The h Bar

Sir Cumference

00:01

@Mithrandir24601 Oh, I said that because I thought I saw you enter the chat

Emilio Pisanty

@Danu ↑ ?

Semiclassical

01:27

@EmilioPisanty yet another reason why going into finance seems like selling my talents to the devil

vzn

03:49

in theory salon, 1 min ago, by vzn

Major Quantum Computing Advance Made Obsolete by Teenager / Hartnett, Quanta

Semiclassical

>For quantum computing, Tang’s result is a setback. Or not. Tang has eliminated one of the clearest, best examples of a quantum advantage. At the same time, Tang’s paper is further evidence of the fruitful interplay between the study of quantum and classical algorithms.

>“Tang is killing [Kerenidis and Prakash’s] quantum speedup, but then in another sense Tang is giving a big improvement and building on what they did. Tang never would have come up with this classical algorithm but for their quantum algorithm,” Aaronson said.

also, here's Scott's post on it:

scottaaronson.com/blog/?p=3880

heather

04:46

ouch, even wikipedia's up to date now:

> 2018 Fields Medal. Birkar's Fields Medal was stolen from him on the same day it was awarded to him.[14]

Sid

09:45

@JohnRennie are you there? I am stuck on transformers

John Rennie

@Sid the film? :-)

Physics Meta

0

Q: What do Phys.SE mean by "Physics"?

Right so I posted a question about something (not really important because the response is the important part). One of my responses (who I believe flagged the article) complained that it was not Physics because it was engineering. Now I know a few engineers and there is no doubt that what they ...

Sid

If we connect a load to the secondary circuit and an input voltage to the primary, then obviously there would be a flux due to the primary input current. Obviously this flux will give rise to an induced emf in the secondary which will give secondary current and by extension another secondary flux

John Rennie

OK ... ?

Sid

Which makes me confused. Because then there would be two flux in the magnetic core Which should cancel each other out by Lenz Law, no?

John Rennie

09:59

@Sid sorry, my phone rang.

John Rennie

10:26

@Sid you still around?

Beta Decay

In relativistic equations can the terms in $c$ be considered to be just a unit conversion. For example, does the $c^2$ in $\frac{1}{\sqrt{1-v^2/c^2}}$ or $E=mc^2$ have any physical significance apart form the conversion factor between geometrised and normal units?

John Rennie

It is a unit conversion, but it's also the maximum speed any object can travel.

So it has physical significance.

Beta Decay

Ah yeah, that makes sense

Thanks

John Rennie

You can show it's the maximum speed from the invariance of the interval.

Sid

10:45

@JohnRennie sorry i was in class

John Rennie

@Sid no problem :-)

harambe is currently asking me something in the Problem Solving room so I'll be brief.

When the secondary is open circuit the primary behaves like an infinite resistance i.e. no current flows.

That's because the voltage applied to the primary creates a magnetic field and this creates a back EMF that opposes the applied voltage.

Since the secondary is open circuit no current flows in the secondary so it has no effect.

When you connect anything to the secondary current now flows in the secondary and this creates a field in the secondary that reduces the back EMF in the primary. That means current now flows in the primary.

Anonymous

12:31

@Sid That secondary reverse flux only reduces the primary flux rather than completely cancelling it

Anonymous

As always, write down the equations to verify!

user1732

@Semiclassical tell that to Jim Simons :P

user1732

12:55

@ACuriousMind do you think it speaks highly of "human nature" to steal anything at such an event?

Emilio Pisanty

@DanielSank have I linked you to this album?

Clase de Tango

►

user1732

\o @EmilioPisanty

Semiclassical

13:32

hi chat

Green

@Semiclassical hi!

Semiclassical

So we've got Green and Blue as active chat participants

where's Red?

danielunderwood

hello

Green

If I have a hydrocarbon gas and want to zap the H-C bonds specifically and leave the C-C and C=C bonds alone, can I just illuminate the molecule with a laser having a wavelength tuned to the vibrational frequency of the H-C bonds?

@danielunderwood hi again! Your info from a few days ago was super helpful.

@Semiclassical I can make a sock puppet to fill out that color triad.

Semiclassical

lol

probably best not to take the joke that far

danielunderwood

13:44

Glad to help!

Semiclassical

14:07

I'm looking at some discussion of Jonsson's double slit experiment (1961) and it takes the wavefunction incident on the slits to be $\psi\sim e^{i (k_1 x+k_2 y)}$ where the slits are at $(0,\pm Y)$

But why is $k_2$ there?

I can see why you need $k_1$ in there: the electrons have to moving towards the screen, after all

the source I have indicates that this implies the waves transmitted from the slits will have group velocities $\pm \hbar k_2/m$, which (aside from the sensible units) I'm not really understanding

Semiclassical

14:25

the other confusing thing is that they claim that, if $k_2<0$, then the wave packets from the slits won't approach each other

bolbteppa

16:09

5

Semiclassical

@bolbteppa my version of that is "hahaha you thought academia was actually a stable career path"

3

Yellow

hi everyone! Iwant to learn python so I can do physical simulations, based on an advice by my physics olympiad teacher. I have no knowledge whatsoever on python and programming in general. Any advices on learning the basics?

bolbteppa

Absolutely

Anonymous

@Yellow Socratica is good

Anonymous

@Semiclassical We've got @Yellow instead ;)

Semiclassical

16:20

works for me

Yellow

@Blue I was thinking on wheter I should convert to purpleism?

Anonymous

@Yellow You should. At least that way you'd have some shade of red in you, completing the color triad in hbar.

Yellow

Maybe I should just be red

But what is Red Blue?

Semiclassical

ah, red vs blue, the eternal conflict

Yellow

arxiv.org/abs/1802.00802

check this out

Semiclassical

16:25

oh hey, SYK model

I've heard of that

Yellow

Did anyone here participated in ipho?

danielunderwood

16:37

@Yellow pick out something you want to write code for and try to write it by googling problems as you run into them. You'll fail a lot, but also learn a lot

Oh and have an IDE with autocomplete

Secret

Have you ever had a weird feeling where you felt like you are not only not feeling anything, your sense of touch, sight, hearing etc. is like something barely above a baseline, but also that it seems like you momentarily seemed to be not registering your surrounding environment as if you are on the brink of losing consciousness without actually fainting?

danielunderwood

@Semiclassical From what I hear, after fighting through getting a postdoc and a professorship, you spend your remaining years as a junior professor fighting for funding. Sounds thrilling

Avantgarde

Huh

Everything alright, Secret?

Secret

(and that the reason you are aware that you are almost not feeling everything as if you never existed because you noticed the emotion level is so low from what you are familar with in normal life)

Uh... It's a weird emotion, I got that after I looked at this artwork (or actually, visited an art gallery which has this artwork)

in The Symposium, Jul 11 at 20:05, by Secret

I don't think I am suffering something because my life is currently pretty usual

But this emotion, which is like numbness on steroids blended with as if you are being on a anesthetic, is very strange

I tried asking my arts and neuroscience professors, they do not know of anything similar

Avantgarde

I don't feel anything unusual looking at this.

Secret

16:44

yeah so do I, but somehow the gallery has some effect on me according to my art professor, as they try to rationalise why I felt that emotion 2 days after when I was asleep

Avantgarde

Interesting

Secret

I wonder... if this emotion is because I spent too much time studying about the metaphysics of nowhere in philosophy things...

danielunderwood

17:26

Do you guys ever get to working with music and your ears finally turn on after the music has changed to something insane?

enumaris

wut

Semiclassical

the case I run into more is if I leave Youtube on autoplay

it'll be playing something i recognize, and so I'll tune out a bit

but eventually I'll tune back in and be like "where the heck did that come from"

danielunderwood

Yeah it's definitely with youtube autoplay. I guess it's because I change what music I listen to a lot

Mine just got to this

Loituma - Ievan Polkka (DOPEDROP x Adryx-G Bootleg)

►

it's...interesting

enumaris

hmmm

Semiclassical

it depends on what video you start autoplay on, of course

if it's part of a series, it'll usually stay on topic

if it's just music, though...yeah, that's subject to change

in other news, I think I just accidentally rediscovered the difference between group and phase velocity for a gaussian wavepacket. while I'm no longer confused, I'm a bit annoyed at myself for being mixed up in the first place

danielunderwood

17:34

And to go alongside that, my next up list has John Denver in it...I think I broke their autoplay

I would have thought that they were the same for a Gaussian. Though I suppose I've never really thought about it

Semiclassical

17:55

@ACuriousMind Got a question for you, based on your knowledge of quantization: What’s the classical phase space of a spinning top like?

I remember that there’s some weirdness but a good reference is eluding me

Sid

@Blue Huh? Think of an ideal transformer. Primary Emf is the same as secondary emf. Then, why wouldn't the secondary flux be the same as the primary flux?

Anonymous

@Sid "Primary Emf is the same as secondary emf." Why should that be true?

Sid

assuming the number of turns are same, of course

Anonymous

@Sid Well, if the number of turns are the same, then the transformer is useless.

Anonymous

A "transformer" is either used for stepping-up or for stepping-down voltage/current

Sid

18:05

Well, okay. Think of it, this way. Secondary flux is produced due to the secondary current, which is in turn produced due to the secondary emf which is induced by the flux(by Faraday's Law). Then, wouldn't the secondary flux be equal and opposite to the primary flux(by Lenz Law)?

Anonymous

@Sid Nope, the "number of turns" is really the key here

Anonymous

Can you write down the equations?

Anonymous

That way you'll understand better

Sid

Equations of? Primary and secondary emf?

Anonymous

Yup

Sid

18:07

That would be $E=4.44*f*(phi_m)*N$

where N= Number of turns.

and blah blah blah

Anonymous

Okay, so $E_1=-N_1\frac{d\Phi}{dt}$ and $E_2=-N_2\frac{d\Phi}{dt}$. By the way, are we considering magnetic flux leakage and winding resistance or not?

Sid

@Blue Ideal transformer bro.

We will go to that later. Not now

So, basically you are saying that since the Number of turns are not equal, the induced flux due to the currents will be different?

Anonymous

So see, in your case $R_1,X_1,R_2,X_2=0$

Anonymous

But when you have placed the secondary load $Z_L$

Anonymous

18:14

The voltage across the secondary circuit's inductor changes, yes?

Anonymous

$E_2$ is the initial emf across the secondary circuit's inductor

Novalium Company

Hello everybody! :-)

@Blue How are you?

Anonymous

Let's actually write down the equation

enumaris

hi

Sid

Changes? There's only a current flowing across the load. Nothing else. KVL says that E2 will be the same as I_2 times Z_L

Anonymous

18:17

@Sid Ah, you're right since $R_2,X_2=0$. But in that case $\phi'=0$ isn't it?

Anonymous

I don't see why you say $\phi-\phi'=0$ by Lenz's law

Anonymous

@NovaliumCompany Hi, I'm fine!

Anonymous

How's school?

Novalium Company

Still summer holiday. 1 Month and 15 days until school.

Sid

@Blue Because? Flux is due to the secondary current, no? And the secondary current is due to the primary flux(or rather secondary voltage which is in turn due to the flux). Hence, by Lenz Law, Secondary flux will oppose the primary flux, no?

Anonymous

18:20

@Sid The secondary current $I_2$ is already flowing in the direction specified by the flux $\phi$

Novalium Company

I've been very busy with the game, but I just finished the main update. Gonna upload it and advertise tomorrow. After that I'll finally have time for the sweetest thing in life, physics and math.

Sid

@Blue Okay, so?

Anonymous

@Sid $\Phi_2'=L_1I_1-L_2I_2$ and $L=\frac{\mu_0 N^2A}{l}$

Anonymous

I don't see any reason why $L_1I_1=L_2I_2$

Anonymous

Remember that here $\Phi_2=L_1I_1$ which was shared by both the primary and secondary coils

Anonymous

18:29

$\Phi_2\mapsto \Phi_2'$

Anonymous

12 mins ago, by Blue

@Sid Ah, you're right since $R_2,X_2=0$. But in that case $\phi'=0$ isn't it?

Anonymous

The second part of this sentence was wrong ^

Anonymous

The net flux through the secondary coil will reduce as soon you place a load

Anonymous

"Hence, by Lenz Law, Secondary flux will oppose the primary flux, no?"

Anonymous

The secondary flux will oppose the primary flux, yes. But there's no reason why it should cancel it out

Anonymous

18:36

Novalium Company

Seeing the chat and I'm just thinking how much I want to advance but so little time...

Danu

> ???
> Profit

Novalium Company

I hope someday I'll achieve your level of knowledge, guys. :-)

Anonymous

$$E_P=-L_P\frac{d(I_P)}{dt}$$ and $$I_P=\frac{V_P}{j\omega L_p}$$

Anonymous

$$\Phi_{\text{shared (per turn)}} = \frac{L_PI_p}{N_p}$$

Anonymous

18:41

$$\Phi_{\text{shared (for Secondary)}} = \frac{L_PI_pN_s}{N_p}$$

Novalium Company

Ok, bye everybody. I'll be back soon with a lot of free time to gain a lot of knowledge :D Thanks for being such an awesome community!

enumaris

o.o

later

Anonymous

@NovaliumCompany See you. Sorry, couldn't chat much today :P

Novalium Company

@Blue No problems! :-)

Anonymous

$$\Phi_{\text{shared (for Secondary)}} = \frac{L_PI_pN_s}{N_p} = L_S I_s$$

Anonymous

18:43

$$\implies I_s = \frac{L_PI_pN_s}{N_p L_S}$$

Anonymous

Anyhow, basically we wanted to show that $L_PI_p-L_SI_s\neq 0$.

Anonymous

Which is quite evident from here, since $N_s\neq N_p$

Anonymous

All we get is $L_PI_p(\frac{N_s}{N_P})=L_SI_s$.

Anonymous

@Sid Okay? So the "number of turns" is important!

Anonymous

This is exactly why a transformer having an equal number of turns in both primary and secondary is useless (due to the reason you yourself mentioned)! Due to Lenz's law the reverse back EMF would cancel the primary EMF.

Anonymous

18:54

BTW:

Anonymous

Isolation transformer

An isolation transformer is a transformer used to transfer electrical power from a source of alternating current (AC) power to some equipment or device while isolating the powered device from the power source, usually for safety reasons. Isolation transformers provide galvanic isolation and are used to protect against electric shock, to suppress electrical noise in sensitive devices, or to transfer power between two circuits which must not be connected. A transformer sold for isolation is often built with special insulation between primary and secondary, and is specified to withstand a high voltage...

Anonymous

@Sid But one important fact to remember is that even if the secondary coil cancels out the shared flux, the primary source will start producing more current, keeping the net shared flux constant.

Anonymous

(which is of course due to the fact that the primary coil will counteract any reduction of flux)

ACuriousMind

@Semiclassical Since the angles as the space of generalized coordinates shoud be $\mathrm{SO}(3)$ (you can specify a rotational position by specifying a reference orientation and a rotation to reach it), that should be the cotangent bundle of $\mathrm{SO}(3)$. Maybe the "weirdness" you remember is that configuration space is a bona fide manifold that's not just a product of circles and $\mathbb{R}$s?

Semiclassical

quite possibly

I think the weirdness I have in mind is that, when you do the cotangent bundle of R^n, you get R^n x R^n

so there's an easy distinction between coordinates and conjugate momenta

Emilio Pisanty

19:09

how is that weird?

Semiclassical

it's not. but my recollection was that this doesn't work for spin

Emilio Pisanty

you want weirdness, I'll show you weirdness

Semiclassical

on the quantum side, that should just be the fact that spin operators don't have canonical conjugates

Emilio Pisanty

-1

Q: Where universe exist?

As we know there are many dimensions exist, like matters are exist in spacetime. Black holes, Worm holes and white holes are exist in universe, But in which dimension universe is existing. Is that dimension have no spacetime, Is universe occupies space, Is universe in higher dimension or not ha...

cosmology universe big-bang spacetime-dimensions

Semiclassical

lol

I prefer bounded weirdness

Emilio Pisanty

19:10

Why you hold this question? — Saif KhAn 21 mins ago

Is this question have no any answers... Please help me. — Saif KhAn 24 mins ago

Semiclassical

All your answer are belong to us.

ACuriousMind

@Semiclassical The same is true for $T^* \mathrm{SO}(3) \cong \mathrm{SO}(3)\times \mathfrak{so}(3)$. The tangent bundle of all Lie groups is trivial, and so is their cotangent bundle.

Semiclassical

Hmm.

Emilio Pisanty

@ACuriousMind watch who you're calling trivial, eh

they say SO(3) doesn't take too kindly to that kind of language

it's been known to come round at night and rotate your bed so it's on the wall and you fall right down as soon as you wake up.

Semiclassical

a sentence which seems pertinent, looking at Altland and Simons, is: "As a consequence, the classical phase space of the system, the sphere, cannot be covered by a global choice of coordinate system."

that's in regard to "The classical action of a spin is one of a massless particle (there is no standard kinetic energy term in Eq. (3.48)) moving on a unit sphere. The particle carries a magnetic moment of magnitude S. It is coupled (a) to a conventional magnetic field via its magnetic moment, and (b) to a monopole field via its orbital motion."

ACuriousMind

19:17

However, this space probably indeed presents a challenge for naive quantization procedures, since you can't really exhibit "position operators" whose collective spectrum "is $\mathrm{SO}(3)$"

Semiclassical

@ACuriousMind that sounds like the right idea, but why is that forbidden?

ACuriousMind

@Semiclassical Well, in naive quantization, you would just seek representations of the CCR for the local coordinates, but Stone-von Neumann says that if you represent the CCR (+Weyl relations, but that is a subtlety naive procedures can't see) you always get the same representation, i.e. infinite continuous spectra $\mathbb{R}$ for all position and momentum operators

Semiclassical

A&S reference an article by Stone which does seem to address this pretty well

(this one, for reference: inspirehep.net/record/262703?ln=en)

ACuriousMind

I mean, that's generally a problem naive quantization has with systems whose configuration space is not a product of circles and $\mathbb{R}$ - you can do the particle on a ring still rather "naively", but if you go to general manifolds where you don't have nice "almost global" charts ($\mathrm{SO}(3)$ arguably has a nice "almost global"chart in the Euler angles) you have no idea what to do

Semiclassical

sounds right

ACuriousMind

19:23

@Semiclassical Oh, I'm by no means saying you can't quantize this system. I'm just saying that you need to actually use a well-thought out quantization procedure instead of "let's morph the Poisson bracket into a commutator and see where we end up" :P

Semiclassical

I'll note that the article of Stone includes the statement that " In this case the configuration space and the phase space coincide. "

@ACuriousMind oh, sure. what I vaguely remember is stuff like geometric quantization

not that I remember it well enough to say anything useful...

ACuriousMind

@Semiclassical Just a difference in terminology. I call the base manifold of generalized coordinates "the configuration space" and its cotangent bundle of generalized coordinates and momenta "the phase space".

Semiclassical

Wouldn't the configuration space and phase space in general coincide, then? (in the sense you're attributing to Stone, I mean)

ACuriousMind

I now occurs to me that I have no corresponding name for the tangent bundle, which is the Lagrangian equivalent of phase space

@Semiclassical ...how could a manifold and its cotangent bundle ever coincide, unless you allow the perversity of the empty manifold?

Oh, I don't know what Stone means by "configuration space"

It may well be that he is using "phase space" for the cotangent bundle and "configuration space" for the space of physical states after possible reductions (due to gauge symmetries)

Semiclassical

he indicates that the configuration space in his case is S^2, though that's not a definition

"By "spin" I mean a system with classical variables $J_i (i = 1,2, 3)$ whose Poisson brackets obey $\{J_i,J_j\} = \epsilon_{ijk}J_k.$"

" The phase space for this system is four-dimensional being $T^*(S^2).$"

...that seems inconsistent

oh, wait

that's for the case of motion of a rigid rod about its centre of mass

he cites some other sources, maybe those will clarify

DanielSank

19:41

@EmilioPisanty Not previously.

Semiclassical

@ACuriousMind I mean, the main upshot of what he's saying seems to be: "There is no globally defined one-form whose exterior derivative is equal to the area form, i.e. it is a generator of the cohomology group H2(S 2, R), so we cannot make a global decomposition into p's and q 's. "

I can't parse what the configuration space / tangent bundle / cotangent bundle (in your sense of the words) are supposed to be in this case, which is confusing

ACuriousMind

@Semiclassical I think you're getting something mixed up here. You initially mentioned the spinning top, which I take to be a system whose generalized coordinates are Euler angles, i.e. the base manifold of generalized coordinates is $\mathrm{SO}(3)$. But then you mentioned $S^2$. The (co)tangent bundle of $S^2$ is not trivial and so there is indeed no such global decomposition.

This is not true for $\mathrm{SO}(3)$, and $S^2$ is not the space of generalized coordinates for a rotating system like the spinning top.

Semiclassical

Yeah, I think I was wrong to say the first one

(which in fact, looking at stone, he specifically calls out in a footnote. derp)

So I really should have been saying S^2 in the first place

ACuriousMind

Ok, then there is nothing mysterious here: If you have a manifold with non-trivial cotangent bundle as your space of generalized coordinates, then there is no global split into coordinates and momenta.

Semiclassical

mmkay

enumaris

19:50

mmkey mmdokey

Semiclassical

So the configuration space would still be S^2, but there's no neat way to write the cotangent bundle as S^2 x (momentum space)

ACuriousMind

Not only no neat way, there is literally no way.

Semiclassical

right

So does that mean there's no sense of conjugate momentum in the first place?

Globally, I mean

(locally there is by Darboux, iirc)

ACuriousMind

There's also no notion of "position" globally :P

Semiclassical

huh

So I guess the equation of 'configuration space' with 'space of generalized coordinates' breaks down

ACuriousMind

19:52

The phase space in this case is just a blob whose points correspond to states of the system, but you can only locally and non-uniquely split these states into generalized positions and momenta

@Semiclassical Hm? The space of generalized coordinates is still $S^2$.

But there is no way to uniquely identify an $S^2$ inside the phase space $T^\ast S^2$, as would be the case if we had $T^\ast \cong S^2 \times \mathbb{R}^2$.

Semiclassical

I guess I'm confused by how to state that in terms of terminology

ACuriousMind

I didn't use enough terminiology? D

Semiclassical

And how to distinguish 'configuration space' from 'generalized coordinates'

ACuriousMind

As I said, different people use different definitions of "configuration space" and you still haven't said what Stone understands by it.

Semiclassical

bleh, fair enough

enumaris

19:55

details...details...

Semiclassical

I guess it again comes down to what 'configuration space' supposed to mean

enumaris

space of configurations

Semiclassical

I don't suppose you have a source with sensible definitions on this?

enumaris

tautology man strikes again!

ACuriousMind

@Semiclassical Again, the problem is that different people mean different things here. I am not under the impression that there is a standard usage of "configuration space". I've seen it used for the space of generalized coordinates, for the tangent bundle (i.e. the space of generalized coordinates and velocities), and quantumly for both the Hilbert space and the projective Hilbert space :P

Semiclassical

19:58

ew

Emilio Pisanty

@DanielSank so? whatcha think?

Semiclassical

guess i gotta find a source which is consistent about it

DanielSank

20:36

@EmilioPisanty Haven't listened yet.

y u no read my paper draft?

Emilio Pisanty

@DanielSank it's a bit too long and too intense than what I have time for at the moment

I really want my current paper to be out the door by Monday or so

DanielSank

21:11

@EmilioPisanty That is a legitimate answer.

Emilio Pisanty

@DanielSank where are you going to send that, by the way?

or is it 'publish' in the not-completely-formalized sense?

DanielSank

@EmilioPisanty arXiv :-P

I guess I could send it to a journal, but no idea which.

Recommendations?

Emilio Pisanty

@DanielSank you should

at least once the document has settled

@DanielSank ... but I don't know where =P

maybe Am. J. Phys. might be interested?

you don't lose anything by making an informal inquire as to whether it's within scope, once you arXiv it

DanielSank

@EmilioPisanty True dat.

Emilio Pisanty

J. Phys. B publishes tutorials from time to time, though you might want to look at others in that family?

J Phys B is AMO physics

that paper is on the edge between that and condensed matter, I guess?

enumaris

21:40

hmm

bolbteppa

23:30

0

Q: SUSY Transform on a Superfield

Given a 4D superfield of the form $$\Phi(x,\theta) = A + \overline{\theta} \Psi + \frac{1}{4} \overline{\theta} \theta F + \frac{1}{4} \overline{\theta} \gamma_5 \theta G + i \frac{1}{4} \overline{\theta} \gamma_5 \gamma^{\mu} \theta B_{\mu} + \frac{1}{4} \overline{\theta} \theta \overline{\the...

supersymmetry

:(

Very defeated

Took like a day to guess at how to get the derivative expression alone

user2236

Ouch!

danielunderwood

Wow that took like a solid second to render the mathjax

Sir Cumference

When you realize the year 2040 is closer than 1995

danielunderwood

23:45

That makes me feel old considering I was born before 1995

only a year, but still

enumaris

23:57

hmmm

Transcript for