The h Bar

12:00 AM

As in, rather literally - you can wait a little longer to get 4 photon generation (still fractions of a second I think) or you can use 2 photons, then have to deal with more loss and so, have to wait a little longer to get a good number of results... Not much difference really

heather

okay

Mithrandir24601

@DanielSank I think that's pretty much how everyone on every [what we call] platform feels

@heather To be a bit clearer, the difference isn't really in adding the odd gate here or there, but in scaling the system up - if you want a few qubits, then not only do you have to deal with loss, but you also have to deal with multi-photon generation, which is slow, while cluster state means that you only have to deal with multi-photon generation

skullpatrol

12:19 AM

Gotta run peeps, cya.

DanielSank

@Mithrandir24601 Chip design is important.

Crosstalk is a problem.

Mithrandir24601

@skullpatrol Bye!

@DanielSank Yeah, I'm not expecting to be any good at it, but it's one of those things that everyone should at least have a go at - we're lucky in that we have a few people who are really good at that sort of thing :)

DanielSank

@Mithrandir24601 E&M simulation software?

I've learned during my time in the field that using such software correctly is a valuable and not-automatic skill.

Mithrandir24601

@DanielSank They actually built the building blocks of some photonic chip design software, which is apparently much easier to use than doing everything from scratch

DanielSank

Oh, photonic chips. Cool.

okey dokey y'all, gotta go.

Mithrandir24601

12:25 AM

Everyone designing chips always gets stressed near chip-run times as it only happens a few times a year and is kind of expensive... Anyway, night!

heather

@Mithrandir24601 I found this paper on arXiv - why does something like that not work for the deterministic photon source?

Mithrandir24601

@heather They're actually working on something that's essentially this in Oxford - it's deterministic in the sense that if an atoms there and you 'do the thing', you get a single photon... But the atom's not where you want to to be all the time

heather

oh.

Mithrandir24601

So, I think it works a little more than a tenth(?) of the time

heather

12:43 AM

@Mithrandir24601 so, you said my idea has already been thought of - what are some of the problems with it? (aside from the standard problem of many gates -> loss)

Emilio Pisanty

Hey, dupe hunters, we've seen this before, haven't we?

1

Q: Why is the angular momentum conserved with relation to this point?

I have this system: It is a given that the mass $m$ rotates around the vertical axis with constant speed, and that $\theta = 30^\circ$ initially. The massless string is then pulled upwards, until the angle $\theta$ becomes $60^\circ$. The problem then asks about some quantities, but a key par...

Mithrandir24601

@heather ::cough:: since when did I say there were problems with it? ::cough:: You'll still have to multiplex it and crystals are big so don't go on chip, so you have to couple it to chip... Or use non-linear effects of a chip to do something similar on chip, then multiplex, or get a source and fab it on the chip or somehow move it onto the chip... <.< >.> <.<

heather

erm...are you coughing so much because this is something you're working on?

ACuriousMind

@EmilioPisanty Have we?

My memory for Newtonian mechanics questions is rather fuzzy

Mithrandir24601

@heather So, you know that my PhD will be simulating non-Hermitian Hamiltonians (as in, systems that aren't time symmetric, such as lossy systems) and I'll also be looking at non-linear optical systems (I've already ordered the textbook) as this is a topic that's apparently related?

Oh, @ACuriousMind - I've just answered one of your questions for my AMA ^

heather

12:52 AM

@Mithrandir24601 sure, i knew about the first part, not the second.

wait, is SPDC non-hermitian?

Mithrandir24601

@heather So part of it will be involving looking at if non-linear processes can be 'perturbed' into the non-Hermitian regime in the hope that it'll improve intuition/knowledge of how photon generation works

(I only found this bit out recently actually)

heather

what do you mean by "perturbed"?

Mithrandir24601

I honestly have no idea

heather

do you have any papers/references i can look at?

i just find your research very cool.

when you explained the matrix representation of measurement and stuff, that was very interesting.

Mithrandir24601

@heather technically, that was a projection :P but yeah, same difference

@heather and I'm slightly overexcited to be honest :)

heather

12:57 AM

@Mithrandir24601 sorry

@Mithrandir24601 pretty sure you can't be overexcited about physics. it's like having too many books: impossible.

Balarka Sen

yeah physics is underexciting

agree 100%

Mithrandir24601

@heather Is there anything in particular you're looking for/interested in?

heather

@BalarkaSen wat, that is not what i said 0_0

Mithrandir24601

In terms of authors, there's CM Bender

heather

@Mithrandir24601 what this "perturbation" might mean, and moving non-linear processes to non-Hermitian regimes.

Balarka Sen

12:59 AM

@ACuriousMind why do you have a thug as your gravatar now

ACuriousMind

@Mithrandir24601 You told me what a non-Hermitian Hamiltonian does, but you didn't tell me what's so special about the PT-symmetric ones ;)

heather

@Mithrandir24601 oh, i just looked up perturbation theory, and i think i get what you mean:

ACuriousMind

@BalarkaSen It is Nines Rodriguez from Vampire the Masquerade: Bloodlines.

Balarka Sen

So a Spanish thug

Mithrandir24601

@heather I just looked at one of the things that might have an answer to this and got "perturbation theory beyond the rotating-wave approximation" in one of the papers. This is super, super-exciting :D

ACuriousMind

1:01 AM

@BalarkaSen Not a thug. An Anarch.

Mithrandir24601

Perturbation theory is just a method used a lot in QM

ACuriousMind

Perturbation theory is just a method used a lot in QM physics (FTFY ;) )

Mithrandir24601

@ACuriousMind My PhD is now about non-Hermitian ones in general :P (I don't think there's a difference in terms of actual physical systems though)

@heather try this: arxiv.org/pdf/1508.07001.pdf

@ACuriousMind I'll answer that in more detail in a few months when I've actually properly looked at applications :P

vzn

3:08 AM

@heather think a bell experiment might not require as many costly components although other experiments might be fun to play with... it does seem to need beamsplitters for a more rigorous one where one wants to measure horizontal/ vertical polarizations simultaneously at each arm...

@NeuroFuzzy awesome, thx :) can you do sep19th tues 1600Z? thats the std mtg time which seems to go better... if so all that is needed is a meta post announcing it...

Alex K Chen

6:23 AM

@vzn Great. Count me in !

That guy's very good.

Emilio Pisanty

@ACuriousMind I think it might've had an @DavidZ answer?

Anonymous

0

Q: What would an intuitive explanation for the E-k diagram?

In all of my solid state books they seem to plot the $E-k$ diagram. But I don't understand why ? What is the physical significance of the $E-k$ diagram ? Please don't give answers like it shows the forbidden and allowed bands. For that we should have been better off plotting $E-r$ (where $...

quantum-mechanics semiconductor-physics

Anonymous

I would be grateful if someone gave me an intuitive explanation for the physical significance of momentum space...

DanielSank

8:06 AM

@ACuriousMind Yeah it's kinda ridiculous how many important physics things are only taught in the context of QM and/or QFT.

Danu

@DanielSank have you seen this?

@DanielSank At least in my degree, it seems like similar techniques are taught in several courses which are aimed at other groups of students (condensed matter, HEP, etc), suggesting that it's just because we don't have much time besides QFT courses or something to teach these techniques...

DanielSank

8:30 AM

@Danu no, but I like it.

user400188

10:01 AM

What to do, when the lab report you had been referencing (for your own report) appears to be wrong?

I have a lab I'm meant to hand in soon, and for it I referenced two other lab manuals. The method in all three (the one supplied to me, and the two I found for reference) are different.

In the theory section of one however; there is a claim that contradicts a statement from the other.

Mithrandir24601

@user400188 Find out if/why it's wrong and give an explanation? If you don't have the time, then ask a lab demonstrator lead demonstrator/lecturer or whoever it is you ask about lab stuff - is that possible?

@user400188 This is a bit more alarming...

user400188

oh, hi @Mithrandir24601

Mithrandir24601

Although depends on when they were written - maybe someone thought they had something, but someone came up with an argument for why it doesn't work and so one of them is just outdated, which is slightly less alarming :P

@user400188 And hi!

user400188

Its kind of weird, because I found the lab manual supplied by the manufacture of the equipment. This is the one that is most likely incorrect, because the two others I can refer to (the one supplied by demonstrator ) and another I found online state the same thing.

Mithrandir24601

@user400188 Ah, manuals supplied by the manufacturer are... Not very good...

user400188

10:07 AM

really? I would have thought that they would be top notch.

Mithrandir24601

Not necessarily always true, I suppose, but what I've used...

Let's take an example - I was writing a program to take data using a bit of equipment called a polarimeter (measures polarisation of light connected via fibre). The manual said that it came with a particular library which meant that I could write programs to control it and take data (as opposed to using the software that came with the thing that didn't do what we wanted). Quite simply, it didn't come with this library, which meant that I had to reverse engineer said library

Of course, a fair chunk of the manual assumed that the library was there, despite the fact that it simply wasn't :/

CooperCape

That just seems like a bit of a dick move on the manufacturers part...

Mithrandir24601

@CooperCape Possibly... Especially considering that when they finally replied to my e-mail several weeks later, they said that it was possible to buy said library

CooperCape

sheesh...

Mithrandir24601

Completely separately to the polarimeter, which cost £6000, of course

CooperCape

10:15 AM

Funded by the department, I assume!?

Mithrandir24601

@CooperCape Well, one of the groups paid for the polarimeter - I managed to get my own library that did the same thing working :) so they didn't need to pay for that

CooperCape

At least there's that... I can imagine much unwanted time wasted, however...

Mithrandir24601

That's experimental physics for you though - nothing seems to works as it should straight of the box

Software for experimental physics is (in my experience) not-so-good :/

That's not to say that doing experimental physics is worse than theoretical physics - they just have different problems :P (and also different fun moments :) )

CooperCape

Yeah I can imagine. I'm not really sure which one I really want to go into rn... What I can say however is that I'm absolutely awful at practical experiments currently - but that just might be school...

user400188

10:31 AM

Cooper, are you a second year student by any chance?
I think I might know you. Your profile pic, along with the short description you gave in your profile align almost perfectly with a person I know.

CooperCape

Nope! Going into my second year of A-levels in... ooh, two days.

(American last year of highschool)

user400188

oh wow. Thats a strange coincidence.

CooperCape

Hey maybe a lotta people like constants...

Mithrandir24601

@CooperCape I was pretty bad at experiments at school, then went to uni and somehow was that one guy that got abnormally accurate results (not the same as getting abnormally high marks :P - I did get good marks for experimental stuff, but there were plenty of people who did better). Only, at the beginning of the year here, I'd lost that knack (maybe as a result of doing mathematical physics as an MSc...).

I may or may not have gained some of that knack back since, but we'll find that out in about half a year or so, so it seems to come and go a bit with me

user400188

It seems to be a common thing. You either remember then perfectly or you use natural units, but never in between.

CooperCape

10:35 AM

@Mithrandir24601 I'll just have to wait and see... good luck :p

@user400188 All those engineers like $\pi=3=e$

user400188

@CooperCape I'm beginning to think you are him. He said the exact same thing to me.

CooperCape

uhhhhh woooww

There's always this classic

Mithrandir24601

Ha! Fine structure constant! 1/140 :D

Anonymous

10:50 AM

@CooperCape lol

user400188

is squaring both sides of an inequality order preserving?

Anonymous

@user400188 No.

user400188

didnt think so

unfortunately I could swear I heard it said by someone just a few days ago.

Ill have to take a look at my notes.

Mithrandir24601

@user400188 It can be if everything is real and $>1$

Anonymous

@Mithrandir24601 I have a condensed matter/ semiconductor question. Would you like to have a look?

Anonymous

10:58 AM

I asked it on the main site but no one answered it yet

Mithrandir24601

(assuming you mean $x>y \implies x^2>y^2$ (Oops, sorry everything $>1$ :P )

user400188

was about that ask about that

Mithrandir24601

I think it's still true if everything's $>0$

::sigh:: Yeah, I was right the first time...

user400188

I think you were haha

Anonymous

If $x-y >0$, either $x+y\geq 0$ or $x+y<0$. For case (i) $x^2-y^2 \geq 0$ (ii) $x^2-y^2 <0$.

Anonymous

11:06 AM

@Mithrandir24601 So, the condition for that to be true is $x+y>0$

Mithrandir24601

@Blue if $x \land y >0$, then how can $x+y<0$?

Anonymous

@Mithrandir24601 I never said that

Mithrandir24601

@Blue Ah, OK - I was putting the assumptions in at an earlier stage :P

user400188

@Blue are you missing a $2xy$ term here? Or is it just me?

Anonymous

@user400188 $(x-y)(x+y)=x^2-y^2$

user400188

11:08 AM

oh I see what your doing now

Anonymous

@JohnRennie Halp. No one answered this question yet :P

John Rennie

@Blue I'm afraid I remember very little from the band structure lectures.

Anonymous

@JohnRennie Oh :/ I'm really struggling with this stuff

Anonymous

Anyway, thanks for having a look

Mithrandir24601

@Blue Take a step back and apply the simple approximation that $E\propto k^2$

Anonymous

11:14 AM

@Mithrandir24601 Okay...I do know that part

Mithrandir24601

(don't worry about the band gaps just yet - we can sort out the change with $k$ first)

@Blue So, you've got a nice, simple graph :)

Give me a minute to check the name of something...

Anonymous

@Mithrandir24601 Yes. I know how to derive and plot the graph. I just don't understand why we are plotting E-k diagrams instead of E-r diagrams. What is the physical significance of $E-k$ ?

Anonymous

And then there are stuff like this ^

Anonymous

Where they plot $k$ along a certain direction

Anonymous

11:18 AM

It seems like they are using math without stating any physical intuition/motivation behind it

Mithrandir24601

@Blue I can't the name of the thing, but anyway :/

Take a step back and don't worry about that yet

We know that the lattice is periodic, right?

So $r\equiv r+dr$

Anonymous

@Mithrandir24601 Are you looking for Brillouin zone ?

Anonymous

@Mithrandir24601 Yep!

Mithrandir24601

@Blue No, that's not it - I was actually looking for the name of a periodicity thing

Anonymous

Bloch theorem?

Anonymous

11:20 AM

Anyway

Anonymous

Go ahead

Mithrandir24601

@Blue So, if you transform this into momentum space what do you get?

Anonymous

Kronig Penney :P

Mithrandir24601

@Blue Actually, I think this is it :)

Anonymous

@Mithrandir24601 Transform what into momentum space ?

Anonymous

11:21 AM

@Mithrandir24601 Hehe :D

Mithrandir24601

@Blue the condition that $r\equiv r+dr$ (maybe this isn't a great way to write it - it might(?) be easier looking at potentials or wavefunctions or energy)

Anonymous

@Mithrandir24601 Not sure how to do that

Anonymous

Could you show?

Mithrandir24601

@Blue So, let's just look in 1D for simplicity of not writing dot products and everything in bold: $\psi\left(x\right) = e^{ikx}u\left(x\right)$

Anonymous

$k \equiv k+\frac{2\pi}{a}$....did you mean that?

Mithrandir24601

11:28 AM

@Blue yeah :) That's what I was going to type next

Anonymous

Okay...go ahead :)

Mithrandir24601

@Blue So, this can also be written as $k\equiv k-\frac{2\pi}{a}$, right?

Anonymous

@Mithrandir24601 Yes

Mithrandir24601

@Blue So, take all points in your graph with $k\geq\frac{2\pi}{a}$ and take off $\frac{2\pi}{a}$

Anonymous

11:32 AM

@Mithrandir24601 Okay?

Secret

> If f is instead a vector field f : ℝn → ℝm, i.e.

f ( x ) = ( f 1 ( x ) , f 2 ( x ) , … , f m ( x ) ) , {\displaystyle {\mathbf {f}}({\mathbf {x}})=(f_{1}({\mathbf {x}}),f_{2}({\mathbf {x}}),\dots ,f_{m}({\mathbf {x}})),} {\mathbf f}({\mathbf x})=(f_{1}({\mathbf x}),f_{2}({\mathbf x}),\dots ,f_{m}({\mathbf x})),

then the collection of second partial derivatives is not a n×n matrix, but rather a third order tensor. This can be thought of as an array of m Hessian matrices, one for each component of f:

Mithrandir24601

@Blue Wait, Oops, I wrote that wrong...

Secret

Second covariant derivative

In the math branches of differential geometry and vector calculus, the second covariant derivative, or the second order covariant derivative, of a vector field is the derivative of its derivative with respect to another two tangent vector fields. Formally, given a (pseudo)-Riemannian manifold (M, g) associated with a vector bundle E → M, let ∇ denote the Levi-Civita connection given by the metric g, and denote by Γ(E) the space of the smooth sections of the total space E. Denote by T*M the cotangent bundle of M. Then the second covariant derivative can be defined as the composition of the two ...

Mithrandir24601

@Blue Take all points with $k>\frac{\pi}{a}$ and take off $\frac{2\pi}{a}$ and add $\frac{2\pi}{a}$ to all points with $k<-\frac{\pi}{a}$

Anonymous

@Mithrandir24601 Alright. Then?

Mithrandir24601

11:35 AM

i.e. All points outside the range $-\frac{\pi}{a} < k < \frac{\pi}{a}$ are equivalent to a point within that range, only with a different energy - these are the different bands, right?

Anonymous

Yes

Mithrandir24601

@Blue If you draw the graph, does that go some way towards explaining why $E$ decreases with $k$ sometimes and increases with others? - You've broken the symmetry of free space and replaced it with a periodic symmetry. As a result of this, all points in momentum space are equivalent to a point in the first Brillouin zone

(Only, with different energies etc.)

Anonymous

I was trying to think of it in a more physical perspective. I was thinking like: Say we are increasing momentum of electron i.e. $k$, then why does $E$ decrease sometime and increase sometime. I could get mathematical interpretation but not the physical one. Now I realize that it always increases but not uniformly (with increase in $k$). But still, I can't understand something: What does it mean to increase $k$ along a certain direction (say [1,1,1]) like in the picture I posted above

Mithrandir24601

Maybe to put things ever so slightly differently - you've broken the symmetry of free space and replaced it with a periodic symmetry. The corresponding conserved quantity with the symmetry of free space is momentum, so you've broken the conservation of momentum

Anonymous

11:48 AM

@Mithrandir24601 I get that. We've broken the conservation of momentum. But I don't know how to interpret a graph like the one above. What do they mean by plotting the Energy with $k$ in a particular direction ?

Anonymous

The graph would stand for something like "As you go along the direction [1,1,1] and increase momentum of electron the energy would first reach a maxima and then a minima and then increase again

Anonymous

That interpretation doesn't make much sense to me

Anonymous

Or I should not think of $k$ as momentum anymore?

Mithrandir24601

@Blue So, you increase the momentum of the electron, but momentum's not conserved in the same way. The electron reaches a certain momentum (at the boundary of the brillouin zone), then absorbs/emits a phonon

(this is the 'momentum's not conserved in the same way' bit)

Anonymous

@Mithrandir24601 Agreed

Mithrandir24601

11:53 AM

So, because of this, it's momentum is now negative (say) and at the opposite Brillouin zone boundary, with a higher energy

Anonymous

Okay

Mithrandir24601

So, you resume increasing momentum, only momentum's negative, so you're now increasing it towards $0$

Hence, what happens to the energy?

Anonymous

@Mithrandir24601 $E\to 0$ ?

Anonymous

From the graph

Secret

[Philosophy]

Good spreads quietly, while evil tend to be vocal

skullpatrol

11:57 AM

[Please wait]

Anonymous

@skullpatrol ?

skullpatrol

Pivotal moment in learning about to occur :-)

Mithrandir24601

@Blue Possibly... It depends

Secret

12:11 PM

blog.rsb.org.uk/the-biology-behind-a-fire-breathing-dragon

Mithrandir24601

@Blue At this point, it's maybe a good idea to look at diatomic chains and learn about acoustical and optical modes, which I suspect will give you what you're looking for (or at least help)

[read: I'm off to make lunch]

Anonymous

@Mithrandir24601 Okay...I'll check them out...I stiil didn't understand what increasing wavevector along a particular direction physically means...but ok

Mithrandir24601

@Blue It's just increasing momentum

Anonymous

@Mithrandir24601 And why would momentum increase along a particular direction? Or do we assume that it increases and plot the E for that assumption?

Anonymous

Or no...perhaps I'm grossly misinterpreting...it's just a graph

Mithrandir24601

12:16 PM

@Blue Oh, it's just the same as any plot - in this case you want to know how changing momentum changes the energy

Anonymous

@Mithrandir24601 Huh...right...I was overthinking

Anonymous

Thanks!

user84215

12:29 PM

in Mathematics, 5 mins ago, by MathematicsAminPhysics

Suppose that $\omega_{ij}=\sum da_{ik}a_{jk}$ . I do not know why a minus signappears in the exterior derivative of the above: $d \omega_{ij}=- \sum da_{ik} \land da_{jk}$

user84215

A space is needed between "sign" and "appears".

0ßelö7

12:41 PM

@MathematicsAminPhysics use Leibniz rule for 1-forms

heather

hello

Mithrandir24601

@heather Rytsas!

user84215

My problem has been solved. Thanks for your answers.

user84215

@0ßelö7 No, I think it does not need the Leibniz rule.

0ßelö7

@MathematicsAminPhysics oh I forgot you're clearly the expert here

user84215

12:55 PM

@0ßelö7 "expert" ???????????!!!!!!!!!!!!!!!!!!!!!!!!!!!!

heather

@Mithrandir24601 rytsas =)

user84215

I know little math and physics (a bit higher than high school math and physics)

heather

@Mithrandir24601 thanks for the link to the arXiv paper by the way, i'm reading it with interest.

Mithrandir24601

@heather I have a nice list of them that I'm planning to read over the next few weeks, as well as a textbook on nonlinear optics :)

heather

nice!

hmm...this is weird.

digikey gave a certain company as the manufacturer of the part, but i cannot seem to find the part on their website.

user84215

1:02 PM

@0ßelö7 I think it only needs to expand the differentials with respect to the bases dx 's. Am I wrong?

0ßelö7

@MathematicsAminPhysics Yes you're wrong and I said exactly what you need to use to get the answer you want.

ııııııııııııııııııııııı

Hi @MathematicsAminPhysics

user84215

@ııııııııııııııııııııııı Hi

ııııııııııııııııııııııı

Why are you studying exterior algebra? (if you're a high school student, and interested in physics)

user84215

@0ßelö7 Can you explain me why I am wrong?

0ßelö7

1:05 PM

@MathematicsAminPhysics No. I bet you didn't even look up the Leibnitz rule for forms.

user84215

I have seen it.

0ßelö7

Sigh...write it here.

user84215

Can you tell me first why my argument was wrong?

heather

@MathematicsAminPhysics do you even know what it is? just say if you don't, and then look the thing up. reading about it will probably tell you why your argument was wrong.

0ßelö7

Your argument isn't wrong, everything in exterior calculus can be done by expanding. You just did it wrong. I'm telling you how to do it easier.

user84215

1:11 PM

Thanks.

0ßelö7

The idea is simply that when you move the $d$ past the $da$, you pick up a minus sign. If that doesn't make sense, you just need to stare at it for a few hours.

Remember the definitions and write down a few examples.

user84215

I do not want to bother any one.

CooperCape

Yo guys I know that the arclength on a metric is $$\int_a^b\sqrt{d\mathrm s^2}$$ in Cartesian coordinates but does this work in spherical/polar etc. as well?

0ßelö7

Jesus, please use correct notation

CooperCape

I'm new to this apologies :/

0ßelö7

1:22 PM

@CooperCape Given a $C^1$ curve $\gamma:[a,b]\to M$, $\int_a^b \sqrt{g(\gamma',\gamma')}\, dt$ is the arc length.

This does not involve coordinates at all

CooperCape

Maybe my misunderstanding is rooted in the fact that I haven't ventured outside Cartesian... oopsie. thanks.

Wrichik Basu

Excuse me, and my apologies for disturbing. Do I lose reputation if I downvote a question?

CooperCape

nop

0ßelö7

@CooperCape you should check that $g(\gamma',\gamma')$ is coordinate-indepedent

Wrichik Basu

@CooperCape thanks.

CooperCape

1:25 PM

Okaaaaay....

Mithrandir24601

@CooperCape For context, @0ßelö7 is a mathematician and so, much more rigourous than most physicists

0ßelö7

@Mithrandir24601 I think every physicist worth his salt can check that.

CooperCape

yeah I'm still tryna work out the feck $\gamma '$ is...

I by no means claim to be a physicist

0ßelö7

the derivative of $\gamma$...

Mithrandir24601

@0ßelö7 The poor OP's not an undergrad yet though

CooperCape

1:27 PM

$\gamma (t)$ is the path?

0ßelö7

@Mithrandir24601 Balarka is like 13

CooperCape

and $\gamma '$ is velocity vector?

0ßelö7

I knew this when I was 15. Age isn't an excuse

@CooperCape yes

CooperCape

okay...

maybe I'm just a lil' slower than you...

0ßelö7

@CooperCape write out $g(\gamma',\gamma')$ in index notation

CooperCape

1:31 PM

$g_{ij}(\gamma _ie_i,\gamma _je_j)$ as a guess but idk man...

0ßelö7

$$g_{\mu\nu}(x(t))\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}$$

CooperCape

Okay right...

Mithrandir24601

@0ßelö7 I agree, but not having learnt it yet is an excuse for not knowing it :P so they're entitled to a bit of slack. At least, until they've learnt it...

0ßelö7

@CooperCape so now go $x\to \tilde x$ and compute this in the new coordinates

CooperCape

Gimme like 2 hours crikey...

0ßelö7

1:37 PM

@CooperCape do you know how $g_{\mu\nu}(x)$ transforms?

CooperCape

Nope...

or what $\tilde x$ means butttt yano...

0ßelö7

it's just another coordinate system. Call it $y$ if you want

what book are you using?

CooperCape

I'm not using any book just collating basic knowledge...

0ßelö7

use a book

Carroll or Zee

CooperCape

okay I'll find a boook

this was the most relevant I found by Carroll?

Then again I don't value my education at £43.12 for a book I probably won't understand for at least the next 1-2 years.

0ßelö7

1:49 PM

"Spacetime and geometry"

It has a red cover

CooperCape

this?

0ßelö7

Yes

CooperCape

kinda looks like the older version of the other one tho

okay cool

I'll... see what I can do to get it...

0ßelö7

The other one is the Indian edition

It might be the same book, I don't know.

CooperCape

Oh right... thanks :p

Mithrandir24601

2:08 PM

@CooperCape Have you tried asking your physics teacher if they happen to have a copy?

CooperCape

Back to school Tuesday - Might be able to ask the library to order one in? otherwise I'll try second hand...

Or I could be a bit naughty and see if it's online in pdf format somewhere but meh...

Mithrandir24601

In my experience, school libraries don't seem to be very good at getting any textbooks outside of the syllabus :/ No harm in asking though! Second hand is usually a good bet, certainly for such a popular textbook

heather

I know I've posted this in here before, but please join and vote the questions on the quantum computing proposal! We only need 9 more questions with a score of 10 or more before reaching the commitment stage

143

Quantum Computing

Proposed Q&A site for computer Professionals, Quantum-Computing, Quantum Software Programmers, Quantum Data Center Architects, and Quantum Computing Engineers.

Currently in definition.

2

Mithrandir24601

I'm totally with @heather on this ^ :D

heather

=)

0ßelö7

2:11 PM

@CooperCape there is

CooperCape

oooh boyooyo

heather

it's very active in terms of followers.

CooperCape

that's £96 I don;t have to spend.

0ßelö7

@heather how is that different from physics?

CooperCape

Can you link me it if you know where it is? If you can't dw I'll do some searching.

Mithrandir24601

2:13 PM

@0ßelö7 It involves physics, maths, computer science and (quantum) engineering, so it's not just physics...

heather

@0ßelö7 quantum computing combines computer science, physics, math, engineering - it's an intersection of multiple topics deserving of its own site.

@Mithrandir24601 beat me by a second =P

0ßelö7

@CooperCape libgen

be careful not to get a virus

heather

besides, from the number of followers it definitely deserves its own site. I mean, bioinformatics was deserving of its own site.

CooperCape

Okay lol... thanks for the tip :p

heather

ditto for quantum computing.

Mithrandir24601

2:14 PM

Although it's worth mentioning that the description in the proposal is... Not great

heather

@CooperCape: not the full textbook but these are some lecture notes

CooperCape

@heather thanks :p

heather

(quite legally obtained)

np

CooperCape

Okay @0ßelö7 the torrent is 'banned by the high court' so I guess I'll have to vpn it... (100% legit)

okay the french government apparently don't care.

0ßelö7

Banned by the high court? What country are you in

CooperCape

2:27 PM

uk... I clicked on the wrong link, however

Downloaded as a .torrent file jsut gotta sort it out now :p

vzn

@Mithrandir24601 yes, notice S is an equality. since when are important physical laws expressed as inequalities? this has been remarked on occasionalliy. in this way it has a thermodynamics-law-like feel to it. ps do you have some objection to discussing this in another room? would like to do sohttps://chat.stackexchange.com/rooms/9446/theory-salon

CooperCape

when the ETA says $\infty$ facepalm

@0ßelö7 It's just given me a file named '1390317a05846205b53ae7eecbf0179e' Does it need to be unpacked or what?

Balarka Sen

top of the morning to you ladies

Prathyush Poduval

@CooperCape go to libgen, it has non-torrent pdf downloads

CooperCape

oh right lol

oopsie

Prathyush Poduval

2:43 PM

@WrichikBasu Yes, you do lose reputation (2 points) to prevent misuse of the tool

Mithrandir24601

@PrathyushPoduval I don't think this is the case for downvoting questions

Prathyush Poduval

Is it only answers then? Thats weird.....

Yeah you're right, I stand corrected

@WrichikBasu physics.meta.stackexchange.com/a/5358/55703

@CooperCape If you are looking for a good relativity book, A.Zee is the best book for beginners. I highly recommend it

The author literally speaks to you through the book.

Balarka Sen

@0ßelö7 why are you giving the explaining the Riemannian length function to a kid

heather

3:02 PM

@DanielSank hello =)

Balarka Sen

@0celo7 "YouTu🅱e" - New YT logo suggestion by an infamous anonymous.

Anonymous

@PrathyushPoduval Well, I learnt something new today. "I stand corrected" means "I admit I was wrong" and not "I was correct and still am". I misinterpreted the phrase when @SirCumference used it last time. :P

Balarka Sen

most probably you sit corrected

Anonymous

lol

Sir Cumference

@Blue :P

Avantgarde

3:19 PM

Childish Gambino - Redbone (Official Audio)

►

Balarka Sen

that's the meme song isn't it

DanielSank

@heather yo

My colleague said your project sounds fun.

Anonymous

@Avantgarde Meh. You chose this dp instead of the amazing skull with the hanging mustache? ;)

yeah haha

@Blue no good? I love nature

Balarka Sen

3:21 PM

you know epic sax guy right

Anonymous

@Avantgarde Just joking :)

Avantgarde

Though that skull was amazing. It's the cover art of an album

@BalarkaSen who?

Balarka Sen

Epic slash ultra sax guy. Google it!

Avantgarde

@BalarkaSen oh yeah, seen that before

heather

@DanielSank =)

Avantgarde

3:31 PM

@BalarkaSen did you like the spacey metal?

Anonymous

@0ßelö7 I hate to say this but it often seems like you're boasting. I admire knowledge but I admire humility even more. Anyhow, I am just trying to give some constructive suggestion. Hope you get my point.

5

DanielSank

^

Prathyush Poduval

@Blue I learnt it from @SirCumference :P

Sir Cumference

@PrathyushPoduval Haha :)

Is it a U.S. saying?

Prathyush Poduval

3:46 PM

It sounds like a classy phrase worth remembering :P

John Rennie