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12:35 AM
@dmckee Are you around?
 
12:52 AM
@SirCumference if 'ton' is >2k, no
Why?
 
I guess I can use comments if I'm iffy on a suggestion
 
The idea is that after 2k got should be clear on what is and is not an acceptable edit
If you're not sure, it's probably better to just leave it in a comment
 
@0celo7 wot wot
 
1:39 AM
Can someone check my numbers here:
1
A: Can you measure temperature using blackbody radiation and a 13GHz receiver?

DanielSankI think this has no chance to work because the power emitted by black body radiation is so low at the frequencies and temperatures you're working with. The power spectral radiance of black body radiation at frequency $\nu$ and temperature $T$ is $$B(\nu) = \frac{2 h \nu^3}{c^2} \left( \exp \le...

I think I may have misplaced a factor of ten thousand...
Come on, someone.
Hey astro people (@DavidZ ?), how is it possible to observe the CMB?
 
2:03 AM
why would it not be
 
2:45 AM
@Slereah The power emitted by a 2 Kelvin source in the radio band is not very high.
At 10 GHz, for an object of one square meter area, you get -150 dBm/Hz radiated into the whole $4 \pi$ solid angle.
Hmmm, I guess that's not crazy.
An off the shelf spectrum analyzer's noise floor is -140 dBm/Hz.
I guess with a half-decent preamp you can measure the CMB at 2 Kelvin.
How do you think of the area of the black body emitter than is the CMB, @Slereah @SirCumference?
 
3:12 AM
@DanielSank help
Can I put this in a PSE answer or is it unprofessional
 
3:36 AM
Why are you contracting a loop?
 
4
Q: Stings in the loop space of timelike curves

SlereahIn Smith's paper on homotopy groups for Lorentz manifolds, he builds the loop space of all timelike loops in the following fashion : Consider all piecewise continuous timelike curves which start and end at point $x$. This include timelike curves with $q$ changes in time orientation (the tangen...

 
4:02 AM
An electron placed in a uniform magnetic field with a strength B that has a velocity v perpendicular to the field will travel in a circle. How would you determine the radius of the circle?
 
@micsthepick set the Lorenz force equal to the cetrifugal force
qvB=mv^2/r
 
r=mv/q/B
 
seems legit
 
@0celo7 Shouldn't it be the centripetal force?
 
@micsthepick Which reference frame do you want to use? Decide that first.
 
4:11 AM
@micsthepick I don't remember/care about the difference
You just set it equal to $mv^2/r$ because $qvB$ is what's making it go round the circle
 
If you are using something like ground frame, then use centripetal force
If you are using the electron's FOR then use centrifugal force
 
@AccidentalFourierTransform Haha you got destroyed
 
Remember that centripetal force is a real force towards the centre which is provided by $q(\vec{v} * \vec{B})$
 
@2017 what is $*$ supposed to be there?
you want $\times$ \times
 
Centrifugal force on the other hand is a pseudo force.
@micsthepick
 
4:19 AM
@2017 Thats what I was thinking
 
@micsthepick You got it now?
:-)
The difference between centripetal force and centrifugal force is very important to understand...I've seen too many people confused about this!
 
 
6 hours later…
10:04 AM
@0celo7 bra-ket notation "fails" precisely because it assumes that the Hermitian conjugate and the adjoint are the same thing
i.e., if you wrote inner products the usual way, you would write (x, Ay); the operator A is acting on y, without ambiguity
and with A^* being the adjoint of A, you could also write (A^*x, y)
but $\langle x \rvert A \lvert y \rangle$ assumes that $A^* = A^\dagger$
the problem has to do with where these operators are defined
and it can actually lead to wrong calculations sometimes
 
10:28 AM
sooo, why are signs made with red light?
i mean why are STOP signs red?
 
why not
 
..
i think its something related to the properties or red light
 
probably not
 
red light dissipates the least
so shouldn't choosing violet be a better idea?
i mean red light bends the least during refraction
idk
why not violet?
it dissipitates more so it might be easier to see..
but it still isn;t easier to see..
idk can someone clear it for me?
 
It might just be semiotics, really
Just that overall red is associated with stopping and green with going
for historical reasons of some kind
 
10:33 AM
@MartianCactus there is no physical reason. For some reason humans regard red as attention grabbing so signs that you really need to see are always red. Green is a more relaxing colour - maybe because it's the colour of plants.
 
why is red attention grabbing?
 
"In the 19th century, rail yards all used their own signal systems. Then, in 1841, railway bigwig Henry Booth pushed to standardize Britain’s color scheme, suggesting red signal “stop,” white signal “go,” and green signal “caution.” Decades later, after a train plowed through a stop signal when a red lens popped off, causing a fake white “go” light and a nasty crash, officials decided to eliminate confusion by giving green an upgrade."
 
its something to do witth the face that red can be distinguished from a distance unlike other colors due to its properties
tho i dont know the property..
 
@MartianCactus no idea. Ask a psychologist.
 
i read somewhere its related to the properties of light..
 
10:35 AM
"Booth’s use of these colors, in turn, seems to have come from a common practice in heavy industry of the era. Many motors and other industrial equipment had indicator lights that were red when the equipment was stopped and green when it was running. Since people were already familiar with this color scheme and its meaning, it makes sense that it was handed down into the railroad industry and then to traffic lights."
"As for why the early industrialists chose red to mean stop and green to mean go, we can only speculate. In color symbolism, red generally evokes danger and makes it a good choice for a warning signal. Green, on the other hand, is calming. While that doesn’t seem to lend itself well to “go,” it may have been chosen for a more practical reason - it contrasts well with red and is highly visible without being harsh or distracting."
 
11:01 AM
Stupid question. I have $\int_{-\infty}^{E_c - E_g}dE \ \sqrt{E-E_c -E_g} \ (1-H(eV-E))$ with H a heavyside step function (its 1 up to eV and 0 after). But I'm not entirely sure how to evaluate an integral of the step function. Intuitively I'd say I get $\frac{2}{3} (eV-E_c - E_g)^{3/2}$ which is only valid for $eV > E_c - E_g$?
 
Integral of a step function is $\int_{-\infty}^b H(x) f(x) dx = \int_{0}^b f(x) dx $
0
Q: Vacuum state in path integral and QFT

SvKarI don't understand the following derivation of ground state: $$ <x_0,t_0|x,0>=\int_{x_0,t_0}^{x,0} \mathfrak{D}x(t) \mathrm{e}^{iS[x(t)]} =<x_0|\mathrm{e}^{i\mathrm{H}t_0}|x>\tag1 $$ After the Wick rotation and $\tau \to \infty$ it become: $$ <x_0|\mathrm{e}^{-\mathrm{E_0}\tau}|0><0|x>\tag2 $$...

>Fraktur D
 
11:18 AM
@Slereah So then I have a sign mistake, if I;m not mistaken
 
11:46 AM
Fraktur D's, but no \rangle's!
 
12:04 PM
@BenNiehoff what? Hermitian conjugate and adjoint are the same thing.
Only physicists use the word Hermitian conjugate
But they mean adjoint
Of course in the unbounded case you have to worry about domains but it's not like physicists give a shit anyway. And that's not a failure of braket notation
@BenNiehoff example?
Oh, I see what you're saying
Yes, it's an issue
But still, I don't know of a real-life example where you get into trouble
I've been looking for one
 
12:21 PM
@0celo7 Yes, exactly, the problem is with the domains of unbounded operators
and there are actually a lot of those!
 
Ok I'm well aware
I misread what you said
 
I read a paper about this once, but alas, I forgot where I found it or what it was called
 
@BenNiehoff I know of a Hamiltonian of the form p^2/2m+V(x), where V is smooth, that is not self-adjoint. But I don't know if that actually causes any problems
It probably doesn't have a basis of bound states
 
yeah, I forget the exact nature of the problems that can be caused
 
@BenNiehoff I'm going to make a resource request on the topic
Can I count on your +1?
 
12:33 PM
if you explain what that means :)
 
@BenNiehoff you go on a date with me
Or upvote
Either one
Or both!
 
I'm not so keen on long distance relationships
 
[Spacetime] I had a weird spacetime which is represented by light cones as follows (where the arrangement of light cones are the same along the vertical direction, and the trend you saw here is continous in all directions)
Now to figure out how to model it and investigate its properties
 
12:54 PM
$$ds^2=-\frac{1}{x}dt^2+dx^2$$
 
1:09 PM
Yo folks
does anyone round these parts know how to use Gaussian for materials chemistry?
 
grr...I have a collaborator who keeps writing $$ ... $$ instead of \begin{equation}...\end{equation}
 
@BenNiehoff suck it up, it's a perfectly OK thing to do
 
it prevents an equation number from appearing
and it defies regex-based search-and-replace :P
 
@BenNiehoff presumably that's intentional on the part of the collaborator
 
@EmilioPisanty If it is a molecule related calculation, I might be able to help, but if it is extended materials like crystals and frameworks, then I am not sure other than I will use DFT to model them
 
1:13 PM
@Secret I need to reproduce the band structure calculations for a monolayer crystal
(via dft)
 
@EmilioPisanty I'm not sure why that comment was met with animosity, but whatever
 
@BenNiehoff no animosity was meant =)
just making random fun of LaTeX purists
 
I also provided a perfectly good collection of macros which they just decided not to use
and yes, I'm a purist about semantic markup :P
 
I found something related to how to perform solid state calculations such as band structures using g09
You also need to use the PBC keyword to build your crystal in the input file
Otherwise it seems to proceed like an ordinary DFT energy calculation
 
well, I'm mostly a purist about semantic markup
I can't stand \left and \right, because they are usually wrong
 
1:23 PM
I use \left \right when I need a big bracket to surround my equations
e.g. $$\left(2^{2^{2^{2^2}}}\right)$$
 
yeah, I use the explicit sizes
\big, \Big, \bigg
because \left and \right tend to overestimate how much size is needed
 
@Secret is g09 free?
paper explicitly mentions it
but the gaussian on the website looks expensive
 
@EmilioPisanty No, gaussian software generally need a license. If you are doing your research in some institution, then usually the physics and chemistry department already had a license from them. If your research is commercial company based, then either your group or the company has to purchase a license.
 
@BenNiehoff why do you always want an equation number
Maybe it's because I'm a math person but I never number them regularly
It looks silly
Put a (*) or something if you want to refer to it on the page
Or refer to the theorem number or whatever in general
 
@Secret I'm in a research institute but I doubt we have a gaussian license
oh well
 
1:28 PM
3
Q: How to run crystal simulation on GAMESS?

AugI know GAMESS is not best suited for crystal simulation but I am not ready to start learning another package as VASP. As I heard GAMESS is able to do some calculations in a "terminated" crystal (I mean using an ion in the periphery instead of repeating it multiple times). Mostly I need to calcul...

 
@Secret non-periodic is probably a deal killer for this
I guess that does rule out both gamesss
GAMESSs
gamesss
that's a fun word
 
yeah crystal and solid state quantum chemistry are pretty much beyond my current expertise. You can also try asking the people in the chemistry chat. Some people there are more familar with the variety of packages avaiable
 
@Secret yeah, chemistry chat is a good idea
 
@0celo7 Well, first of all, to have an equation without a number, one can use \begin{equation*}...\end{equation*}
but secondly, physics papers are not written in the Definition, Lemma, Theorem style of math papers
 
@BenNiehoff what's wrong with \[ \] like everyone uses
 
1:38 PM
so, it's more important to have equation numbers
and lastly, even if I don't need to refer to an equation in a paper, there's always a chance that someone else might want to
 
@BenNiehoff maybe one should fix that :P
Sounds like a bug
 
I mean, physics isn't math
I've never proved a theorem in a paper; I've only solved problems
\[ \] is at least preferable to $$ $$
but, if you have \begin{equation}, then it's easy for me to change to \begin{align}, or \begin{gather}, etc.
 
@0celo7 who's this 'everyone' that uses the awkward notation?
@BenNiehoff That's the main point, yeah
 
TeX.se
 
In any case, I've definitely had this situation happen a lot, where someone else decided that some expression in their paper was not important enough to have an equation number, and yet I wanted to talk about that equation...it's kind of frustrating
 
1:43 PM
@0celo7 they're all weirdos there
 
What the hell is gather?
 
well, a large fraction of them
 
gather collects multiple lines, but just centers them rather than aligning
 
@BenNiehoff or indeed you yourself when talking about your paper later on down the line
I've had it happen plenty of times
I learned my lesson
equation numbers on everything displayed
 
yep
I thought it looked stupid at first, but I've learned my lesson
if it's important enough to put in displaystyle, it's important enough to reference
 
1:46 PM
\begin{equation} \label{eq:solve}
blahblahblah
\end{equation}
ok it does not work on chat
 
no, MathJax is specifically looking for dollar signs
and in chat, it would be silly to use \begin{equation}
but any decent LaTeX editor has tab-completion, so you don't actually have to type out all of these verbose commands
 
@BenNiehoff out of curiosity, what editor do you use?
 
I never got tab completion to work
 
I use TeXstudio
 
So I stopped numbering equations
 
1:53 PM
I used to use TeXworks, but it didn't play nice with Windows 10 per-monitor display scaling
 
@BenNiehoff why do you use malware?
 
huh?
 
@BernardoMeurer see above.
@BenNiehoff windows is malware
 
2:12 PM
oh, come on
I would run Linux if it worked on the SP4 with ALL hardware features available
and no bizarre battery consumption effects, etc.
 
@BenNiehoff $H=p^2/2m-x^4$ is not essentially self-adjoint on $C^\infty_0(\Bbb R)$
@BenNiehoff I want to know if someone doing a real-life problem (making a pharmaceutical or something) ran into some subtle functional analytic issue and it was a big deal
 
I highly doubt any of quantum field theory makes contact with pharmaceuticals, let alone the subtleties of the domains of unbounded operators
 
@BenNiehoff wat
I don't care about quantum field theory
that field is a mess I wouldn't touch with a mile-long stick
 
haha
 
I'm sure there's quantum mechanics used in biochemistry
 
2:22 PM
I'm sure the only kinds of places where you'll see a problem are things like lasers and condensed matter systems
and collider experiments
 
Yeah quantum chemistry is a thing
Dunno if there's any quartic interaction though
Let's see
 
Doesn't have to be quartic, just something
@BenNiehoff doesn't have to be a domain issue, I'd be interested in any functional analytic issue
For instance, there are Hamiltonians for which $1=\sum |n\rangle\langle n|$ fails
 
Aha!
 
there are quartic interactions in chemistry models
 
2:25 PM
there you go!
chemistry is wrong!
 
maybe aspirin doesn't actually work and it's a placebo
 
clearly
 
@Slereah so have you given up on that homotopy paper?
 
Well i'm not doing it right now
I might get back to it later
But I feel I'll have to just prove everything myself
I guess I'll try to see what happens if I try to use loops without stings
 
Time to buy some books.
oh no, my bank wants me to put in my password instead of my fingerprint
who knows what that is
 
2:27 PM
@0celo7 what will a complex energy eigenvalue even mean if that hamitonian fails to be self adjoint?
 
imaginary molecules
 
@Secret not self-adjoint does not mean the eigenvalues are complex
likewise, self-adjoint does not mean there are any eigenvalues
damn physicists always get that wrong
 
Isn't $x^4$ just divergent or something
 
Hermitian operators always means they have real eigenvalues, no? and self adjoint are the same as hermitian operators unless you go to infinite hilbert spaces?
 
@0celo7 paracetamol always ever
 
2:30 PM
@Secret Indeed, except that Hilbert space is infinite-dimensional, usually
 
@Secret the spectrum is real, but the spectrum and point spectrum (eigenvalues) don't always agree
if your operator has compact resolvent, they agree
99% of the operators in quantum mechanics have compact resolvent
that's why everything works out nicely, despite everything being wrong
how can I take that author's name seriously
 
I knew another guy named Gerd, I think it's not too uncommon
 
@BenNiehoff the last name
 
Grubbs' catalysts are a series of transition metal carbene complexes used as catalysts for olefin metathesis. They are named after Robert H. Grubbs, the chemist who first synthesized them. There are two generations of the catalyst, as shown on the right. In contrast to other olefin metathesis catalysts, Grubbs' catalysts tolerate other functional groups in the alkene, are air-tolerant and are compatible with a wide range of solvents. For these reasons, Grubbs' catalysts have become popular in synthetic organic chemistry. == First-generation catalyst == The first well-defined ruthenium catalyst...
 
that picture is too thin to click, @EmilioPisanty
 
look what I found
 
@Slereah get a better mouse
 
my Apple ® trackpad works just fine
 
2:34 PM
still too thin to read
 
click to magnify
@BalarkaSen btw, how are you doing
 
also this
black magic, that stuff
never ceases to mystify me
 
@0celo7 Not bad.
 
@Secret Finite dimensional spaces are too easy because all the operators are continuous and you can pick bases and whatnot
all the topologies are equivalent, everything is a Banach space
all subspaces are closed
it's just too nice for things to go wrong
all operators are compact
@EmilioPisanty I fail to see what you're so amazed by
 
@0celo7 so all the arrows and stars and check marks in the site come from one single image file
that gets shifted and rescaled by css
 
2:39 PM
is that really more efficient than just having 20 small images?
 
@0celo7 yes
 
> Are you an author of research articles?

Take our author survey for a chance to win a €300 / $300 giftcard!
Dammit
 
as I understand it, it's forced by considerations on the load on the server
i.e. you don't want it to be giving the client each little arrow over and over
 
@0celo7 time to write a paper
 
so why not give the client all 20 images and store that somehow
 
2:40 PM
you want a single cached sprites sheet that you can just send over, and ideally tell the client itself to cache
... or something like that
 
I hate my life. On my circuits test I wrote .4-.686 = .583
How????
 
rip
 
@BenNiehoff unrelated, but if you do $H=H_0+\lambda V=p^2/2m-\lambda x^4$, the perturbation series has zero radius of convergence
 
@0celo7 that's one ugly hamiltonian
 
Yeah I don't know what it would correspond to
very strong repulsive potential
 
2:44 PM
@0celo7 probably nothing?
It doesn't even have well-defined classical propagation for all times
 
it's called the "anharmonic oscillator", and it's somewhat important
 
you get to $x=\infty$ too fast
 
@EmilioPisanty so that's well-known?
 
@BenNiehoff presumably that'd have a plus sign?
 
How do I do a resource rec that doesn't get closed instantly?
 
2:45 PM
well, sign of lambda wasn't specified
 
@0celo7 I'm pretty sure it's in the Classical Symptoms of Quantum Illnesses paper
 
but, phi^4 theory is basically an anharmonic oscillator at every point in spacetime
 
quantum field theory is wrong
all of it
 
yup, it is
 
yes, but it produces such great answers!
 
2:46 PM
if it is correct, then it has nothing to do with what physicists actually do
 
you can't argue with 11 significant digits
 
@BenNiehoff yeah that's the mystery
 
and yes, the fact that the perturbation series with a quartic potential has zero radius of convergence is discussed in the opening chapters of Zee
 
Zee is at home, but I don't remember that
 
and there are good arguments that the perturbation theory of QED has zero radius of convergence as well
 
then why does it work?
 
asymptotic series can appear to converge for arbitrarily long before they begin to diverge
 
what is an asymptotic series anyway
I'm trying to download Zee but the FBI must be attacking the Russian servers
 
it's a series whose first N terms can give you a very good approximation, but which formally diverges if you take N to infinity
 
2:51 PM
@BenNiehoff can you give a simple example?
 
for some N, one reaches a turning point where more terms make the approximation worse rather than better
a simple one is say you want to integrate e^{-x^4} over the real line
 
ok?
 
you can do this with a perturbation series by introducing a parameter, i.e., e^{-\lambda x^4}, and expand in that parameter
then integrate term by term
 
that series will always diverge, no?
 
yes
 
2:53 PM
I mean, how are any of the terms supposed to be finite
 
but, plot the first N finite truncations of it, and you will see what I'm talking about
 
you're integrating polynomials over R
 
no, polynomials times something like e^{-x^2}, I don't remember every detail
 
but...then it's not a taylor expansion in lambda
 
yes it is...you need to expand around lambda = 1
 
2:55 PM
huh
 
I think I am describing it wrong, but this is the example from Zee
 
wtf I don't remember that at all
libgen is broken
 
you can integrate each term, but their sum does not converge
 
argh
 
but, if you plot the first N terms, as a function of N, then you will see them begin to approach the value of the original integral for a while, and then start to drift away
 
2:58 PM
do you know where in Zee it is?
 
I think it may have been Chapter 1
Zee's QFT book, of course
 
yes, I'm looking now
 
it's really quite amazing, and worth plotting in Mathematica
 
is that what you're thinking of?
 
yes
 

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