The h Bar

John Rennie

07:12

@ACuriousMind Sounds like a good time to light a spliff :-)

Little Alien

Is it right that electric and magnetic field come hand in hand so that I cannot have static magnetic field without having perpendicular electric field in that element of space?

John Rennie

@Danu as in the flavours in lemon peel are all from oils that are water insoluble?

Secret

No you can have a static magnetic field without any electric field. The idea that electric and magnetic fields goes hand in hand is that a magnetic field is basically an electric field in a moving frame

or more accurately, both electric and magnetic fields are part of the electromagnetic field tensor

thus you can transform from one into the other

Little Alien

Secret

that's an electromagnetic wave, it is one of the many forms the electromagnetic field tensor can take

John Rennie

07:22

@LittleAlien that is a depiction of a propagating electromagnetic wave.

That is one particular type of electromagnetic field, and it does always have both an electric and magnetic field at right angles.

Secret

in an electromagnetic wave, as solved in maxwell equations, the E and B field are indeed always orthogonal to each other. However not all electromagnetic field have to go hand in hand like this

John Rennie

However there are many other different types of electromagnetic fields and they don't necessarily have electric and magnetic fields at right angles.

Little Alien

@JohnRennie How can you have magnetic field alone in some element of space if to propagate, you need the electric counterpart there?

John Rennie

@LittleAlien when I say a propagating electromagnetic wave this is a specific type of wave. Other fields can be time dependent without looking like an EM wave.

Secret

A time independent field is not propagating. Since a static magnetic field is time independent, it is unchanged with time at all points in space, thus no electric field will be induced

Similarly for static electric fields

John Rennie

07:27

@Secret I think Alien's point is that the magnetic field must have been created at some point and while being created it was time dependent.

Little Alien

exactly

Secret

Well in that case, for an electromagnetic, the moment you turn it on the magnetic field does slowly rise, and thus by lenz law there's an induced electric field that opposes it. This give rise to the inductance of a material

John Rennie

@LittleAlien But a magnetic field being created, e.g. when you turn an electromagnet on, doesn't look like an EM wave and doesn't have the same properties as an EM wave.

Secret

However magnetic fields can also be produced by the spin of particles and for that I am not sure if there is a notion that says there is a point they came to existence suddenly

John Rennie

@Secret right, but the alignment of multiple spins can change with time to give a time varying magnetic field.

Secret

07:30

yup

John Rennie

The point is that an EM wave has a specific type of space and time dependence. Other time varying fields won't have the same properties.

user228700

This is PSE, not MSE, I know, but does any of you know the formula for the sum of the squares of the terms of an A.P? I searched everywhere, to no use. I derived it and all, but I'm not getting the correct answer, I think. Should I venture into the MSE chat maybe or do you know, somehow?

John Rennie

$\sum 1^2 + 2^2 + 3^2 ... +n^2$ ?

Secret

$\sum_{x=0}^n x^2 = \frac{n(n+1)(2n+1)}{6}$ (I need to check again how to derive it though)

user228700

No no, that's just for the natural numbers. I'm talking about a²+(a+d)²+(a+2d)²+...+(a+(n-1)d)²

user228700

07:36

@Secret Using that, I derived a formula but I dunno if it's right...

John Rennie

Offhand I don't know. Sorry.

user228700

^a being the first term of the A.P and d being the common difference.

user228700

@JohnRennie OK, no problem. Guess I'll go and introduce myself over at the MSE chat :P

user228700

Do you know what it's called? The general chat room for Math?

user116211

@KaumudiHarikumar Mathematics

Secret

07:38

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

7

39

user228700

Thanks, guys :-)

Secret

I got:

$$\sum_{n=1}^m(a+(n-1)d)^2$$
$$\sum_{n=1}^ma^2+2a(n-1)d+(n-1)^2d^2$$
$$\sum_{n=1}^ma^2+2ad\sum_{n=1}^m(n-1)+d^2\sum_{n=1}^m(n-1)^2$$
$$\sum_{n=1}^ma^2+2ad\sum_{n'=0}^{m-1}n'+d^2\sum_{n'=0}^{m-1}n'^2$$
$$ma^2+2ad(m-1)+d^2\frac{(m-1)m(2m-1)}{6}$$

user228700

@Secret Oh! You did it a lil' different than I did(thereby teaching me something more!) but the result is the same, thank you loads!

Secret

The good thing is that this is not an infinite series, else you are not allowed in general to bubble out the summation symbol like that

Also shifting indices is a useful skill in simplifying series. You will need such skill when you need to solve some differential equations with pen and paper in more higher up physics stuff beyond high school

user228700

07:55

@Secret What do you mean "bubble out the summation symbol"?

user228700

And also, the second term...are u sure it's correct?

user228700

Shouldn't it be $adm(m-1)$?

Secret

Notice how from step 2 to step 3, the summation sign goes form one into three summation signs. This works because summation is distributive over addition (as long you don't pull the n terms out of the summation that is)

The second term is obtained by using the perfect square formula $(x+y)^2$ and let $x=a$, $y=(n-1)d$. As usual of what happens when you expand $(x+y)^2$, you get that factor of 2 for the middle term

user228700

Yes, but when we sum $n`$ from 0 to $m-1$, we get $m(m-1)/2$...

Secret

ooops, yeah, that's a careless mistake... Let me fix it

user228700

08:01

@Secret Sure, no problem :-) Thank you!

Secret

then indeed the 2nd term is $2ad(m-1)m/2=ad(m-1)m$

FrancescoS

Hi @ACuriousMind . Do you know something about analyticity of the S-matrix?

user228700

08:21

@Secret Can you please answer another math doubt over in the math chat?

Secret

ok

Secret

08:57

[Classical mechanics musings] if $\{F(q,p),q\}=\partial_q F$ and $\{F(q,p),p\}=-\partial_p F$, that is, momentum and coordinates are generators of the translation of each other, then what is motion?

Danu

09:12

@JohnRennie Science! Nice :D

I did not realize that.

Secret

Well, those oils while smell good, are often bitter

I learnt that when I ate an orange rind

John Rennie

09:43

@Danu I confess I don't know what the flavours in lemon peel are, but that sort of chemical trends to be terpene derived.

Secret

yeah, the fruit oil ones generally smell fruity, but they tend to taste bitter

Johnrennie, we knew the vector potential cannot be uniquely defined due to the presence of a gauge freedom in the form of a gradient of an arbitrary vector field, but can the difference in the vector potential be physically measured?

that is, is there an analogous concept for vector potential similar to "electric potential difference" that can be detect by experiments?

Danu

10:24

@JohnRennie So, should I try to (very softly?) fry the lemon peels in the oil that I use for the garlic---before putting the garlic in?

Also, you were right:

Limonene

Limonene is a colorless liquid hydrocarbon classified as a cyclic terpene. The more common d-isomer possesses a strong smell of oranges. It is used in chemical synthesis as a precursor to carvone and as a renewables-based solvent in cleaning products. Limonene takes its name from the lemon, as the rind of the lemon, like other citrus fruits, contains considerable amounts of this compound, which contributes to their odor. Limonene is a chiral molecule, and biological sources produce one enantiomer: the principal industrial source, citrus fruit, contains d-limonene ((+)-limonene), which is the (R...

John Rennie

@Danu traditionally you finely grate the lemon peel to get what's known as zest. Then just mix the zest in with your food.

Zest (ingredient)

Zest is a food ingredient that is prepared by scraping or cutting from the outer, colorful skin of unwaxed citrus fruits such as lemon, orange, citron, and lime. Zest is used to add flavor ("zest") to foods. In terms of fruit anatomy, zest is obtained from the flavedo (exocarp) which is also referred to as zest. The flavedo and white pith (albedo) of a citrus fruit together makes up its peel. The amounts of both flavedo and pith are variable among citrus fruits, and may be adjusted by the manner in which they are prepared. Citrus peel may be used fresh, dried, candied, or pickled in salt. ��2...

Danu

@JohnRennie Yeah, but I like using big pieces.

I know what zest is (obviously?)

John Rennie

@Danu I tend not to assume any knowledge of cooking because i have next to none myself. I only know about lemon zest because my mother uses it to make exquisite lemon biscuits.

Danu

OK :)

Secret

(On Susskind) Everything is straightforward so far until electromagnetism started kicking in

Danu

10:27

We have a very strong tradition of cooking in my family

John Rennie

I suspect frying lemon peel would spoil the flavour by boiling off the oils in it.

Danu

@Secret Aharonov-Bohm effect

@JohnRennie Oh, really?

Damnit, chemistry

So I guess I have to deal with zest...

John Rennie

I've eaten curry that has diced lemon peel in it, and that worked OK. If you want a really nice lemon flavour then I think you should go the zest route.

Danu

I really like the big chunks, visually

John Rennie

Include a few big chunks for the cosmetic effect and use zest for the flavour?

Danu

10:30

hehe

Yeah, maybe

Secret

I think I need to write the main equations from both susskind's books into one single A4 paper, because I had a feeling I don't 100% understand what I have read and do as otherwise it should have already been committed to memory

John Rennie

@Secret: the gauge freedom for the magnetic vector potential is very different to that for the scalar electric potential. The vector potential can have any arbitrary function of space (but not time) added as long as the curl is zero.

Secret

I kinda realised, it is like as if you set a zero for your reference frame except this "zero" is a rather crazy looking vector field

John Rennie

@danu: what's your view on ...

9 hours ago, by 0celo7

Because they are cute and delicious

... venison?

@Secret physics.stackexchange.com/questions/189913/…

Danu

@JohnRennie Venison is nice.

Too expensive to eat regularly.

John Rennie

10:36

It seems to fashionable in the UK at the moment to eat venison sausages and burgers, but I've tried a few at various different price points and I have to say I don't think venison makes good burgers or sausages.

Danu

I think that that's kind of a waste of the nice meat

Secret

@JohnRennie Hmm, Well, reflecting on what Danielsank said ages ago, a potential (scalar or vector) seemed to be mainly a very convenient way to wrap all those local interactions into one single function

Danu

Especially burgers

John Rennie

My mother makes a beef bourguignon type dish with venison and that is delicious. But I think the meat is too dry to work in sausages or a burger.

Danu

dirty?

Oh, dry haha

It's been too long ago that I ate it to have a real opinion.

John Rennie

10:38

Damn, I can't type today :-)

@Secret I'd be inclined to move straight to the electromagnetic four-potential because that's Lorentz covraiant so it's simpler in a way.

Secret

Danu what traditional dishes do you and your family cooked?

user228700

10:52

It's always very strange to watch shows like MasterChef because we use almost literally none of the ingredients that the rest of the world uses(exceptions; salt, pepper, chilli powder etc.)

user228700

And we make completely different sorts of food! Most of which don't even involve meat, like at all! That's probably why we use such diff. ingredients, hm. Anyway; food probably has nothing to do with physics :P

user116211

Man, there are exciting terms in Bourbaki:

user116211

The latest I found is Magma product.

ACuriousMind

11:43

@JohnRennie That type of smoke doesn't seem to trigger these detectors anyway :P

user218912

11:59

@ACuriousMind I figured out the dimensions of the delta function in mass dimension for the ETCR.

user218912

but how do I deal with the commutator?

user218912

just plug in the dimensions and solve?

ACuriousMind

@FrancescoS Not much. What do you want to know?

@IceLord What's there to "deal with"?

You now know the dimensions of all quantities occuring in the relation, don't you?

(except for $\phi$)

user218912

I think

ACuriousMind

So what is your problem?

user218912

12:04

idk what the dimensions of $\Pi^0 = $

user218912

because from its definition

user218912

you're differentiating wrt. to $\partial_\mu \phi$

user218912

is it just $d+1$?

ACuriousMind

Are you guessing or deriving that?

user218912

I derived it

ACuriousMind

12:07

How?

user218912

well I know the dimensions of $\mathcal{L}$ which is $d$

user218912

and the derivative is +1 to that.

user218912

am I wrong?

ACuriousMind

Yes, you are wrong.

user218912

xD

ACuriousMind

12:08

Because "the derivative is +1 to that" was for the derivative w.r.t. $x^\mu$.

user218912

oh I see.

user218912

so for this derivative

user218912

you would have to use the definition of the derivative again?

ACuriousMind

Sure

John Rennie

@ACuriousMind I wouldn't know of course :-)

user218912

12:16

@ACuriousMind is it $d+2$?

ACuriousMind

How did you derive that?

FrancescoS

@ACuriousMind I am looking for a simple and readable proof of the analyticity of the s-matrix. The Weinberg's one is too much detailed

user218912

@ACuriousMind from the definition of the derivative but idk if I did it right.

ACuriousMind

@JohnRennie Indeed, of course not :)

@IceLord When I ask "how did you derive that" I expect something more specific than "by using the definitions" :P

user218912

so I did

ACuriousMind

12:17

I can't help you if you don't show me what you actually did

user218912

$lim_{h \to 0} \frac{f([\phi] + 1 + h) - f([\phi] + 1)}{h}$

ACuriousMind

@IceLord I have no idea what that is supposed to be.

user218912

difference quotient

ACuriousMind

@FrancescoS I don't know of any besides the one in Weinberg, sorry

FrancescoS

@ACuriousMind thanks anyway ;)

ACuriousMind

12:20

@IceLord I'm not an idiot, I know that's a difference quotient. But you didn't tell me how that is in any way related to determining the dimension of $\partial \mathcal{L}/\partial (\partial_\mu \phi)$.

0celo7

@ACuriousMind Since I'm trying to prove a certain topology is second-countable, and I'm defining this topology via subbasic sets, should I show that it is generated by a countable number of subbbasic sets?

user218912

@ACuriousMind alright so I did $\frac{\partial}{\partial(\partial_\mu\phi)}d$ now to figure out what dimension $\frac{\partial}{\partial(\partial_\mu\phi)}$ has I did the difference quotient $\frac{f([\partial_\mu\phi] + h) - f([\partial_\mu\phi])}{h}$

ACuriousMind

@0celo7 I haven't done enough countability proofs to know which method will likely succeed, sorry

@IceLord What are those brackets doing in the argument of $f$? What is $f$, anyway?

user218912

@ACuriousMind idk how you wanted me to use the difference quotient.

0celo7

What are you doing here

ACuriousMind

12:26

I don't know why you're making this so complicated - the derivative is the limit of a ratio, so you just have to figure out what dimension $\mathcal{L}/\partial_\mu\phi$ has.

@IceLord Note that, once again, you have not actually answered the question I asked. When I ask "What is $f$?" I want to know what $f$ is. "I don't know what you wanted me to do" is not an answer, and helps neither of us.

user218912

because idk what $f$ is

ACuriousMind

But you wrote it down!

0celo7

Yikes ACM going hard

user218912

because I didn't know what to write.

ACuriousMind

How can you write something down and not know what it is?

user218912

12:29

it's just a function

ACuriousMind

Ok, but...why? Why do you think you needed a random function there? Why are the "dimension brackets" inside the argument?

user218912

@ACuriousMind how am I supposed to find the dimension of $[\phi]$ when the dimension of $\Pi$ requires it?

user218912

@ACuriousMind because I'm doing a dimensional analysis derivative

ACuriousMind

@IceLord Aha, now that's a better question!

user218912

@ACuriousMind do I just use the ratio in the ETCR?

user218912

12:30

and solve for $[\phi]$

ACuriousMind

@IceLord I don't know what that's supposed to mean

user218912

do I do

ACuriousMind

You can express the dimension of $\pi$ in terms of the dimension of $\phi$. Then you plug that into the relation and solve for the dimension of $\phi$.

0celo7

ETCR?

user218912

$[[\phi], \frac{d}{[\partial_\mu\phi]}]$ @ACuriousMind ?

ACuriousMind

12:32

@IceLord I don't know what that's supposed to mean.

user218912

:|

user218912

what that means is I am writing down the ETCR and plugging in the dimensions of the arguments.

ACuriousMind

Yes, that was the general idea how you are supposed to use the CR to get the dimension of $\phi$.

user218912

@0celo7 equal time commutation relation.

0celo7

@IceLord *favepa

that was supposed to be face palm

ACuriousMind

12:35

However, writing a commutator of $[\phi]$ and some other weird thing doesn't make any sense because $[\phi]$ is just a number, you can't take a commutator of it

0celo7

But my phone had other ideas.

user218912

@ACuriousMind yep that's what I thought.

user218912

but I had no other ideas

user218912

@0celo7 :(

ACuriousMind

I don't know what's so hard here: Write the relation out as $\phi\pi - \pi\phi = \delta$. Take the dimension on both sides and solve for $[\phi]$.

user218912

12:39

@ACuriousMind that's exactly what I was doing

user218912

but you said it doesn't make sense.

ACuriousMind

No, you did not say that

user218912

taking the dimensions on both sides is the same as using the dimensions in the commutator, isn't it?

0celo7

@ACuriousMind I'm missing the point of the compact open topology. I'm assuming you can't help?

ACuriousMind

@IceLord No, because the dimensions are just numbers so the commutator is zero.

Note that the dimension of $A-B$ is not $[A]-[B]$.

user218912

12:40

@ACuriousMind oh

ACuriousMind

@0celo7 As I said, I haven't really encountered it, so no

user218912

$[\phi]\frac{d}{[\phi] + 1} \frac{-d}{[\phi] + 1}[\phi] = -1$ ? @ACuriousMind

ACuriousMind

@IceLord I have no idea what that's supposed to mean.

user218912

I plugged in the dimensions into the definition of the commutator

John Duffield

@dmckee : that's not very nice, and nor is it true.

user218912

12:45

we had $[\Pi] = \frac{[\mathcal{L}]}{[\partial_\mu \phi]}$

user218912

right?

ACuriousMind

No

The dimension of $\Pi$ is the dimension of the ratio, but the dimension of a ratio is not the ratio of the dimensions, i.e. $[A/B] \neq [A]/[B]$.

user218912

:(

user218912

we had $[\Pi] = \big[\frac{\mathcal{L}}{\partial_\mu \phi}\big]$

ACuriousMind

Try and figure out what $[A/B]$ is in terms of $[A]$ and $[B]$ first.

Then figure out what $[A-B]$ is. Then use that to solve the problem

John Duffield

12:50

@ACuriousMind : "I say that because there’s a standing joke in our house, that I’m the only one who can change a light bulb. But somehow it isn’t funny. If you selected a hundred people at random and tested their technical and scientific knowledge, I think the average score would be lower than that of a comparable group from fifty years ago. Yes, we’re more specialist these days, and some things are more difficult to understand..."

user218912

@ACuriousMind okay thanks I will try to.

user218912

I have class now though, so I'll be back in a few hours hopefully with a derivation.

0celo7

@JohnDuffield What did he say?

ACuriousMind

@IceLord Uh, okay. It's not much of a derivation through, you just need to know how units work :P

That is, if you find yourself writing more than one line to show anything here, you're probably going the wrong way about it.

0celo7

Whom should I ask about the CO topology? Analysis or topology prof?

John Duffield

12:58

@Secret : purple is not in the spectrum.

Secret

No that's the RGB approximation of violet, you cannot make true violet on a computer

ACuriousMind

@0celo7 I'd expect both to know, I guess

John Duffield

@0celo7 : he couldn't change a battery in a smoke detector. He covered it in tape. Tsk.

@Secret : oh, OK.

Uh, a conference call. Bye.

DanielSank

13:16

Hi, everybody.

ACuriousMind

Hi @DanielSank

DanielSank

I'm feeling a big push on my linear algebra document coming up.

The one I think you and @0celo7 said was crap :)

You guys are wrong though, so it's cool.

ACuriousMind

I don't remember judging anything you've written, and I only remember reading that "too obvious" paper, but no linear algebra thingy

Might be I forgot about it if I really thought it was crap, though :P

DanielSank

@ACuriousMind heheheh

@ACuriousMind Oh you read the non-RWA thingy? What'd you think?

ACuriousMind

@DanielSank Yeah, you wanted someone to read it to see whether the exposition was clear or something. I didn't see anything wrong with it and found it understandable, but since I'm not versed in your field I couldn't say anything about its importance or impressiveness.

user116211

13:27

@0celo7: Linear Algebra: Hoffman, Kunze has just been delivered ;))

user116211

Price?

user116211

Just 3 dollars ;)))

DanielSank

@MAFIA36790 how?

@ACuriousMind groovy

Xasel

@MAFIA36790:IS that beginner's book..I mean can it be used by HS students?

JUST 3 DLLARS?

jyotishraj thoudam

13:45

0

Q: Double Slit Experiment/Transition of Classical to Quantum problems in Probability Addition in "An Experiment on bullets"

The First Picture is taken out from the Book The Character of Physical Law By Richard Feynman And the second picture is from his own The Feynman Lectures on Physics. Both figures correspond to the An experiment with bullets topic which appear on his lecture. In the first lecture $N_1$ , $N...

DanielSank

@Danu I think I'll write a blog article about why "setting hbar to 1" is stupid language ;)

ACuriousMind

@jyotishrajthoudam Why did you post that here?

jyotishraj thoudam

I find people posting their questions here. Is it not in the policy? Sorry then @ACuriousMind

ACuriousMind

@jyotishrajthoudam Well, we don't really have a policy of things you may post in chat (aside from the overall Be Nice policy), but I don't usually see people just posting their own questions without any additional commentary

I also don't see why you posted that. If you just wanted more people to look at your question then I think that's not what chat is for.

Emilio Pisanty

Hey folks

jyotishraj thoudam

13:52

Ok, I won't next time

Emilio Pisanty

any of y'all notice this lately?

physics.stackexchange.com/users/10851/chris-white

ACuriousMind

@EmilioPisanty Yes

Sep 15 at 12:48, by ACuriousMind

@JohnRennie At least you've managed to stay room owner longer than Chris, so maybe you're overestimating your divisiveness ;) (CW's account is gone entirely btw, for those that hadn't noticed)

Emilio Pisanty

@ACuriousMind I guess job searches made him do it?

It's a real shame, that is.

ACuriousMind

@EmilioPisanty I'm pretty sure it's a (over?)-reaction to him getting banned on chat.

Because right after that happened, his profile text was changed to "delete me" and he never returned.

Emilio Pisanty

@ACuriousMind That's pretty strong stuff.

I think his entire SE profile is gone, too

6

A: How should I interpret an almost-but-not-quite-alphabetical author list?

How should I interpret this authorship convention? The non-alphabetic people are primary contributors, with their own ordering, while the alphabetic ones all worked together to contribute something that lead to the paper. It might be impractical or undesired to rank authors within the latter...

that one was his

Danu

13:58

@DanielSank As we discussed (and agreed on!) before, it's unpedagogical. For research level stuff it's perfectly reasonable.

I find it unnecessary, at best, to say "stupid"

Xasel

@ACuriousMind:Is he a popular memeber here
?

Emilio Pisanty

@Xasel Judge for yourself

web.archive.org/web/20160722072644/https://…

Xasel

@EmilioPisanty: hmm..quite popular then why he left?

Emilio Pisanty

@Xasel I'm completely in the dark about it.

Xasel

Well then we may move on from this toppic then

0celo7

14:13

@ACuriousMind Prof. Dr. Elven Lord said my $\Delta$ closed $\Leftrightarrow X$ Hausdorff proof was "excellent", but my PhD level boundary proof was just "good" :(

DanielSank

@Danu "stupid" gets attention

ACuriousMind

@0celo7 I don't know what I'm supposed to do with that information

0celo7

you're bad at this

Emilio Pisanty

@DanielSank I'm curious as to why exactly you think it's that bad

or rather what you think people should say instead

(but I can also wait for the blog post)

Danu

@DanielSank That's a "stupid" reason :P

0celo7

14:23

@Danu Do you have any experience with the compact open topology?

Danu

No.

Balarka Sen

14:42

@0celo7 What's this boundary proof?

user218912

@ACuriousMind to find the dimension of $[\frac{\mathcal{L}}{\partial_\mu \phi}]$ do I use the fact that $\frac{\partial \mathcal{L}}{\partial\phi} = \partial_\mu \Pi$

ACuriousMind

@IceLord That's not a fact.

user218912

xP

user218912

it's wrong to do it that way anyway.

user218912

sorry.

0celo7

14:53

@BalarkaSen Boundary is disjoint from interior, union of boundary and interior equals closure

I spent a lot of time on it, gave a very thorough proof

user218912

@ACuriousMind is this problem supposed to be really simple?

ACuriousMind

@IceLord Yes.

user218912

feels bad.

user218912

sigh...

user218912

@ACuriousMind are you sure $[\frac{A}{B}]$ isn't equal to $\frac{[A]}{[B]}$?

0celo7

14:59

Pretty sure it is!

ACuriousMind

@IceLord Yes.

0celo7

@ACuriousMind Explain

ACuriousMind

If $[A/B] = [A]/[B]$ were true, then dividing something of mass dimension $n$ by somethign of mass dimension $1$ would still have mass dimension $n$. But the ratio of two masses is obviously dimensionless.

user218912

brb

Danu

@0celo7 Really?

Don't forsake your physics knowledge completely :P

0celo7

15:03

@Danu Yes, it's obviously true. I don't know what ACM is on about.

If something has mass dimension $n$ and you divide by mass dimension $1$, you get mass dimension $n-1$.

That's what $m^n/m^1=m^{n-1}$ says.

ACuriousMind

@0celo7 Fully and utterly correct. So why do you say that $[A/B] = [A]/[B]$ is true? The $[\cdot{}]$ denotes taking the exponent of the mass unit.

0celo7

Oh, what kind of shitty notation is that?

Of course it's not true then.

Danu

lmao

It was pretty clear to me (and indeed, obviously wrong :P)

David Z

@ACuriousMind that'd be important to clarify because a lot of other people (like me) use the bracket notation to mean something else

0celo7

@DavidZ Thank you.

ACuriousMind

15:07

@DavidZ We've used it in that way in this discussion in the last two days (see e.g. $[\phi] = \frac{1}{2}(d-2)$ further upthread).

David Z

Ah, well I haven't been following, so if everyone involved is clear on what $[\cdot]$ means, sure, carry on.

ACuriousMind

I'm aware that in units where there's more than one kind of unit it's usually just taking the units of the expression

@0celo7 Don't spoil :P

0celo7

He hasn't figured that out?

ACuriousMind

I'm...not sure

But his last question would indicate not.

0celo7

Yeah.

@ACuriousMind Ok, all roads lead to this 1939 paper by Steenrod or this obscure functional analysis text on Lie groups.

ACuriousMind

15:12

@0celo7 All roads for what? The Lie-ness of the isometry group?

0celo7

yeah.

The Steenrod paper assumes a lot of stuff about the compact open topology though.

As does Petersen. I'll ask my topology prof if we can cover CO later in the semester.

If not...dunno what I'll do.

Kobayashi-Nomizu have some sketches.

@ACuriousMind The idea is this: orbits of the isometry group are submanifolds. The charts on these submanifolds give rise to charts on the isometry group, which can be smoothed to give a smooth atlas.

I'm sure no one has bothered to check that the multiplication and inversion are smooth, however...

@ACuriousMind The author of the Lie groups book is like Montzipillion. Amazing name.

Secret

16:14

Ok that's all for today. I have completely underestimated the content in Susskind's classical mechanics

I thought I can just fill in half of my A4 paper and left the other half for QM so I can compare them side by side

[Worldbuilding musings] (Red=Steps that might break in a back to the future time travel scenario)

yuggib

@DanielSank of course it is not 1...I almost always take it to converge to zero

Balarka Sen

@0celo7 okie

yuggib

@ACuriousMind he got banned here? wasn't he a room owner?

user116211

16:31

@Xasel HS doesn't limit what books you would read; that's simply your count.

user116211

@DanielSank amazon.in/Linear-Algebra-Kenneth-Hoffman/dp/8120302702/…

John Rennie

Homework, or am I being excessively zealous?

0

Q: Does changing resistance of resistor connected parallel to voltmeter change the output voltage of it?

Here is the circuit. http://i.imgur.com/cdxC8TJ.png My own logic says that the voltage output should stay the same even if the resistance of the resistor is increased because in parallel the voltage is equal.

homework-and-exercises electricity electric-circuits

user116211

@JohnRennie For me, it's hw.

Xasel

16:51

@MAFIA36790:I mean is there any prerequisite for this book other than my high-school mthematical background

@JohnRennie : ALthough it seems to be HW but atleast he is try to put his own reasoning and asking for suggestions (whether it's correct or no..I guess he is self-studying) rather than asking for fully-fledged cake

user116211

17:19

@Xasel Just only started to read the book; but at the first sight, it seems to be self-contained.

0celo7

@DanielSank I said it was not rigorous.

That's not "crap"

ACuriousMind

17:35

@yuggib Yes, he was. Then he made Not Nice comments about a certain user and got suspended for a day. After that he never came back to chat and evidently decided to delete his account.

0celo7

@ACuriousMind Welp, neither know how to show it.

ACuriousMind

@0celo7 Well, I expect them to know about the compact-open topology, not that they carry the proof of the Lie-ness of the isometry group in their heads

This is something that begs to be buried in one of these proof graveyards :P

ACuriousMind

Does someone know whether the HNQ algorithm prefers questions from newly launched sites? I'm seeing a disproportionate number of Esperanto questions lately

Physics Meta

1

Q: Why are these off-topic questions off topic?

Today I've seen a bunch of examples on the site of a certain type of question. In order to avoid priming responses, I'm not going to give this type of question a name. Instead, let me show by example what I'm talking about: How to get a constant force parallel to the inclined? Would someone hel...

discussion scope

0celo7

@ACuriousMind Oh god, not that.

I first need to do the 9 part preamble exercise

My advisor's first reaction to what I was doing "that's not a very nice proof"

ACuriousMind

17:43

lol

0celo7

@ACuriousMind I'm half of the mind to make my senior thesis a resurrection of the geometry proof graveyard.

dmckee

@ACuriousMind I believe that it does privilege new site to some degree. But I don't have a link to back up that claim.

0celo7

Holonomy group is a Lie group, isometry group is a Lie group, Frobenius integrability for PDEs, etc.

Obliv

Hey does anyone know how $v_{avg} = \int_0^{\infty} v P(v) dv$ where $v_{avg}$ is the average velocity of a molecule of a gas, $P(v)$ is the speed distribution function, this integral returns an average velocity? What is $v$?

ACuriousMind

@dmckee Yeah, I couldn't find anything specific on meta either, the official algorithm doesn't include site age

I question the value of putting specific questions about a language there even more than I usually question the value of HNQ, though :P

DanielSank

17:48

@0celo7 You were literally ridiculing the document for using notation in some way you didn't like.

0celo7

@DanielSank Doubtful, doesn't sound like me.

I believe ACM ridiculed your notation for the zero vector.

DanielSank

@DavidZ Subtle meta post.

0celo7

But that's about it.

ACuriousMind

@0celo7 Have you heard yourself?

DanielSank

@0celo7 A yes, it was that.

0celo7

17:49

@ACuriousMind Yes, every day.

ACuriousMind

You complain about "shitty" notation rather often :P

0celo7

@DanielSank I defended that notation.

DanielSank

@ACuriousMind, you're found guilty.

ACuriousMind

@0celo7 Oh, right, was @DanielSank the heretic who wrote $\lvert 0\rangle$ for the zero vector? :P

0celo7

Yah.

DanielSank

17:50

@ACuriousMind I think it was.

That was rather foolish now that I think about it...

At the time I was hoping for feedback on the content of the document though, so I was miffed that all anyone could talk about was that one stupid notational mishap.

0celo7

I would complain about the notation if you wrote a thing on geometry @DanielSank

But I'm sure there would be worse sins than notation ;)

@ACuriousMind Why quotes?

ACuriousMind

@0celo7 Because I don't always agree it's indeed shitty :P

0celo7

@ACuriousMind Example?

Bass

Confusing $|0\rangle$ with $0$ is one of my achievements, too.

0celo7

Some smartass grad student in QM asked why my prof was writing the vacuum state all the time when he was writing the axioms of Hilbert space

David Z

17:57

@DanielSank No it's not.

ACuriousMind

@Bass Well, DanielSank didn't confuse the two, he decided to write $\lvert 0\rangle$ for the zero vector. I don't remember whether there was an occurence of a vector labled by $0$ it could have been confused with, but probably not

0celo7

@ACuriousMind I think we agree on notation for the most part.

ACuriousMind

@0celo7 Yes, that's why I said "not always".

Mathematics

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$Mathematics$ Mathematics