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3:55 AM
@EmilioPisanty @ACuriousMind
0
Q: Drive frequency for second order quantum transitions

DanielSank$\newcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\newcommand{\bra}[1]{\left \langle #1 \right \rvert}$ $\newcommand{\braket}[2]{\left \langle #1 | #2 \right \rangle}$ $\newcommand{\bbraket}[3]{\left \langle #1 | #2 | #3 \right \rangle}$ Summary How do we show, from first principles, that...

 
4:24 AM
in Mathematics, 33 mins ago, by johnny09
Hi everybody, I created a new room called Social Behavior Mathematics (Game theory). If anyone is interested please join the room and say hi. Cheers.
 
5:08 AM
@DanielSank you've already written the time integral
Looks to me like it's just plug it in and calculate
Don't be calculation-shy =P
I'll try and give it a go if I have time
 
5:26 AM
@EmilioPisanty I'm generally not calculation-shy.
I just like asking about calculations that I think others have probably already done.
 
5:41 AM
::shots fired::
 
@user1732 wut
 
/joke
:-)
 
6:15 AM
@DanielSank I mean, the matrix elements come out of the integral
You're just left with two cosines times the exponential
The first integral is trivial and resolves to a sinc function
The second one is probably also trivial
 
6:53 AM
 
7:07 AM
@EmilioPisanty Yes, the first one is easy and the second is not-too-hard.
The result isn't particularly enlightening yet.
 
7:51 AM
Time must be measured with respect to some changing event. In a void world, time can't be measured at all.
 
8:12 AM
but it's strange that people generally deem there is an uniform time ticking against which we have to race for everything.
my master advisor once told me in the past, there was no clock (or calendar, I can't quite recall now) and people made appointment based on some events.
 
@CaptainBohemian you need to distinguish between time as a coordinate and the flow of time. Time as a coordinate certainly exists but it's far from clear whether the flow of time is anything other than a trick of the way the human brain works.
 
@JohnRennie but even time as a coordinate requires you to make a foliation of spacetime.
 
@JohnRennie It's just $U_{\nabla t, \lambda}$!
 
@CaptainBohemian you can always construct a local coordinate system with a time coordinate. You may or may not be able to construct a global coordinate system.
@Slereah I don't understand your notation. What are $U$ and $\lambda$?
 
8:28 AM
Vector flow :p
Flow by the vector field $\nabla t$ with parameter $\lambda$
 
I actually have been confused by time, I treated events occurring 3 years ago like what just occurred, and what occurred 6 to 7 years ago like what happened 1 year ago.
 
that's just getting old
 
that's because I'm in a world not distracted by anybody.
I can observe the moon phase change to track time.
when I was in a world surrounded with people, I had never done that.
then I understand why people who are engaged with abundant socialization usually consider 1 year is very long.
after studying relativity, I started to doubt everyone experiences time uniformly.
 
 
1 hour later…
9:48 AM
@DanielSank there you go
0
A: Drive frequency for second order quantum transitions

Emilio PisantyYour final expression factorizes quite nicely as \begin{align} \braket{f}{\Psi'(t)} = & - \left( \frac{V_0}{\hbar} \right)^2 \sum_n \bbraket{f}{\mathcal{O}}{n} \bbraket{n}{\mathcal{O}}{i} \\ & \qquad \times \int_0^t \mathrm dt' v(t') \exp\left(i(\omega_f-\omega_n)t'\right) \\ & \qquad \quad \...

it's exactly the same as for the energy-conservation condition in first-order perturbation theory
just a bit more involved
when in doubt, plot them functions
 
10:05 AM
@JohnRennie what would it even mean for there to be a flow of time? What exactly is the difference between that and it being a trick of the mind?
 
I find once you stop encountering new experience, your brain keeps old memory like what just happens no matter how long it actually happens if you can somehow trace it.
 
10:23 AM
when I sleep, I experience time dilation--I can make a dream which seems to span several days while I actually only sleep several minutes--and I only feel a new day starts after a long enough sleep.
 
11:06 AM
58
Q: What is time, does it flow, and if so what defines its direction?

John RennieThis is an attempt to gather together the various questions about time that have been asked on this site and provide a single set of hopefully authoritative answers. Specifically we attempt to address issues such as: What do physicists mean by time? How does time flow? Why is there an arrow of ...

↑ I'd recommend going through this before asking further ;-)
 
@EmilioPisanty although the issue of whether time flows is kind of fudged
 
@JohnRennie still.
 
@EmilioPisanty if I'm honest I only wrote that question to provide a launch pad for the twin paradox question.
 
 
1 hour later…
12:46 PM
@JohnRennie is this better?
also, I see @KyleKanos disagreed with you.
Also, it's on HNQ
Obviously
 
1:03 PM
@EmilioPisanty Yes, thanks. Annoyingly I clicked the approve button then discovered it needed another approve vote so I couldn't correct the capitalisation. It's one of the odder features of the SE that while I can single handedly make an edit or reject and improve an edit I can't single handedly approve an edit.
 
Hi, I'm here for the scheduled chat session.
 
welcome
 
@Nick that's not for three hours yet
 
I did disagree. Didn't think it was worth the approval when there were other fixes that could be done (and fixing on Mobile sucks)
 
16:00 UTC
 
1:06 PM
Oh, I'm UTC+5:30 xD So, that's 9:30pm for me!
 
@KyleKanos shrug :-) You're right though. I should have checked for other changes before approving the edit.
@Nick you're in India?
 
Possibly. But to be fair, I reject a lot of edits
 
I'm actually busy with learning cryptography and image processing. I have tests on those tomorrow. Any application of those to physics?
@JohnRennie Yup, southie.
 
@Nick Kerala? Tamil Nadu?
 
@JohnRennie KL :) We survived a flood recently.
 
1:08 PM
Latest stats say I've rejected 1008 & approved 1478, improving another 231.
 
@Nick Ah yes. I have a friend in Kochi and she said the floods were very bad around there.
 
@JohnRennie My car literally swam like a submarine. It was both epic and seriously scary to be inside.
 
@KyleKanos hmmmmm. Where are those stats?
 
@Nick wow, when you said you survived a flood you meant that quite literally!
 
@EmilioPisanty click into the latest suggested edit you've performed & select the more button at the top
Says 688 approved, 229 rejected & 174 improved for you
 
1:13 PM
@KyleKanos if someone has made the effort to suggest an edit I feel a little unkind rejecting it. It's just a pain that if I accept their edit then try to improve it. I have to wait for a second reviewer vote.
 
@JohnRennie It was just 10 minutes, and it was only a foot above us. We could see the water make waves above us as we moved past.
Then, our car made it back to land. We were so close but it stopped midway. Some people came and pushed it a few inches onto the unflooded road. but really, a lot of people had it worse. I'm still helping with relief efforts.
 
what kind of car?
 
@JohnRennie that's reasonable, but I'd rather people take more time to make more improvements than doing just one quick thing they noticed.
 
@user2646 The green gangsta one from Transformers 2: Revenge of the Fallen
 
:-D
 
1:16 PM
Chevy Beat :) Green just like in the movie.
 
@KyleKanos sounds about right
4
Q: Could a real-life X-Wing fly in Earth's atmosphere?

Alan BFrom an aerodynamic point of view, could a full-size aircraft of X-Wing design fly in Earth's atmosphere? Assuming you were free to add control surfaces here and there, could the wings in open configuration create lift ?

Our star editor's sole PSE post
It's pretty good, tbh
 
@EmilioPisanty thanks, although I did read that already actually. I've been thinking about writing an answer to it from a more thermodynamic point of view. I noticed @JohnRennie mentioning the difference between time as a coordinate and the flow of time but wasn't sure exactly what the difference is.
 
@Stuart I dunno about John, but I'd be fairly annoyed if I got a chat question about something I've already written about, and the commenter had already seen it but neglected to mention it.
Just sayin'.
 
@JohnRennie What's the mathjax code for adjoint of matrix?
 
1:32 PM
@Abcd Not sure. There's a MathJax summary here though it doesn't specifically mention matrix adjoints.
 
@JohnRennie Found it in a question its \operatorname(adj) (but thats tooo long)
@JohnRennie can you please explain why $A (\operatorname{adj}A)= (\det{A})I_n$
 
@EmilioPisanty sorry about that, I should have mentioned it. I only meant to ask what was meant by the words. Sorry for the inconvenience @JohnRennie
 
@Abcd don't know, sorry.
 
@JohnRennie Okay nvm I am done. Actually, the ij notation is very confusing! I got the proof when I actually wrote down the matrix and saw!
 
1:49 PM
@Abcd that's more typically labelled as the matrix 'adjugate' nowadays, with the 'adjoint' name being older and liable to confusion with the concept of 'adjoint operator'
 
Okay
@Semiclassical Are you aware of determinants' properties?
 
depends what you mean. at the level of matrices, yes.
at the level of linear transformations, less so
(the latter is the more fundamental/geometric perspective)
 
Can you tell me why:
"Sum of the products of the elements of any row or column with the cofactors of the corresponding elements of any other row or column is 0"
 
well, it'd better be like that in order for A(adj A) = (det A)In
 
Yes I know I used that property to prove that^
 
1:53 PM
right. hmm
 
But I have no clue how that determinant property is proved. Can you tell me the reason/proof?
 
ah
Let's start with a special case to get a feel for it. Simplest nontrivial one is $n=3$
 
Okay
@Semiclassical What to do with it?
 
distracted, sorry---back now
taking A=(a_{ij}) and writing adj A = C^T, the fact that A*(adj A)=(det A)I_3 will require (for instance) $\sum_{k=1}^3 a_{1k}c_{2k}=0$
or, writing that out more fully:
$$-a_{11}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}+a_{12}\begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33}\end{vmatrix}-a_{13}\begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32}\end{vmatrix}=0$$
Which we can conveniently re-express as a single determinant:
hmm, ignore the last line. I don't trust it right now
I think the better thing to focus on is that the determinant is multilinear, i.e. it's a linear function of each of its columns
 
2:13 PM
@Semiclassical Not sure I understand what that means
 
1 hour ago, by Nick
I'm actually busy with learning cryptography and image processing. I have tests on those tomorrow. Any application of those to physics?
So, anything?
 
@Nick image processing is definitely a big deal in various applications of physics, e.g. astronomy
 
@Semiclassical Ah, thank you!
 
and to the extent that image processing is connected with Fourier transforms, it's definitely related to physics
cryptography is not so directly related. but to the extent that it's about transmitting information in some (encoded) form, it's natural to talk about Shannon entropy
 
My courses are Digital Image Processing (DIP) and Cryptography and Network Security (CNS). Hoping STEM fields like Physics do have good applications for them.
 
2:16 PM
and that connects you to thermodynamics/stat mech
 
@Semiclassical This is so interesting to know.
 
also, quantum cryptography is a thing
@Abcd well, suppose you think of a matrix as a row of column vectors e.g. $A=(a_1,a_2,a_3)$ with $a_1=(a_{11},a_{21},a_{31})^\top$ and so forth
 
@Semiclassical Yeah, I'll have to look at my previous year's math courses for that. My teacher reffered me to Kreyzig. Can't believe I have to sit through that again.
@Semiclassical oh, that is neat!
 
@Semiclassical ok
 
one can then show that $\det (\lambda a_1+v,a_2,a_3)=\lambda \det(a_1,a_2,a_3)+\det(v,a_2,a_3)$
i.e. det is a linear function of its first column
if you buy that, it shouldn't be surprising that the same is true for its other columns as well
and as such the determinant is a multilinear function, i.e. it's a linear function of each of the column vectors
(same is true for the rows)
What that means for the present case is that the following is true:
 
2:22 PM
@JohnRennie Top of the line phones (like top of the line most stuff) dod get you more powers and features at less weight than a merely good model. It's just that you pay in terms of price/performance.
 
$$-a_{11}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}+a_{12}\begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33}\end{vmatrix}=\begin{vmatrix} -a_{11} a_{12}+a_{12}a_{11} & a_{13} \\ -a_{11}a_{32}+a_{12}a_{31} & a_{33}\end{vmatrix}$$
 
But I think you're right: most people toting them want you to know what they've got.
 
@dmckee if you can point to specific differences between a premium phone and a mid ranger like a OnePlus 6 or the Huawei equivalent then please do. I am unconvinced any significant difference exists.
 
I used to live on the trailing edge of the technology curve, but now that I have some money I like to buy middle of the line (or sometimes the marked down top-of-last year's line) to get as much goody as I can without breaking the bank.
 
@JohnRennie I'm increasingly convinced that the price premium does help ensure a much more consistent working.
 
2:26 PM
@JohnRennie When I look inside any given manufacturer's line I find that you generally get some subset of more memory,more pixels, slightly faster processors, less weight, thinner package or more battery power (thought the faster speed and improved display may eat that up and actually reduce the expected battery life).
 
and then the fact that the upper-left element vanishes means we can expand that determinant to get $$a_{13}(a_{11}a_{32}-a_{12}a_{31})=a_{13}\begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32}\end{vmatrix}$$
which cancels the third term in the original expression to give zero.
 
I currently use a ZTE axon 7 (roughly equivalent to the OnePlus 6, I think) and it's pretty good. But there's a longuish series of glitchy behaviour which I'm fairly sure wouldn't happen to a top-end flagship.
the most annoying of which is a software focusing stabilization problem which causes the camera to bug out for hours at a time.
 
So that validates the fact that that weighted sum of determinants is zero. But I'm not sure how satisfying that is...
 
@dmckee the premium phones are certainly better, but my point is that they are incrementally better and the increment is sufficiently small that for most people the improvement is undetectable.
 
Whether a NameBrand (tm) Premium Phone XXVVII is better than a good phone from Solid Manufacturer Without the Prestige logo is a different problem.
@JohnRennie Sure. The price/performance curve is steep.
 
2:28 PM
There's also the fact that, while premium brands are typically more reliable, you can still have some major problems. C.f. the whole Samsung battery debacle
 
Any way. Back to the salt mine. Tata!
 
7
Q: For what reason do we define center of electron's mass and charge coincide?

J CTraditionally for a free electron, we presume the expectation of its location (place of the center of mass) and the center of charge at the same place. Although this seemed to be reasonable for a classical approximation (see: Why isn't there a centre of charge? Lagerbaer), I wasn't sure if it's...

any takers for a better title for that one?
it's on HNQ so that has an added relevance.
 
Okay, wtf
I'm looking at the algebra I did above, and what I initially wrote as the sum of cofactors identity was simply wrong
but somehow it still summed to zero
wtf
never mind, I see why it's fine
 
@Semiclassical we're glad we could help.
 
Anonymous
@Semiclassical Are you referring to the explosive batteries in some of their faulty models? One of my pet peeves about Samsung and most other smartphone manufacturers is that they started making their batteries non-removable....which renders one of the most effective methods of rebooting useless (i.e. opening the battery and putting it back in, when the phone hangs) :P I actually don't know what advantages non-removable batteries have...and if the reason is substantial enough to justify it...
 
2:40 PM
@Blue yeah. exploding batteries in a premium phone does not make for a good reputation
seems like a pretty good summary of my part of the conversation
 
@Semiclassical Reminds me, I learnt engineering chemistry in my first year of collge. We had a whole topic on Li batteries.
Just a few exothermic equations out of balance, and kaboom! you get an explosion. Pretty stuff.
 
alternative title: stock price plunge compilation
 
@Semiclassical I like that side-effect. Hate the catastrophe that causes it though.
Speaking of stock plunge, heard about Tesla's dip? Apparently, Musk smoked a blunt live on a podcast. Investors weren't happy.
 
2:51 PM
jeeze, the one starting at 2:18
 
Anonymous
@Nick Why do you care about applications in physics? :P Quantum cryptography is a thing...but it's a lot more computer science/pure math and a lot less physics. But yeah, as SemiC said, there are lot of techniques in digital image processing which were possibly inspired from physics.
 
@Nick the suspenseful part of that: seeing multiple phones on screen and wondering which one of them is going to explode. (though the most disturbing one really does have to be where it explodes while next to someone's head, and they collapse right after)
 
@Blue I just felt like making what I'm doing relavent to this chat, that's all. I'll possibly code up something neat by tonight.
@Semiclassical I remember that from somehwere else. He survives that.
 
good
i mean, i'd hope they wouldn't post a video of someone getting killed on screen
 
Anonymous
@Nick Almost everything other than physics is relevant here, don't worry. And if you want to be a programmer that's especially welcome. :P
2
 
2:58 PM
but it's definitely not clear from that video
 
@Blue Aw thank you. I'm only good with Python and C at the moment. That's all I've ever needed and will need in colleg life. Maybe a sprinkle of Jave sometimes but beyond that is just a hype circus.
@Semiclassical I thought he died too. But check it out, a girl named Katy survived a musket to the face. She recently got a full face transplant. 40th in the world known to have undergone that, and at 21 (my age), she's the youngest. She recieved the face of a 30 yr old woman named Adrea who donated her body to science.
 
Even Malala Yousofzai survived a shot to the head the Taliban gave her. It's suprising how many documented cases of head injury survivors there are in the world right now.
@Semiclassical ok, that's more than I knew. thnx
 
Anonymous
@Nick I see. Any plans for the near future btw? Applying to grad school?
 
Anonymous
3:14 PM
Now who starred that....
 
I and a friend have just managed to repair his motorcycle. Now there's applied physics for you :-)
2
 
3:33 PM
@Nick I think someone mentioned astronomy, but I believe image processing is also used a decent bit in biophysics to capture data. I think it can be useful for getting data out of a large number of images in general
 
@JohnRennie I just wrote an excellent highly-applied-physics answer.
 
@EmilioPisanty which one was that?
 
@JohnRennie the HNQ, of course
the mass-and-charge-coincide one
 
The centre of charge question? That's not applied physics.
Calculating the electron EDM is right up there in the more abtruse reaches of high energy physics.
 
3:47 PM
@Blue I should apply MS but I haven't started yet. You just reminded me, the date to write the GATE exam is getting closer. I ought to register soon. It's the next major thing after JEE.
 
Anonymous
@Nick Umm, wait, you're applying for MS as well as appearing for GATE? That's pretty ambitious
 
@JohnRennie ah, I see the tongue-in-cheek detector is off ;-)
 
It's a good answer though :-) One day I will correlate your answer frequency with your publication frequency. I'm sure your answer rate peaks just after a paper is accepted, presumably as a stress relief :-)
 
@JohnRennie I suspect it's rather the opposite
 
gateprep dies
 
3:49 PM
answer rate peaks just before a paper is submitted
May 27 '17 at 18:08, by Emilio Pisanty
user image
I mean, you can certainly pick out my thesis-writing months in my answer rate
 
@EmilioPisanty Well, you did something very specifically assumption-y there.
You only used one value of $n$.
That's probably the approximation we were looking for :-)
 
@DanielSank doesn't matter
 
Yes, it does.
 
it's linear in $n$
the total amplitude will be the interference of all of those processes
 
True, but the plot you showed is an absolute value and would look completely different if you had more than one $n$.
@EmilioPisanty Correct.
 
3:52 PM
there is a non-resonance assumption, in that you shouldn't be on-resonance for any $n$, which translates to the requirement that there shouldn't be any significant population on the intermediate levels
that's the "virtual level" talk
if you do get into such a resonance, then the energy-conservation spiel does go away
the formalism you want holds when all of the relevant $\omega_n$ are far away from $\omega_p$
(assuming $\omega_i=0$)
where "far away" is in terms of both the lifetimes and the pulse bandwidth
 
@EmilioPisanty ...and when one of the the $n$ is much closer to $i$ and $f$ than any of the other $n$.
 
@DanielSank not necessarily
you could have a cluster of $\omega_n$'s at a reasonable distance from the midpoint
 
orly?
 
@Blue nah, it's a last ditch effort at achieving something this late in life.
 
@DanielSank the distance from $\omega_n$ to $\omega_f$ and $\omega_i$ isn't that relevant. unless you're worried about population transfer to $n$ instead of $f$, I guess, but then you're twisting the formalism
if that situation worries you, then the normal way to phrase it would be to have multiple $|f⟩$s instead of having an $|n⟩$ close to $|f⟩$
the in-built assumption is that the $|n⟩$s will be close to the midpoint
they're only relevant in that they provide support for the "virtual levels"
 
3:57 PM
@EmilioPisanty Yes.
 
and yes, the total amplitude is an interference of those virtual-level processes
but that holds only in terms of the matrix elements
the two-photon-resonance energy-conserving sinc function is the same for all of those processes and it factorizes out
 
@EmilioPisanty Ah, that's a good point.
 

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