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Jim
12:00 PM
So, I could do one, but I don't make promises about the quality
 
@DanielSank cc @vzn ^
 
@ACuriousMind I did literally every quest and explored everything in the Orchard and am not even level 5
 
@Jim Would one of our chat sessions work for you or would you prefer a different timeslot?
@0celo7 That's completely fine, iirc
 
Ghouls and Drowners are impossible
How do I fight them
 
What difficulty are you playing on?
 
Jim
12:01 PM
@ACuriousMind chat sessions are at the worst possible time for me every time
 
The second one
 
@0celo7 Hm, then they shouldn't be so hard. What you need to learn in contrast to many other games is that the prime directive of Witcher combat is don't get hit - you are squishy and will always die in a few hits
Quen is your friend for that
If you invest in Igni, it can reliably ignite enemies and keep them occupied with that instead of attacking you
Otherwise, dodge, dodge, dodge - I usually spend more than half of the early combats rolling around trying to hit them without getting hit
 
@ACuriousMind yeah with ghouls I find their attack to be very hard to dodge
It's just a little slap, hard to see coming
And they're always en masse
 
Well, I'll not say that ghouls aren't the most annoying enemy early on. That's where Quen - get hit without taking damage - and Aard - knock them down and then insta-kill them - are pretty useful.
@Jim What sort of time would be better?
 
I haven't used Quen. I'll take a look
 
Jim
12:11 PM
@ACuriousMind Well, around this time is best for me, otherwise it depends on the day
 
@Jim Well, I'd say the time is pretty much up to you if the chat sessions don't work for you, but we usually announce the AMAs something like 2 weeks or more in advance on meta (also to collect questions).
 
Jim
@ACuriousMind how long is the AMA?
 
@Jim An hour, usually
 
As long as you want to talk :P
 
Jim
so then 8:30am ET on any given Tuesday or let's say noon on a Thursday
whichever is better
 
12:16 PM
ET?
 
Jim
eastern time
 
Ok, I'll pass the message^ along to @DanielSank
vzn hasn't been in here for awhile
 
@LasVegasRaiders Uh, why? Though he likes the AMAs (I think), we don't really have anyone in charge of them.
 
Jim
so who announces them?
 
Usually vzn or Daniel
 
Jim
12:19 PM
I see
 
@Jim We usually have the speaker themselves post a short intro on meta asking for questions after the timeslot is decided, see e.g. physics.meta.stackexchange.com/q/7782/50583, physics.meta.stackexchange.com/q/8981/50583, physics.meta.stackexchange.com/q/9613/50583 for previous AMAs.
 
They'll contact you.
 
Jim
@ACuriousMind asking for questions? Hey, I'll set aside an hour to answer any questions someone wants to ask me (or at least, give a reason why I won't answer it) but I'm not begging the question. I'll probably just say "space is big" and that'll go for an hour or two
 
@Jim We've usually had a mixture of spontaneous questions and those collected on meta beforehand, it's worked quite well
You don't have to do anything beforehand except to tell us who you are
 
Jim
@ACuriousMind you mean talk about Jim? THAT I can do!
 
12:24 PM
(not breaking anonymity, but what kind of things you work on or know or are interested in)
 
@Slereah has been looking for somebody to talk about black holes with @Jim
 
Jim
@LasVegasRaiders it's been a few years since I really delved into black holes. They've changed what's new
 
I am reading "Physics from Symmetry" by Jakob Schwichtenberg. so far pretty enjoying it. but I have ppl on the Physics Forum complained the author was an undergraduate.
but I think they also mentioned they haven't read it
 
Jim
@Shing if the content is correct and easy to understand, who cares who it comes from?
 
anyone here has read it? and would like to share any opinions on it?
@Jim true
 
12:32 PM
@Jim Well, the problem with introductory books is that if you're the sort of person who's reading that to learn it, you have a pretty hard time judging whether it's correct or not.
 
Jim
@ACuriousMind true, hopefully some reviews address the correctness and don't just focus on the qualifications of the author
Alright, just give me a day to nail down my schedule and I'll write up an AMA introduction
 
Ok, looking forward to it :-)
 
Jim
then we'll all get to do something that doesn't happen often; listen to me ramble on about how great I am
 
Suppose you have the joint state of two systems averaged: $$\sum_{n}p_n \rho_n \otimes |n \rangle \langle n |$$ where $p_n$ is the probability of being in that joint state. Why would it follow that we can diagonalize the states $\rho_n$ without affecting the total von Neumann entropy of the joint state?
 
12:40 PM
Hey hey
 
Is it simply because if $U$ is unitary then $U \otimes I$ is unitary on tensor products and von Neuamnn entropy is invariant under unitary operators?
 
Jim
Side note: I'm finding way too much personal amusement from the thought that "Jack" is short for "John" and so your name can be interpreted as "John John". Seriously, that's not funny, but I'm enjoying it
 
or "Jack Jack"
 
Jim
^ indeed
 
"John" always makes me think of Johnathan from Roller Ball.
 
Jim
12:53 PM
okay, that's a bit odd
 
John a than, John a than
 
Jim
terrible grammar
plus, it's two names: Jo Nathan
the h is invisible
 
actually what is the relation between the name Steven and Stevenson?
 
Jim
stevenson is the son of steven
 
lol is that really the origin of stevenson?
 
Jim
12:56 PM
yes, really
last names are not very creative
Stevenson is an English language patronymic surname meaning "son of Steven". Its first historical record is from pre-10th-century England. Another origin of the name is as a toponymic surname related to the place Stevenstone in Devon, England. Notable people sharing this surname include: Adonis Stevenson (born 1977), Canadian boxer Alexander Campbell Stevenson (1802–1889), American politician and physician Alexandra Stevenson (born 1980), American tennis player Ben Stevenson (born 1936), British ballet dancer and director Carter L. Stevenson (1817–1888), American soldier Charles Stevenson (1908...
 
lol
 
#jim_for_AMA , BOOK IT MODS (@ACuriousMind @DavidZ @Danu)
 
Jim
@Fawad I would be unsurprised if in a few years there was a similar post reading "#jim_for_pres, ....."
 
1:01 PM
Star it so mods will see
 
Jim
#jim_for_pres
2
let's star that one too
 
President?
 
Jim
yes
 
If trump can then why can’t you
 
Jim
I figure I'm at least as good as the current one
 
1:03 PM
@Fawad What do you mean by "book it"?
 
Make a booking.
 
Jim
plus, I've also got ties to another country that would make it illegal and possibly treasonous for me to be president
 
Jim said he needs a bit to figure out the best time, what am I supposed to do? (Or Danu, for that matter, who isn't even a physics mod and I'm not sure why you'd ping him)
 
@ACuriousMind hey. Thanks for responding to my message. Can you arrange AMA of Jim?
 
Jim
@ACuriousMind yeah, I've got a note in my calendar to write up an intro tomorrow. I'll announce a good date then as well
 
1:06 PM
I thought mods help each other. If you don’t know how to arrange AMA then danu can give advice
 
@Fawad I thought that's what I've been doing here just now. What specifically do you want me to do?
 
They do help each other @Fawad thanks for the enthusiasm :-)
 
@ACuriousMind remove my question ban
 
That will never work pal @Fawad
 
@Fawad 1. I can't do that (as in, I literally can't, moderators have no power over the automatic bans). 2. That has nothing to do with the AMA or "arranging" it. I was asking what you wanted me (or other mods) to do when you said we should "book it".
 
1:10 PM
Try asking for improvement guidelines @Fawad
 
@LasVegasRaiders I know. Just showing mind what he can’t specifically do for me.
 
3 messages moved to Trash
 
:O
that's strike 2 pal
 
Which strike? I don’t see
 
@LasVegasRaiders What does that mean, and who are you talking to?
 
Jim
1:16 PM
me, him and I are bowling
 
You have bowling in Canada? I though you all curled ;P
 
Jim
we can do two things
bowling is like summer curling
 
Baseball
 
Jim
because shuffleboard is stupid
 
True that^
@Fawad a mind is a terrible thing to waste
 
Jim
1:30 PM
and a waist is a terrible thing to mind
 
^indeed
 
@LasVegasRaiders acuriousmind
 
@Jim wot
You're not a natural US citizen
And Trump is the president from now on
 
Jim
@0celo7 read my 4th next message
 
1:47 PM
When Jim is here it's like the good old days again.
All we need now is Sophie :P
 
@Justwinbaby she couldn't handle all of the porn we have
And Chris is still gone
But I guess you forgot about him.
 
Jun 23 at 3:03, by Secret
May 2 at 17:13, by Bernardo Meurer
Please come back Chris White
Jun 23 at 3:03, by Secret
Apr 22 at 9:30, by Secret
Mar 27 at 22:27, by ACuriousMind
I am beginning to believe @BernardoMeurer is stuck in some kind of bizarre time loop that involves Chris White and free software.
Actually...
that time loop is broken a long time ago already
 
2:11 PM
How to write the set of matrices
$M$, $\mathfrak M$, $\mathcal M$ or $\mathscr M$
 
2:32 PM
@Slereah Mathbb M
Or mathrm Mat
 
vzn
2:44 PM
@Jim hi / @LasVegas thx for interest/ memory of AMAs. the next regular mtg is in 7 days tue jul 11 1600Z. Jim, wanna do that one? its enough time for a (meta) announcement but youd have to post asap. @Emilio was our last invitation, hasnt gotten to his announcement yet. chat.stackexchange.com/rooms/info/71/the-h-bar?tab=schedule physics.meta.stackexchange.com/questions/7783/…
 
@vzn If you read the rest of hte chat log, you'll discover that chat session times aren't good for Jim and that he'll figure out till ~tomorrow when exactly would fit his schedule best
 
I'm a doctor not a physicist, @Jim
 
vzn
@ACuriousMind ok thats fine @Jim you can pick the time in the meta post if you like, plz post announcement to chat & cc me
 
@ACuriousMind Do you know if we simply define the von Neumann entropy of a joint system say $V = \sum_n \rho_n \otimes |n \rangle \langle n |$ as $-\sum_n\lambda_n \text{ln} \lambda_n$ where $\lambda_n$ are the diagonal elements of a dialgonalized $V$?
 
@Slereah you are a physician?
 
2:57 PM
[Testing]
Hmm
> I would rather not.
*Trollface*
 
@JohnJack Why would we define it specifically for joint systems if the ordinary definition ($\mathrm{tr}(V\ln(V))$) works perfectly well? That said, what you wrote is exactly the same as that, so I don't understand the question.
 
@ACuriousMind So your answer is yes?
 
user228700
@JohnR: Maybe just me but I'm taking a lot longer to complete the next levels. I'm still on level 18...
 
In most games the levels get harder as you go on.
 
user228700
Yeah, yeah...
 
3:00 PM
How close to the end are you? Is it time to buy Portal 2 yet? Was the cake a lie? :-)
 
@JohnJack No, what I'm saying is that 1. I don't understand why it matters that this is a joint system and 2. For any system, $\mathrm{tr}(\rho\ln(\rho)) = \sum_n \lambda_n \ln(\lambda_n)$ where the $\lambda_i$ are the eigenvalues (or "diagonal elements of its diagonalization")
 
@ACuriousMind Yeah but it becomes trickier when working with tensor products so I'm just confirming. So you would have to diagonalize each block matrix of $V$ to find $\lambda_n$ ?
 
@JohnJack What do you mean by the block matrices? Why would you not just diagonalize $V$? If you have $V = W\otimes X$, then you could diagonalize $W,X$ to make it easier, but in the general case where $V = \sum_i W_i\otimes X_i$ (assuming the $W_i,X_i$ are what you mean by blocks), I don't see how that would make anything easier
 
@0celo7 I will heal you
 
3:23 PM
@ACuriousMind From time to time it becomes easier to consider in the way I mentioned.
For each block $\rho_n$ I know that the entropy is invariant under unitary transformation $U_n$ i.e. $S(\rho_n) = S(U_n \rho_n U_n^{\dagger})$. Further if we define $U_n \otimes I$ then $S(U_n \rho_n U_n^{\dagger} \otimes |n \rangle \langle n|) = S(\rho_n) + S(|n \rangle \langle n |) = S(\rho_n)$.

But, do you see though how for $V = \sum_n p_n \rho_n \otimes |n \rangle \langle n |$ we have that $S(V) = S(
\sum_n p_n U_n\rho_n U_n^{\dagger} \otimes |n \rangle \langle n |)$. Since $S$ is not a linear operat
 
@JohnJack You've just applied the unitary operator $U_n\otimes \mathrm{Id}$ to $V$, so since you know entropy is invariant under unitaries, that shows it's true.
 
@ACuriousMind It doesn't show it for the case of a series of $n$. It shows it for the one case I mentioned where you have $\rho_n \otimes |n \rangle \langle n|$. That's what I'm asking. It seems you would need to invoke linearity to extend to the series case.
 
OH, wait, is the claim really supposed to be that there's a different $U_i$ for every $\rho_i$?
 
@ACuriousMind Yeah
@ACuriousMind That's my claim. Each $\rho_n$ would need a different $U_n$ to diagonalize it (possibly need a different $U_n$).
 
What reason do you have to believe this is true?
 
3:33 PM
@ACuriousMind How else would you diagonalize each $\rho_n$ than to find a unitary $U_n$ for each $n$? The claim from the text is that "you can diagonalize each $\rho_n$ without changing the total entropy of the joint state $V$"....
 
@JohnJack Do they really mean that you can diagonalize them all simultaneously? The "each" could also just mean that you can choose each of them and diagonaize it, but not more than one.
 
What's the proof that the orthogonal group is a Lie group, anyway
 
@Slereah Preimage theorem.
 
@Slereah parametrize via Euler angles, maybe?
 
3:45 PM
Sounds tough to do in arbitrary dimension and signature
 
@Slereah then yeah, probably via the preimage theorem
 
@EmilioPisanty Okay, but that's horrible :D Much easier to exhibit $\mathrm{O}(n)$ as the zero level set of the smooth map $A\mapsto AA^T - \mathrm{Id}$ and use the preimage theorem.
 
but why do you care about O(d) for d>3?
=P
 
@EmilioPisanty Spacetimes of arbitrary dimension, obviously
 
@ACuriousMind I think they do because to diagonalize each of $\rho_n$ at a time you would need a $U_n$ for each of them right? And so we would still get $S(V) = S(
\sum_n p_n U_n\rho_n U_n^{\dagger} \otimes |n \rangle \langle n |)$ in order for the statement "you can diagonalize each $\rho_n$ without changing the total entropy of the joint state $V$" to make sense. We are diagonalizing $V$ so all $\rho_n$ need to be diagonalized.
 
3:48 PM
Plus as you well know spacetime is an 11 dimensional manifold
 
@EmilioPisanty I wonder how much the decision weighed
 
it is true
If the earth was round we would slide off
2
also it would roll away
 
still, I have yet to come across any one of those that's better than Concave Earth theory
 
didnt you guys know? the earth is a Möbius strip
 
3:55 PM
I guess the flat Earth thing is effectively a religion. People choose to believe it then cast around for evidence to support that belief.
 
Well like a lot of conspiracies like that it's rooted in deep distrust of the Authority
How can u believe anything the Man says
 
> if the earth was flat, cats would have pushed everything off of it by now
2
 
It's always surprised me how people believe scientists constitute any form of authority. They clearly haven't met many scientists.
 
also of course the flat earth is 1) not too easily testable by a random shlub and 2) any argument that doesn't require to move a lot requires some finesse
 
@JohnJack Ah! We have that $S(\sum_n p_n V_n) = S_\text{Shannon}(\{p_n\}) + \sum_n S(V_n)$, where I wrote $V_n = \rho_n\otimes \lvert n\rangle\langle n\rvert$. And $S(U_n V_n U^\dagger_n) = S(V_n)$, which shows the claim.
Hm, wait, that's only an inequality in general and I forgot the conditions to have equality...
 
4:00 PM
@ACuriousMind So consider the orthogonal group as a subset of the matrix space $\Bbb R^{n\times n}$?
 
@Slereah Yes
 
That makes sense yeah
 
Jim
@vzn Nah, tuesday at noon doesn't work for me. Is it possible for me to just announce a random time and hope everyone shows up then?
 
lets do the ama now
 
@ACuriousMind Thank you.
 
4:02 PM
whats your stance on abortion?
should drugs be legalised?
 
@EmilioPisanty Why the %(@&* not?
 
is chemistry a science?
 
@Jim I expect that is possible.
 
@Jim Yes, if you do so sufficiently in advance (I'd say 2 weeks or so)
@AccidentalFourierTransform Stop trolling :P
 
4:07 PM
@ACuriousMind That would resolve it it seems. Where did you find that result? There are so many of these results but that one seems very important.
 
@JohnJack I think it's not true in general, sorry
 
Oh Damnit
 
I am currently trying to download the Lois Lane comics
Hard to do
Not a lot of seeds
 
@ACuriousMind does high RAM latency explain me sucking at fighting ghouls?
 
@AccidentalFourierTransform That does look like the kind of explanation our flat-earth agnostic friend @0celo7 would be deeply interested in !!
 
4:14 PM
@0celo7 perhaps u just suck
For ghouls I would advise getting a good hand to hand weapon, btw
Especially a sustained damage one
The minichainsaw thing is very good for that
 
@0celo7 No.
 
Since ghouls just throw themselves on it
 
@JohnRennie - Thanks a lot for the immensely useful discussion the other day. Maybe I'll bother you again some time when I have more substance to talk about!!
 
@Slereah Uh...what game do you think we're talking about?
 
@ACuriousMind Fallout 4?
 
4:16 PM
@0celo7 Not really, also, I don't know what "high" RAM latency is
@Slereah Nope, TW3
 
o
Are you doing a lot of witching
Maybe flying around on a broom
Stirring a big cast iron pot
 
@ACuriousMind It states further that it is irrelevant that each of the $\rho_n$ will have a different eigenbasis since $V$ is block-diagonal where each $\rho_n$ is one of the blocks. I don't really see how that helps.
 
@JohnJack Oh! Well, of course, that helps a lot
If you have a matrix of the block form $\mathrm{diag}(\rho_1,\dots,\rho_n)$, then the matrix that diagonalizes $\rho_1$ is of the form $\mathrm{diag}(D_1,0,\dots,0)$, that that diagonalizes $\rho_2$ is of the form $\mathrm{diag}(0,D_2,0,\dots,0)$ and so on
So all these matrices commute, hence you can diagonalize all the $\rho_i$ simultaneously
 
also, if you write $D_i=(D_i+1)-1$, you can conclude that $D$ need not be Toeplitz
 
4:32 PM
@ACuriousMind So your $D_{1}$ is some unitary matrix such that $D_{1}\rho_1 D_{1}^{\dagger}$ is diagonal?
 
@JohnJack Yes.
 
@TheDarkSide you're welcome. I'm always happy to chat about physics, though I must admit that I wasn't always sure exactly what we were chatting about :-)
 
Well, I was borrowing from your colloidal expertise to shovel knowledge over my ignorance !!
 
@ACuriousMind I know that two commuting matrices can be simultanously diagonalised...In this case, is the idea that since the $diag(0,0,D_i,0,)$ commute with each other they diagonalise the blocks seperately hence we can apply them all at once in any order since they only affect one block at a time?
 
4:50 PM
If we have a oscillating system, in general the particular solution describes the system in a steady state, while the homogeneous solution describes the transient effect, so it appear to me that when having a system with having start conditions as in steady state the hom. sol. is $0$, I'd appreciate a quick approval from someone more experienced, thanks :)
 
7
Q: Is it viable to wall modern cities against "zombies"?

X22T7VLet's say the city covers about 40 square miles (104 square kilometers), that machinery and some oil refineries are still usable (for construction vehicles), and that there is a lot to salvage in way of constructing walls. In addition, let's say one would have to be athletic to outrun infected hu...

^ lol what ?
 
why do you need to wall cities against zombies
Zombies are like humans except worse at everything
They're like slow children
 
How will military defense kill what's already dead?
 
REMOVE THE HEAD OR DESTROY THE BRAIN
 
> If you want to build a wall around a city after a zombie apocalypse, it's to keep the zombies in not out.
So, zombies can't climb walls?
 
4:55 PM
No
Zombies are very dim
 
So basically, corpses can walk but not climb?
 
they can barely walk
they suck
 
I have a problem with these zombies acting as per their convenience.
Comment posted. Wrath invited.
What if zombies climb the wall? If corpses can walk, why can't they climb? — The Dark Side 1 min ago
 
@Slereah slow? Have you not seen World War Z?
 
@JohnRennie No thanks I'd rather see blood in my urine
2
 
5:31 PM
@ACuriousMind 75ns
 
6:03 PM
> The whole point to mathematical physics is that it is a self-contained kind of structure which can make its own internal predictions. As Wigner has emphasized, the most amazing thing about mathematical physics is that it describes the world at all -- that's a miracle, something we don't understand.
 
I have no idea what will a physical law be like if it cannot be written in mathematical terms
 
Here's what I have done:
Then I tried to equate the two equations of $x$ to solve for $t$
But that didn't work.
Can someone please help me out?
@DawoodibnKareem Thanks a ton for replying to my queries brother!
 
Sorry @Abcd i'm afraid I have important business to take care of right now
Like posting this
 
@DawoodibnKareem @JohnRennie I got the answer when I took the Range as Range + 5 cos 15
@Secret Can like poles repel and opposites attract be represented mathematically?
@Slereah ok
 
you need to go to QFT to do that
 
6:15 PM
@Secret QFT?
 
quantum field theory
 
ok
 
6:30 PM
@ACuriousMind I overclocked my RAM
What have I done
@BernardoMeurer the speed increased by 1ns...
@JohnRennie I can't figure out the speed issue
 
user228700
2 days ago, by Kaumudi. H
I don't like biscuits...
 
user228700
@JohnR: I take this back. I quite like these.
 
Are space filling curves actually space filling
Or do they leave out a set of measure 0
 
@Slereah - This one's for you:
5
Q: Every Fourth Wormhole

paramesisAn entry in Fortnightly Topic Challenge #33: Surface Geometry Mazes

It has no answer as of now.
 
6:45 PM
@Slereah what a picture
 
@Slereah they are subjective.
Subjective.
Surjective
 
Isn't all math subjective in the end
 
7:05 PM
Objectively subjective depending on your perspective
(also, there is no "end")
 
@Fawad FWIW vzn is the one who manages the AMAs. Mods have nothing to do with that. You're barking up the wrong tree ;-)
 
Hi @DavidZ
 
Hi
 
How are you?
 
Fine
 
7:19 PM
@ACuriousMind Eek I have to fight one of the Wild Hunt??
 
Ted is chatting in the math room @0celo7
(if you have any questions)
 
@0celo7 Ah, Keira's quest? :D
 
@LasVegasRaiders Why would I have questions for him?
@ACuriousMind Yeah. Using Quen is the real powerplay in this game
attack, dodge, get hit, roll away, use quen, repeat
 
Well... he does have over 30 years of university math teaching experience.
Hmm, I wonder if he would be interested in doing an AMA?
 
@Abcd Brother, there'll always be a little trial and error with a problem like this. Here's how I'd solve it. First, calculate the height of the ball after it's travelled 110m. Use that height to estimate which bench it will hit, and whether it will hit the vertical or the horizontal part of the bench. To help you estimate, bear in mind that at the angle the ball is travelling, it will go approximately 3m horizontally for every 4m vertically. (continued)
Then, if you estimated that it would hit the vertical part of a bench, calculate the exact height that the ball is at once it has travelled the right horizontal distance, to confirm your estimate. If you estimated that it would hit the horizontal part of a bench, calculate the exact horizontal distance that the ball has travelled once it has reached the height of the bench, to confirm your estimate.
If you estimated correctly, then al-hamdu Lillah, you have your answer. If not, then re-estimate based on how far off your calculations were, and do the calculations again.
 
7:44 PM
@ACuriousMind is "An Invitation from Keira" a booty call?
 
¬¬ damn exponential scaling
 
@0celo7 Yeah, at least at the beginning it is
@0celo7 Go and find out! (Better not post pictures of the result here, though ;P)
 
TMI on the "booty calls"
So Ted is too busy right now for an AMA...
 
8:08 PM
@Kaumudi.H Har Cookie Mein Kayi Smiles ROTFL
 
@ACuriousMind what does adrenaline actually do in this game?
 
@0celo7 Not much (slight damage increase I think) until you take some talents that use it - e.g. you can take one that lets you cast signs with adrenaline when you don't have stamina.
 
@ACuriousMind the bloody baron seems like a decent guy
 
@0celo7 In the interest of avoiding spoilers, I will decline to comment on the baron at all until you have completed the main story in Velen fully.
 
8:42 PM
@ACuriousMind I can't figure out how to craft Samum
wiki says it's a default formula but it's not in my alchemy menu
 
@0celo7 And you didn't already craft it? What bombs are in there?
 
@ACuriousMind Just Grapeshot.
 
Hm.
That's odd
 
@ACuriousMind just killed the Satanic Chicken
 
8:57 PM
@0celo7 What's that?
 
9:13 PM
I have to say that the question buried in the faux proof here is actually both subtle and interesting.
Moreover the answers posted before it was closed (presumably in part due to the OP's attitude) failed to hit the key point.
The key point is that the "proof" assumes uniform circular motion, but when the shape changes various mass elements trace out non-circular paths.
The insidious part being that the motion is uniform and circular before and after the shape alteration.
Very distracting.
 
@dmckee I...actually think that's not very subtle, but you're right that somehow every answerer managed to completely miss the point
 
@ACuriousMind the shrieker
cockatrice I think
 
Btw, I think that question just got closed because it fails to actually be a question - there's an implied "What did I do wrong?", but really no clear question that one could answer
 
@dmckee Nice, thanks.
 
@ACuriousMind - I've finally had the time to update my placeholder answer, physics.stackexchange.com/a/342932/52112 . Could you undelete it?
 
9:19 PM
@dmckee Although, I still have questions. Part of the reason I was unsure about the ability to experimentally realize the rapid measurement was I never really understood what the measurement rate (in the sense that matters for observing these short time effects) was to begin with.
 
@ACuriousMind Well, subtle at the level of the intro course. I wouldn't expect a first year student to spot it and would be pleased if any did.
 
For example, let's take measuring radioactive decay with a geiger counter. I have a sample of e.g. Radon next to a geiger counter for a couple hours so I can later histogram bin the decays and try to determine a decay law - what was the measurement rate, in the sense that matters for these short time violations?
the inverse dead time of the geiger counter seems an unsatisfactory answer
 
@GPhys Depends a lot on the setup, but typically you latch signals at a granularity somewhere between tens of nanoseconds and milliseconds, though each latch will render the detector dead for tens of nanoseconds or more.
 
@DavidHammen That practice of posting a placeholder answer that isn't even an answer is, well... I'm sure there's a more appropriate term which I can't think of right now but I'm going to call it "scummy". Since ACM is handling this I leave it to him to decide what to do in this case, but personally I would not undelete the answer, and I'd advise not doing that in the future.
 
I seem to recall that the QZE experiment probed processes with a relaxation time on order of seconds and probe the state on the millisecond scale, which is easier than trying to poke something every nanosecond.
 
9:23 PM
@DavidHammen I just did since you already put the work in (but don't post other placeholders in the future and expect me to do that again), I just have one question: If you think the question is clear enough to answer it, why haven't you voted to reopen it?
 
But I think the problem here is that a Geiger counter doesn't reach out and measure the source: it just waits for something to come to it.
 
@David Z -- I saw four votes to close on what I saw as a good question. I knew the question would be closed due to overly itchy close fingers by the time I had time to answer it. (It's the Fourth; I had grilling that needed to be attended to.)
 
Those QZE experiment probe the system they are diddling: the experimenter poke the system rather than waiting for the system to poke the instrument.
 
@ACuriousMind -- On it.
BTW, Happy Fourth of July to all!
 
::lights sparkler::
 
9:36 PM
There was an interesting 4th of July filler story on NPR yesterday, an interview with a chemE professor from Purdue. He toyed with inventing explosives as a kid. Those experiments eventually resulted in a trip to the hospital. His Dad gave him a sparkler and told him not to play with anything more dangerous until he knew what he was doing.
 
@DavidHammen Noted, but that doesn't change anything I said. (Or perhaps more clearly, I stand by what I said even after considering your response.)
 
@DavidHammen Yes. Even peeling BlackCats apart for the powder can lead to stupid, stupid behavior. My delinquent buddies and I decided to build a small bomb with the powder. Just to see how big a boom it would make, you understand; not to blow anyone up.
We built in in a glass peanut butter jar and figured a 3/8" (AKA 1+ cm) plywood partition would be an adequate safety barrier.
Then we just had to peer over the top of the barrier when we set it off.
We got lucky, but on seeing that some of the shrapnel went nearly through the board we wised up.
 
@dmckee -- All of the interesting people I know did incredibly stupid things when they were young. It's amazing we lived.
 
@dmckee can I prepare the system in a high enough temperature to observe the low time (QZO) decrease too?
 
9:53 PM
@GPhys Not a clue. This isn't really a subject I study, just something I went on a reading spree about once, and then being satisfied started ignoring again.
 
I backed myself into an interpretational corner with quantum mechanics and I'm having a crisis, but maybe you can help me ask the right questions
 

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