (in the case of an increase) and less are leaving
so, we can interpret this, as more trajectories go towards the same region, then leave it
and since a trajectory represents the time evolution of the system,
in other words, the different states the system is in, during a certain period of time
then by having a change in density, in the case of an increase
that means that more states are having nearly the same energy
am I correct until now ?
what is the physical meaning of this density not being constant ?

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

nlab has nothing on the worldfunction
Most important of biscalars!

@ACuriousMind Well if you would divide with the nr. of states per unit volume, you would get the density as a probability, no?

depends on your definition of the phase space density, I guess, yes

11:01 AM
+ for a system's time evolution, it's state does change, but the pdf shouldn't necessarily change
Like, a trajectory in phase space, is the change of the state of the system over time, but the pdf can be (in phase space) constant
So it doesn't change, but the state of the system changes
so why do you say : "he density represents the state of your system - if it changes, the state of your system changes" ?
Am I missing something here?

@imbAF that (the pdf being stationary) would be the case if your statistical system is in equilibrum
that's a very special assumption, "most" systems are not in equilibrium

then we can't speak about trajectories in phase space in that case
or can we?

every point in phase space lies on a trajectory
I don't see what the existences or shape of the density matters for that

One second, let me try and properly explain
at least what I understand
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

11:08 AM
No I am about statistical currently
So let's continue
A point moving in a trajectory in phase space, is physically the system changing it's states over time, time evolution of the system, correct ?

yes

Before I continue one more clarification
In phase space we have all the possible states a system can be in
right?
and to be more precise all the possible microstates a system can be in*

Yes

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?

1. Of course, it's $\mathbb{R}^{6N}$
2. I don't understand the "it's phase space is one" clause

11:13 AM
in the sense that
how do I explain this
whether the system is in eq. or not
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
I hope I somehow tried to explain what I need

that's not how equilibrium works

The equilibrium just means that the macrostates are constant

All individual states - all "single points" - always move along their trajectories

unless they have zero momentum I suppose :p

there might be very rare states that don't move at all because their solutions to the e.o.m. are constant, but that essentially doesn't happen

11:18 AM
does moving along a trajectory means physically the system is changing states over time?

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

Depends what you mean by "state", but that will not change the macrostate, no
if at equilibrium

yes macrostate doens't change, but it contains a mulitplicity

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

so a bunch of microstates belonging to it, i get that

11:20 AM
At equilibrium you still have particles bumping around and changing momentum

yes
Then, what is the physical meaning of the point (which I assume represents the system) moving along a trajectory ?
Cuz we say the point moves along the trajectory, and we don;t give a physical meaning to it
And until now, I was under the impression that the point = system, trajectory=time evolution of it, but apparently no?

the point is a possible state of the system
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

Ok
and what is a trajectory?

It's still the same as before - the path a state traces out in phase space as it evolves in time
I'm not sure why the notion of trajectory should change just because we have to look at more than one state at once

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

11:30 AM
well, I mean, that's still true - but in statistical mechanics, it's just one possible state the system can be in
the "cloud of points" of course evolves into a "cloud of trajectories"
you have to be careful with the notion of "state" here: When I say state, I mean microstate, i.e. a point in phase space

yes

the density, the "cloud of points", that's the macrostate

ofc

but when we stay stuff like "An equilibrium state is a state that doesn't change in time", that's a macrostate we mean there

one second cuz we are getting into something finally

11:33 AM
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 density , when we integrate the density

you still get the probability that the system is in that region
And: "the region" can be a shorthand for the state/density in the case where you assume the density is uniform
i.e. you might define a density by specifying a region and just taking the density that's uniform in that region and zero outside

"you still get the probability that the system is in that region" what does this mean physically ?
probability of the system being in the macrostate?

no, the "probability to be in a macrostate" is 1 :P

xD
brain fart right now

11:38 AM
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

in one of these MACROSTATES?
right?
macro states

no, what we integrate over are sets of points, i.e. microstates

but you are using density

we have a function $\rho(x,p)$, that's the macrostate

which is the macrostate

11:41 AM
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"

macorstate is the multiplicity of microstates

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"

I understand the concepts macro and micro
very good actually
what I am rn trying to understand is
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
and I just can't get things to click in my mind
wait
lets for a moment say that the system is isolated

@imbAF that's the definition of the macrostate, kind of
what do you think the "density" $\rho(x,p)$ represents if not a probability density?

one second
cuz I think I got all the integrating and probability and microstate
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
this is how I understand all of it

11:53 AM
$\rho$ must satisfy $\int \rho(p,q) dp dq = 1$ i.e. it's a normalized probability density

there is no "probability of a system being in a macrostate"
At any point in time, the system has a single macrostate, a single $\rho$
its probability to be in any other macrostate is 0, its probability to be in that one macrostate is 1

yes

I think you are confusing the basic notions of probability densities with thermodynamic arguments like "systems maximize their entropy"
because I can't explain why you think "the macrostate with the highest multiplicity is the one" otherwise
that's not some universal law of how macrostates work, that's just the fundamental assumption of thermodynamics in equilibrum

why wouldn't it be?
When you reach equilibrium
why would the system be in a macro state that compared to other macro states has smaller multiplicities?

yeah, that's an assumption - that you reach equilbrium, and that the equilibrium macrostates will be uniform

11:56 AM
will be uniform?
what you mean by that?

you're implicitly assuming here that every microstate in a given macrostate will be equally probable
otherwise the notion of "multiplicity" isn't really a good one

that would be the case for an isolated system right?

a probability density where this is the case - everything not zero is equally probable - is called uniform

which is not the case for every system, ofc
aha ok
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

12:01 PM
yes
so rho is the macrostate, yes?

yes

And in equilibrum rho is all over phase space constant, yes?

no
it's a non-zero constant over some region, and zero elsewhere
the region it's non-zero over is the region of microstates "belonging" to that equilbrium macrostate, e.g. those with a certain energy

yes
it's what I was about to write
but I wasn't sure
and what when it's not constant all over phase space
this is a non equilibrium case right?
or non constant over a region*
Idk which would be the corrent
non constant over phase space or over a region

@imbAF yes, exactly
that wouldn't be equilibrium

12:13 PM
so rho is a changing macrostate of the system
over time

so if you're just doing equilibrium thermodynamics you can essentially forget about $\rho$ entirely, all you need is the set of allowed microstates

yes, but I was dealing with the Liouvill equation

@imbAF yep, and in equilibrium it's just $\rho = 1/V$ where $V$ is the volume of the set of allowed microstates

YES
would that imply that the microstates, of this volume (in eq.) have the same probability
or it's the case only for the micro canonical ensemble
i.e isolated system
?

well...the different ensembles have different ideas about what the phase space is and what a macrostate is defined by
but I think they all agree that microstates are equiprobable in equilibrum

12:16 PM
Ok

you just won't have that the set of microstates is defined by a constant energy if you're not microcanonical

I mean even in the microcanincal
the microstates have energies between E and E+dE
then you average them out
the energies
and you get the energy of the macrostate
which in the case of an ideal gas we use U=3/2 nRT
to find it
or am I mistaken?

you're right, and yet you will come to talk about a thermodynamic (macro)state being defined by its energy
don't think too hard about it :P

xD

classical statistical mechanics is a mess, really

12:19 PM
no no no hahahaha
I hate it
I find quantum statistical so much better
@ACuriousMind
I have 2 final questions, I know I spend a lot of your time
but for me it was worth it cuz I got somewhere

# Truely have no idea what I am even thinking about that "physics question" huh?
I digress

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

corrent
but I do have an exam

my condolences

and I can't avoid this crap until I CAN AVOID and never return back to it
thx
So my 2 questions
well actually one
since rho =1/v and integrating over V gives you 1
So I derived the liouvill eq.
And now i am considering equilibrium
before i ask the question I want to say this
equilibrium=macrostate=certain multiplicity= certain nr. of microstates= nr. of points in phase space
this is the chain of thoughts for me
anyway
eq. then \partial \rho / \partial t =0
which means the poission brackes
{\rho,H}=0
then In my script i have this
/rho = \delta (H(p,q) - E)
where e is a constant value of energy
where does this come from
the equation with delta

12:27 PM
okay, let's go back to "$1/V$"
as we said, the function is 1/V on one region and 0 outside of it

yes

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 $E$"

yes

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

ok let me decipher what you just said
ok
since delta has that behavior that goes to infinity for 0 and then 0 for the rest
we use it
but my question was more about
what type of reasoning is made
when we jump from the fact that the poison brakces are zero
to that delta expression
or there is none

12:31 PM
well, there need to be some additional assumptions :P

aha ok

the zero Poisson bracket just means your $\rho$ doesn't change in time

I am digging to deep again
my stupid brain, always asking why for everything, thx to my OCD apparently xD

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.

yes

12:33 PM
there's no "reason" for this, it's just the definition of equilibrium/microcanonical ensemble

of course
and not really a big deal
one last thing tho, regarding the liouville theorem
rho is macrostate as we said and as it makes sense
and it's called
density function
am I correct?

sure

and density of states
is the same thing
as rho
or is something else?
because we called rho as density and it represents the macro state
but but in liouvilles theorem the following is said
"The density of states in an ensemble of many identical states with different initial conditions is constant along every trajectory in phase space"
This iswhy I preemptively was asking you about what we call \rho
and in the statement we use the terms "Density of states"
and for \rho we use the terms "density function", or macro state

tbh I never really understood what people wanted to express with that phrasing of Liouville's theorem

yes
the theorem,what it says I can't make sense of it

12:41 PM
in terms of $\rho$, the whole "theorem" is just $\dot{\rho} = \{H,\rho\}$
not a very interesting or surprising statement :P

Ok
so forget the statement and let's focus on the mathematical expression

I think it's a relic from "ancient" times before we had modern Hamiltonian mechanics or statistical mechanics

Aha
ok
but density of states is not rho which is density function / macrostate?
Or also that's the case
but becuase the statement it's a relic from "ancient" times

ah, wait, now I remember

oh no

12:43 PM
the "constant along every trajectory" comes from this: Consider $\rho(q,p,t)$

ok

now take a solution to the equations of motion $q(t),p(t)$ and plug it in there: $\rho(q(t),p(t),t)$
this is now only a function of $t$, and if you take the time derivative of this it's zero

wait
if it's a function of t when you take the time derivative
why would it be zero?

that's what Liouville theorem says!

its like f(x)=3x+1 is a funciton of x and it's derivative w.r.t x is zero
but its 3

12:45 PM
I'm not saying every function does this
I'm saying a $\rho$ that's a solution to $\dot{\rho} = \{H,\rho\}$ does this

aaaa this function does this
Ok
and D.O.S =/= \rho ?

nah, they mean the same
in this case
careful, there's another meaning of "density of states" in the context of statistical mechanics :P

yes
but you see
how pronunciation and badly spoken physics
can throw you off so much
and the worst is that people don't pay attention to what they say when they explain stuff
But thank you for your time
I was able to progress quite a lot in my understanding
of classical statistical mechanics

1:07 PM
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

yes, they're all completely inside the region
that's very easy to see if the region is defined by constant energy since energy is constant along a trajectory

energy is constant along the trajectory?
I mean I get it that it comes from the liouvile

energy is conserved!

aaa yeah
thats the macro state
while microstates might change
the macro state has a well defined energy
or no?

I mean, in the microcanonical ensemble it has
but I was making a more general statement - on trajectories (solutions to the equations of motion), energy is a constant because of energy conservation

1:13 PM
See I was forgeting the fact that constant energy is the case for eq

you don't need statistical mechanics or any assumptions for that

1

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 $c$ ...

blessed

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)

yes yes you mentioned that above
ok, but I guess the trajectories inside the volume follow the same path
like over writing
if i can express it like this

1:17 PM
are you asking whether the trajectories are closed/periodic or not?

yes

that doesn't need to happen for them to stay in a "region" since nothing forbids your region to extend to infinity

periodic and do not intersect
ok, but if the region is finite

whether or not the orbits are closed depends on the specific Hamiltonian

aha
Sorry for asking so many questins, but as I am reading/studying different questions pop up in my mind

1:26 PM
it's fine, those are pretty understandable confusions

1:37 PM
But you seem to pretty much know then well, abstractly and their physical meaning
what book would you recommend ?

1:50 PM
Nov 30 '19 at 17:05, by ACuriousMind
@yuvrajsingh I haven't read many textbooks so I'm about the worst person in here to recommend any books on basic topics.

lol

2:27 PM
Do mathematicians typically call it the "inverse exponential map" or the "log map"

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