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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
 
Okay thanks
 
 
6 hours later…
6:13 AM
morning
 
 
1 hour later…
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
 
@Ishwaran Hi Ishwaran. We deal with JEE problems in the Problem solving room. Shall I move your question there?
 
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
 
@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
 
@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
 
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
 
how exactly is the worldline of a "Rindler observer" defined mathematically
 
@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
$a = x^{-1} e_x$?
 
8:50 AM
yeah, I don't think you can guarantee that this covers the entire manifold
 
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
 
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
 
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
 
9:11 AM
@JohnRennie yes john... please
 
Moved
 
Also saying that I know what happens on $\Sigma$ is even an exaggeration, I don't know what happens on the entire universe!
 
 
2 hours later…
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
 
 
1 hour later…
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
 
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 ?
 
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?
 
Depends, but the basic one is normal ordering
otherwise you have to go through renormalization if things are too complicated
 
@DIRAC1930 you...don't, really, at least not in canonical quantization
 
I thought normal ordering just deals with the ordering ambiguity?
 
you just act as if there's no problem and then you have to do renormalization
 
How come renormalization doesn't exist in non-rel theories then?
 
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
 
@DIRAC1930 that's a bold claim
 
be bold I say
 
the basic idea of renormalization is neither special to relativistic theories nor special to quantum field theories
 
Okay thanks
 
it's just something that happens in perturbative theories when your perturbation series diverge, see e.g physics.stackexchange.com/q/144220/50583
 
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
 
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?
 
The example which comes to mind is a non-relativistic Fermi gas with Coulomb interaction
 
also you can just use like non-quantum examples
just look at distribution algebras
 
If you write down the Hamiltonian naively in momentum space, you get a 1/q^2 divergence
 
Oh right there's also like a classical example
The energy functional of a point particle in EM
 
2:40 PM
If you want to get rid of that, you can introduce a positive charge background and a finite screening length
 
@ACuriousMind How do you deal with model spaces that aren't vector spaces in Cartan geometry
 
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
 
@Slereah I'm not sure what gave you the impression I know anything about Cartan geometry :D
 
@ACuriousMind Doesn't hurt to ask!
idk who else to ask
 
In such a way that, upon setting the screening length to zero, you get the same Hamiltonian but w/o q=0
 
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
 
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?
 
Maybe?
Let's hope
 
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
 
@Semiclassical Nothing ever goes well when you remove obviously irrelevant terms
 
I guess this is more like regularization than renormalizatin tho
No infinite series etc
 
2:46 PM
also for the fancier structures there are cone structures, whatever that is
for the fancy causal structure and its light cones
 
@Slereah Why doesn't non-rel QFT use distributions?
 
@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?
 
which one - $G$ or $H$?
 
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
 
Why is the voltage source the most difficult to "control" in an RC circuit?
 
2:57 PM
Although maybe the problem is that G-structures are a type of Cartan geometries, but maybe not all of them are?
 
@Slereah yes
 
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
 
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
 
but it doesn't act on the tangent space directly
What do you call them?
 
@Slereah no, it's a Cartan connection but not a G structure since the $G$ in $G/H$ doesn't have the right form
 
3:08 PM
Those damn physicists
 
@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 ?
 
$T^{-1} = \partial_E S$ is independent of any process
it's just the definition of temperature as a state function
 
really?
 
really. :)
 
Cuz when I derived it, I took the case where dU = Tds - pdV
and as we know
 
3:20 PM
what do you mean, you derived that?
 
$\delta Q = TdS$ and $\delta W = pdV$ are valid only for reversible
 
that's the statistical definition of temperature :P
 
see
dU = Tds - pdV +\mudN
 
what you did was probably showing that this statistical definition coincides with the "naive" thermodynamic concept of temperature for some particular process
 
then you have ds= T^-1 ( dU + pdV - \mudN)
 
3:22 PM
I (probably) understand what you did
 
then from here you can see that $(\frac {\partial S}{partial U})_{V,N} = T^-1$
 
I'm just saying that $T^{-1} = \partial_E S$ is how we define temperature in statistical mechanics as a state function
 
Ok
 
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"
 
I thought it was the case when we consider the reverisble process
aha
so a confirmation basically
 
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
 
I see
temperature is a state variable right?
 
(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)
 
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
 
well, what you did is the motivation to define temperature like this
 
I think you do have to assume everything changes smoothly, so free expansion of a gas is out
 
3:26 PM
but the definition itself makes no reference to any process anymore
 
Otherwise the derivative doesn’t make sense
 
so it's just a state function
 
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
 
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
 
how does that of the environment increase ?
aha
true lol
 
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
 
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
 
Sure. Need some equation of state
 
yes
but the process is not an isothermal one and also not a isentropic one right?
 
I don’t think you’ve said enough to know either way
 
well for once, if we are concerned with Nernst heat theorem , we know two things for certain that T changes and S also changes
 
3:41 PM
Could keep temp fixed while lowering entropy if work is being done on system
Ah
 
But I agree, in order to find an expression about \Delta T you need more information
 
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
 
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
 
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
 
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
 
3:51 PM
I think you do run into a contradiction here. See en.wikipedia.org/wiki/Third_law_of_thermodynamics#Specific_heat
 
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
 
@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"!
 
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
 
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
 
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
 
3:58 PM
I think you're just confused about how the math works
 
What do you mean?
Usually the types of processes I have done are either isotropic/adiabatic/isentropic
 
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
 
ok
how would you find the heat take from the system
 
it might not be particularly easy to compute this integral, but there's no conceptual problem
 
when it drops in entropy and also changes temperature
 
4:00 PM
well, you need to describe the actual path through state space
because heat is not a state function, after all
 
no it is not
 
so just saying "these things change" is not enough information
 
but if the process is reversible you can use the $\delta Q = T dS$
expression
yes
I am aware
that you need additional information
 
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
 
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?
 
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
 
@DIRAC1930 "bounded below" means that a function is never below that bound for any input
 
So if the classical field is bounded below, that means that the quantized field must be too right?
 
yes, but classical fields aren't bounded below
 
4:50 PM
The classical field isn't the important part, it's the Hamiltonian
 
classical fields are just (perhaps compactly supported) smooth functions
they're not "smooth functions that never exceed the value X"
 
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
 
Why do people use scalar Yukawa theory then?
 
Why not
It's very much bounded from below
 
@DIRAC1930 that's a weird question
 
4:53 PM
At $0$, if properly renormalized
a good value to be the bottom of the scale
 
are you implying Yukawa theory is not bounded below?
 
You can check whether a theory is bounded from below via a physical process called "look at it"
 
No I'm just asking a question because at first glance the interacting part of the Hamiltonian seems to approach $-\infty$ as $\phi\rightarrow -\infty$
 
The potential energy of Yukawa is like $\phi^2$
 
@DIRAC1930 sure, but it's linear so it's dominated by the $\phi^2$ mass term, which is bounded below
 
4:55 PM
a very bounded from below quantity
 
@ACuriousMind Ah yes, I never thought about that
 
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
 
But what if $\psi^\dagger \psi \phi$ dominates the $\psi^\dagger \psi$ and $\phi^2$ terms?
 
what do you mean "what if"
the boundedness of a function is not a question of imagination
 
If ifs and buts were candies and nuts
there are too many orthogonal groups
 
5:00 PM
@ACuriousMind You didn't consider all the possible terms though
 
I can think of like 5 or 6 groups that are somewhat orthogonal
 
$(\psi^\dagger\psi +1)\phi + \phi^2$ 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.
 
ah, yes, it's not bounded below in $\psi$
 
oh no
 
thing is, perturbative QFT only expands around a vacuum anyway, so stable local minimum is quite enough
 
5:12 PM
not true for Lorentz manifolds apparently, though
 
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
 
Is that related to the uuuuh uniformization theorem
 
the Higgs famously has (or had, before we detected it) a range of possible masses where the running coupling just makes the SM unstable
 
Apparently the model space for Lorentzian conformal structures is the Einstein universe?
I barely remember that one
nobody cares for that poor thing
 
@ACuriousMind Okay thanks
 
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?
 
5:41 PM
If a particle is characterized by a z- component of Spin, can it also have an x component ?
 
that's the wrong question
what do you mean by the particle "having" an x-component of spin?
 
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?
 
sure
it's a Stern-Gerlach experiment, what you find is that the beam splits in half
 
but in my example
yeah I thought that
Nothing is said about an external magnetic field
 
what would that matter?
 
5:52 PM
because it should?
 
sure, S-G does the measurement via an external magnetic field, but how the measurement works is completely irrelevant
 
Ok
but the ideal was that we are having this set up
and
 
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
 
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*
 
@imbAF why would the result of a measurement of x-spin be a |z,+> state
 
5:54 PM
Honestly I don't understand the question
if the spin component of Z is +1/2
 
|z,+> is your input state for the measurement, it's not the result
measurement changes the state!
 
if
the state is a superposition or mixed
how would you change a eigenstate?
 
|z,+> is not an eigenstate of x-spin, is it?
 
no
But at the same time the components of spin do not commute
 
"However, the structure group of gr(TM) = TM is reduced to G0 = CO(r,s)."
Have mercy Cartan
What is even CO
 
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
 
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
 
@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
 
@ACuriousMind Obviously you've never been in a lab
that's what you get most of the time
 
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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