The h Bar

0celo7

00:03

I GOT IT

I HAVE THE PROOF

@ACuriousMind please check it when you see this

0celo7

00:16

I don't care that it's sloppy

I have convinced myself that it's true

Secret

04:27

Hmm, if I understood your proof and figure correctly, it seems by having connected components ending on the boundary, it prevents the existence of a few points (x,y) to lie in the open subset of $\Gamma$ hence preventing such subsets to not be connected components themselves...

3075

06:24

why am I awake.

Bass

11:23

hi guys, what are the reasons that $C\lvert\gamma\rangle=-\lvert\gamma\rangle$ and $C\lvert 0\rangle=\lvert 0\rangle$? I understand that their charge parity must be $\pm 1$ because both the vacuum and the photon are not charged, but why $+1$ for the vacuum and $-1$ for the photon?

Slereah

Because the vacuum has no charge at all?

Also the vacuum is invariant under the Lorentz group

So it should be that $CPT \vert 0 \rangle = \vert 0 \rangle $

Which should be obvious enough that this is true also for $P$ and $T$

Apparently the stress energy tensor doesn't diverges for the vacuum of the Gödel metric

0celo7

12:12

@Slereah do you know what the proof of the square lemma is

Slereah

Step 1 : draw a square

Step 2 : ???

0celo7

...

ACuriousMind

12:27

@0celo7 That appears to be some kind of correct argument

0celo7

@ACuriousMind so you get the jist?

Slereah

You know

I wonder if it would be possible, for a theory with a known classical solution, to transfer that solution to the wavefunctional

ACuriousMind

@0celo7 Well, I also think I got the jist from what Milnor wrote, so I'm not really the best judge here

0celo7

@ACuriousMind :(

What jist

Slereah

Can you easily transfer the solution to a PDE to the solution of a functional differential equation

0celo7

12:34

He didn't write anything

@ACuriousMind I have a reasonably correct proof now but I still don't know what Milnor wrote

@ACuriousMind I'm not asking to whine or be annoying, I'm legitimately curious what jist you saw

because you got one half of the proof, yeah, but not the other

ACuriousMind

12:53

@0celo7 Uh...how do you think I'm able to explain that in any other words than the inadequate one we already tried?

David Z

@ChrisWhite I think there are such people, they just work at Google.

0celo7

@ACuriousMind Because I now understand it?

user116211

13:21

4

A: What Do We Get From Having Higher Generations of Particles?

Dear Chad, I thought you were an atheist. Most atheists tend to realize that many things that exist in the Universe have no "purpose". The existence of the Universe has no "purpose" that may be scientifically demonstrated. Even if life could exist in a Universe with 1 generation of quarks and le...

user116211

Was the first line really necessary?

user116211

@yuggib: o/

yuggib

\o

@Slereah in which sense?

Slereah

If I have the solution for an equation like $$\frac{\partial}{\partial x} f(x) = a$$

Can I transfer that solution to an equation of the form $$\frac{\delta}{\delta \varphi(x)} f[\varphi(x)] = a$$

Obviously for more complex equations

yuggib

what do you mean by "transfer"

Slereah

13:27

Well are the solutions going to be of the same form

With some transformations can I use them as solutions

yuggib

probably not

Slereah

Unfortunate

yuggib

for instance, the initial conditions are different

what did you want to do that for?

Slereah

The Sine Gordon equation

yuggib

what for

not in which model

Slereah

13:29

To write out explicitely the Hilbert space

since apparently it's Complicated

yuggib

?

it's called problem of quantization

Slereah

It is possible to write out explicitely the wavefunctions of QFT states using functionals

But it is rarely done

For that reason, I suppose

yuggib

the problem is to identify the measure in the space of wavefunctionals

Slereah

Hm, true

yuggib

the space of wavefunctionals is something like $L^2(\mathscr{S}'_{\mathbb{R}}(\mathbb{R}^d);d\mu)$

Slereah

13:32

Since the Hilbert product is $$\int \mathcal D \varphi(x) \Psi[\varphi(x)] \Psi^*[\varphi(x)]$$

yuggib

it is the measure $d\mu$ that gives "problems"

Slereah

I guess you need to define it, as well

yuggib

for the free case, it is a Gaussian measure

for the interacting case....a mess

Slereah

So I hear

yuggib

no explicit form

Slereah

13:33

I try to read Jaffe once in a while but then my eyes kinda glaze over

yuggib

it is obtained as a limit of something using the Feynman-Kac formula

Slereah

I say let's all go back to Bohmian mechanics

Then there will be no Hilbert space

yuggib

that does not solve any problem of course

Slereah

Can't know until we try!

I'm not quite sure how the QFT of Bohm works

yuggib

the point is that we are not so able to cope with the non-commutative probability theory given by the algebra of CCR/CAR observables set in infinite dimensional classical spaces

Slereah

13:47

IIRC it still uses particles

No fields involved

0celo7

@Slereah I think so?

Maybe with some integrals thrown in

I know Weinberg does it at least once

Slereah

mb

0celo7

I don't have his book here

He does a Gaussian functional solution to something

yuggib

@Slereah as long as you want to reproduce the results of non-commutative probability theory, there is not much hope of obtaining something simpler I am afraid

0celo7

check chapter 9 of his first book

Slereah

13:50

Do you happen to know what the measure for $\phi^4_2$ looks like, btw

I can never understand what Jaffe is on about

0celo7

@ocelo7 : it is so that $F_{\mu\nu} \to \Omega^{2}(x)F_{\mu\nu}$. If I would know this explicit rule, I would not ask the question here. — Name YYY 2 mins ago

Sass!

Heyo

Slereah

Isn't the reason that the EM field is conformal invariant just that the Maxwell equation doesn't use the metric at all

It's all differential forms bullshit

ACuriousMind

@Slereah It does use it (it's hidden in the Hodge star/index positions)

Slereah

Oh yeah

0celo7

@ACuriousMind Is a symmetry of an action a symmetry of the EL equations

Slereah

13:59

$$\star d \star F = j$$

Or something

ACuriousMind

@0celo7 Not necessarily

0celo7

@Slereah real men use $\mathrm d^\dagger$

ACuriousMind

@0celo7 See this question for more info.

Slereah

I wonder if there's any calculations done for the QFT in CTC spacetimes by what's his name

F-algebra or whatever he calls it

0celo7

@Danu Could you do this?

yuggib

14:08

Let $d\mu_0(\varphi)$ be the Gaussian measure with mean zero and covariance $(-\Delta+m_0^2)^{-1}$ on $\mathscr{S}'(\mathbb{R}^2)$. Then denote $$V(h)=\lambda\int_{\mathbb{R}^2}:\varphi^2(x):h(x)dx\; ,$$
where $h$ is a space(time) cutoff. This defines a measure (essentially via the Feynman-Kac formula) $$d\mu_h(\varphi)=\frac{e^{-V(h)} d\mu_0(\varphi)}{\int_{\mathscr{S}'(\mathbb{R}^2)} e^{-V(h)} d\mu_0(\varphi)}$$

then the euclidean measure $d\mu$ is defined as $d\mu=\lim_{h\to 1}d\mu_h$

Slereah

Errrr

0celo7

#rekt

Slereah

Isn't that already what I would have guessed

yuggib

and the measure for the wavefunctionals is the one restricted to time zero fields

Slereah

What did he prove with that?

That it existed?

yuggib

14:09

@Slereah yep

Slereah

Oh well

So much for that

yuggib

and that the theory satisfies Osterwalder-Schrader i.e. Wightman axioms

Slereah

By the way

Jonah J. Jameson and the Goombas have in the book something about the limit of the measure $d\mu_t$, of a path integral at time $[-t/2, t/2]$, with $t\to \infty$, is unique

Is that related to the fact that the vacuum expectation value is just a path integral with no boundary

Bass

14:55

@Slereah Well but the photon has not charge too. And still its eigenvalue is $-1$.

Slereah

Is it though?

Shouldn't it be 1

Since it is a real field

C parity −1

Hm

Odd

Bass

The CPT argument makes sense, but only if you know that CPT is always conserved, right?

Slereah

If you don't have a shitty theory, CPT is always conserved

I forget the exact theorem but basically for any reasonable Lorentz invariant theory, CPT is conserved

Bass

Yep but if you want to check if your photon theory violates CPT, you need to know the C parity of the photon. You cannot deduce the C parity from CPT, because then, CPT is trivially conserved.

Slereah

Oh wait

Bass

15:00

@Slereah Why? -1 is just a phase change.

Slereah

Yeah after checking that isn't a problem

Obliv

What's the difference in proving a left or right group action? $g_1(g_2 * a) = (g_1g_2) * a$ is this only for right group actions?

ACuriousMind

@Bass The photon charge parity is deduced from observing that the current changes its sign under charge conjugation, so for the coupling term $A_\mu J^\mu$ to be invariant you need that $A_\mu$ also changes sign

This is then inherited by the modes of $A_\mu$, which create the photon states, and hence by the photon states

@Obliv What do you mean "proving a group action"? What you've written there is a left action, the right action would be written $(ag_2)g_1 = a(g_2g_1)$.

Obliv

oh.. okay. Thanks @acuriousmind

Bass

@ACuriousMind I see. And the vacuum? Must it be $C\lvert 0\rangle=\lvert 0\rangle$ for a consistent theory, or is -1 also possible?

ACuriousMind

15:07

@Bass My argument above already presupposes an invariant vacuum - If both $A_\mu$ and the vacuum changed their sign, the photon states $A_\mu$ creates from the vacuum would not

I'm unsure whether it would be in principle possible to have the vacuum have parity -1, but I suspect not.

Bass

Oh of course. So $C\lvert 0\rangle=\lvert 0\rangle$ is just a postulate, and we could as well say that vacuum has -1 parity and the photon has 1? The current would still change its sign, since it's made up of two particles..

@ACuriousMind Ok I see.

thanks

0celo7

I still don't know what $|0\rangle$ is

QFT is such bullshit

Bass

@0celo7 It's nothing :P

ACuriousMind

^that :D

0celo7

You either get physics crap that makes zero sense and sounds silly or you get legit PhD level math

yeah keep joking

QFT being shit is why I didn't become a physicist

Obliv

15:11

you don't have to learn qft to become a physicist lol

0celo7

Yeah you do

Any interesting physics

interesting/grant worthy

Slereah

@0celo7 It is the Lorentz invariant state

0celo7

@Slereah I don't understand how the Lorentz group acts on states

Slereah

Read Weinberg on it mb

Chapter 2 is all about Lorentz actions on states

0celo7

I did read Weinberg

Weinberg explains fucking nothing

And Weinberg is unreadable

No one understand the cluster decomposition principle

Slereah

15:14

It's pretty equation heavy

ACuriousMind

@Slereah A mathematician currently giving a course on the unitary representations of the Poincaré group has told me he has no idea what the fuck Weinberg is doing there :D

Slereah

heh

0celo7

@ACuriousMind language

All QFT books are garbage honestly

Slereah

What about Glim and Jaffe

It's all math

u should love it

0celo7

And when you read "good" notes they don't bother explaining anything that's confusing in the garbage books

I'm bad at math @Slereah

Took me 4 days to figure something out that ACM got instantly

Obliv

15:15

@acuriousmind when you connect a positive terminal to a negative terminal in a circuit, does the information travel instantaneously, so that electrons begin to flow from the positive to the negative terminal?

ACuriousMind

@Obliv If only there were a wonderful Q&A site on the internet where that question certainly has been asked and answered...

Obliv

If only

0celo7

@ACuriousMind not everyone has your search skills dammit

Slereah

The vacuum state is just the lowest energy eigenstate of the Hamiltonian operator

at least for the usual theories

Obliv

physics.stackexchange.com/questions/215205/… got it

0celo7

15:17

why does it have to have a lowest

ACuriousMind

@0celo7 Well, it appears Obliv had not much trouble finding a fitting question :P

0celo7

@ACuriousMind I couldn't

Slereah

Because if it's unbounded from below u get bad things

0celo7

what's another word for enjoy

Slereah

You can have theories which are not bounded from below, btw

ACuriousMind

15:20

cherish?

Slereah

They have no vacuum state then

Or well they do I guess

But it is not the lowest energy

0celo7

@ACuriousMind mmm, does one cherish company?

Slereah

Does $\phi^3$ have a vacuum state?

I suspect it do

ACuriousMind

@0celo7 Mmmh...relish perhaps?

0celo7

@Slereah $\phi=0$ probably

@ACuriousMind did you google "enjoy thesaurus"

Bass

15:22

@Slereah which theories don't have a vacuum state?

Slereah

Well some theories don't have a lowest energy state

But I think they still have a vacuum state

Some even have several vacuas

Bass

example?

Slereah

$\phi^3$

Bass

@Slereah does not have a vacuum?

Slereah

Fermions with commuting field operators

It is unbounded from below

Still has a vacuum tho

Bass

15:24

@Slereah Ok, right.

But $\phi^3$?

Slereah

I think it's the same

ACuriousMind

@Slereah That's not a theory, the spin-statistics theorem forbids that.

Slereah

I know it's not bounded from below, but probably still a vacuum state

Not sure tho

Bass

$\phi$ being a scalar, right?

Slereah

yes

@ACuriousMind Sure it's a theory, it's just a garbage theory :p

Bass

15:25

Why does $\phi^3$ not have a vacuum?

Slereah

If you compute the stress energy tensor you have a term $\propto \phi^3$

ACuriousMind

@Slereah Well, but it's not a consistent QFT, so it's a garbage example for your claim "you can have theories"

Slereah

which means you can make the energy arbitrarily negative

Bass

Isn't $\phi^3$ just the interaction term of a free scalar theory? And the free theory has always a vacuum, after all its Fock space is constructed from the vacuum.

Slereah

As said

It probably has a vacuum

Bass

15:27

@Slereah Arbitrarily negative $\neq$ no vacuum.

Slereah

As in, a state invariant under Lorentz transform

Bass

ok

ACuriousMind

@Bass What? The exponent is cubic, not squared

Slereah

But this isn't the lowest energy state

I think any odd power of the field will also be like that?

Although IIRC it gets non-renormalizable pretty quick

Like $\phi^6$ or so in 4D

Bass

@ACuriousMind I mean, the full lagrangian is something like $\mathcal L = \mathcal L_{\mathrm{free}}+\mathcal L_{\mathrm{int}}$, and $\mathcal L_{\mathrm{int}}=\phi^3$. And in the free theory we just have $\phi^2$.

John Rennie

15:29

I feel like this should be closed, but I'm not sure exactly what the close reason should be ...

0

Q: Is running in a zigzag helpful in avoiding being shot by an arrow?

Two people A and B start at a certain point. B runs away from A to another point. A shoots arrows at B. Is it helpful for A to run in a zigzag rather than in a constant direction? To be more precise: Suppose A starts and remains at $(0,-1)$, and B starts at $(0,0)$ and has to get to $(0,...

Bass

@Slereah $\phi^n$ with $n>4$ is non-renormalizable.

Slereah

yeah sounds about right

0celo7

where is Duffield

Slereah

But $\phi^3$ is also garbage

So basically it's only $\phi^4$

and also maybe $\phi^4$ doesn't make sense either

Basically QFT is garbage

ACuriousMind

@Bass Yeah, so? That you can construct a free vacuum doesn't mean that there is an interacting vacuum.

Bass

15:31

So in $\phi^3$, there are operators acting on the vacuum that lower its energy?

0celo7

Mod alert!

Bass

@ACuriousMind Oh I see. Didn't know that.

0celo7

@Emrakul what brings you here

ACuriousMind

Doesn't $\phi^3$ have some sort of catastrophe with those tadpoles?

Slereah

Raising operators are only for free theories!

@ACuriousMind Yeah

The vacuum can radiate

Hence why things go downhill pretty fast

ACuriousMind

15:32

@Bass Renormalizability may not be all that it's made out to be - if QFT is just an effective description at "low" energies (like in string theory), there's nothing inherently wrong with non-renormalizable QFTs

Slereah

But then what is the real theory!

Aaaaah

ACuriousMind

Renormalizability becomes crucial when you want that the QFT is a valid description at all scales

@Slereah Well, that's what nobody knows (yet)!

Slereah

(it's probably not)

Is there a not garbage theory, though

Like... LQG?

LQG sounds alright

It has an effective cutoff

ACuriousMind

@Slereah Sometimes I think that although QFT is probably not the last word, it's too good for its own sake concerning the accuracy of predictions - we don't really get any hints as to what the "real" theory is from experiments currently

Slereah

Well obviously yes

ACuriousMind

15:35

Because they all just tend to agree with QFT

Which is good, but also kinda boring

Slereah

But it would be nice to have a theory that makes sense

user61230

@0celo7 Physics! Poking around chat, honestly.

0celo7

@ACuriousMind So, how does Fubini's theorem for $(n-1)$-dimensional vertical slices imply it for $(n-k)$-dimensional ones

I've heard it's by induction on $k$ but I don't get it

I have proved it in the $k=1$ case but I don't see what the induction step is

Slereah

Prove it for (n-k-1), imho

ACuriousMind

@0celo7 Ugh, I don't want to do integration theory now

Slereah

15:38

My favorite trick to make cutoffs for QFT is the QUANTUM FOAAAAM

0celo7

@ACuriousMind not integration

Slereah

I've seen papers about making cutoffs with just a very dense wormhole gas at Planck scales as a cutoff

0celo7

sets of measure zero

ACuriousMind

@Slereah "just a very dense wormhole gas"...

0celo7

I'm trying to complete the proof of Thm 7.9 @ACuriousMind

Slereah

15:40

Yeah that is how it is described

ACuriousMind

That's not what I know as Fubini, and I like measure theory even less

Slereah

arxiv.org/abs/1006.5181

0celo7

@ACuriousMind This follows from the usual Fubini theorem according to Lee

It's the form of Fubini needed for Sard's theorem

this sucks

I guess I have to make a bunch of vertical slices $V^k_c$ or something

Then the induction step is that Fubini with $V_c^k$ implies Fubini for $V_c^{k+1}$??

Obliv

a center of a group of matrices doesn't have to be a symmetric matrix, right?

user54412

@Chris White I really don't understand why people here behave like short minded . If you see other stackexchange sites like mathstackexchange ,there are many similar questions with huge reputation like question1, question 2 ,question 3 and many more — pandu 3 mins ago

user54412

15:47

People keep making this argument. Do they not realize the logical conclusion is that we should just become a porn site? After all, that would get the most traffic.

7

0celo7

I would support physics porn

It's an election year, I support strippers. - 2 Chainz.

Actually I think that song was from 2008, so

It's an election year, I support strippers. - Titty Boy.

Obliv

@0celo7 was listening to this youtube.com/watch?v=oiQiTCxSaHA probably in my top 5 albums i've heard

i wonder if you'd like it

ACuriousMind

@ChrisWhite They are not making the argument because they want anything (good or bad) for the site. They are making it because it would mean their own question doesn't get closed, which is all they see.

0celo7

@ACuriousMind Hmm, I need help with this

Not from a measure theory PoV but induction

When I assume it's true for the $n-k$ dimensional slices do I get to assume that $A$ is measure zero?

does that come with it?

well that might be a stupid question

is this a trivial induction?

John Rennie

@ChrisWhite I used to attempt a reasoned argument. These days I smile sweetly and click the Close link.

John Rennie

16:09

Re this question:

0

Q: C-parity violation evidence

I know about the CP-violation experiments from the 60's and the P-violation from the 50's. But, is there a similar experiment which displays (perhaps historically in the same way as the experiements of Wu and Christensen) the violation of Charge conjugation alone?

particle-physics nuclear-physics symmetry

Can you have just charge conjugation? Wouldn't that mean having the same particle(s) with opposite charges?

Slereah

Charge conjugation is just the antiparticle?

ACuriousMind

@JohnRennie What do you mean? You can certainly apply just the charge conjugation to a process and ask whether that's a symmetry

John Rennie

@Slereah I thought the particle and anti-particle were related by CP ...

Young

@dmckee You can let your fellow moderator @Qmechanic know he's a real hero for crushing the spirits of a 13 year old girl by suspending her account ... she really loved being on the site and doing research to answer questions. Did I vote up some of her answers? yes ... Where they good answers? yes ... Did she have 100s of rep points that were from other users? YES

Slereah

Well it will be an antiparticle with different parity, jeez

It's like $C a_k \approx b_k$

Or something

There might be a sigma 2 in there

0celo7

16:15

@Young oh?

which user is this

John Rennie

@Young Qmechanic wouldn't have done that alone, it would be a decision agreed amongst all the moderators.

0celo7

Mod abuse is nothing new to this site

Young

Heather

good job to all of them then

0celo7

"voting irregularities"

Can someone translate this please?

Young

My account was deleted entirely from physics

0celo7

16:17

good lord

Young

@Qmechanic @DavidZ @dmckee Good job inspiring the next generation ...

Slereah

I say the next generation can take a hike!

0celo7

@Young don't bother, they have no feelings.

ACuriousMind

@Slereah Um, I think charge conjugation doesn't change the parity

Slereah

I dunno

I don't have my books here

ACuriousMind

16:20

That's the reason there is "C-violation" by the weak interaction - it sends the interacting lefthanded neutrinos to the inert lefthanded antineutrinos

Slereah

Do they even exist

John Rennie

@Young "voting irregularities"? Perhaps you can explain what young Heather was up to and why suspending her for seven days is an excessively harsh punishment.

0celo7

not all sets of measure zero satisfy the Fubini theorem, do they

certain hyperplanes don't

lines

ACuriousMind

@Young Since the site does not verify any information about its users, we cannot treat users that claim to be "13 year old girls" different from those claiming to be emeritus professors. And if the account was suspended for voting irregularities, there must be something in the database that looked to the mods as if someone was abusing the voting system.

We've had troubles with sockpuppets voting each other up in the past, so they tend to be rather harsh on such irregularities.

Young

She was on the site a ton and answered 14 questions in 2 days, having a blast researching questions. I up-voted most of her posts, but she had 100s of other rep points

John Rennie

16:24

@Young that's not what we asked you

Slereah

I'm a 13 year old emeritus professor

Young

Well, I'm not a sockpuppet

0celo7

@Young Proof?

You're guilty of that until proved otherwise in my book

Young

Ok then I don't understand the quest @JohnRennie

ACuriousMind

@0celo7 Stop trolling.

Young

16:25

I assume the irregularities were my votes

0celo7

@ACuriousMind I'm not.

I'm trying to prove a theorem and it's not working

Slereah

Maybe the theorem is wrong

John Rennie

@Young so you know nothing about the reasons for her suspension then? You're just making assumptions?

ACuriousMind

@Young I don't think anyone gets suspended for such "serial upvoting". Usually the votes are just reversed, and maybe the voting account suspended.

0celo7

@JohnRennie I like how you instantly assume the mods are just.

Young

16:26

@JohnRennie She showed me the email from @Qmechanic and it claimed she was using sockpuppets

and my account was deleted from physics

She has done nothing else

0celo7

How is one supposed to prove that one is not a sockpuppet?

ACuriousMind

Ah...you're sitting at the same IP address, and if you never did anything but vote on questions where she posted, then that looks exactly as if she was using a sockpuppet to bolster the reputation.

Young

@ACuriousMind well she did

yes we're in the same house

I posted my own questions and answers as well but wasn't nearly as active as she was

John Rennie

Ok, so the mods saw mutual upvoting from two accounts using the same IP address.

That looks very like someone using a sockpuppet.

Young

Ok well she responded to @Qmechanic and explained and he said take it up with SE and she is

Qmechanic

16:31

Hi @Young, Thanks for the feedback. I understand if you are upset, but venting it in the hbar chat room is not constructive. Instead, to resolve your Phys.SE account and/or complain about Phys.SE moderators contact the SE team directly.

Young

@Qmechanic sure it was constructive I got your attention and now you know what a great job you did ... have fun in your high tower

John Rennie

@0celo7 I made no such assumption. Young says one thing, the mods don't say anything at all, so I have no way of knowing who is right or wrong. My point is simply that because a user unknown to me joins the chat and accuses the mods of foul play doesn't mean that user is right.

@Obliv: arrows don't travel in zigzag lines!

Slereah

Well according to Zeno they don't travel at all!!!

0celo7

Zeno was a smart man

Obliv

@johnR I meant a random straight path

David Z

16:40

I echo what Qmechanic said. No more discussion on that issue here.

Obliv

between the bounds of the zig-zag path. I.e if it oscillates between x = -1 and x = 1 then the arrow travels straight at some x in between

0celo7

hmm, but if we take $V_c$ to be $A$

then the induction fails

and I have an explicit example for why

John Rennie

@Obliv well yes, if you fire the arrow in a random direction you have a non-zero chance of hitting the randomly moving target. Even so, if I were the target I wouldn't be running in a straight line.

Obliv

If it's a long distance away, it's pretty much random and you might have a better chance running straight as you'll cover more distance to avoid further fire.

John Rennie

OK we'll try the experiment. I'll get my bow and you get your running shoes :-)

3

Obliv

16:48

@JohnR If you've ever played Rust, h1z1 or any other survival game, at a distance the first shot is usually meant to calibrate your next shot. so, if they're not an amazing shot you can get away with running straight for the first shot I think.

actually no.. that's if they're running in a straight path that's not parallel to you. If they're running parallel to you then forget it.

0celo7

@ACuriousMind I literally need help with the induction step, I don't even know what it means to use induction here

@ACuriousMind I don't want your help with the measure theory bits, I just don't even know what the induction step looks like

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