The h Bar

Silly Goose

05:16

Shweg

uhoh

07:47

5

Q: Can terrestrial infrared telescopes see through clouds or haze, sometimes at least?

My answer to Could UV-A imaging sensor reasonably see a total eclipse in progress through clouds? suggests that while clouds blocking visible light observation of the (partially) eclipsed solar disk would also likely block near UV viewing, there's a chance that there might be some transmission in...

Ryder Rude

08:11

hi

nickbros123

08:21

asking bing gpt to write tikz code for random illustrations works surprisingly well

let me try chat gpt

Minsky

14:06

cute article about light scattering, blue sky, blue sea and so on scientificamerican.com/article/why-does-the-ocean-appear

Minsky

14:17

another one loc.gov/everyday-mysteries/physics/item/why-is-the-ocean-blue

Monty

In this link web.stanford.edu/~peastman/statmech/…, at section `the maxwell-boltzmann distribution' the object they are looking to calculate is the probability of our system (which they call $A$) to be in a macrosate with energy $E_A$. They start by saying that $A$ is non-isolated, and that they can "use a simple trick to extend our analysis to non-isolated systems."

However I dont understand what "analysis" they are referring to that works for isolated systems but not non-isolated systems (i.e. up to section 2.5 where in these notes can they calculate $P(E_A)$ for an isolated system and why wouldnt that work for a non-isolated system?).

ACuriousMind

@Monty $P(E_A)$ is not for an isolated system - I think you misread what the setup here is: The system $A+B$ is isolated, $A$ alone is not because it interacts with the heat bath $B$

Monty

I guess they are referring to the idea that you can calculate probability $P(E_A)$ by using whatever statistical ensemble you chose. However why does this only work for isolated system?

@ACuriousMind no I agree here $A$ is not isolated.

ACuriousMind

then I don't understand what the question is

Monty

@ACuriousMind Sorry, I am asking : If the system $A$ IS isolated, how would you calculate $P(E_A)$?

and if you tell me how, then WHY doesnt this work if $A$ is not isolated?

ACuriousMind

14:27

@Monty The question is: from what givens do you want to calculate this?

or, rather: The notion of $p(E_A)$ that you're computing for this $A+B$ case makes no sense at all for an isolated $A$

it's the probability for $A$ to be in a state of energy $E_A$ when the full system $A+B$ has energy $E_T$

Monty

I am saying forget the full system

ACuriousMind

when $A$ is isolated, this makes no sense - there is no "full system", and there is no probability "when the full system has energy $E_T$"

Monty

pretend you just have an isolated system, forget $B$

How does one go about obtaining $P(E_A)$, or we can just call it $P(E)$ now since $A$ is isolated

ACuriousMind

if we were being really pedantic, we'd have to write the probability as $p(E_A\vert E_T)$ to express that this really is a conditional probability depending on the energy of the full system

@Monty what is your definition of $p(E_A)$ in this isolated case?

Monty

thats what I am trying to figure out

It will depend on the statistical ensemble you choose I imagine

ACuriousMind

14:31

the answer is that there is no analogous notion for an isolated system

for isolated systems, you just have what the text discusses before this section: All microstates of a given energy are equiprobable

so you might write $p(E_A) = \frac{1}{\Omega_A(E_A)}$ or something like that, but really that's a completely different probability from the $p(E_A\vert E_T)$ you're computing for the system in a heat bath

Monty

yeh but you can still ask for the probability OF a given macrostate

ACuriousMind

@Monty no, there is no such thing as the probability of a macrostate

the macrostate is the probability distribution

you can't have a probability over probability distributions :P

Monty

So, for an isolated system, should I think of a macrostate, i.e. a given energy as something I perceive

ACuriousMind

You should always think of a macrostate as being defined by observable (i.e. "macroscopic") parameters

Monty

ok sure

so the probability part enters when we ask about the microstatew

ACuriousMind

14:39

the probability distributions that thermodynamics deals with are: Given a macrostate, i.e. a list of observable parameters, what is the probability of a given microstate of the system to be realized

there are different kinds of systems and different kinds of macrostates: Your isolated systems are "the microcanonical ensemble", you specify the macrostates by the energy of the system and get that all microstates with that energy are equally probable

Monty

ah but there is also The canonical ensemble

ACuriousMind

the system $A+B$ in the heat bath is the canonical ensemble, you specify the macrostate by the total energy/the temperature $\beta$ of the heat bath, and you get the Maxwell-Boltzmann distribution as the probability for the microstates of $A$ - a microstate with energy $E_A$ has probability $\mathrm{e}^{-\beta E_A}$ to occur

Monty

wait!

I need to get the first bit straight in my head

(thank you btw)

ACuriousMind

the text you are reading is right now constructing this canonical ensemble for you. If you like the ensemble language more, the "let's move from isolated to non-isolated systems" is "let's move from the microcanonical to the canonical ensemble"

Monty

so depending on the kind of macrostate you fix, there is usually a `correct' choice of ensemble to use

e.g. fix energy and you use the micro canonical ensemble

ACuriousMind

14:45

really, "kind of macrostate" and "ensemble" are effectively synonyms

Monty

ahh

and the micro canonical ensemble assigns equal probability to each microstate

however I read before the canonical ensemble maximises entropy

ACuriousMind

@Monty to those with the specified energy of the macrostate, yes. don't forget that it assign zero probability to those microstates that have a different energy

Monty

@ACuriousMind ok thanks. (this is difficult for me since I learnt math and havent done any physics for years and years, and have studied SDEs so usually time is playing a role in my work, but that is not the case here - yet!)

i appreciate your help.

ACuriousMind

oh, if you're a mathematician you should just think of a macrostate as a probability distribution on the phase space of microstates, and there's special kinds of macrostates labeled by a few macroscopic variables that are particularly useful in physics. What the text you're reading is doing right now is deriving the form of those special probability distributions from physical arguments

Monty

Ok super

So why now in the non-isolated case does it make sense to talk about $P(E_A)$

are we no longer taking the energy of $A$ as a given?

ACuriousMind

14:54

well, if we take the more mathematical view, in both cases what we're really looking at are probability distributions $P(x,p)$ on the phase space of particles with positions $x$ and momenta $p$

in the microcanonical case, we fix $E_A$ and then we get the distribution $P(x,p) = \frac{1}{\Omega_A(E_A))}$

Monty

yup

ACuriousMind

in the canonical case, we fix the heat bath temperature $\beta$ and we get $P(x,p) = \mathrm{e}^{-\beta E_A(x,p)}$, where $E_A(x,p)$ is now a function returning the energy of a given tuple $(x,p)$

the pitfall is that $E_A$ is a constant in the first case, but not in the second, and that for some reason physicists love to write the second probability as $P(E_A)$

Monty

ohh

so, in the canonical case we are saying that we observe the temperature to be given as $\beta$, and then ask for the distribution of microstates with such a temperature? - AND this will depend on $E_A$, BUT $E_A$ actually changes over such microstates for which the temperature is $\beta$

ACuriousMind

actually, to be really consistent, we should write the microcanonical distribution as $P(x,p) = \frac{1}{\Omega_A(E)}\delta(E_A(x,p) - E)$, with $E$ as the constant energy and $E_A(x,p)$ now in both cases consistently a function

Monty

Yup but it is zero unless at the energy observed (which you write as $E$)

ACuriousMind

15:01

That's why I put the $\delta$ in there

Monty

yehyeh I was just repeating in words

you have been very clear!

I will read through this non-isolated case now

Monty

15:34

@ACuriousMind you are right it seems really strange to write this probability as $P(E_A)$!!

and quite misleading !

naturallyInconsistent

16:41

apparently you dont have, now

Claudio Menchinelli

lol

you are way too quick

@naturallyInconsistent anyways I forgot to tell you something

naturallyInconsistent

covers ears

Claudio Menchinelli

I think your computations from last time are not correct according to a mathexchange user

naturallyInconsistent

which computation?

Claudio Menchinelli

basically they gave me an answer after some time and it contradicts what you said, namely: $$\delta_{ij}\delta_{jm}\delta_{mj}x_ip_j = 3\mathbf{x}\cdot \mathbf{p}$$

it should be $x \cdot p$

according to him

naturallyInconsistent

16:45

I checked that before, and it agrees with Sakurai, and it agrees with the extremely well-known connection between the Laplacian operator and the radial squared momentum and the angular momentum operators. It cannot be wrong.

Claudio Menchinelli

he said this basically: $$i = j = m \Rightarrow \sum_{i,j}\delta_{ij}\delta_{jj}\delta_{jj}x_ip_j $$

naturallyInconsistent

What I would say, however, that Sakurai wrote it in a form that is extremely atypical and that is why I did not recognise it as correct in the first step.

@ClaudioMenchinelli if i=j=m then there is no ij sum

nickbros123

@ClaudioMenchinelli whats the difference between $\delta_{mj}$ and $\delta_{jm}$ here?

Claudio Menchinelli

nothing

they dont matter in fact

but there's no 3 factor apparently

naturallyInconsistent

@nickbros123 Kronecker delta are symmetric, so they must be the same. They must both be identity matrices

Claudio Menchinelli

16:46

It's just a dot product

nickbros123

this thing is summed over all 3 variables is it?

Claudio Menchinelli

1

Q: Sum with multiple Kronecker deltas

I have a problem dealing with the following expression: $$\sum_{i,j,m}\delta_{ij}\delta_{mj}\delta_{jm}x_ip_j$$ which I know it should yield the following result: $$ 3\mathbf{x}\cdot\mathbf{p} $$ I was thinking in terms of matrices and thought that because we have that $$\sum_{j,m}\delta_{mj}\del...

linear-algebra summation kronecker-delta

naturallyInconsistent

The multiplication of identity matrices gives back identity matrix. The trace of the identity matrix is 3.

Claudio Menchinelli

@nickbros123 yes I omitted it

@naturallyInconsistent I guess you have more authority but nobody seems to disagree with his answer

nickbros123

mse answer seems about right

naturallyInconsistent

16:52

I just pointed out the mistake in the M.SE, where it is not the answerer that is wrong, it is you that asked it wrongly.

ACuriousMind

@ClaudioMenchinelli The indices here are weird: Why are there more than 2 $j$?

Remember the laws of index notation as physicists usually use it: Each side of an equality must have the same free indices (indices occuring once) and no index must occur more than twice

naturallyInconsistent

Indeed. In my own personal notes rendering this expression, I had up and down indices, despite the fact that I didn't learn the correct expression to deal with the Hodge dual (inside the cross products).

ACuriousMind

also, your separate claim that $\sum_{j,m} \delta_{jm}\delta_{jm} = I^2$ is wrong and you should have noticed that because the l.h.s. is a number and the r.h.s. is a matrix

The thing about index notation is that it's easy to make mistakes, but it's also usually easy to notice that you've made one if you just very mechanically check these rules

Claudio Menchinelli

I have inverted the indices somewhere wait

naturallyInconsistent

oh, I think index notation is one huge way to help meow prevent mistakes

Claudio Menchinelli

16:56

no it's my mistake

It’s this big sum here that created problems

nickbros123

@ACuriousMind perhaps hes talking about the sum of diagonal elements of $I^2$

ACuriousMind

@nickbros123 then he should be explicit about this, and write $\mathrm{tr}(I^2)$

which would be correct, and equal to 3

Claudio Menchinelli

no it's my mistake

naturallyInconsistent

Actually, it is my mistake from when I typed it into this chat.

Claudio Menchinelli

well physics was saved again today

naturallyInconsistent

17:02

The way Sakurai derived it is just very awkward. If he wanted to get us to make fewer mistakes, he would have tried a more symmetric presentation, so that following the argument is easier.

Claudio Menchinelli

should I delete my question on M.SE at this point

naturallyInconsistent

no point deleting.

ACuriousMind

@ClaudioMenchinelli you can't delete it, it has received upvoted answers

naturallyInconsistent

lol

nickbros123

out of context that mse question page looks extremely funny

Claudio Menchinelli

17:04

who the hell upvoted it lmaooo

didnt even notice that

I must say it was like 3 am when I was doing this indices stuff

and tears were making it hard to see clearly

:P

nickbros123

"....You typed the original question wrong..." i dont know why im laughing at this lol

Claudio Menchinelli

I'm getting roasted simultaneously on both sites

I guess they commute

naturallyInconsistent

17:18

Anyway, I would have written it as $$\begin{align}L^2&=(\vec x\times\vec p)\cdot(\vec x\times\vec p)=\epsilon^{ijk}x_ip_j\delta_{k\ell}\epsilon_{\ell mn}x_mp_n\\&=(\delta^{im}\delta^{jn}-\delta^{in}\delta^{jm})x_ip_jx_mp_n=(\delta^{im}\delta^{jn}-\delta^{in}\delta^{jm})x_i(x_mp_j-i\hslash\delta_{mj})p_n\\&=x^2p^2-i\hslash\vec x\cdot\vec p+i\hslash\delta^{jm}\delta_{mj}\vec x\cdot\vec p-\delta^{in}\delta_{jm}x_i(p_nx_m+i\hslash\delta_{nm})p_j\end {align}$$

oops, that was $\epsilon^{\ell mn}$

and $\delta^{jm}$

Claudio Menchinelli

too bad I'm not familiar with tensor notation. I should probably learn it on my own at this point

naturallyInconsistent

Well, with this much work on it, you are already somewhat there

Claudio Menchinelli

that's good news then

Slereah

raw.githubusercontent.com/nameiwillforget/hegel-in-mathematics/…

Paper on the nlab Hegel thing

Slereah

17:38

it is not super more specific than Schreiber's own tho

Slereah

17:54

God was Hegel paid by the page

nickbros123

is smooth conditions assumed on equation of state for a thermodynamic system warranted?

im asking this because the isotherms for water dont rly look smooth

Slereah

Odds are pretty good that the condition is there for being nice

nickbros123

defo

afaict thermodynamics lives on implicit function theorem

and that needs C^1 i think

Slereah

18:11

> In continuity, therefore, magnitude immediately possesses the moment of discreteness — repulsion, as now a moment in quantity. Continuity is self-sameness, but of the Many which, however, do not become exclusive; it is repulsion which expands the selfsameness to continuity. Hence discreteness, on its side, is a coalescent discreteness, where the ones are not connected by the void, by the negative, but by their own continuity and do not interrupt this self-sameness in the many

I'm gonna need an even clearer paper I think

nickbros123

@Slereah ..sir, this is a wendy's..

ACuriousMind

lol

Slereah

ncatlab.org/nlab/show/Hegelian+taco

Mr. Feynman

18:27

In the RN (non extremal case) you have two event horizons at $r=r_\pm$. After your cross the outermost one, you are forced to cross the innermost horizon, but then since $r$ is spacelike again inside and the singurality is timelike, you don't have it in your future (actually it's repulsive). At any rate, you can bounce off it and get outside of the innermost horizon. How come you can cross an event horizon backwards?

I know that these coordinates are not really the best, but they should work outside and inside even though not at the horizon itself

Maybe I should ask: why is the innermost horizon an event horizon if you can cross it inward and outward?

This question possibly explains the issue better

> According to Carroll, if you go into the first one, you will fall until you reach the second one, at which point you are free to move around. From here you can choose to avoid the singularity and cross the Cauchy horizon again (I'm not sure I get how, isn't the horizon a null surface?

Slereah

@Mr.Feynman You can't reach null infinity from there

although I guess that's more the outer horizon that's the boundary there :p

But I think the name is mostly historical

Mr. Feynman

@Slereah So is Carrol being a little sloppy by calling it an event horizon?

Slereah

I think it's a trapped surface?

Which is sort of what many people mean by "event horizon"

Mr. Feynman

How is it trapped though? If I cross it, I can bounce off the singurality and get back outside

Slereah

it is trapped in that you have to cross it

Trapped surfaces don't necessarily require you to not be able to get out

ie. a black hole that eventually evaporates still has its "event horizon" as a trapped surface

Mr. Feynman

18:40

Ok, that's something I didn't realize. The definition I know for trapped surfaces is in terms of outgoing and ingoing light rays both converging

Slereah

There are many types of trapped surfaces

Mr. Feynman

So do you confirm that the one I wrote is a different notion of trapped surface?

Maybe more naive

Slereah

That's just my guess

I cannot confirm

Mr. Feynman

Slereah

The many trapped surfaces

Mr. Feynman

18:46

In the picture above you can see the part I'm talking about

And there is an additional question, which is certainly related: why is it timelike but reversed after crossing it back?

I mean, if I write the metric it's exactly the same as the one I had the first time I crossed it

How would I translate in equations that "reversed"?

Slereah

Also looking at the Penrose diagram, can you really cross the same horizon?

from the looks of it you just cross another different outer horizon

Mr. Feynman

In the Penrose diagram that's what happens. Basically you enter BH and exit WH

Though I'm confused as to how that should be clear from the paragraph

Is this understandable only going through the explicit calculations of the Kruskal coordinates and Penrose compactification?

Mr. Feynman

19:20

@ACuriousMind since you said a couple of times you had many pure math courses, I wonder which ones. Do you have an approximate list? :P

ACuriousMind

@Mr.Feynman uh, sure: real analysis, differential geometry, complex analysis, algebraic topology; linear algebra, abstract algebra (mostly ring theory), algebraic geometry, sheaf cohomology (which was about deriving Verdier duality); functional analysis, Lie group theory, representations of the Poincaré group (which more generally developed Mackey's and Harish-Chandra's theory of representations)

5

Slereah

"In 19th century German, the word "Ring" could mean "association", which is still used today in English in a limited sense (for example, spy ring), so if that were the etymology then it would be similar to the way "group" entered mathematics by being a non-technical word for "collection of related things"."

Dammit

It's another word meaning "a bunch"

Mr. Feynman