But when have we disagreed?

Ah I see. (Confusing the two is an achievement of mine anyway)

@Bass I think everyone has done that at some point

Not me. Never. Nope.

There goes my dream. I thought I achieved something.

@Bass I'm sure you're special in some other way ;P

@ACuriousMind I don't consider the [] thing a disagreement.

You're simply wrong...it's a different matter.

@MAFIA36790:DO share your reviews after crossing the 51st or 57th page(if you find it suitable for beginnerr..I wil buy it)

@IceLord That is correct!

@DavidZ I was, most assuredly, joking.

@IceLord Flagging...

@IceLord Wait a moment, why is there a $-3$ on the r.h.s.? The dimension of the delta function also depends on $d$.

@ACuriousMind yeah I just failed to realize that "0" is a perfectly reasonable name for a vector which is not the zero vector.

@ACuriousMind you're a girl

@IceLord "inverse length" would have mass dimension 1, since "length" has mass dimension -1, so I don't know what that means

@IceLord What is the $d$ in your equations then?

@0celo7 hm?

Your avatar is a gril

@IceLord Then I don't understand why you're writing $d$ and not 4 :P

@IceLord No, I'm not saying it's 1!

Did you read the problem correctly?

Reread the problem and answer and make sure you computed the same thing

@ACuriousMind what was your last profile pic and what is it now?

He is a girl now

So we can practice getting you a date @Obliv

also on your profile it says "interested in non-perturbative QFT" @acuriousmind why is that? Isn't perturbative more accurate to reality?

Ok @Obliv

@0celo7 I don't think a female A.I. would be a good practice partner.. no offense acm

Wait another moment. Why $[\Pi][\phi]$

Supoose you see @ACuriousMind in Whole Foods

@0celo7 You around?

How would you ask her out?

I'm now confused as to how one is supposed to get the dimension from the CR at all. If I do it the correct way, $[\phi]$ just drops out.

@BernardMeurer it's Kat's birthday

@IceLord $[AB]\neq[A][B]$.

@0celo7 I sent her a happy birthday text first thing in the morning :)

Thanks though

@ACuriousMind we've been over this.

Had my first day of uni today guys :)

That IS true.

I wouldn't pick up girls at whole foods first of all. But probably something along the lines of "are those melons fresh?"@0celo7

@0celo7 Not in the meaning we're currently using. Deal with it.

@bernard nice how did it go? you're a real boy now

Ok @Obliv I will show you how it's done

@Obliv It was pretty good, actually

@0celo7 I want to ask a question about how to prove a number is irrational on MathSE, which tags are appropriate?

Ask ACM, he will tell you how it works.

@IceLord Ooops

Ah, no, that was right

@IceLord It's not that exactly

If I multiply two quantities, then I have to add their mass dimensions

Lol

Shitty notation. Confirmed.

@0celo7 Dood halp

@BernardMeurer Something like <irrational numbers> tag maybe 8^)

I'm in class.

maybe number theory?

what number is it?

@0celo7 24

@Obliv But what is the branch of maths that worry about which number belongs in where? (I, R, C, N,...)

@IceLord What do you mean, "but in exponents"? The idea is that we have a single mass unit $m$, and $[\text{number}\cdot m^k] = k$, right?

@0celo7 You'll see, I don't want to spoil the surprise

Yeah, so $[AB] = [A]+[B]$.

@bernard my best guess is number theory

@Obliv the last one was Kreia/Darth Traya from KotoR 2 and this is Triss from The Witcher series.

Terrible notation.

@Obliv Alrighty

@ACuriousMind Say it with me.

@Obliv No , "perturbative" means that you use an expansion in a small parameter and hope that the first terms in that expansion are close enough to the full result. "non-perturbative" essentially means "exact" in this case

@IceLord No, $[a][a] = [a]^2$, but $[aa] = 2[a]$.

Lololo

Just replace the $[\cdot]$ mentally with $\ln$ to see how it works.

@IceLord I will explain this properly later.

I'm pretty sure that one can't deduce the field dimension from the CR because the r.h.s. is the same dimension regardless of field type, but one must be able to write them down for all fields in all dimensions. Are you sure the problem states that one can get the dimension from the Cr alone, @IceLord?

@ACuriousMind :^Another witcher fan spotted

@ACuriousMind What? The delta function changes by dimension.

So you should be able to get it from the CR.

@0celo7 I said the r.h.s. is the same dimension regardless of field type, not regardless of (spacetime) dimension.

We have $[\Pi] = d-[\phi]-1$ and $[\delta] = d-1$. So $[\Pi\phi] = d-1$ regardless of $[\phi]$.

Hm. Do I recall Fermions have a different mass dimension than scalars

@0celo7 Yes, 3/2 for a 4D fermion

Ok...I agree with you then.

But isn't the dimension of pi coupled to that of the field?

Yes, it's $d-[\phi]-1$, as already states multiple times

Yeah. So solve the equation for phi!

@0celo7 What equation?

the. CR

Holy crap

My phone has stopped autocorrecting.

If you try to evaluate $[\Pi\phi]$ you get $d-[\phi]-1+[\phi] = d-1$ even without using that the r.h.s. is the $\delta$ function

Aha.

So you can't solve for $[\phi]$ because it just drops out

Then I agree with you, it seems unfeasable to get it from the CR.

@ACuriousMind Yes I can do basic PhD algebra

So, use the Lagrangian. What's the big deal?

None, except that it sounded to me as if @IceLord was doing a problem that stated one can get the dimension from the CR alone.

Doubtful.

He's probably reading the problem wrong.

@Obliv "Are those melons fresh?"???

@ACuriousMind So...suppose I have a subbasis. I want to find a refinement. What's a necessary and sufficient condition that it generates the same topology?

Lol @ACuriousMind

That's why he's single :P

@0celo7 Hm, I don't see any nicer condition than it just also being a subbasis.

Wait guys would the internal energy of a gas actually be $E_{int} = \frac{f}{2}nRT$ where $f$ is the number of degrees of freedom?

Works every time.

@ACuriousMind I would go with "I'm in the market for some roast beef, I hear you're got some"

@0celo7 Is that a line to hit on a man or a woman? :P

I hear you are got some, girl.. <- smoothest dude around

@Obliv yes

Besides, I'm in general not sure if people going grocery shopping appreciate being hit on

Germans generally don't take well to attempt to interrupts their efficient shopping trips ;)

@acuriousmind as you increase the temperature of such a gas, how can you check to see if its gaining degrees of freedom?

and yeah you'd never want to be in between a german and their mission

@ACuriousMind whichever is DTF...

@0celo7 don't let your gf hear that.

@Obliv Uh, I don't think any gas gains degrees of freedom what heating up. The number of d.o.f. is usually a result of the shape of the molecule

@0celo7 down to...fiddle?

@ACuriousMind No. Functional analysis.

@MAFIA36790 Triss from The Witcher series.

@0celo7 Do you know how to answer that?

It's for forming a functional analysis study group.

@0celo7 For which you need a supply of roast beef, apparently

@acuriousmind imgur.com/a/cLBNk

Snacks.

wrong person

Holy shit, why is there a massive firework going on outside?

Drunk Germans.

@Obliv Oh, yes...well, for that you determine the lowest energy states belonging to that kind of excitation and then ignore them until you are at energies where those can actually be excited

@0celo7 1. It looks pretty professional. 2. You can't buy fireworks this time of the year here

@acuriousmind to celebrate my birthday. it was 2 days ago but I'll let it slide since time is backwards in non-american countries

@acuriousmind so it's probably not a good idea to measure it by a graph that relates the molar specific heat to the temperature of the substance then?

@ACuriousMind 1. Drunk professionals. 2. Wtf?

@0celo7 The only regular occasion to which fireworks can be bought in stores is around New Year's Eve. Although I guess that some services might deliver them year-round.

Would be pretty pointless, though, since fireworks outside of New Year's Eve are forbidden without a special permit, so you have to employ professionals regardless

amzn.com/1118473507

Does anyone have this to sell?

The internet does...

For all kinds of prices :P

Link

Oh man my brain is failing. How does $nC_v dT = -pdV \to pdV + Vdp = nRdT$ using the ideal gas law $pV = nRT$

Oh I figured it out. $R = C_p - C_v$ was used

Gosh, it's hard work explaining how a magnet works.

LOL, have you seen this? In QED "photons play the part of the boomerangs pushing opposite-charged particles together".

@johnD is that incorrect?

@Obliv : yes, it's incorrect. Virtual particles are virtual. They only exist in the mathematics of the model.

oh I suppose it wasn't specifically defining the photons to be virtual. Kind of misleading then

@MAFIA36790 I wonder if asking dumb questions is allowed.. technically it will choose the 8 top voted questions <|8O)

@Obliv Of course asking "dumb" questions is allowed, but they'll likely not get top voted :P

Also, note the "We reserve some editorial control in the selection of the questions and may opt not to select a question that is tangential or irrelevant to moderation or the election"

@acuriousmind do you want to be a mod? I can try rigging the votes for you if you continue answering my trivial questions throughout my time in undergrad.

aw boring

@Obliv I doubt you have the ability to rig votes :P

Does campaigning for you on the streets count? If not, let's just say I know some people.. @bernard

You don't even have enough rep to vote in the election (150 required).

@Obliv Ehm, everything counts. i need the books & I'm broke :)

...what?

Ah, wait

he mentioned me I thought he was talking about the book

Sure I'll be a mod

let's burn down Rome

Lol i don't even have 150 rep this sucks

Oh wait

You want me to rig the elections for ACM

Ah, sure

@BernardMeurer Your reading comprehension seems to be below average today :P

@ACuriousMind I'm tired :(

Oh I know @acuriousmind just set up a bunch of 150 rep bounties on a lot of algebra questions i'll answer with multiple accounts. That way I'll rig the votes and I'll get practice 8D win win

@Obliv : your wish is my command.

@Obliv Sure. I'm certain nobody will notice that.

Tsk, some people don't half tell some lies to children. People who should know better.

@acuriousmind Even if they do notice you can threaten to ban them from this chatroom as you are a room owner. You have far more power than you think ACM

you practically outrank mods.

@Obliv ...what?

I'm not a room owner

oh you aren't? I thought you and JR were

Just because I'm always here doesn't mean I own the place :P

@ACuriousMind hehe you're a girl

@johnD oh that's what you meant.. thanks :D even though that question and answer were garbage. Also, are you only annoyed with the usage of the word photon or are you also annoyed with the description of the exchange of said photons?

@Obliv : I don't have any issue with photons. But I do have an issue with the notion that the electron and the proton are forever chucking photons at one another. It just isn't true. When they move together they "exchange field" such that the resultant hydrogen atom doesn't have much of an electromagnetic field. But hydrogen atoms don't twinkle.

@0celo7 Given that you already believe I'm Alicia Keys, I'm not sure why this is surprising to you

@ACuriousMind What? Why would I believe that.

@BernardMeurer Help!

I'm going to buy more books

stop me please

@0celo7 Buy me some

No

Does anyone known if there's a commonly used name for quantum scattering where the wavelength of the incoming plane wave is much much larger than the size of the scattering potential?

@ACuriousMind I wanna kum...quat.

@IceLord How am I supposed to tell if you're doing something wrong if you haven't shown me what you did? However, using $\dot{\phi}=\Pi$ I get $[\phi] = \frac{1}{2}(d-2)$ as desired.

What seems to be the issue?

l.h.s. is $[\phi]+1$, r.h.s. is $d-[\phi]-1$, simple algebra gives the result.

@ACuriousMind With which Lagrangian?

wait what

@0celo7 Hm? What has the Lagrangian do to with this?

what are the units of $\Pi$?

@0celo7 We determined them to be $d-[\phi]-1$ far earlier.

@IceLord Yes...take $\dot{\phi}=\Pi$.

You don't need the CR at all if you use that equation.

@ACuriousMind Christ. Are you given the Lagrangian?

Or are we still committed to using just the CCR?

Why not compute this using the Lagrangian?

It's what, $S=\int (\partial\phi)^2d^dx$?

@IceLord Well, "it" is stupid, then. You can't determine them from the CR. If you use $\dot{\phi}=\Pi$, then you can just directly take the units of that equation instead of using the CR

So that's like $1=2[\phi]+d-2$

Is that right?

What is the actual answer?

@0celo7 We did the "determine it from the Lagrangian" before this one.

@ACuriousMind Was my wild guess anywhere close?

Crap, I got $d+3$.

What is the mass dimension of action?

Find a better prof. :P

Hmm, I'm bad at this.

what's the mass dimension of $dx$

is it $1$ or $-1$

Ok, then I'm right

Good.

@IceLord Did you tell him his record is still intact?

ACM & DS used BCH.

Or did you have the idea?

But...how else are you supposed to do it? I don't know QFT but I know one uses BCH to solve that problem.

Speaking of that. @ACuriousMind I'm taking credit for a proof you showed me on a handout in analysis. Sorry.

@0celo7 DS showed an alternative way expanding the exponentials and $\langle n \vert m\rangle = \delta_{nm}$

@0celo7 Which one?

@ACuriousMind Finite dimensional subspace of normed vector space is closed.

Ah, I vaguely remember that one

My prof showed it in class today, took the whole class. He used linear functionals or something.

I gave him the one-line proof you told me and that's being put on a handout with some other stuff.

@ACuriousMind Yes. I was trying to show that the Lie algebra of a normal subgroup is an ideal.

One needs to show that $f(t)\in\mathfrak h\implies f'(0)\in\mathfrak h$.

This follows from $f'(0)$ being the limit of difference quotients, which lie in $\mathfrak h$.

@ACuriousMind I told him the proof wasn't mine, but he insisted on giving credit.

Shrug.

@0celo7 I don't think of proofs as things one possesses. Very few of the proofs I know sprang entirely from my own mind

@ACuriousMind Yeah...remember my really quick way of proving $[0,1]$ is connected? My topology prof gave the same proof word for word.

@ACuriousMind What's a proof that sprang from your own mind?

I have some: holonomy splitting, Lorentzian Stokes thm

I do.

I find physics that doesn't relate to real life more boring though.

Hi all! Is anyone here familiar with QED? I have a question, and it'd be nice to talk it through before posting.

Question: The Breit-Wheeler process describes the interaction between two photons which results in the creation of an electron / positron pair. By specifying the parameters m and e when we define the Lagrangian for QED, it feels like we're inherently "over-specifying" this process, since m and e are quantities that should arise out of it.

Is the information in the \gamma + \gamma system insufficient to determine e^+ + e^- on its own? Do we need to know the outcome in order to describe the process?

@LukeBurns 1. You are correct in so far that mass and charge are inputs to QED and not derived. However, there is no ground for saying that they "should" arise out of it. 2. Yes, you cannot predict what exactly will come out of a $\gamma+\gamma$ process, just like you can very rarely predict the exact result of a (quantum) measurement in general.

@IceLord Moderately.

It was Hawking & Ellis that really did it.