The h Bar

DIRAC1930

12:15 AM

Does an internal degree of freedom mean that its associated operator commutes with $\hat{q}$ and $\hat{p}$?

@DIRAC1930 yes, but that's not the definition (the definition is just "something other than a spacetime symmetry"), that's the Coleman-Mandula theorem

DIRAC1930

Okay thanks

Slereah

6:13 AM

morning

Slereah

7:41 AM

@ACuriousMind Riddle me this, Batman

Consider some Cauchy hypersurface $\Sigma$

On it is some kind of initial vector field from which stems curves

And all over the spacetime, a vector field $a$ exists, such that $\nabla_{\dot{\gamma}} \dot{\gamma} = a$

Under what conditions does this form a foliation of the manifold

John Rennie

@Ishwaran Hi Ishwaran. We deal with JEE problems in the Problem solving room. Shall I move your question there?

ACuriousMind

8:36 AM

@Slereah what exactly is the foliation here?

if you have a bunch of integral curves of a vector field, that's a foliation iff the vector field is everywhere non-zero

Slereah

@ACuriousMind the foliation defined by every curve going through $\Sigma$

@ACuriousMind I am thinking that maybe there could be pathological behaviours like if the acceleration is unbounded then there is the formation of a horizon

And then the set of curves doesn't cover the entire spacetime

ACuriousMind

@Slereah but how are you extending these curves to all of $M$

just saying that they pass through $\Sigma$ doesn't define the entire curve

Slereah

It would be a problem for a single curve, but I don't know about all those curves

Like imagine that through the surface, there passed all Rindler observers

Does it cover the entire spacetime

None of them would reach timelike infinity, but otoh they would cover the entirity of null infinity I think

ACuriousMind

how exactly is the worldline of a "Rindler observer" defined mathematically

Slereah

@ACuriousMind as I said, passing through sigma, an initial vector, and an acceleration vector field

For a Rindler observer for instance let's say everyone with initial velocity $\partial_t$ and uuuuh

what's the acceleration of a Rindler observer again

$a = x^{-1} e_x$?

ACuriousMind

8:50 AM

yeah, I don't think you can guarantee that this covers the entire manifold

Slereah

Well famously a single Rindler curve has a horizon, but otoh, if the curves cover the entire Cauchy surface, I think maybe it can work?

They will all have different horizons, which will possibly span the entire spacetime

But that's a maybe

I remember that there's a paper about foliating spacetimes with Hamiltonian flows, I guess it's the same problem here

Just turn the acceleration vector field into a "force"

but otoh that would require $a \propto \nabla U$ I think

ACuriousMind

well, here's a formulation that actually foliates the manifold: If there is a non-zero vector field $X$ such that $\nabla_X X = a$ and $X\vert_\Sigma$ is your initial vector field, then the integral curves of this vector field are your Rindler observers and form a foliation of the manifold

Slereah

Unfortunately I can't predict the future, I only know what happens on $\Sigma$!

Ah well, let's look into it

btw I think a possibly neat example of a foliation that doesn't admit an orthogonal complement one is the case of a random spacetime

Every spacetime admits a line element field, which I think defines a foliation, but otoh they may not admit a foliation into spacelike hypersurfaces

Like Gödel's spacetime

I guess all those line element fields define a fleet of observers

Physics problem, there is a mysterious field that has descended upon the town

Ishwaran

9:11 AM

@JohnRennie yes john... please

John Rennie

Moved

Slereah

Also saying that I know what happens on $\Sigma$ is even an exaggeration, I don't know what happens on the entire universe!

Slereah

11:46 AM

you know there's a thermodynamical cycle for every type of thermodynamical variable

ie pressure/volume, magnetization/magnetic field, polarization/electric field

I wonder if there is one for the chemical potential and also for gravitational potential energy

possibly tidal generators are an example but I'm not sure

Slereah

1:04 PM

"For instance, we could choose a Klein model $M_0$ that has the same topology than $M$"

Ouh la la

Bespoke tangent bundles

imbAF

2:00 PM

What is one (or some) physical interpretations of the Sackur–Tetrode equation ?

Can one be, the fact that entropy is expressed as a function of state variables, which makes it a state variable too?

Or is there another physical interpretation ?

DIRAC1930

2:27 PM

When canonically quantizing for example $\phi(X)^2$ for a real field $\phi(X)$, how do we avoid squares of operator-valued distributions at the same point which I think are undefined?

Slereah

Depends, but the basic one is normal ordering

otherwise you have to go through renormalization if things are too complicated

ACuriousMind

@DIRAC1930 you...don't, really, at least not in canonical quantization

DIRAC1930

I thought normal ordering just deals with the ordering ambiguity?

ACuriousMind

you just act as if there's no problem and then you have to do renormalization

DIRAC1930

How come renormalization doesn't exist in non-rel theories then?

Slereah

2:30 PM

Nah

Normal ordering is just you removing $\delta(0)$ from the theory

@DIRAC1930 1) it does 2) it doesn't use distributions

ACuriousMind

@DIRAC1930 that's a bold claim

Slereah

be bold I say

ACuriousMind

the basic idea of renormalization is neither special to relativistic theories nor special to quantum field theories

DIRAC1930

Okay thanks

ACuriousMind

it's just something that happens in perturbative theories when your perturbation series diverge, see e.g physics.stackexchange.com/q/144220/50583

Slereah

2:32 PM

you can even do it for regular boring functions

renormalization isn't too hard to understand when you remember that distributions can also be expressed as a series of weakly converging function

You just change that series to another series with a substracted term

It's everywhere finite, and it converges to something finite

Even though if you look at "the last term" of the series, it would be like substracting an infinite term

imbAF

While we reduce the Temperature of the system to absolute zero, according to Nernst heat theorem the entropy of the system is reduced or tends to go to zero, but from the 1.st law we know that the entropy can never be smaller then zero, therefore while the system has an entropy reduction the enviroment must have an entropy increase, so how exactly is the 1st law true while we reduce the temperature?

Semiclassical

The example which comes to mind is a non-relativistic Fermi gas with Coulomb interaction

Slereah

also you can just use like non-quantum examples

just look at distribution algebras

Semiclassical

If you write down the Hamiltonian naively in momentum space, you get a 1/q^2 divergence

Slereah

Oh right there's also like a classical example

The energy functional of a point particle in EM

Semiclassical

2:40 PM

If you want to get rid of that, you can introduce a positive charge background and a finite screening length

Slereah

@ACuriousMind How do you deal with model spaces that aren't vector spaces in Cartan geometry

Semiclassical

If you account for the energy of said background and it’s int retraction with the (negative) Fermi gas, you find that the 1/q^2 part gets cancelled

ACuriousMind

@Slereah I'm not sure what gave you the impression I know anything about Cartan geometry :D

Slereah

@ACuriousMind Doesn't hurt to ask!

idk who else to ask

Semiclassical

In such a way that, upon setting the screening length to zero, you get the same Hamiltonian but w/o q=0

Slereah

2:42 PM

Every Cartan geometry stuff just talks all the time about how the model space is put in isomorphism with the tangent bundle and how the geometry is a subgroup of $GL(n)$ but also some of them are not vector spaces at all?

They're kind of the tangent bundle, but it's not really said how you're supposed to deal with it

Sometimes it's the projective structure of the tangent bundle, sometimes it's the one point compactification of the tangent bundle

I guess the first one you use $PGL(n)$, but how do you deal with the tangent bundle at infinity

ACuriousMind

from a quick nLab reading, what's identified with the tangent spaces is not the model space $G/H$ but its tangent space $\mathfrak{g}/\mathfrak{h}$, which is a vector space because it's a Lie algebra, no?

Slereah

Maybe?

Let's hope

Semiclassical

So you can either do a long calculation to justify getting rid of the problematic q=0 term, or you can get rid of it on the grounds of this divergence ‘obviously’ being irrelevant

Slereah

@Semiclassical Nothing ever goes well when you remove obviously irrelevant terms

Semiclassical

I guess this is more like regularization than renormalizatin tho

No infinite series etc

Slereah

2:46 PM

also for the fancier structures there are cone structures, whatever that is

for the fancy causal structure and its light cones

DIRAC1930

@Slereah Why doesn't non-rel QFT use distributions?

Slereah

@ACuriousMind they do say that the group has to be a subgroup of GL though

Which I don't know if that's true here?

ACuriousMind

which one - $G$ or $H$?

Slereah

Lemme see

It's a bit hard because the discussion is split between the article on G-structures and the cartan connection article

I don't know if notations are consistent across it

Qwin

Why is the voltage source the most difficult to "control" in an RC circuit?

Slereah

2:57 PM

Although maybe the problem is that G-structures are a type of Cartan geometries, but maybe not all of them are?

ACuriousMind

@Slereah yes

Slereah

That could be it

that helps but only somewhat, because there are still relations between Cartan geometries where the base model is $R^n$ and ones that aren't though :p

Since some structures imply other structures

Also is "conformal structure" a G-structure?

It has structure right in the name

ACuriousMind

the G-structures are Cartan geometries (turns out I know what this is, just not under that name) where you have $H$ as a subgroup of GL and $G= \mathbb{R}^n\rtimes H$, nlab says this here

Slereah

but it doesn't act on the tangent space directly

What do you call them?

ACuriousMind

@Slereah no, it's a Cartan connection but not a G structure since the $G$ in $G/H$ doesn't have the right form

Slereah

3:08 PM

Those damn physicists

imbAF

@ACuriousMind when we say that $\frac 1 T = \frac {\partial S}{\partial E}$ this is the case in which we can write $\delta Q = TdS$ and $\delta W = pdV$, which is the case for a quasi static process/ reversible process, is my assumption correct ?

ACuriousMind

$T^{-1} = \partial_E S$ is independent of any process

it's just the definition of temperature as a state function

imbAF

really?

ACuriousMind

really. :)

imbAF

Cuz when I derived it, I took the case where dU = Tds - pdV

and as we know

ACuriousMind

3:20 PM

what do you mean, you derived that?

imbAF

$\delta Q = TdS$ and $\delta W = pdV$ are valid only for reversible

ACuriousMind

that's the statistical definition of temperature :P

imbAF

see

dU = Tds - pdV +\mudN

ACuriousMind

what you did was probably showing that this statistical definition coincides with the "naive" thermodynamic concept of temperature for some particular process

imbAF

then you have ds= T^-1 ( dU + pdV - \mudN)

ACuriousMind

3:22 PM

I (probably) understand what you did

imbAF

then from here you can see that $(\frac {\partial S}{partial U})_{V,N} = T^-1$

ACuriousMind

I'm just saying that $T^{-1} = \partial_E S$ is how we define temperature in statistical mechanics as a state function

imbAF

Ok

ACuriousMind

what you did was showing that for a process where we have $\delta Q = T\mathrm{d}S$, this indeed is temperature "as we know it"

imbAF

I thought it was the case when we consider the reverisble process

aha

so a confirmation basically

ACuriousMind

3:24 PM

but that doesn't limit the validity of the definition $T^{-1} = \partial_E S$ to such processes - this definition is intended as a state function for all thermodynamic states and doesn't depend on any particular idea of process

imbAF

I see

temperature is a state variable right?

ACuriousMind

(for one, note that nothing in that definition involves anything to do with a process - S is a state function, E is a state function, so T is a state function, too)

imbAF

yeah

but the way we found it was as I explained

since it wasn't explicitly said in my lecture

and I assumed that it was the case for the rev. process not of irrev. for example

ACuriousMind

well, what you did is the motivation to define temperature like this

Semiclassical

I think you do have to assume everything changes smoothly, so free expansion of a gas is out

ACuriousMind

3:26 PM

but the definition itself makes no reference to any process anymore

Semiclassical

Otherwise the derivative doesn’t make sense

ACuriousMind

so it's just a state function

imbAF

Yes

I know, same with entropy

you find that it depends from volume change in rev processes

but even if the same process is irreverisble the entropy change will have the same value

regardless the path we took to reach the state

I had this question done earlier, can you confirm whether my understanding is correct

While we reduce the Temperature of the system to absolute zero, according to Nernst heat theorem the entropy of the system is reduced or tends to go to zero, but from the 1.st law we know that the entropy can never be smaller then zero, therefore while the system has an entropy reduction the enviroment must have an entropy increase, so how exactly is the 1st law true while we reduce the temperature?

how does the entropy of the enviroment increase while evidently that of the system decreases ? Can you give an answer, or you need more informations about how the temperatur drop happens, whether we have volume change etc etc

Semiclassical

What’s the contradiction? Entropy of system goes down, entropy of environment increases more

Heat will flow out of the system , so entropy of environment increasing is hardly unexpected

imbAF

how does that of the environment increase ?

aha

true lol

Semiclassical

3:32 PM

I mean, you could also have work being done at the same time

But if the system entropy goes down then heat must flow out

imbAF

Yes

but one thing

if we assume that the process is quasi static

so that we can express \delta Q = TdS and \delta W = pdV

we can calculate Q as the integral of TdS

but obviously we need to express T in another way

since it's not a constant

since T of the system drops

Semiclassical

Sure. Need some equation of state

imbAF

yes

but the process is not an isothermal one and also not a isentropic one right?

Semiclassical

I don’t think you’ve said enough to know either way

imbAF

well for once, if we are concerned with Nernst heat theorem , we know two things for certain that T changes and S also changes

Semiclassical

3:41 PM

Could keep temp fixed while lowering entropy if work is being done on system

Ah

imbAF

But I agree, in order to find an expression about \Delta T you need more information

Semiclassical

If you assume an ideal gas and take temperature to zero, I don’t see why you couldn’t hold the either pressure or volume fixed

It’s compatible with the equation of state at least

imbAF

and if the process was rev how would you find the heat

since you would need to integrate this \delta Q = TdS

but T ain't constant

It's a very hypothetical exercise, that is missing many informations, so forget it

Semiclassical

I’d guess that you’d get different heats if you assume constant pressure vs constant volume

There’s no work being done in the second case, so dU=T dS in that case

imbAF

dU=3/2nRdT

and from here you find how \Delta T is related to \Delta S

I guess

@Semiclassical I have one question regarding spin

Semiclassical

3:51 PM

I think you do run into a contradiction here. See en.wikipedia.org/wiki/Third_law_of_thermodynamics#Specific_heat

imbAF

I see

but this is the case when we go to zero

I merely was talking about simply reducing the temperature

to a certain point

not striving for zero

and then calculating different changes in entropy of the system, heat given and taken, Temperature

But you need more information about the type of the process that we are observing

ACuriousMind

@imbAF what's the problem with integrating $T\mathrm{d}S$ for varying $T$?

you can integrate $f(x)\mathrm{d}x$ along a path, even though $f$ "varies"!

imbAF

Nothing, you simply get how \Delta S changes with \Delta T

but

if you need to find the heat

then on the right side you have two variables 1 is temperature and 2 is entropy

ACuriousMind

but both are state functions

and you have a path through state space

integrating a state function along a path through state space is perfectly possible, it's just a line integral

imbAF

the thing is

ok

what I asked was not a good thing to do, since you need more information regarding the type of process you are having/ what changes in the system etc

yes

so you have Q = \int T dS

but T also changes same as S

ACuriousMind

3:58 PM

I think you're just confused about how the math works

imbAF

What do you mean?

Usually the types of processes I have done are either isotropic/adiabatic/isentropic

ACuriousMind

let's say we're in $\mathbb{R}^2$, coordinates $x,y$ and we have functions $f(x,y)$ and $g(x,y)$. Then $f\mathrm{d}g$ is a perfectly fine 1-form you can do line integrals over

$T\mathrm{d}S$ is exactly the same

imbAF

ok

how would you find the heat take from the system

ACuriousMind

it might not be particularly easy to compute this integral, but there's no conceptual problem

imbAF

when it drops in entropy and also changes temperature

ACuriousMind

4:00 PM

well, you need to describe the actual path through state space

because heat is not a state function, after all

imbAF

no it is not

ACuriousMind

so just saying "these things change" is not enough information

imbAF

but if the process is reversible you can use the $\delta Q = T dS$

expression

yes

I am aware

that you need additional information

Slereah

4:35 PM

What does a conformal structure select, anyway

Is it like a direction field?

IIRC it's the projective orthogonal group or somethnig

It's weird to call it "orthonormal" considering that there's no notion of distance involved

well I guess it's orthogonal

get those angles

DIRAC1930

If in real life, a classical field takes on finite values, does that automatically mean that when quantized, the interacting Hamiltonian with linear terms in that field is bounded below?

Slereah

I mean it is to be hoped for, but no, I don't think that has any bearing on it

The basic example of a theory that isn't bounded from below is commutative spinors

a fairly unremarkable theory in the classical regime

ACuriousMind

@DIRAC1930 "bounded below" means that a function is never below that bound for any input

DIRAC1930

So if the classical field is bounded below, that means that the quantized field must be too right?

ACuriousMind

yes, but classical fields aren't bounded below

Slereah

4:50 PM

The classical field isn't the important part, it's the Hamiltonian

ACuriousMind

classical fields are just (perhaps compactly supported) smooth functions

they're not "smooth functions that never exceed the value X"

Slereah

Because otherwise you get the Hella Jeff effect

System is at energy $E$, radiates some energy away, now it is at energy $E - \Delta E$, it radiates some more

etc etc

and then it turns out that the universe explodes

See $\phi^3$ theory if you enjoy that gag

DIRAC1930

Why do people use scalar Yukawa theory then?

Slereah

Why not

It's very much bounded from below

ACuriousMind

@DIRAC1930 that's a weird question

Slereah

4:53 PM

At $0$, if properly renormalized

a good value to be the bottom of the scale

ACuriousMind

are you implying Yukawa theory is not bounded below?

Slereah

You can check whether a theory is bounded from below via a physical process called "look at it"

DIRAC1930

No I'm just asking a question because at first glance the interacting part of the Hamiltonian seems to approach $-\infty$ as $\phi\rightarrow -\infty$

Slereah

The potential energy of Yukawa is like $\phi^2$

ACuriousMind

@DIRAC1930 sure, but it's linear so it's dominated by the $\phi^2$ mass term, which is bounded below

Slereah

4:55 PM

a very bounded from below quantity

DIRAC1930

@ACuriousMind Ah yes, I never thought about that

ACuriousMind

polynomials of even degree are bounded below (or above if they have a negative sign in the top degree), the lower degrees don't matter for boundedness

DIRAC1930

But what if $\psi^\dagger \psi \phi$ dominates the $\psi^\dagger \psi$ and $\phi^2$ terms?

ACuriousMind

what do you mean "what if"

the boundedness of a function is not a question of imagination

Slereah

If ifs and buts were candies and nuts

there are too many orthogonal groups

DIRAC1930

5:00 PM

@ACuriousMind You didn't consider all the possible terms though

Slereah

I can think of like 5 or 6 groups that are somewhat orthogonal

DIRAC1930

$(\psi^\dagger\psi +1)\phi + \phi^2$ I guess you did

But if $\psi^\dagger \psi$ shoots up to infinity faster than $\phi \rightarrow -\infty$ won't you get the whole thing going to $-\infty$?

In David Tong's notes

The scalar Yukawa theory has a slightly worrying aspect: the potential has a stable local minimum at $\phi= \psi= 0$, but is unbounded below for large enough $-g\phi$. This means we shouldn’t try to push this theory too far.

ACuriousMind

ah, yes, it's not bounded below in $\psi$

Slereah

oh no

ACuriousMind

thing is, perturbative QFT only expands around a vacuum anyway, so stable local minimum is quite enough

Slereah

5:12 PM

not true for Lorentz manifolds apparently, though

ACuriousMind

the whole boundedness thing is much more complicated anyway because running couplings (or "the renormalization group") can make your theory unstable even if it started off stable in the classical Hamiltonian

Slereah

Is that related to the uuuuh uniformization theorem

ACuriousMind

the Higgs famously has (or had, before we detected it) a range of possible masses where the running coupling just makes the SM unstable

Slereah

Apparently the model space for Lorentzian conformal structures is the Einstein universe?

I barely remember that one

nobody cares for that poor thing

DIRAC1930

@ACuriousMind Okay thanks

imbAF

5:22 PM

If in an experiment, the spin of the individual particles in the x-direction is measured on a beam of particles with spin 1/2 in the positive z-direction.

How does this work?

then what's the result of this experiment, what are the states?

Should the state be an eigenstate, whose probability is equal to them square of the coeffiecient

with with this is multiplied?

imbAF

5:41 PM

If a particle is characterized by a z- component of Spin, can it also have an x component ?

ACuriousMind

that's the wrong question

what do you mean by the particle "having" an x-component of spin?

imbAF

I know that you cannot speak about the spin

like it's a vector

I am a bit rusty in this part of QM, I haven't revised it in a long time

But I was able to deduce that If in an experiment, the spin of the individual particles in the x-direction is measured on a beam of particles with spin 1/2 in the positive z-direction, is a way of saying

that we are having a mixture of states

or am I understanding the problem wrong?

ACuriousMind

sure

it's a Stern-Gerlach experiment, what you find is that the beam splits in half

imbAF

but in my example

yeah I thought that

Nothing is said about an external magnetic field

ACuriousMind

what would that matter?

imbAF

5:52 PM

because it should?

ACuriousMind

sure, S-G does the measurement via an external magnetic field, but how the measurement works is completely irrelevant

imbAF

Ok

but the ideal was that we are having this set up

and

ACuriousMind

like, in order to tell me what velocity something has you don't need to know what brand of radar cannon I'm using to measure velocity

imbAF

I am asked, which is the result of the experiment, which are the states present

and i sad |z,+> and |x,+>

Are the states for which we are concerned

@ACuriousMind hahaha nice analogy

said*

ACuriousMind

@imbAF why would the result of a measurement of x-spin be a |z,+> state

imbAF

5:54 PM

Honestly I don't understand the question

if the spin component of Z is +1/2

ACuriousMind

|z,+> is your input state for the measurement, it's not the result

measurement changes the state!

imbAF

if

the state is a superposition or mixed

how would you change a eigenstate?

ACuriousMind

|z,+> is not an eigenstate of x-spin, is it?

imbAF

no

But at the same time the components of spin do not commute

Slereah

"However, the structure group of gr(TM) = TM is reduced to G0 = CO(r,s)."

Have mercy Cartan

What is even CO

imbAF

5:57 PM

so if the system is in a particular spin state, which is corresponds to a spin component

let's say the system has a spin in the positive z-direciton

if you try to measure the x-spin

you wouldn't get anything

Slereah

Orthogonal group

In mathematics, the orthogonal group in dimension n, denoted O(n), is the group of distance-preserving transformations of a Euclidean space of dimension n that preserve a fixed point, where the group operation is given by composing transformations. The orthogonal group is sometimes called the general orthogonal group, by analogy with the general linear group. Equivalently, it is the group of n×n orthogonal matrices, where the group operation is given by matrix multiplication (an orthogonal matrix is a real matrix whose inverse equals its transpose). The orthogonal group is an algebraic group and...

mama mia

Will they never run out of orthogonal groups

ACuriousMind

@imbAF what do you mean "you wouldn't get anything"

a measurement is a measurement

"you don't get anything" is not a valid measurement result

Slereah

@ACuriousMind Obviously you've never been in a lab

that's what you get most of the time

ACuriousMind

the possible results of measurements are the eigenvalues of the operator you're trying to measure

stop trying to think intuitively and just apply the rules of QM

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