The h Bar

Jim

12:00

So, I could do one, but I don't make promises about the quality

Las Vegas Raiders

@DanielSank cc @vzn ^

0celo7

@ACuriousMind I did literally every quest and explored everything in the Orchard and am not even level 5

ACuriousMind

@Jim Would one of our chat sessions work for you or would you prefer a different timeslot?

@0celo7 That's completely fine, iirc

0celo7

Ghouls and Drowners are impossible

How do I fight them

ACuriousMind

What difficulty are you playing on?

Jim

12:01

@ACuriousMind chat sessions are at the worst possible time for me every time

0celo7

The second one

ACuriousMind

@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

0celo7

@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

ACuriousMind

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?

0celo7

I haven't used Quen. I'll take a look

Jim

12:11

@ACuriousMind Well, around this time is best for me, otherwise it depends on the day

ACuriousMind

@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?

ACuriousMind

@Jim An hour, usually

Las Vegas Raiders

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

ACuriousMind

12:16

ET?

Jim

eastern time

Las Vegas Raiders

Ok, I'll pass the message^ along to @DanielSank

vzn hasn't been in here for awhile

ACuriousMind

@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?

Las Vegas Raiders

Usually vzn or Daniel

Jim

12:19

I see

ACuriousMind

@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.

Las Vegas Raiders

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

ACuriousMind

@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!

ACuriousMind

12:24

(not breaking anonymity, but what kind of things you work on or know or are interested in)

Las Vegas Raiders

@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

Shing

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?

Shing

anyone here has read it? and would like to share any opinions on it?

@Jim true

ACuriousMind

12:32

@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

Las Vegas Raiders

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

Las Vegas Raiders

lol

John Jack

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?

Slereah

12:40

Hey hey

John Jack

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

Las Vegas Raiders

or "Jack Jack"

Jim

^ indeed

Las Vegas Raiders

"John" always makes me think of Johnathan from Roller Ball.

Jim

12:53

okay, that's a bit odd

Las Vegas Raiders

John a than, John a than

Jim

terrible grammar

plus, it's two names: Jo Nathan

the h is invisible

Shing

actually what is the relation between the name Steven and Stevenson?

Jim

stevenson is the son of steven

Shing

lol is that really the origin of stevenson?

Jim

12:56

yes, really

last names are not very creative

Stevenson

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...

Las Vegas Raiders

Rollerball - Jonathan, Jonathan

►

Shing

lol

Fawad

#jim_for_AMA , BOOK IT MODS (@ACuriousMind @DavidZ @Danu)

Las Vegas Raiders

:-D

Jim

@Fawad I would be unsurprised if in a few years there was a similar post reading "#jim_for_pres, ....."

Fawad

13:01

Star it so mods will see

Jim

#jim_for_pres

2

let's star that one too

Fawad

President?

Jim

yes

Fawad

If trump can then why can’t you

Jim

I figure I'm at least as good as the current one

ACuriousMind

13:03

@Fawad What do you mean by "book it"?

Las Vegas Raiders

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

ACuriousMind

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)

Fawad

@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

Fawad

13:06

I thought mods help each other. If you don’t know how to arrange AMA then danu can give advice

ACuriousMind

@Fawad I thought that's what I've been doing here just now. What specifically do you want me to do?

Las Vegas Raiders

They do help each other @Fawad thanks for the enthusiasm :-)

Fawad

@ACuriousMind remove my question ban

Las Vegas Raiders

That will never work pal @Fawad

ACuriousMind

@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".

Las Vegas Raiders

13:10

Try asking for improvement guidelines @Fawad

Fawad

@LasVegasRaiders I know. Just showing mind what he can’t specifically do for me.

ACuriousMind

→ 3 messages moved to Trash

Las Vegas Raiders

:O

that's strike 2 pal

Fawad

Which strike? I don’t see

ACuriousMind

@LasVegasRaiders What does that mean, and who are you talking to?

Jim

13:16

me, him and I are bowling

ACuriousMind

You have bowling in Canada? I though you all curled ;P

Jim

we can do two things

bowling is like summer curling

Las Vegas Raiders

Baseball

Jim

because shuffleboard is stupid

Las Vegas Raiders

True that^

@Fawad a mind is a terrible thing to waste

Jim

13:30

and a waist is a terrible thing to mind

Just win baby

^indeed

Fawad

@LasVegasRaiders acuriousmind

0celo7

@Jim wot

You're not a natural US citizen

And Trump is the president from now on

Jim

@0celo7 read my 4th next message

Just win baby

13:47

When Jim is here it's like the good old days again.

All we need now is Sophie :P

0celo7

@Justwinbaby she couldn't handle all of the porn we have

And Chris is still gone

But I guess you forgot about him.

Secret

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

Slereah

14:11

How to write the set of matrices

$M$, $\mathfrak M$, $\mathcal M$ or $\mathscr M$

AccidentalFourierTransform

yes

0celo7

14:32

@Slereah Mathbb M

Or mathrm Mat

vzn

14:44

@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/…

ACuriousMind

@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

Slereah

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

John Jack

@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 $V$?

0celo7

@Slereah you are a physician?

Secret

14:57

[Testing]
Hmm
> I would rather not.
*Trollface*

ACuriousMind

@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.

John Jack

@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...

John Rennie

In most games the levels get harder as you go on.

user228700

Yeah, yeah...

John Rennie

15:00

How close to the end are you? Is it time to buy Portal 2 yet? Was the cake a lie? :-)

ACuriousMind

@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")

John Jack

@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 $\lambda_n$ ?

ACuriousMind

@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

Slereah

@0celo7 I will heal you

John Jack

15:23

@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…

ACuriousMind

@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.

John Jack

@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.

ACuriousMind

OH, wait, is the claim really supposed to be that there's a different $U_i$ for every $\rho_i$?

John Jack

@ACuriousMind Yeah

@ACuriousMind That's my claim. Each $\rho_n$ would need a different $U_n$ to diagonalize it (possibly need a different $U_n$).

ACuriousMind

What reason do you have to believe this is true?

John Jack

15:33

@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 $V$"....

ACuriousMind

@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.

Emilio Pisanty

physics.stackexchange.com/questions/342935/…

it seems OP changed his mind

Slereah

What's the proof that the orthogonal group is a Lie group, anyway

ACuriousMind

@Slereah Preimage theorem.

Emilio Pisanty

@Slereah parametrize via Euler angles, maybe?

Slereah

15:45

Sounds tough to do in arbitrary dimension and signature

Emilio Pisanty

@Slereah then yeah, probably via the preimage theorem

ACuriousMind

@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.

Emilio Pisanty

but why do you care about O(d) for d>3?

you're probably in the wrong place, bro

=P

Slereah

@EmilioPisanty Spacetimes of arbitrary dimension, obviously

John Jack

@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.

Slereah

15:48

Plus as you well know spacetime is an 11 dimensional manifold

John Rennie

@EmilioPisanty I wonder how much the decision weighed

Emilio Pisanty

Flat Earth PROOF: Spirit Level Flight Experiment

►

the Earth is flat

Slereah

it is true

If the earth was round we would slide off

2

also it would roll away

Emilio Pisanty

still, I have yet to come across any one of those that's better than Concave Earth theory

AccidentalFourierTransform

didnt you guys know? the earth is a Möbius strip

John Rennie

15:55

I guess the flat Earth thing is effectively a religion. People choose to believe it then cast around for evidence to support that belief.

Slereah

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

AccidentalFourierTransform

> if the earth was flat, cats would have pushed everything off of it by now

2

John Rennie

It's always surprised me how people believe scientists constitute any form of authority. They clearly haven't met many scientists.

Slereah

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

ACuriousMind

@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...

Slereah

16:00

@ACuriousMind So consider the orthogonal group as a subset of the matrix space $\Bbb R^{n\times n}$?

ACuriousMind

@Slereah Yes

Slereah

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?

AccidentalFourierTransform

lets do the ama now

DanielSank

@ACuriousMind Thank you.

AccidentalFourierTransform

16:02

whats your stance on abortion?

should drugs be legalised?

DanielSank

@EmilioPisanty Why the %(@&* not?

AccidentalFourierTransform

is chemistry a science?

DanielSank

@Jim I expect that is possible.

ACuriousMind

@Jim Yes, if you do so sufficiently in advance (I'd say 2 weeks or so)

@AccidentalFourierTransform Stop trolling :P

AccidentalFourierTransform

._.

John Jack

16:07

@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.

ACuriousMind

@JohnJack I think it's not true in general, sorry

John Jack

Oh Damnit

Slereah

I am currently trying to download the Lois Lane comics

Hard to do

Not a lot of seeds

0celo7

@ACuriousMind does high RAM latency explain me sucking at fighting ghouls?

The Dark Side

@AccidentalFourierTransform That does look like the kind of explanation our flat-earth agnostic friend @0celo7 would be deeply interested in !!

Slereah

16:14

@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

DanielSank

@0celo7 No.

Slereah

Since ghouls just throw themselves on it

The Dark Side

@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!!

ACuriousMind

@Slereah Uh...what game do you think we're talking about?

Slereah

@ACuriousMind Fallout 4?

ACuriousMind

16:16

@0celo7 Not really, also, I don't know what "high" RAM latency is

@Slereah Nope, TW3

Slereah

o

Are you doing a lot of witching

Maybe flying around on a broom

Stirring a big cast iron pot

John Jack

@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.

ACuriousMind

@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

AccidentalFourierTransform

also, if you write $D_i=(D_i+1)-1$, you can conclude that $D$ need not be Toeplitz

John Jack

16:32

@ACuriousMind So your $D_{1}$ is some unitary matrix such that $D_{1}\rho_1 D_{1}^{\dagger}$ is diagonal?

ACuriousMind

@JohnJack Yes.

John Rennie

@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 :-)

The Dark Side

Well, I was borrowing from your colloidal expertise to shovel knowledge over my ignorance !!

John Jack

@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?

Felix.C

16:50

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 :)

The Dark Side

7

Q: Is it viable to wall modern cities against "zombies"?

Let'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 ?

Slereah

why do you need to wall cities against zombies

Zombies are like humans except worse at everything

They're like slow children

The Dark Side

How will military defense kill what's already dead?

Slereah

REMOVE THE HEAD OR DESTROY THE BRAIN

The Dark Side

> 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?

Slereah

16:55

No

Zombies are very dim

The Dark Side

So basically, corpses can walk but not climb?

Slereah

they can barely walk

they suck

The Dark Side

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

John Rennie

@Slereah slow? Have you not seen World War Z?

Slereah

@JohnRennie No thanks I'd rather see blood in my urine

2

0celo7

17:31

@ACuriousMind 75ns

AccidentalFourierTransform

18:03

> 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.

Secret

I have no idea what will a physical law be like if it cannot be written in mathematical terms

Abcd

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!

Slereah

Sorry @Abcd i'm afraid I have important business to take care of right now

Like posting this

Abcd

@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

Secret

you need to go to QFT to do that

Abcd

18:15

@Secret QFT?

Secret

quantum field theory

Abcd

ok

0celo7

18:30

@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.

Slereah

Are space filling curves actually space filling

Or do they leave out a set of measure 0

The Dark Side

@Slereah - This one's for you:

5

Q: Every Fourth Wormhole

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

grid-deduction mazes three-dimensional

It has no answer as of now.

bolbteppa

18:45

@Slereah what a picture

0celo7

@Slereah they are subjective.

Subjective.

Surjective

Slereah

Isn't all math subjective in the end

Las Vegas Raiders

19:05

Objectively subjective depending on your perspective

(also, there is no "end")

David Z

@Fawad FWIW vzn is the one who manages the AMAs. Mods have nothing to do with that. You're barking up the wrong tree ;-)

Las Vegas Raiders

Hi @DavidZ

David Z

Hi

Las Vegas Raiders

How are you?

David Z

Fine

0celo7

19:19

@ACuriousMind Eek I have to fight one of the Wild Hunt??

Las Vegas Raiders

Ted is chatting in the math room @0celo7

(if you have any questions)

ACuriousMind

@0celo7 Ah, Keira's quest? :D

0celo7

@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

Las Vegas Raiders

Well... he does have over 30 years of university math teaching experience.

Hmm, I wonder if he would be interested in doing an AMA?

Dawood ibn Kareem

@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.

0celo7

19:44

@ACuriousMind is "An Invitation from Keira" a booty call?

Emilio Pisanty

¬¬ damn exponential scaling

ACuriousMind

@0celo7 Yeah, at least at the beginning it is

@0celo7 Go and find out! (Better not post pictures of the result here, though ;P)

Las Vegas Raiders

TMI on the "booty calls"

So Ted is too busy right now for an AMA...

Dawood ibn Kareem

20:08

@Kaumudi.H Har Cookie Mein Kayi Smiles ROTFL

0celo7

@ACuriousMind what does adrenaline actually do in this game?

ACuriousMind

@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.

0celo7

@ACuriousMind the bloody baron seems like a decent guy

ACuriousMind

@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.

0celo7

20:42

@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

ACuriousMind

@0celo7 And you didn't already craft it? What bombs are in there?

0celo7

@ACuriousMind Just Grapeshot.

ACuriousMind

Hm.

That's odd

0celo7

@ACuriousMind just killed the Satanic Chicken

ACuriousMind

20:57

@0celo7 What's that?

dmckee

21:13

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.

ACuriousMind

@dmckee I...actually think that's not very subtle, but you're right that somehow every answerer managed to completely miss the point

0celo7

@ACuriousMind the shrieker

cockatrice I think

ACuriousMind

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

GPhys

@dmckee Nice, thanks.

David Hammen

@ACuriousMind - I've finally had the time to update my placeholder answer, physics.stackexchange.com/a/342932/52112 . Could you undelete it?

GPhys

21:19

@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.

dmckee

@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.

GPhys

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

dmckee

@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.

David Z

@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.

dmckee

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.

ACuriousMind

21:23

@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?

dmckee

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 Hammen

@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.)

dmckee

Those QZE experiment probe the system they are diddling: the experimenter poke the system rather than waiting for the system to poke the instrument.

David Hammen

@ACuriousMind -- On it.

BTW, Happy Fourth of July to all!

dmckee

::lights sparkler::

David Hammen

21:36

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.

David Z

@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.)

dmckee

@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.

David Hammen

@dmckee -- All of the interesting people I know did incredibly stupid things when they were young. It's amazing we lived.

GPhys

@dmckee can I prepare the system in a high enough temperature to observe the low time (QZO) decrease too?

dmckee

21:53

@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.

GPhys

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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