The h Bar

0celo7

3:06 AM

A$AP Ferg - Let It Go (Explicit)

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@IceLord maybe this will inspire you

user218912

@0celo7 inspire me to do what?

user218912

@knzhou can you help me please?

user218912

@0celo7 why can't you just show me the steps so I learn it?

user218912

then I'll be able to do it on my own in the future.

user218912

I literally wasted like 3 hours on this trivial thing.

user218912

3:18 AM

I could just ask my TA but it's a weekend.

0celo7

Ask him on Monday.

user218912

why do you refuse to tell me?

0celo7

because I'm mean

user218912

:|

user218912

you always get help from acm and balarka.

user218912

3:22 AM

so return the favour to me.

user218912

plz :)

Bernard Meurer

@IceLord No such thing as returning a favour to a third party

user218912

yes it's called treating people how you would like to be treated.

0celo7

I don't always get help

ACM regularly refuses to help me

user218912

@0celo7 that's because he doesn't know.

0celo7

3:24 AM

then I don't know

user218912

2 hours ago, by 0celo7

cuz I can do it in my head

0celo7

I'm a pathological liar

@BernardMeurer dude did you sleep yet?

Bernard Meurer

@0celo7 Having trouble sleeping tonight

user218912

SOMEONE PLEASE HELP ME SO I CAN MOVE ON

0celo7

@BernardMeurer need help with QM?

I mean chemistry :)

user218912

3:27 AM

@0celo7 so you help him but not me?

0celo7

Yes.

user218912

why?

0celo7

6 mins ago, by 0celo7

because I'm mean

that's my default answer from now on

user218912

...

user116211

And that's it, now the cooling fan is not working properly; the system automatically shuts down in 15s ._.

user116211

3:32 AM

I've to check for HP cooling fans; the price is quite low though :|

user218912

how am I supposed to figure out how to do this.

user218912

it's not worked out in any book.

0celo7

Because someone taking a graduate level class is supposed to be able to do freshman calculus.

user218912

@0celo7 they don't use indices in freshman calculus.

Bernard Meurer

1 min ago, by IceLord

it's not worked out in any book.

^ Never say that

0celo7

3:34 AM

@BernardMeurer He might be right -- it's so trivial no one has bothered to write it down.

user218912

instead of attacking me can you just help so I can move on please?

Bernard Meurer

@0celo7 Lol, you're a meanie

user116211

Anyways, yesterday learnt Lagrange's undetermined $\lambda$ method; it was awesome.

Bernard Meurer

@MAFIA36790 Your name is so weird

because you're like this super nice and sweet guy

but your name is mafia which was a hardcore murderous organization

I'm always like 'lol'

user116211

It increased the degrees of freedom from $n-m$ to $n+m; $ that's hot!!

user116211

3:37 AM

@BernardMeurer ._.

0celo7

@IceLord I'm not attacking.

user218912

@0celo7 yes you are.

0celo7

Ok, then I am. Still not helping

user116211

@IceLord, Did you show "every single step"?

0celo7

Lol! It's in Shankar.

user218912

3:40 AM

@0celo7 what page...

0celo7

No, you read the whole thing, remember?

Let's check Zee...

user218912

@0celo7 I did but I don't remember all of it.

0celo7

It's trivial enough that neither Zee nor Shankar explain it.

So think some more about it...

user218912

@0celo7 since when does shankar use indices?

user218912

I know how to do it with ordinary derivatives...

0celo7

3:44 AM

lol

you do not understand index notation

user218912

then teach me please

0celo7

show me with ordinary derivatives

user218912

I just did ^

0celo7

where

user218912

52 mins ago, by IceLord

$-ike^{-ikx}$

0celo7

3:51 AM

the general case is a straightforward generalization.

user116211

WHATT!!

user116211

There is another candidate O.o

user218912

@0celo7 i don't see it so please tell me.

knzhou

@IceLord Oh, I see the problem. You're asking 0celo7 for help.

Okay, the rules for dealing with indices are actually pretty straightforward, but they can be tricky to figure out what they are the first time around.

What you should do is expand everything out. Apply the summation convention everywhere, until you have an expression with no indices. (Use 1 spatial dimension to keep it from getting too big.)

Then just differentiate normally.

Rewrite the result with indices.

Then you'll see the rule.

user218912

@knzhou thanks, let me try it

0celo7

3:55 AM

@knzhou Ha-ha.

@IceLord Do you understand now?

user218912

not quite.

user218912

this is what I did

user218912

$$\partial_\nu \partial_\mu \epsilon^\mu e^{-ik\cdot x}= \partial_\nu \partial_0 \epsilon^0e^{-ik\cdot x} + \partial_\nu \partial_1 \epsilon^1 e^{-ik\cdot x} + \partial_\nu \partial_2 \epsilon^2 e^{-ik\cdot x} + \partial_\nu \partial_3 \epsilon^3 e^{-ik\cdot x}$$

0celo7

christ

ok

user218912

is that wrong?

0celo7

4:07 AM

what next

@IceLord no

user218912

hmm

user218912

I take $\partial_\nu$ of each term?

0celo7

no

user218912

I take the component derivative of each exponential?

0celo7

yes

user218912

4:13 AM

so i.e. $\partial_\nu \partial_0 \epsilon^0 e^{-ik\cdot x} = \partial_\nu [\epsilon^0 \partial_0 e^{-ik\cdot x} + e^{-ik\cdot x}\partial_0\epsilon^0]$?

0celo7

I thought $\epsilon$ was constant

user218912

yeah it is

user218912

so that whole thing becomes 0

0celo7

...what

no

I'm amazed

You really don't know what the chain rule does!

user218912

wait

user116211

4:27 AM

@DanielSank o/

user116211

Okay, I'm not getting why a ringoid doesn't need to be closed under its operation.

user116211

@0celo7, could you tell me?

0celo7

what the hell is a ringoid

user116211

Ringoid is an algebraic structure with a set and two binary operations; one being distributive over the other and vice-versa.

0celo7

why?

who on earth needs that

user116211

4:30 AM

@0celo7 ._.

user116211

I don't know, actually ;/

user116211

I'm not getting why it is not defined to be closed under the operation.

0celo7

there is nothing to get

it's an arbitrary definition

user116211

okay.

user218912

4:43 AM

@0celo7 are you sure it's not $-i\epsilon^0 k_0e^{-ik\cdot x}$ ?

0celo7

That looks right

user218912

that's what I said before...

user218912

I just forgot the i

0celo7

Nope.

Correct.

user218912

and the exponential

0celo7

4:43 AM

Only took you like 5 hours.

user218912

big deal...

user218912

okay I get it now

user218912

phew

user218912

5 hours wasted

user116211

@IceLord Chill with some cold ~~beer~~ coffee , dude ;)

user218912

4:44 AM

@MAFIA36790 never drank before.

0celo7

@MAFIA36790 wtf why would he drink o.O

user218912

@MAFIA36790 it's 12:45am.

user116211

@0celo7 Because he worked for 5 hrs on a single thing. He needs to be chilled.

user218912

no I need to be disciplined for being so dumb.

user116211

@IceLord So? I drink coffee in 3AM also.

0celo7

4:46 AM

Alcohol is the devil's medicine

user116211

@0celo7 Coffee ;/

user116211

anyways, got an exciting book:

0celo7

It's the Lord's treasure

user116211

amazon.com/introduction-algebraic-structures-Azriel-Rosenfeld/…

user116211

But does it ship to my country....

user116211

4:47 AM

hmm.

user116211

Apparently, this is the only book which defines ringoid.

user116211

My library doesn't have it, damn ;(

0celo7

makes you wonder why the hell you're trying to learn about them...

user116211

Because I'm learning algebraic structures?

0celo7

a ringoid does not have enough structure to actually be interesting.

user116211

4:50 AM

:(

user218912

I gotta say I like 0celo7's helping style, by spending 5 hours on that dumb thing I tried so many different combinations of the derivative with indices on this sheet of paper that I now know how indices work finally. if he just told me straight up I probably wouldn't learn how they work for future occurrences because I would forget since I didn't do it myself.

user218912

so yeah thanks :)

user116211

@IceLord 0celo7 is great ;))

user116211

Ah! @0celo7, when the operations are closed, that is the respective binary operations separately along with set from semigroups, then the ringoid becomes semi-ring.

DanielSank

5:13 AM

@MAFIA36790 \o

0celo7

@MAFIA36790 but...why??

user116211

@0celo7 Because semigroup is a magma which is closed.

user116211

Of course under the respective operation.

0celo7

ok, these terms do not appear in serious mathematics texts...

semigroup does

user116211

ohh.

Secret

7:45 AM

Most world problems are similar to the philosophy of asking a good question and answering them. One need to really get their hands dirty in make the connection clear and highlight the important nodes, and the other need to dig into the network to find out what is important

Block by block the nanocrystals assemble

Secret

8:03 AM

Stereotypes, our brain attempt to find patterns among things that vary like crazy, is known to harm a lot of things from sexualisation to workforce productivity

I wonder what happens if it survives and still communicating with us...?

Some basic nuclear geochemistry on Mars

If we think something is ours, even when handing them to another person, we are more likely to orient it in a way such that it is less helpful for the receiver. This does not happen if the object is perceived to be owned by someone else

user128101

8:19 AM

@JohnRennie, Hi, why did you reopen my question after supporting ACM's point of view? Can you answer it?

John Rennie

@user104372 The answer is no, which is what Lubos answered. I don't think there's any more to be said. I voted to reopen because I don't think the question was unclear. NB I didn't reopen the question single-handedly, I was just one of the five people who voted to reopen.

user128101

9:02 AM

@JohnRennie, what happens when it collides with a spinning particle, why can't be there any change/transfer of AM, an actual rotation? that is the question. Do you have time? do you want to discuss briefly the train tunnel paradox?

Secret

9:18 AM

Einstein's Relativistic Train in a Tunnel Paradox: Special Relativity

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the 2nd and 3rd example weird me out, guess I need to do some calculations

Ladder paradox

The ladder paradox (or barn-pole paradox) is a thought experiment in special relativity. It involves a ladder, parallel to the ground, travelling horizontally and therefore undergoing a Lorentz length contraction. As a result, the ladder fits inside a garage which would normally be too small to contain it. On the other hand, from the point of view of an observer moving with the ladder, it is the garage that is moving, so it is the garage which will be contracted to an even smaller size, thus being unable to contain the ladder. This apparent paradox results from the mistaken assumption of absolute...

Ok fine, they never said that the trapped ladder survive being blown up

user128101

@Secret, if we substitute unlikely doors that close faster than c, and people who see objetcs travelling at c, and use timegate timers that can accurately measure picosecond, don't we get same time for an observer Or at rest and one Ot on the train? What is the problem there, why do they need SR?

Secret

do you mean two different doors when you say "unlikely doors"?

user128101

I mean that is silly to imagine that one can see a thing moving a t c and that doors can close so fast, that perspective distorts the picture

user128101

user128101

If they use timegates at both ends of tunnel they get same T =8 at the end of tunnel and = 1.25 microseconds when the back of train leaves the tunnel

Secret

9:34 AM

In principle you can factor the time to close the door in if the door cannot close near the speed of light. In the tunnel's frame, the train is travel close to c. Since the doors may not close very fast, they have to start closing some time before the train as seen by them is actually contained in the tunnel. So they can still close at the same time in the tunnel's frame, except that now there will be a delay in the closing time to factor that the doors need time to close.

Alternately, we may replace the doors with actual laser barriers, that way they can always close in time because lase…

user128101

Why do you want to close doors? if you electronic time GATEs that sent impulses to Or and Ot they will record same time (and length of train) to pass the tunnel, is that so?

user128101

why do they need SR? why should the train be shorter to Or?

user128101

Is the picture clear?

user128101

I am following another way of reasoning

user128101

There is no contraction

user128101

9:40 AM

according to electronic measurements

Secret

So in your picture you are saying at soon the front of the train passes through the Ot timegate, a time is recorded, and then as the rear left Or timegate, another time is recorded and then you use the difference of these two times to calculate the length of the train by speed of train x time difference?

user128101

Or measures the time his cell has been disturbed by train and gets 12.5 microsecond,( at 0.8 c length is 3 km: no contraction), if there is another cell at the end of tunnel he records 8 mus, and same times records Ot when he gets an impulse from the cell at the back and at the front of the tunnel.

user128101

and they get the measurements simultaneously

user128101

is there any problem with that?

user128101

The both get same measurements for the length of the train 3Km, and they get it simultaneously (well Or gets it 8mus after Ot)

Secret

9:52 AM

Ok so you are saying:
There are two cells Ot and Or in the tunnel, one at the front and one at the rear

When front of train passes through Or, or gets a reading 12.5 (some units) seconds. Now the front of the train reaches Ot and Or get a reading of 8 (some units) seconds. At the same time, Ot records a time of some k (some unit) seconds. Using 12.5 and 8 for Or, k and (some time recorded) for Ot, both Or and Ot both agree that the train is 3km long based on its known speed, am I correct in the understanding of your logic?

user128101

Yep, Or and Ot are "observers" and they both get a reading from timegates at both ends of tunnel on their clock

Secret

Ok, this is ok
Or and Ot are at rest wrt each other, thus they are in the same frame thus they will agree on the length of the train they measure, and also the synchronisation of their clocks

But now here's the kicker:

user128101

No Or is on the platform (r = rest) and Ot (t=train) is on the train

Secret

Oops sorry, let me rephrase:

user128101

10:06 AM

Recap: there is a cell /timegate at both ends of tunnel B and F. Or is on the platform and records the time B is interrupted by train = 12.5 microseconds, he also get a signal from F when train is at front of tunnel T = 8 microseconds. Driver of train Ot gets a signal from B when he enters the tunnel and a signal from F when he leaves it and records same time as Or = 8 microseconds.

user128101

Secret

10:20 AM

Can you tell me the times of both Ot and Or when the train is at the locations 1,2,3,4? I think I still kinda confused on how the measurement is done, other than Ot and Or seemed to take the time measurement at the same time depending on where the train is?

user128101

10:33 AM

The timegate B is at the entrance of the tunnel, I had to adapt a piture I found, that is misleading when the train enters the tunnel time is 0, when the back of the train passes B time is 12.5, when the front of the train reaches F time is 8 when the back of the train passes B time is again 12.5 I'll try to correct the picture.

Secret

Try to draw your own using shapes in e.g. power point, that might make it easier to illustrate the sequence of measurements Ot and OR carried out and how/when they rely the signal to each other as the train passes through various important points

user128101

Is it clear now?, is it better if a post a question on the site?

Secret

Let's see...
Or made a measurement $t_{Or1}$ when the train head is right at the entrance of the tunnel, At the same time, Ot made a measurement $t_{Ot1}$. After the train head reached the end of the tunnel, Ot made a measurement $t_{Ot2}$, and at that same time, Or made the other measurement $t_{Or2}$. Now using $t_{Or2}-t_{Or1}$ and $t_{Ot2}-t_{Ot1}$, and multiply with the train's known velocity of $v$, one gets length of train $(t_{Ot2}-t_{Ot1})v=(t_{Or2}-t_{Or1})v=L$ where $L$ is the rest length of the train. Thus both Ot and Or agree on the train is of length $L$

user128101

Yes, nobody detects contraction.SR is not needed, there is no paradox. Do you concur?

Secret

10:53 AM

Ok I think I understood now: Here's the kicker that inform us something like SR is needed:

For the sake of argument, we assume that the measurements of 1 and 2 are indeed simutaneous. (Readers don't need to worry about me somehow breaking relativity here by allowing a basically superluminal signal here. One way we can assume is that momentarily, Or and Ot are comoving at the instant when they are doing the measurement)

Now experimentally, when they carry out experiments like these, what they found instead is that $(t_{Or2}-t_{Or1})v=(t_{Ot2}-t_{Ot1})v=L' < L$. That is, the measured length…

user128101

No, you are going astray, what you have to consider is the time recorded by B and F, not the time when they receive the message. That is the basic mistake. The clocks of Or and Ot bothe measure 0 when train hits B, then when both clocks measure 8mus,the cell F records time 8, Ot reads it immediately, Ot gets the message some times later, if depends on how the message it is sent, He can receive it after an hour, but the measurements still reads 8.

user128101

On the other hand Or doesn't have to wait for reading fron T because his cell will notify when the back of the trains leaves his timegate T = 12.5, after a seconf an hour or a day he'll get the reatinf from cell T (I'll change it to P)

user128101

Secret

11:19 AM

In that case, what B and F have measured is the length of the train in Or's frame (because the tunnel is stationary wrt Or). Meanwhile Ot, being on the train all the time, is statonary wrt the train. When the BF readings are compared to the rest length of the train, it is still shorter by some factor thus once again both the tunnel and Or basically saw the train as somewhat shorter than from its rest frame version

no it's all good, both B and F are located at the tunnel, not one on the trian and one on the tunnel (But anyway, (even if we are not supposed to compare between two frames))Even in such a case of subtracting one reading from one frame with another reading from another, the length obstained is still shorter than the rest legnth)

user128101

You lost me! you are taking for granted that there is some problem, where there is none. Timegates are at rest, but the receiver on the train is moving, it gets a signal from B,F and P. Ot on the trainknows that it took him 8 mus to cross the tunnel. Or gets same measurement.Where is the paradox? the problem?

Secret

The problem is that all the lengths get by Or and Ot (which agree because they both get 8 mus for train to pass tunnel) and B F (which agrees with Or) are all shorter than the rest length of the train, thus relativity is needed to explain that shortening

If no relativity is needed, the mus measured must multply with the speed of the trian to recover the length of the train for Or, Ot and BF, but this is not what is experimentally observed

which means if there is no relativity, the mus has to be some value k > 8 such that $kv=L$

user228700

12:13 PM

@MAFIA36790: Hi :-) The movie is alright; a true Dhoni fan will definitely love it. Also, my friend has only the version of Atkins with Paula :/

user128101

@Secret, "all the lengths get by Or and Ot (which agree because they both get 8 mus for train to pass tunnel) and B F (which agrees with Or) are all shorter than the rest length of the train," where do you get that from?

user116211

@KaumudiHarikumar okay.

John Duffield

@Secret : they're nearly always correct. Now and then there's a few speculative things, but not as much as you'd think. If you'd like to pick one and try to point out where it's incorrect, please do.

user228700

@MAFIA36790 It doesn't seem like ionic equilibrium is covered by this book...is it?

user116211

@KaumudiHarikumar yes.

user228700

12:26 PM

@MAFIA36790 Ugh. Too many chapters. Will go fish for it...

user228700

Which edition did u follow..?

John Duffield

@Slereah : no, the Physics party. As opposed to the Moonshine party.

@0celo7 : that's because you're realising that JD is right.

user116211

@KaumudiHarikumar I have the third edition of Atkins and fifth edition of Atkins & Paula.

Slereah

I don't think cider qualifies as moonshine

user228700

@MAFIA36790 Oh. Part 1 looks like this:

ACuriousMind

12:31 PM

@Slereah I'm not sure if what you're concocting counts as cider ;)

user228700

Slereah

Well it's applejuice and yeast

If I get vodka I'd be very surprised

user116211

@KaumudiHarikumar It's different than mine.

John Duffield

@dmckee : would you care to give an example of that?

user228700

@MAFIA36790 Is it entirely different? Did urs have ionic equilibrium too?

user116211

12:33 PM

@KaumudiHarikumar Yes, separate chapter.

user228700

@MAFIA36790 Oh, crap. Is it available online? Do u have a soft copy or something..?

user116211

@KaumudiHarikumar I bought it from Amazon; so sure it is available online.

user228700

I checked and only the versions heavily ladden with biology are available. And no, I meant to ask if the PDF is available online.

user116211

@KaumudiHarikumar No idea.

user228700

@MAFIA36790 OK, thanks.

user128101

12:37 PM

@JohnDuffield, can you tell where is the paradox?

Xasel

interetsing:)

What happens when divide by zero on mechanical calculator Facit ESA-01

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Secret

Guys, I don't think our curent technology level anyone can answer that question by user104372, cause it appears that there are no direct experimental check for legnth contraction. We always inferred it from time dilation results (because in the framework of relativity both occur together). The muon experiment came close but still indirect because its measuring time of flight, not how much of the atmosphere the muon have zipped through

23 mins ago, by user104372

@Secret, "all the lengths get by Or and Ot (which agree because they both get 8 mus for train to pass tunnel) and B F (which agrees with Or) are all shorter than the rest length of the train," where do you get that from?

user116211

@Slereah You can get sealed glass with vodka of 150-200 ml in shops of Russia; cool ;))

user116211

Slereah

I remember those when I was in the Czech republic

except they were shots of fake rum

John Duffield

12:40 PM

@user104372 : this is also known as the ladder paradox or the pole and the barn paradox. The pole is longer than the barn, and the barn is longer than the train. It points to an issue with length contraction. It doesn't work the way you think. There are no paradoxes.

@DanielSank : whenever you see a complex number there's always a rotation somewhere.

user116211

@Slereah Did you meet Lumo?

user116211

@Slereah ._.

Slereah

So it turns out there are more than one people there

user116211

;P

user128101

@JohnDuffield, I do not think, I just made a picture with data. Is there any mistake? Why do they use SR?

user128101

12:43 PM

can you answer physics.stackexchange.com/questions/264222/… ?

Secret

Facit NTK Mechanical Calculator: Division by zero, decimal markers

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John Duffield

1:02 PM

@user104372 : I can't see any mistake. The mistake is in claiming that the train is longer than the tunnel and the tunnel is longer than the train. It isn't true. The train is longer than the tunnel.

@user104372 : yes I can. You have to remember that the electron is a "spinor". A spinor is often depicted as a Mobius strip. There are two rotations, around the major axis, and around the minor axis.

Note however that the electron is not some strip. To get a better concept, you need to "inflate" the strip as if it was a bicycle inner tube, and animate it:

That's the ring torus from Adrian Rossiter's torus animations. If you keep on "inflating" the torus the ring gets fatter and fatter until it looks like a horn torus:

Take it all the way and you end up with the spindle-sphere torus. In a way it has two spins, though I'd rather say it has a single compound spin.

This thing doesn't look like it's rotating, but it is.

Note that the electron isn't some ball. It's field is what it is. It has no outer surface.

ACuriousMind

Just for the record: A spinor is never depicted "as a Möbius strip" (the picture taken from the Wikipedia article explicitly says in its description that it's the vector field on the strip that's analogous to the spinor, not the strip itself). Tori are completely unrelated to spinors. The reason Möbius bands and spinors share some features is merely that both are related to certain $\mathbb{Z}/2\mathbb{Z}$ quotients.

John Duffield

1:19 PM

The S-orbital has no outer surface either.

See the mathspages article on Dirac's belt: "An object must be translated around the loop twice in order to be restored to its original position and chirality. In this sense a Mobius strip is reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be rotated through two complete rotations in order to be restored to their original state. "

Secret

@ACuriousMind We knew that total angular momentum is conserved in a system with (forgot condition). Since from Einstein de haas one can show that spin is a type of angular momentum. If I had some electron in some state $a\lvert 1\rangle+b\lvert 0\rangle$ colliding with a nanoparticle (where its consituents don't matter and basically

act like a billard ball with nonzero classical angular momentum), how do these two momenta interact during the collision as the magnitude of $\vec{S}$ cannot be changed by any known interaction (only the polarisation direction of the spin vector can) ? Does that mean spin and classical angular momentum can only interact via the direction only, and cannot affect the magnitude of either of them?

Slereah

Which is exactly what @ACuriousMind said

The spinor group is $S^3/Z2$, but that's harder to draw

ACuriousMind

@Secret I dont understand the question. If you don't have a magnetic field or anything other coupling to the spin, why do you think spin is relevant in that interaction at all?

Secret

Because we knew that spin is a type of angular momentum. Naively I would expect something with spin angular momentum colliding with another thing with a classical angular momentum can exchange momentum in a way to conserve total momentum?

John Duffield

@Secret : see the Wikipedia Einstein-de Haas article: "the Einstein–de Haas effect demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics".

Secret

1:27 PM

JD<-R, A->ACM: Yes, which is why naively speaking if I fire a stream of electrons to a neutral ball that is actually rotating, I will soon expect something like the ball rotation to speed up or slow down?

Or do spin need something (likea magnetic field) to couple it to ordinary angular momentum in order to interact with it?

Secret

Let me elaborate:

John Duffield

@Secret : the important point is that it's quantum field theory. It's like, everything is field. A wave is a field variation, or a field is a standing wave. When you spin a billiard ball, fields are interacting. When you perform Compton scattering, fields are interacting.

The Dark Side

Ahh! Back on this site after a long time. Election time !! Good luck "bar"mates - @ACuriousMind, @tpg2114, @Jim and the two other gentlemen/women :)

Secret

Consider a case where there are two classical billiard balls rotating on their own axes and then heading for a collision. Using conservation of angular momentum, we can easily calculate the angular momenta of each ball after the collision, thus finding that they has sped up or slow down their rotation

The Dark Side

Although there's perennial confusion as to which Jim !!!

user116211

1:35 PM

2016 Physics Stack Exchange Election

Open Discussion for physics.stackexchange.com/election/3

The Dark Side

@MAFIA36790 Hi @MAFIA36790. OK. You mean we are not supposed to talk about it here?

user116211

@TheDarkSide All the election-related query should be aired in the above room; not here.

user116211

And yes, long time @TheDarkSide o/

Secret

@ACuriousMind Now consider instead of having two balls rotating on their own axes on a collision course, we have instead a stationary billiard ball rotating on its own axes being fired upon by a beam of electrons. Since spin is a type of angular momentum, why we don't see in general the billaird ball's rotation got affected by being bombarded by the electrons as would be expected by conservation of momentum?

The Dark Side

:)

OK. I'll transfer the messages over there. :)

ACuriousMind

1:38 PM

@Secret But observe that two actually rotating bodies have a mechanism to change each other's rotation: At the contact point, they exert forces on each other, which leads to torques, which changes the orbital angular momentum. They don't magically exchange angular momentum, they have a well-defined interaction for that. There's no such interaction term in your electron-collides-with-billiard-ball case.

If there is no interaction term that can exchange orbital and spin angular momentum, both will seperately be conserved.

Secret

Ok I see.

John Duffield

@Secret : imagine firing a stream of billiard balls at the Earth. You aren't going to see much of a change to the Earth's rotation.

ACuriousMind

^Red herring - it's a true statement, but in this case it's simply true because the amount of angular momentum in a billard ball is much too small to cause a noticable change in the Earth rotation. Nevertheless, the change is there.

Secret

@JohnDuffield Size is not an issue here, my question is about the existence of the outcome, ACM have answered that part nicely while you have only point out that fields need an interaction term to interact

Secret

Next Q ready in a few sec

user128101

1:41 PM

@JohnDuffield,@Secret can you write an answer at the question and explain after a collision?

Secret

Do the interaction term of spin always related to electromagnetism, that is, in order to couple spin to ordinary angular momentum, the interaction is electromagnetic in nature? I have seen examples like that such as spin orbit coupling in molecules and also spin recession in magnetic fields, where both have interaction terms of EM nature. But is EM the only way spin can interact with something else?

Slereah

No, other forces can have spin interactions

But EM is the one we most commonly see

Secret

what are some well knwon non EM spin interactions, slereah?

Slereah

Well the other gauge forces

Weak and strong

0celo7

@JohnDuffield no, I'm pretty sure you're wrong about everything

John Duffield

1:46 PM

@Secret : when we're talking about electrons, spin is always related to electromagnetism. But note that when you put your hand on your desk, your hand doesn't go through your desk because of electromagnetism.

@0celo7 : you're wrong about that too.

0celo7

@ACuriousMind You can't prove that...

ACuriousMind

@Secret The question is a bit ill-posed. In the case of magnetic fields and such, we are treating the EM field classically. That doesn't work for the other forces, and in the full quantum field theoretic description, the spin of objects can "change" in every interaction (since the actual particle states don't really exist during the interaction to begin with, it doesn't really make sense to speak of "change" here), but total angular momentum will be conserved.

0celo7

@ACuriousMind In a principal bundle we have the trivialization map $\pi^{-1}(U)\to U\times G,p\mapsto (\pi(p),F_U(p))$. For the life of me I cannot figure out how to write down that map's inverse :/

Secret

I see, make sense guys, thanks

user128101

@JohnDuffield, wow, fancy that! 7 downvotes on that question!

user128101

1:51 PM

taking turns!

user128101

blind rage!

0celo7

@ACuriousMind I'm thinking $(x,g)\mapsto F_U^{-1}|_{\pi^{-1}(x)}g$, but that's super awkward.

user128101

@Jim, thanks for opening my question, can you edit it or write an answer?

Secret

31 mins ago, by Slereah

The spinor group is $S^3/Z2$, but that's harder to draw

Ramblings: I can draw $S^3$ quite ok. Once I learn quotient space, I might be able to actually draw this thing

John Duffield

@user104372 : sympathies, IMHO that sort of thing is not good.

ACuriousMind

1:55 PM

@0celo7 I have no idea what $F_U$ is

Or did you simply define it as the second component there?

Slereah

"A function $^*f:^*A \to ^* \Bbb R$ is continuous at a point $a\in ^*A$ if for every infinitesimal $\varepsilon \in \theta$ we have $^* f(a + \varepsilon) \approx\ ^*f(a)$"

neat

ACuriousMind

If so, then the inverse map can be defined by fixing $p_0(x)$ as the point with $F_U(p) = 1$, and then defining $(x,p)\mapsto p_0(x)g$.

Secret

Can an electron have two spins? user104372 need to edit the question and the title to brought out that focus in v2 of the edit. However given that Lubo have answered the question in v1 so satisfactorily, I have no idea how an edit can be carried out to focus v2 without making Lubo's answer become disconnected with the question

Secret

"And no I cannot write an actual answer, because the answer is imperfect and incomplete given my current level of knowledge, and I don't like seeing myself writing imperfect answers"

"I only write an answer if it is certified to be able to fully account every potential subquestion that can arise from it"

"Hijacking maths for a bit: Suppose the Space of question and subquestion of a given question is a topological space, then my answer must be a base of this topological space else it is not an answer"

(No, questions don't actually form a topological space, bleh)

Slereah

why not

Sentences can form a metric space

With the levenstein distance

John Duffield

2:08 PM

@user104372 : I could, but I'm afraid I'm going to politely decline I'm afraid.

Slereah

Oh no

Polynomial functions aren't continuous in the infinite range

Secret

Questions has something that is not just about the difference between isolated characters, Daniel and I suspect that questions have a logical network structure, ACM confirmed this observation. But there's something ... else about a question that determine its properties... and I am not sure if there's a math way to formulate that

Slereah

Since $(\omega + 1/\omega)^2 \neq \omega^2$

Secret

One thing we knew about question is that there's a set called answers whose size depends on the nature of the question

and answers themselves are a complex breed

Both questions and answers have a dependence on what we called reality of which we have no complete way to simulate it

and unknwons in general don't follow statistical correlations thus our prediction on what might out there are gurentee to be almost always off from when we actually discover them

0celo7

@Slereah what?

@Secret yes they do

literally any set can be made into a topological space

Secret

2:23 PM

Ok, that's unexpected from a math ramble...Well then, this answer, then will need to be a very special collection then, if its union is going to generate all the questions and subquestions of a givne question

In which case it is not easy to tell what topology we need to use for them given that we don't know everything about one deciding factor that is tied to questions ans answers: Reality

John Duffield

I've finished my chapter/article/piece on magnetic monopoles. This is the last line: "Magnetic monopoles do not exist. If they did, there would be no such thing as charge". As you can imagine, my next chapter is on charge.

Secret

I have been wasting too much time on too interesting tangents, in the past few hours, I should be reading Munkres... ugh...

Music of the rising carbon dioxide

0celo7

@ACuriousMind What is $(x,p)$?

I want a map on $U\times G$

ACuriousMind

@0celo7 Meant to be $(x,g)$, obviously (if you examine the r.h.s.)

0celo7

@ACuriousMind What is the $p$ in $F_U(p)=1$?

That equation has many solutions

Do you mean the $p\in \pi^{-1}(x)$ s.t. $F_U(p)=1$?

ACuriousMind

2:38 PM

@0celo7 I said you pick $p_0(x)$ as the point with $F_U(p) = 1$. If you can't figure out what that's supposed to mean, bad luck.

0celo7

@ACuriousMind But there are many such points!

ACuriousMind

Yes, there are. I nevertheless believe it's reasonably clear what I intended to do.

0celo7

Did I guess correctly with the third line?

Danu

@user104372 What question?

0celo7

@ACuriousMind Bishop-Crittenden write that one can factor the map $p:G\to \pi^{-1}(x),g\mapsto pg$ as $p=(\pi\times F_U)^{-1}(x,L_{F_U(p)})$. I think that makes sense, but then computing $\mathrm dp_p$ seems impossible...

John Duffield

2:46 PM

@Danu : this question.

0celo7

Correction: $\mathrm dp_e$.

user218912

3:22 PM

lol I woke up at 11.

0celo7

There's a dude just walking in circles holding an empty cup

Oh my god

It's the crazy dude from physics last year!

Secret

Last night I had a time travel dream

It's not very complicated in terms of the actual time travel itself, but causality gets a bit... weird
Basically the story boils down to an organisation trying to stop an apocalpyse thus they send an agent back to a moment in time. The agent has a letter written by someone which summarise everything happened in history
including the 9 key events that the agent must suceed on at least one of them in order to avert the apolcalpyse
The agent, however is clumsy, and failed to change the events one by one until the 8th one

during that he saw the content of the letter changed for each failure, and also making subsequent missions harder or even impossible depending on how he failed at them
Eventually, at the 8th event, he battled the big bad, dream ends here

user218912

@0celo7 I remember him.

2016 Physics Stack Exchange Election

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