Unless you also have population being pumped in from the outside. Is this the case in your simulations?

@bolbteppa Hmmm...the notion of "twist" in the book I linked might not be the same as that in the PDF you linked

@MarkMitchison well, that would depend on the initial rho, correct? Note, when I did these simulations, I had a wrong initial rho, but both times, it converged to 1/7...

@TanMath No, I am asking whether there is a term in the master equation describing population entering the system on a certain site. It is not to do with the initial state. If there is only loss and no gain, then the population on site 3 must vanish after long times. This is common sense, really, if you think about it for a bit.

@MarkMitchison nope.. I am using the same model as in environment assisted transport...

@ACuriousMind I'm almost sure it is despite the scary notation which honestly confuses me haha

@MarkMitchison so, what would cause the prolonged plateau at 1/7?

@TanMath I don't know, but I assume it is an error.

Are you sure you have the correct sign in all your anti-Hermitian terms?

But physically it is clear: if population is being continually sucked out of the system and not replaced, eventually it will all be gone.

@MarkMitchison like which one?

@MarkMitchison originally i thought that too, then I saw the results...

@TanMath I have no idea, you need to check your code. If you have a sign wrong then you might be inadvertently pumping population into the system somewhere.

@0celo7: ?

@MarkMitchison well, solving the lindblad equation is done by an external module.. I only have to put the hamiltonian, collapse operators, initial rho, and a list of times to evaluate rho(t) at...what part of that could go wrong?

could the hamiltonian being wrong affect the results? I am pretty sure the collapse operators are correct...

@TanMath That could go wrong in many, many places. Especially if you don't understand what the module is doing. But assuming the module is not faulty, the error is in your code. You need to carefully go through and debug.

@MarkMitchison ok.. so it can't be the initial rho causing all of this? I am pretty sure the module is not faulty..

@TanMath Generally one expects dissipative master equations in a complex system to have a unique steady state, independent of the initial state. In the presence of symmetries in the master equation, it is possible to have a degenerate manifold of steady states, so that the final state does depend on the initial state.

However, it is clear that no steady state of that master equation will feature non-zero population on sites which have a constant loss rate.

@MarkMitchison so it could just be that because my initial rho is wrong, I am getting a non-zero steady state?

@TanMath No, I am saying it is not to do with the initial state.

@MarkMitchison oh, ok.. could it be the hamiltonian? I am pretty sure my collapse operators are correct...

@TanMath I do not know, it is your job to find out now...

But you might want to check if that module can handle non-Hermitian Hamiltonians.

It might be doing something simple yet incorrect like renormalising the state after each time step.

@MarkMitchison BTW, have you used python? it is a python module called qutip FYI

@bolbteppa I think the notion is not the same. The PDF you link leads to a non-central extension of the Lie algebra, while "my" twist is just real 2-cocycle - this classifies central extensions.

@TanMath No, I use MATLAB for simulations like this.

Although python and MATLAB are very similar

@MarkMitchison oh, ok.. I can see why.. I used python because I was able to find this module...

But I'm not a python expert, I've barely used it

@MarkMitchison but my system hamiltonian is hermitian

@TanMath The model you are trying to copy has a non-Hermitian Hamiltonian, see Eqs. 6 & 7.

@bolbteppa: As far as I can see, the $0\to A_\text{Lie} \to A_\text{Lie} \rtimes \mathfrak{g} \to \mathfrak{g} \to 0$ extension is not central.

@TanMath If you do not add these anti-Hermitian terms then there is no loss acting on the system.

Only dephasing

@MarkMitchison so I add them into my system hamiltonian?

Yup

@MarkMitchison but I don't know the recombination and trapping rates..

@ACuriousMind I don't really appreciate what that means tbh

@TanMath Sounds like you have some work/research to do then

@bolbteppa "appreciate" - you don't understand what it means, or you don't know why I'm telling you this?

@MarkMitchison plus it was not explicitly stated that those two terms are in the system hamiltonain, in fact, the first equation defines the system hamiltonian so I wasn't sure...

@TanMath "augment" means "add to"

@MarkMitchison oh, I see... I wonder where they get the recombination and trapping rates...mainly the trapping rates...

@TanMath Have you read the caption to figure 2?

They state $\kappa_3 = 1~$ps$^{-1}$

@ACuriousMind that non-central extension in my pdf does not apply to the next sentences where he constructs the twisted UEA from $U(g)$

@MarkMitchison but what about $\kappa_1 , \kappa_2 , \kappa_4, \kappa_5, \kappa_6, \kappa_7$?

@TanMath I imagine they are equal to zero. One frequently assumes that only one of the sites is significantly coupled to the RC

@MarkMitchison you know what? you're right!

@bolbteppa Oh, but isn't what he described there just another way to say the $A$-twisted UEA is $U(A_\text{Lie} \rtimes \mathfrak{g})$?

@MarkMitchison and now the recombination rate is 1 exciton/ns ?

@TanMath I am busy now. Sounds like you can work this out for yourself if you read the paper carefully. Good luck.

@MarkMitchison ok.. thanks for the help!

@ACuriousMind I'm not sure but from browsing it looks like the discussion on that page up to the non-central extension is talking about the (un-complexified) lie algebra derived from the lie group, note the lie algebra from a unitary lie group is not unitary hence one needs to complexify, I think this is what he does when he says "In a similar fashion" and twists...

goodbye

@bolbteppa I'm just saying that what I linked is useless because it describes the case $A = \mathbb{R}$.

@ACuriousMind I think your book is taking the complexified Lie algebra coming from a Lie group and forming the twisted UEA of the complexified lie algebra, while my pdf is taking the (complexified? At least in section 2.7 it is complexified!) lie algebra of a Lie group and then twisting it by Gamma matrices?

@bolbteppa Yeah, the difference is what we are twisting by - my book twists by real numbers, yours by gamma matrices/some other algebra elements. The latter method is a more general notion of twists than the former.

@FenderLesPaul SRiGF came already!

@Huy ::clicks, clacks::

@Huy my clickity clack machine

I got the brown switches

They have the same activation pressure as red, but honestly they're a little loose

also I'm bottoming out like a noob

@ACuriousMind you don't have books

Advice to everyone: Don't google "bottoming out" :P

@ACuriousMind it's a mechanical keyboard thing

@0celo7 Read the whole discussion, we dropped the quotes around "my" somewhere :P

@0celo7 dope dude

told you it's fast

yeah but they said 8-10 days when I made the order

I guess that's the max?

they overshoot so murricans don't sue

the return address is right here in TN

I should sue @ACuriousMind

@ACuriousMind what finger do you hit alt with?

@0celo7: Do you like it?

@Huy yes

(thumb for me)

it's very different

Great, glad I didn't make a bad recommendation.

Yes, but you'll get used to it in less than a week.

I think the overall size of the keys is different

I'm hitting keys that I didn't hit before

I felt like their spacing was different.

I used Thinkpad keyboards before, and those were lot closer.

I never liked the membrane keyboards at school

And do you think browns were the right switches for you?

I always thought Apple laptop keys were the best

but man this feels amazing

@0celo7 Thumb

@Huy browns are like reds but quieter, right?

I think thinkpad keyboards are the best typing laptop keyboards

that's what I got from that guide

it's actually really loud, I can't imagine what blue must be like

they are quiter but also have a little tactile bump when pressing them down I think?

just a very little one

fiddles

yeah

maybe

tomshardware.com/reviews/…

you can see it clearly in this animation

what's causing it

browns were cheaper anyway

sure

I found them on a site that doesn't charge tax

they're great too

any halfway decent mechanical keyboard is better than membrane I think

yeah gotta run

laters

@Huy thanks a million bud

would have never thought of this otherwise

np, I hope you'll like the mouse as well !

still have not ordered

I read some bad things about the mouse button

take your time :)

oh ok, then maybe let's look for another one this weekend :P

cool

also I heard you can't reprogram the buttons properly, but the thread was very confusing

like you can't change the DPI buttons to something non-DPI

hm, that would be odd, but yeah better check it out in advance

you can also see how expensive corsair mice are over there, I've never tried them before but I'm sure they're really good as well

I just started with some cheapish logitech models and then sticked with it, but my next one probably won't be a logitech anymore

(not because they're bad but because I wanna see what others are like)

@ACuriousMind Is a finite cylinder technically a manifold?

The sides are not homeomorphic to $\mathbb{R}^2$, right?

@0celo7 No, it's a manifold with boundary.

Good

BTW: the DJIA dropping ~500 pts today has hurt Apple pretty badly (of course, it hit everyone badly)

Beginning of the end for them?

Well Verizon's choice to eliminate 2-year agreements (which greatly subsidize the cost of new phones) likely isn't going to help Apple sales any (who's going to drop $600 for the phone when they can get cheaper ones that do the same thing?)

Totally eliminate?

Dang...

That's what I have.

Yep. Starting Aug 13, they've eliminated the 2-year plan for month-to-month plans

So when my contract runs out, what then?

They have data plans whose prices depend on the GB allowance

1 GB is \$30, 3 GB is \$45, and so on

They even go up to like 12 GB/month for \$80

But for phones, they still sell them, but at full cost

So that \$600 iPhone 6 isn't going to be \$150 anymore, it'll be \$600

wow

And that's due up front

why did verizon decide to do that

they figure they won't lose that much business?

They did some sort of market study

I think T-Mobile switched to that a few years ago and it's not hurt them much

@ACuriousMind what's a good quick way of seeing the necessity of a non-Euclidean sig?

@0celo7 Necessity for what?

why does the squared line interval have a minus, essentially

I know one way of seeing this

I would like to see what you think about it

The Minkowski metric is the unique metric preserved by Lorentz transformations, and the Lorentz transformations follow essentially from the assumption of a maximal speed.

how are you defining the Lorentz transformations?

All transformations that can be written as products of boosts and rotations (and time reversal and parity).

@ACuriousMind this seems really hard to verify

@0celo7 It's a lengthy bit of shifting matrices around, but it's not really hard.

You don't need any theorems or something, it's just computation

hmm, maybe this book will motivate it better

I know what you're saying, I just don't like that argument

@0celo7 so the way Landau and Padmanabhan explain it is that you have, for a light ray, $c^2dt^2 = dx^2 + dy^2 + dz^2$ i.e. $ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 = 0$, and then Padmanabhan actually beautifully proves that $ds^2 = ds'^2 \neq 0$ for the infinitesimal separation of two events in two frames by taking a Taylor expansion of one in the other, which establishes the importance of $ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$ in special relativity

There is a youtube video of this too I can find

From this you have Lorentz transformations as those transformations preserving $ds^2$, hyperbolic rotations...

I don't know anything other than this, if you have a nicer way to think about it

Note this idea comes immediately from the postulates of relativity and derives what a Lorentz transformation is

The computation and an amazing discussion is in the first (or at least the first 3) video of this course youtube.com/watch?v=EotEgl8MMaw based off his book and Landau

@bolbteppa Do you know of Einstein's moving mirror thought experiment?

Zee shows $\Delta t^2-\Delta x^2$ is invariant, it does not depend on which inertial frame you are in

My linear algebra assignment: learn LaTeX

I guess prof. Dydak is upset at all the ignorant engineers using Word equations

I had a look at Zee's example in the start of III.II, I think that example only shows the existence and invariance of the interval when $ds^2 = 0$, then in the next chapter he starts using it when it's not 0 as if it's invariant in those cases too...

Really?

I'll take a look at it.

@bolbteppa did you read footnote 3?

@MarkMitchison Now, since the hamiltonian and dephasing rate are in cm^- natural units (I think), I need to convert the recombination and trapping rates to cm^-1 in natural units?

page?

173

I see no reason why $\Delta s^2=0$ in his example.

(He does not either, but that's less important.)

Okay cool, I need to think about how to explain what is iffy about what Zee does, but instantly I can see his example only holds for light rays, while the spacetime interval is defined to be invariant for any kinds of events

Zee seems to think that's exactly what Einstein did as a patent clerk.

haha, so Zee's footnote warning is about just disregarding all logic when talking about the spacetime interval for light, and says you need Einstein's clock, but I think his example only holds for light between two spacetime events and only holds because he sneakily chose the same starting and ending point of space, while Landau/Padmanabhan's proof holds for all cases for all kinds of interactions

What page in Landau?

Section 2 of his CFT book (vol 2) but you should read section 1 as well, if his derivation in section 2 is a bit too tough then Padmanabhan's book and video does it out a lot better

Only about 4/5 pages

Hmm. That proportionality argument is sketchy...

Check Padmanabhan as well, he will clean any issues up

"Math interpreted in physics enriches the math." - Dr. Dydak

"I suggest you sue your high school for negligence. They should have taught you good physics."

"Somebody messed it up. Well I'll try to correct it but life is hard with the students doing work."

@0celo7 Are you reading "Dr. Dydak's aphorisms" or what?

@ACuriousMind Yes, they're called linear algebra lectures.

@ACuriousMind What do you know about the Routhian?

@0celo7 Till a few minutes ago, I didn't know it existed.

@ACuriousMind yay ignorance buddies

@0celo7 ::fist bump::

stream!

someone is streaming?

dude I should stream The PoE Adventures

wtf is that

OK. Let me finish Chap. 1 in SRGF. Then let me eat dinner.

@ACuriousMind school him pls

@FenderLesPaul I'll be ready by 8 because that's when the dining hall closes :)

::sigh::

I was about to make a comment on that.

I would have regretted it.

Thank you.

(removed)

(removed)

@0celo7 alright bud

@0celo7 Padmanabhan justifies the proportionality argument ($ds^2 = ads'^2$ in the equation between 2.4 and 2.5) by Taylor expanding $ds^2$ as a function of $ds'^2$, i.e. $ds^2(ds'^2) = b + ads'^2$ so that $b = 0$ because of them being equal for a light ray

@bolbteppa I'll accept that.

The rest is here books.google.ie/…

Nice!

@ACuriousMind Now that's convincing.

@0celo7 It's a good physicists' argument.

(Meaning a mathematician would stop this explanation right at "Taylor expansion" with an exasperated "Wait, what?")

The mathematician can go you know what himself.

And why would he do that?

He can go to rigorously derive many results himself, yes :)

@ACuriousMind No one likes a smartass.

@0celo7 One reason would be because equality of Taylor expansions does not mean equality of functions (There are functions with Taylor expansion 0 about a point which are non-zero, for example).

@0celo7 I'm not a smartass then, because I'm universally well-liked ;)

@ACuriousMind That was a rhetorical question, the answer is: because he's a prick and likes arguments that no one can follow.

@ACuriousMind Define universally.

@0celo7 Oh. :D

@FenderLesPaul Is it just me or does Gourgoulhon never motivate the whole Lorentz signature bit?

@0celo7 Definition of "I am universally well-liked": When you pick a random person that knows me, the odds that they will like me are far larger than that they will dislike me. As the number of persons that know me grows to infinity, the probability one of them dislikes me goes to zero.

That's cheating, there are not infinitely many people.

And as we all know, the chance of picking one out of $x$ as $x\to\infty$ is 0!

Your first point is valid, your second is not.

@0celo7 I suppose he doesn't yeah but it's meant to be like a 2nd read in SR

according to the preface

so maybe he assumes you've seen it already

Carroll does not even explain it.

Carroll isn't really meant to be an intro to SR either though