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Bernardo Meurer
Bernardo Meurer
10:00 AM
#marriagematerial
 
user228700
@BernardoMeurer Haha, nice! :-) I know literally no one in those places though.
 
John Rennie
John Rennie
Are Indian students expected to get part time jobs while at university?
 
Avantgarde
Avantgarde
Nope.
 
user228700
^
 
John Rennie
John Rennie
So basically you have no money until you graduate and get a job?
 
user228700
10:04 AM
@JohnRennie Basically. Well, I will have what little money my parents give me for the most basic of my needs but I will try and save even that!
 
Dawood ibn Kareem
Dawood ibn Kareem
@JohnRennie No, she has Rs 5000. Concentrate!
 
user228700
@DawoodibnKareem x'D
 
John Rennie
John Rennie
@Avantgarde I used to be a Rush fan back in the Caress of Steel, 2112 and farewell to Kings era, though in the 80s they left me behind a bit.
 
Avantgarde
Avantgarde
I agree with that sentiment, John
 
Secret
Secret
I also have second thoughts about the notion of "propagating CTCs" It might be possible to have something like that without changing the spacetime topology (or at least not violating the uniqueness of values of any point in spacetime). Consider the following diagram and my guess on a possible metric that describes it:
 
Avantgarde
Avantgarde
10:05 AM
2112 and Farewell to Kings were so nice
Hemispheres too
 
John Rennie
John Rennie
@Kaumudi.H you need to be careful about living too meanly as a student. It's hard work and you do need to spend a bit on treats for yourself or it'll become too much of an ordeal.
 
user228700
Dammit, I've already done everything I could possibly do while procrastinating revision. Now what?
 
John Rennie
John Rennie
Why do you have to revise?
 
Dawood ibn Kareem
Dawood ibn Kareem
@Kaumudi.H Don't worry about the bragging. Nobody here will resent you for it. I'm glad I saw that bit before you edited it out.
 
user228700
@JohnRennie But Darjeeling!
 
user228700
10:07 AM
@JohnRennie Not now, before!
 
user228700
@DawoodibnKareem ...why?
 
John Rennie
John Rennie
@Kaumudi.H ah, I see. You mean you've already explored all the options within your budget in the days when you were desparately avoiding doing any work for the exams :-)
 
user228700
@JohnRennie Exactly :-P I have nothing to do now.
 
user228700
Oh, wtf, u removed that.
 
John Rennie
John Rennie
OK, poor tapestry :-)
 
Dawood ibn Kareem
Dawood ibn Kareem
10:09 AM
@JohnRennie LOL
 
John Rennie
John Rennie
In the financial sense I mean
 
user228700
x'D Yes.
 
John Rennie
John Rennie
When I was your age I spent all my free time hacking computers
 
user228700
@JohnRennie Eh :-P I actually think I'll get started on cleaning my desk and drawers...
 
user228700
Dusty and filled with many books I won't need anymore...
 
John Rennie
John Rennie
10:10 AM
I once wrote a basic CAD system on a Vax 11/780 - in Vax assembler!
 
user228700
@JohnRennie I have no idea what any of those words mean :-/
 
Dawood ibn Kareem
Dawood ibn Kareem
See, Kaumudi, that's bragging!
 
John Rennie
John Rennie
You youngsters!!
 
user228700
I only understood that u used assembly language; and even of that bit I'm not sure :-|
 
John Rennie
John Rennie
VAX-11
The VAX-11 is a discontinued family of minicomputers developed and manufactured by Digital Equipment Corporation (DEC) using processors implementing the VAX instruction set architecture (ISA). The VAX-11/780 is the first VAX computer. == VAX-11/780 == The VAX-11/780, code-named "Star", was introduced on 25 October 1977 at DEC's Annual Meeting of Shareholders. It is the first computer to implement the VAX architecture. The VAX-11/780 central processing unit (CPU) is built from transistor-transistor logic (TTL) devices and has a 200 ns cycle time (5 MHz) and a 2 kB cache. Memory and I/O are accessed...
 
user228700
10:12 AM
Wokay, I see...
 
user228700
(only sort of)
 
John Rennie
John Rennie
A 5MHz CPU. 5MHz!!!!
 
Slereah
Slereah
That computer isn't mini at all!
 
John Rennie
John Rennie
It was compared to the IBM 3081 that Cambridge had in the 80s :-)
 
user228700
@JohnRennie You computer nerds!!!
 
Slereah
Slereah
10:15 AM
please, this is a minicomputer
 
John Rennie
John Rennie
IBM 308X
The IBM 308X was a line of mainframe computers, the first model of which, the Model 3081 Processor Complex, was introduced November 12, 1980. It consisted of a 3081 Processor Unit with supporting units. Later models in the series were the 3083 and the 3084. The 3083 was announced March 31 and the 3084 on September 3, both in 1982. The IBM 308X line introduced the System/370 Extended Architecture. All three 308X systems, which IBM had marketed as "System/370-Compatibles," were withdrawn August 4, 1987. == IBM 3081 == The initial 3081 offered, the 3081D, was a 5 MIP machine. The next offering, the...
 
Slereah
Slereah
MSDOS watch with tiny floppy disks
This is the future
 
Dawood ibn Kareem
Dawood ibn Kareem
@JohnRennie Let me guess. You had to walk 40 miles in the snow to deliver your punched cards.
 
John Rennie
John Rennie
Reduced power consumption, 23 kilowatts - its predecessor consumed 60kW!!!
 
user228700
@DawoodibnKareem x'D
 
John Rennie
John Rennie
10:17 AM
@DawoodibnKareem I did use punched cards in my first year at university
In fact I wrote an adventure game mostly using punch cards
 
Slereah
Slereah
what was it about
also how were loading times
 
John Rennie
John Rennie
On the IBM3081 it was pretty fast. But it was a text only adventure like Colossal Cave
 
Slereah
Slereah
The mini computer craze I remember was the electronic agenda in the 90's
Every businessman needs one!
To write your business on it
It was completely useless
It basically offered no benefit over a paper agenda
 
John Rennie
John Rennie
I've still got my Psion S3.
I don't know if it still works. It hasn't been turned on for 25 years.
The Psion was good for its day and it was genuinely useful. I used to write up notes on it while on the train, then transfer them to my PC.
 
Slereah
Slereah
Graphics!
Also pagers were a big things
90's pagers
to send your 90's text messages
"For example, in the case of the bosonic open string theory in 26-dimensional flat spacetime, a general element of the Fock-space of the BRST quantized string takes the form (in radial quantization in the upper half plane)"
help
 
Dawood ibn Kareem
Dawood ibn Kareem
10:26 AM
@Slereah Ha! I had one that only did numbers.
 
Slereah
Slereah
I think I need to learn about BRST quantization
 
user228700
@JohnR: What dyou think:
 
user228700
 
Dawood ibn Kareem
Dawood ibn Kareem
There is no harm in applying. If you don't enjoy the job, you can quit.
 
John Rennie
John Rennie
Data entry is a spectactularly boring job. My brother did it for a while. But if it's only to raise some money in the summer then I'd say go for it.
 
Dawood ibn Kareem
Dawood ibn Kareem
10:29 AM
But in the future, when you are looking for a better job, that would be a nice thing to have on your CV.
 
user228700
I have only about two months though, I hope that won't be a problem...
 
user228700
I'll ask my parents when they're back and decide what to do!
 
Dawood ibn Kareem
Dawood ibn Kareem
It's good to get into a rhythm of working and earning money from a young age, no matter how menial the work you do.
If I were your father, I'd say, yes, do it.
 
user228700
I guess, yeah. I better do some more searching to check for better jobs for me here...
 
Dawood ibn Kareem
Dawood ibn Kareem
As a way of procrastinating your revision still further?
 
user228700
10:32 AM
@DawoodibnKareem Nah man, I'm on vacation!
 
Secret
Secret
If I had not mistaken... the idea would look something like this:
 
manshu
manshu
@Kaumudi.H then why work? :p
 
Secret
Secret
 
user228700
@manshu I'm trying to save up so that I can visit Darjeeling as soon as possible :-) Besides, there's next to nothing to do at home, despite the copious amounts of work I put into making a comprehensive to-do list (while procrastinating :-P)
 
Secret
Secret
So the idea is that in O's frame, taking a xyzt volume with height in t corresponding to the period of the CTC, and translating this xyzt volume every 1 proper second, one should end up with a CTC basically slowly moving in the x direction
(For O' frame which is moving at some velocity wrt O, the whole spacetime diagram will be distorted accordingly, thus all timelike observers should agree that the trajectory of the CTC is timelike)
(O wait sorry, mistake, forgot to add the bump function to localise the CTC)
 
manshu
manshu
10:35 AM
There are still too many places to visit for me. Best of luck. :)
 
Secret
Secret
$$ds^2=-\Omega_{[t-R,t+R]} R\sin \theta dt^2+vdx^2+dy^2+\Omega_{[t-R,t+R]}R\cos \theta dz^2$$
The bump function ensures the CTC only exists at a radius $R$ away from the line t=vx and that the CTC smoothly fade into flat spacetime as we move further away from said line
 
Slereah
Slereah
$\Omega$ isn't a bump function
I assume you mean $1 + \Omega$
Otherwise the metric becomes degenerate
 
Secret
Secret
err, what symbols should I use for a bump function that is nonzero only between t-R and t+R?
the standard bump function is defined to be nonzero for all $|values| \leq 1$
so I guess I need to shift it somehow... not sure about the notation
 
Slereah
Slereah
Omega is fine, but if it's a bump function your metric will be $ds^2 = vdx^2 + dy^2$ outside of the support
 
Secret
Secret
ah yes, then $dt^2$ will be missing, right
 
Slereah
Slereah
10:49 AM
Also there are no CTCs in this metric
 
Secret
Secret
Hmm... guess I will need to try again much later... The diagram illustrates roughly what I had in mind for the observer O, however I need to figure out how to account for the - sign in the time component to construct a metric so that it includes a CTC that is on the zt plane (if that makes sense)
and that CTC will be "moving" in the x direction at some velocity v (it looks that way when viewing against flat spacetime)
 
Slereah
Slereah
If the metric tensor is diagonal in $\Bbb R^n$ there are never any CTCs.
 
Secret
Secret
so I am guessing if I want a CTC on the zt plane (as seen by O, and transformed accordingly for other observers) will require adding the cross term $(some function)dtdz$ into the metric?
 
Slereah
Slereah
yes
also actually checking that there are CTCs
Find a piecewise timelike curve that is closed
Hm, this paper has the opposite problem of the usual Lagrangian problem
Instead of overusing functional derivatives as $$\frac{\partial L}{\partial \phi}$$, he underuses them and writes $$\frac{\partial S}{\partial \phi}$$
Or does he
 
Secret
Secret
$S$ is an action?
 
Slereah
Slereah
11:04 AM
yes
 
Secret
Secret
If I recall, field lagrangians are usually integrated wrt the fields $\phi$ themselves to give the action. Not sure if that rule still holds for string theory
 
ACuriousMind
ACuriousMind
11:24 AM
@Secret No, you integrate a Lagrangian density over spacetime to get the action. Integrating "with respect to the field" would be a path integral.
 
Slereah
Slereah
@ACuriousMind Is the gauge function just a projection of orbits in the associated bundle to a single point of the orbit
the function $\mathcal{F}^\alpha (\phi) - f^\alpha = 0$
 
ACuriousMind
ACuriousMind
Yes, the idea is that a "perfect" gauge function selects precisely one representant from each gauge orbit
 
Slereah
Slereah
IIRC we rarely use such perfect gauge function
 
ACuriousMind
ACuriousMind
If you have "residual gauge symmetry" it selects more than one (you might say it selects one "suborbit", which is now an orbit under the residual symmetry), and more generally Gribov ambiguities often prevent such choices from being globally possible
 
Slereah
Slereah
Lorentz gauge is kind of overnumerous still, right?
If we have residual gauge symmetries but the path integral still converges, are they all equivalent?
 
ACuriousMind
ACuriousMind
11:31 AM
The path integral doesn't like residual gauge symmetries, it should be completely fixed.
 
Slereah
Slereah
So the spooky ghosts fix the gauge completely?
 
ACuriousMind
ACuriousMind
Otherwise you're "overcounting" the physical states, and the path integral is supposed to integrate over physically distinct configurations. I don't think that convergence is really the reason we gauge-fix the path integral - reducing it to an integral over physically distinct configurations is.
@Slereah They should, yes. There are different ways to do this, some "gauge-fixed versions" average over many possible gauge choices, some make a particular kind
 
Slereah
Slereah
Do the spooky ghosts fix it completely in classical field theory, too?
Seems weird because for EM, the spooky ghosts are independant of the field
 
ACuriousMind
ACuriousMind
That's not what they do in classical FT, no
They fix the gauge in the path integral because you introduce them via the Faddeev-Popov trick explicitly to rewrite the FP determinant that comes from the gauge fixing.
 
Slereah
Slereah
Alright
Is there any big problem about residual gauge in classical field theory?
 
ACuriousMind
ACuriousMind
11:36 AM
If you do BRST and also startintroducing them on the classical level, the purpose of the introduction of the ghost is quite different
They "fix" the gauge only in the sense that the BRST cohomology on the algebra of both ghosts and "normal" dynamical variables computes precisely the gauge-invariant observables, but no gauge choice is made
Then, in the path integral, there still appears something called a "gauge-fixing fermion", which ends up yielding the same terms that the fP trick gave
(At least, that's what I recall from Henneaux/Teitelboim without looking it up)
 
Slereah
Slereah
Apparently a full gauge for EM is $\nabla A = 0$ and $\chi = 0$
$\chi = 0$ fixes a section of the principal bundle I guess
 
ACuriousMind
ACuriousMind
What is $\chi$?
 
Slereah
Slereah
The gauge $A \to A + d \chi$
 
ACuriousMind
ACuriousMind
The condition $\chi = 0$ makes no sense to me, then.
 
Slereah
Slereah
1
Q: Residual Gauge Freedom

Gaurav KatochHow are we still left with one Residual Gauge Freedom in the choice of Electromagnetic Potential after having already exploited the Gauge Freedom once. As is mentioned in Halzen and Martin Section 6.9. Doesn't choosing another scalar to induce Gauge Transformation spoils the Lorenz Gauge condit...

electromagnetism gauge-theory degrees-of-freedom gauge
Sayeth this guy
 
ACuriousMind
ACuriousMind
11:42 AM
You need to pose all gauge conditions as a position on the actual fields. If I hand you a random configuration $A$ with $\nabla A =0$, how are you going to decide whether it fulfills $\chi = 0$ or not?
 
Slereah
Slereah
What's the classic "full" gauge condition for EM?
Is it fixing one of the component?
I sometimes see fixing of $A_0$ popping up
 
ACuriousMind
ACuriousMind
Yeah, $\nabla_i A^i = 0, A_0 = 0$, for instance, is a complete gauge
 
Slereah
Slereah
Thx
I guess there is some implicit gauge fixing when you solve the Maxwell equation
 
ACuriousMind
ACuriousMind
There should be others, I can't recall any off the top of my head, though
@Slereah You usually fix a gauge to get a unique solution and to make certain annoying terms vanish, yes
 
Slereah
Slereah
The initial conditions of the Maxwell equation is the "gauge fixing" I suppose
 
ACuriousMind
ACuriousMind
11:48 AM
Which terms are the "most annoying" of course depends on your application
 
Slereah
Slereah
well, the residual gauge fixing
Since for a fixed $A_0$ you'll get a unique $A(t = 0)$ I guess
 
ACuriousMind
ACuriousMind
> [1] private correspondance
Well played :P
 
Slereah
Slereah
I suspect that the single worldline is gonna be some totally delocalized momentum state
Basically a single Fock state $| p \rangle$
 
ACuriousMind
ACuriousMind
Yeah, the resulting Hilbert space is traditionally exhibited as being spanned by momentum states
 
Slereah
Slereah
Although I'm not sure that makes sense
Since those states are typically not normalized
Maybe it is if we consider a single state, but I dunno
Also is there a "string Fock space"?
Since the worldsheets are basically Feynman diagrams are there asymptotic string spaces that are just string number occupation states
Vacuum + 1 closed string + 1 open string + 2 closed string etc etc
 
ACuriousMind
ACuriousMind
11:59 AM
@Slereah Yes, although it has more than one sort of excitation. You should see that after stepping through the quantization of the string when whatever you're reading discusses the string spectrum
@Slereah No, not like that
 
Slereah
Slereah
Yeah from what I've seen there's the mass and angular momentum sort of excitation
 
ACuriousMind
ACuriousMind
The states do not correspond to a "number of strings". You build the string spectrum on a fixed number of strings.
 
Slereah
Slereah
Well yes, there is the spectrum of individual strings, but is there also states for every possible number of strings?
 
ACuriousMind
ACuriousMind
Then you have things like "vector-like excitation on string a", "massless scalar excitation on string c", etc.
@Slereah Well, since the strings don't interact with each other at that level, it's just the direct sum of the states of the individual strings
The main important difference in spectra you can have is open vs. closed, and different GSO projections.
 
Slereah
Slereah
But the full path integral isn't on individual strings, though
It's the sum over every interpolation between every boundaries
 
ACuriousMind
ACuriousMind
12:02 PM
No, it's not. There's a rule for how to insert the states from individual strings as vertex operators, though
 
Slereah
Slereah
Oh
I guess I need to read more strings
At least it's starting to make more sense
 
ACuriousMind
ACuriousMind
In the end, there will be a rule for how to compute the "string S-matrix" that tells you how to compute the scattering amplitude for the states X,Y,Z on string configuration A turning into states U,V,W on a possibly different string configuration B
 
Slereah
Slereah
I tried to look up some string field theory but it's some spooky shit
The wavefunction equation was nonlinear
That's illegal
 
ACuriousMind
ACuriousMind
And the marvelous thing that convinces some people that there is something interesting here is that the tree-level computations of that S-matrix look precisely like the amplitudes one gets from the 10d SUGRA theories.
 
Slereah
Slereah
Hm, sugar theory
The dimension of the target space in the string theory is the linear sigma gauge, right?
 
ACuriousMind
ACuriousMind
12:06 PM
So string theory may be seen as "higher order corrections" to these theories of quantum gravity, and unlike their perturbation series, the string perturbation series is renormalizable/"UV-finite"
@Slereah I do not understand the question - the dimension of the target space is not necessarily fixed prior to quantization, and it's just the number of the scalar fields $X^\mu$ on your worldsheet
 
Slereah
Slereah
Yeah, but those fields are related by a linear sigma model, no?
 
ACuriousMind
ACuriousMind
Once you quantize, you find that bosonic ST is only consistent with 26, and superstring theory is only consistent with 10 dimensions.
 
Slereah
Slereah
Gauge invariant under the Poincaré group of the target space
Hence why there are $n$ fields
The fields $X^\mu$ are invariant under $\Lambda^\mu_\nu X^\nu + a^\mu$
At least for a flat target space
 
ACuriousMind
ACuriousMind
@Slereah There's no gauge invariance - the Poincaré symmetry is global, not local
 
Slereah
Slereah
Ah right
How does string theory deal with a non-topologically trivial target space?
 
ACuriousMind
ACuriousMind
12:11 PM
Which is why it is an interesting accident that ST produces graviton-like states whose amplitudes correspond to a SUGRA where the Lorentz symmetry is local.
@Slereah What's there to deal with - then you don't have full Poincaré symmetry, obviously, but it's just some global symmetry to begin with, it has no particular importance for string theory. The story of what ST on non-trivial targets does is basically the story of compactifications, which is very interesting and also very broad by now
 
Jim
Jim
This is a public service reminder: One more day until the first annual Physics Pun Day. I hope you have all been working on your best awful puns for tomorrow. I know I (should) have
 
Slereah
Slereah
Alright, thanks
That should be enough string theory questions for today
2
Q: Lorentzian path integral for string theory and causality

SlereahIs the Lorentzian path integral in string theory well defined, as opposed to the usual Euclidian path integral that is commonly used for simplicity? The path integral is roughly $$\sum_{\mathbf{\mathcal M \in Top}}\int \mathscr DX \mathscr Dg\ e^{i\lambda_{\chi}[\mathcal M ]}\exp{\lbrack i \int ...

general-relativity string-theory metric-tensor causality wick-rotation
except that one, of course
I'm pretty sure string theory is a swindle
 
ACuriousMind
ACuriousMind
Aren't you pretty sure all of physics is a swindle? ;)
 
Slereah
Slereah
Well most of it is full of cheats
Are any string path integral even solvable
Except maybe for the trivial ones
open to open and closed to closed
 
ACuriousMind
ACuriousMind
@Slereah People computed them to compare them to other amplitudes, so yes. It's kind of a Herculean effort, though, you'd have to look up the actual papers that did that because I don't think many other people have ever drudged through that.
 
Slereah
Slereah
12:23 PM
I'll bet
The integral over $X$ is probably fine but the integral over $g$ sounds like a hassle
And that's without all the weird gauge
 
ACuriousMind
ACuriousMind
There's also the issue that computing the higher order corrections in flat space - or any other target space - is pretty pointless since we don't know yet which of these to choose to get predictions that in principle could relate to the real world.
So much of ST is focused on the phenomenology of the effective QFTs you get for different target spaces in order to know which one to choose
 
Slereah
Slereah
The solution may give hints to generalities
Who knows
Hey @yuggib
 
yuggib
yuggib
@Slereah o/
 
Slereah
Slereah
Why were you bothering me about $\Pi = -i\delta / \delta \phi$ when that's 100% valid
I checked, the definition of $\Pi$ can be expressed that way or the other way
The only difference is the measure you use
 
yuggib
yuggib
what is $\Phi$ in the representation given by the measure where $\Pi= -i\delta / \delta \phi$?
are they unitarily equivalent?
 
Slereah
Slereah
12:29 PM
The same I think, but then the product of two states has a different measure
 
yuggib
yuggib
@Slereah this does not make sense
 
Slereah
Slereah
One has a gaussian measure and the other I forget what
Lemme see
 
yuggib
yuggib
chatjax does not work me today
 
Slereah
Slereah
$\phi = x_k$, $\pi = -i \partial_{x_k}$ for $L^2(\Bbb R^n)$ and $\phi = x_k$, $\pi = -i \partial_{x_k} + i x_k$ for $L^2$ with the measure $d\mu_k = \pi^{-1/2} e^{-x_k^2} dx_k$
Sayeth Simon & Reed p. 229
The first one is the representation that a lot of QG stuff uses
 
yuggib
yuggib
but it does not hold in infinite dimensions
 
Slereah
Slereah
12:42 PM
Hm
 
yuggib
yuggib
in infinite dim, there is no Lebesgue measure
 
Slereah
Slereah
Oh, that's why
Does it mean that Rovelli's book is all wrong
 
yuggib
yuggib
or that it is all physical
 
Slereah
Slereah
same thing
 
ACuriousMind
ACuriousMind
::sobs quietly::
 
yuggib
yuggib
12:43 PM
if there were a lebesgue measure in infinite dimension, everything would be trivial :-D
 
Slereah
Slereah
for instance p. 132
What is the dealio
It's not the only author to use that convention
Are the results equivalent?
It's a difference of $- \phi^2$ in the Hamiltonian
 
yuggib
yuggib
I don't really know what Rovelli writes
 
Slereah
Slereah
He uses $$\pi(x) = -i \hbar \frac{\delta}{\delta \phi(\vec x)}$$
 
yuggib
yuggib
The fact that he uses it, does not make it correct
 
Slereah
Slereah
yes, but
why does he, is the question
Do the results coincide if you use the two different conventions?
 
yuggib
yuggib
12:51 PM
well, because heuristically it gives nice formulas that probably work more or less well
 
Slereah
Slereah
It seems like if you added $+ i \phi(\vec(x))$, it would shift the mass term to $m^2 - 1$
 
yuggib
yuggib
it is not a matter of conventions, it is a matter of one thing making sense (mathematically), the other not
 
Slereah
Slereah
Which sounds important enough
 
yuggib
yuggib
it is more a problem of what you want to do. If you want to represent the canonical commutation relations, then they have to be satisfied
 
Slereah
Slereah
I guess I should check by hand if they give the same observables or not
I don't doubt that it's more mathematically correct, I just wonder if they give the same physical observables
 
yuggib
yuggib
12:53 PM
without defining a measure for your $L^2$ space, it is going to be difficult to tell
 
Slereah
Slereah
That never stopped physicists :p
 
yuggib
yuggib
if you want to do physics, then you can avoid to care altogether
 
Slereah
Slereah
Well I want to do it rigorously, but if everyone does not do it, I need to see why that work, and how that translates to as rigorously
 
yuggib
yuggib
it "works" in the measure on which it gives correct predictions (so I doubt it "works" for LQG)
 
Slereah
Slereah
Hm
Let's see if Rovelli uses the correct operator for LQG at least
He writes the momentum operator of the 3D gravitational connection as $$\frac{\delta}{\delta {A^i}_a(\vec x)}$$
Was LQG a scam all along???
 
Moses
Moses
1:05 PM
@ACuriousMind Hi, can you maybe see how we obtain the following integral, where $|x' \rangle$ are eigenstates of the position operator $x$:
$$\int_{-\infty}^{\infty}e^{-(x'-c)^2}|x' \rangle \langle x'| dx' = e^{-(x-c)^2}?$$
 
Slereah
Slereah
1:16 PM
Maybe I should stop trying with Simon & Reed and check the authors in the Notes instead
Simon & Reed gives me a headache
 
user228700
@JohnR: YHM! (Erm, I mean a message on gchat)
 
SBM
SBM
hello
 
Slereah
Slereah
Apparently the original paper is "Tensor algebras over Hilbert spaces"
Let's check it out
"There are many other realizations of Q space where the "points" differ, but the algebra of measurable sets and the measure are the same."
Hm
When you have a graded Hilbert space, does it imply some selection rules?
 
heather
heather
hello
 
Slereah
Slereah
$$f \to -i \sqrt{2c} \frac{\partial f}{\partial \xi_\mu} + i \frac{1}{\sqrt{2c}} \xi_\mu f$$
I think that's the one
 
Avantgarde
Avantgarde
1:40 PM
sup @heather
 
Slereah
Slereah
good old Jaffe
although that book isn't complete
 
Avantgarde
Avantgarde
mathematically rigorous treatment of QFT. I like it
always a pain to get used to new notation
 
yuggib
yuggib
@Slereah yeah, that and the nelson paper "Free Markoff field" that came later
anyways, with Segal you can't go wrong
 
Slereah
Slereah
Seems like a headache too, but I'll try :p
That domain seems to not have a middle ground
Either it's physicists that do it all on gumption or math people that do it all abstractly
What happens exactly if you try to use $\delta/\delta \phi$ on the $L^2$ space with gaussian measure?
Both obey the CCR so the problem isn't from there
 
yuggib
yuggib
@Slereah there is probably a problem with self-adjointness
 
Slereah
Slereah
1:50 PM
Ah, could be
Hm, what would it be
 
yuggib
yuggib
you change the scalar product so the "integration by parts" (or the infinite dimensional analogue) would change to take into account the different measure
 
Slereah
Slereah
I think I get it
There's a similar thing that happens with the QM operator for momentum in curved space
$p = \partial_x$ isn't self adjoint because of a change in the measure
You have to switch to $p_a = \partial_a + \Gamma_a$
I guess I should check if for Rovelli, $(\Psi, \pi \Psi)$ is real
And then check for $\pi + i \phi$
 
ACuriousMind
ACuriousMind
@Moses Since the l.h.s does not contain any $x$, but the r.h.s. does contain $x$, that integral does not seem to make any sense. In case the $x$ on the r.h.s. is an operator, that there is just the continuous analogue to a matrix being diagonal in its eigenbasis. Take care though that most mathematicians wouls still recoil in horror from that integral :P
 
Slereah
Slereah
@yuggib thanks
 
Emilio Pisanty
Emilio Pisanty
any mathematica'ers around?
0
Q: Why don't Part, Apply and other standard tools work with Region objects

Emilio PisantyConsider the following discretization of a region object: region = DiscretizeRegion[Disk[], MaxCellMeasure -> ∞] Head[region] InputForm[region] which returns something like this: By all appearances, this has created an object region with Head MeshRegion and which contains a bunch of poi...

expression-manipulation regions core-language
I am super confused
 
Secret
Secret
1:59 PM
[Random] A worldbuilding scenario of falling down (Newtonian) indefinitely but bounded:
The displacement $S$ is given by $S=C\tanh (t),t>0$ and $S=C\textrm{arctanh}(t),t<0$

Therefore as $t$ progress, the object fell further, but the displacement decreases thus eventually the object can fall no further

The equation of motion is clearly time symmetric, thus reversing time at $t=t_0$, the object retrace the path it takes to get back to the starting point. This means, if the object have been falling indefintely close to the cutoff for an extended period of time, when time is reversed, it will…
 
yuggib
yuggib
@Slereah no prob ;-)
 
Secret
Secret
(NB If you want $S$ to be a smooth function, then the following differential equation has to be solved: $f'(x)f(-x)-f(x)f'(-x)=0$, which so far I failed)
 
Slereah
Slereah
I guess the question is also "how is quantization performed"
Should the canonical momentum be replaced by the self adjoint operator?
Ashtekar variables aren't real
Quantum gravity is a headache
Who knew
 
heather
heather
why is it required that resource recommendation questions be community wiki?
 
Slereah
Slereah
Because many people can contribute to ressource recommendations
 
heather
heather
2:05 PM
how do bounties work on them then?
 
Slereah
Slereah
Not a clue
Don't worry too much about your internet points though, I'd say
 
heather
heather
because i answered a 200 point bountied resource recommendation question, and it was made community wiki.
 
ACuriousMind
ACuriousMind
@heather I think it was a compromise partly based on the idea that voting on recommendations will not reflect the quality of an answer, but merely how much people like the resource being recommended, and that people shouldn't derive rep just from recommending popular material for which not actual expertise is needed.
@heather I am 95% sure bounties completely ignore CW status.
 
heather
heather
@Slereah what, not care about meaningless points? =/
@ACuriousMind ah, okay.
 
Slereah
Slereah
quite easy for me to say since I have more than you
 
ACuriousMind
ACuriousMind
2:08 PM
@Slereah ...you don't.
 
Slereah
Slereah
Well, more than @heather, not more than you
 
ACuriousMind
ACuriousMind
No, I meant that heather has more than you
Look it up :P
 
Slereah
Slereah
Ah yes
Misread
Well even more reason to not care then!
 
Secret
Secret
0
Q: Do black holes recapture the CMB?

Joe CBasically if I shined a flashlight at a black hole, would I cause it's Hawking radiation temperature increase by more than the temperature of the light I shine at it, at any time during the life of that black hole? Or when CMB photons fall into the black hole are they returned to the universe ho...

thermodynamics cosmology black-holes temperature cmb
Hmm, I am not sure if during the epoch when all black holes started to evaporate, would the universe be temporally in a nonequalibrium state due to the high temperature of hawking radiation compared to CMB before settling into heat death...
(assuming the universe expands forever of course, as the above scenario obviously cannot hold in a closed universe nor a big rip)
[Yet another random thought] Entropy of a big rip well defined...?
 
BAYMAX
BAYMAX
2:32 PM
Hi
Any example of a Chaotic system whose state space structure is not a fractal?
 
Slereah
Slereah
The government
 
BAYMAX
BAYMAX
Physical systems?
@Slereah
 
Moses
Moses
@ACuriousMind So would the discrete equivalent be that for say some operator $A$ we can write $A = A \sum_{a'}|a' \rangle \langle a' | = \sum A | a ' \rangle \langle a ' | = \sum a'
|a '
\rangle \langle a '|$ where the RHS is the operator in matrix diagonal form?
 
ACuriousMind
ACuriousMind
@Moses yep
 
Secret
Secret
> Take care though that most mathematicians wouls still recoil in horror from that integral
 
Avantgarde
Avantgarde
2:39 PM
@Slereah lol
 
Secret
Secret
Ah, no wonder why I cannot make sense of it (except my possible guess it might be somehow related to the gaussian approximation of dirac delta)
 
Moses
Moses
@ACuriousMind I see, and further since operator $\hat{x}$ commutes with and function of $x$ (which can be expressed as a power series) we have that $|x \rangle$ diagonalizes operator $e^{c\hat{x}}$...?
 
ACuriousMind
ACuriousMind
@Moses yep
 
Moses
Moses
@ACuriousMind So we think of the RHS as an infinite matrix then?
 
ACuriousMind
ACuriousMind
It's an operator, so yes (but, again, don't tell the mathematicians about that :P)
 
Moses
Moses
2:47 PM
@ACuriousMind :) This is what makes physics books so hard to work through, but I guess you only have so much time for rigor...
 
Mithrandir24601
Mithrandir24601
@JohnDoe I've never heard of 'resolving power' in that context before, but to me it sounds like it might have something to do with making a measurement and how good your device is at giving the correct results. I may be wrong though. Have you ever heard of von Neumann measurements? That's essentially the track I'm thinking along. I'll be better able to talk about it in a couple of hours, when I'm back in my room
 
Secret
Secret
Ah ok, that makes sense (that is why I like to add those hats when thinking about operators in quantum, especially for things like the RHS to remind myself it is an operator and not just some numerical function)
i.e. I will be less confused if it looks like this:
$$\int_{-\infty}^{\infty}e^{-(x'-c)^2}|x' \rangle \langle x'| dx' = e^{-(\hat{x}-c)^2}$$
 
Mithrandir24601
Mithrandir24601
@JohnDoe I.e. there's a finite probability that your measurement device tells you that it measured one thing, when it actually didn't, but I'm not sure if this is what 'resolving power' means here or not...
You can also get e.g. 'photon number resolving', but that doesn't seem to make sense with what you've written, at first glance
 
Slereah
Slereah
3:35 PM
@ACuriousMind some dude now says that it is the point particle theory
I would help if you could shed some lite on that private correspondence of yours. I see no reason why you shouldn't identify 1-dimensional Polyakov action theory with the relativistic particle. Take a look at my path integral derivation of the Klein-Gordon propagator from the 1-dimensional Polyakov action: solenodonus.com/file/path-integral-for-point-particle.html — Solenodon Paradoxus 26 mins ago
 
ACuriousMind
ACuriousMind
@Slereah See my comment, I think everyone involved needs to be more precise what they actually mean by a "point particle theory" :P
 
Slereah
Slereah
Well in my case I mean RQM
And I think he does too
His link is the path integral of a relativistic point particle
Which IIRC kinda gives KG but formally it diverges
Due to the Problems
 
ACuriousMind
ACuriousMind
@Slereah As you can see his link regularizes, so what's the issue?
 
Slereah
Slereah
Yeah, I think you can regularize it away for scalar fields
It's equivalent to the Dirac sea IIRC
 
ACuriousMind
ACuriousMind
So, uh, what's your question exactly, then?
 
Slereah
Slereah
3:44 PM
Same as before, i suppose
Is the worldline of one particle formalism equivalent to RQM or something else?
RQM should be localizable, I think
I wonder how well the relativistic point particle + Dirac sea behaves on GR
 
Slereah
Slereah
4:19 PM
Relativistic point particle is such a hack, really
Take an unbounded from below Hamiltonian and fill it with garbage
infinite garbage
 
Secret
Secret
4:36 PM
I would like a quantum gravity that make absolutely no reference to particles except at the non relativistic limit
 
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