The h Bar - 2017-05-14 (page 4 of 6)

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4:00 PM

@BernardoMeurer I don't understand.

Bernardo Meurer

@JaimeGallego I have your phone number now, lol

867-5309, sister's name is Jenny

Bernardo Meurer

And I have your address

"867-5309/Jenny" is a 1981 song written by Alex Call and Jim Keller and performed by Tommy Tutone that was released on the album Tommy Tutone 2, on the Columbia Records label. It peaked at #4 on the Billboard Hot 100 chart and #16 on the Billboard Top Tracks chart in May 1982 (see 1982 in music). The song caused a fad of people dialing 867-5309 and asking for "Jenny". == Creation == Lead guitarist Jim Keller, interviewed by People in 1982, said "Jenny is a regular girl, not a hooker. Friends of mine wrote her name and number on a men's room wall at a bar. I called her on a dare, and we dated for...

Then come here and we may have some tea

meh i have work to do. prob better then being in this chat

Bernardo Meurer

4:03 PM

You have an iPhone?

Wtf is going on

No, I don't

Bernardo Meurer

I am trying to find out your sister's age

Because I'm bored

Fun Microwave Games - a book that all the family can enjoy

Bernardo Meurer

Hmm, I wonder how much does a hamster cost

user228700

@JohnRennie What, erm, what?

4:07 PM

@BernardoMeurer Not much if you buy in bulk, which you have to because there isn't much meat on a single hamster.

user228700

Jesus Christ, why?

Bernardo Meurer

I wonder how hamster tastes

Can you grill'em?

@Kaumudi.H It was referenced in an article on string theory that I was reading.

0celouvskyopoulo7

yes

0

Q: Mathematica stuck in "running" for every calculation I attempt, kernel error

I downloaded Mathematica this morning to do an implicit polar plot. I attempt to run the code Manipulate[ ContourPlot[ Evaluate@With[{r = Sqrt[x^2 + y^2], \[Theta] = ArcTan[x, y]}, {r Cos[Pi \[Theta]/(Pi - 2 chi)] - Exp[1/(r^2 Sin[Pi \[Theta]/(Pi - 2 chi)]^2- 1)] == 0, r^2 == 1}], {x, -2, 2}...

I will sell my soul for upvotes, holy crap people

user228700

@JohnRennie Yes, but hamsters?! :'-( Poor hamsters.

4:09 PM

I'm sure no-one is really planning to microwave hamsters.

user228700

Ah, that's a relief. One can never be too sure these days...

0celouvskyopoulo7

I will do it if it fixes my issue

However guinea pigs are eaten in Latin America, and I know this for certain because my mother ate some when she was there.

user228700

:'-(

Bernardo Meurer

@0celouvskyopoulo7 Give me Shankar

I upvoted

4:10 PM

They are cooked like a spatchcock chicken.

user228700

@JohnRennie Please stop :'-(

Noted.

0celouvskyopoulo7

@BernardoMeurer you have to come to knoxville I'm afraid

Bernardo Meurer

Tsc

I'm downvoting

0celouvskyopoulo7

why?

4:11 PM

@Kaumudi.H Keep in mind that guinea pig were kept as meat animals.

It make sense for subsistence agriculturalists to have low cost livestock.

user228700

@JohnRennie Oh, I wasn't being too serious or anything.

As David says, they are kept like livestock. It's not a case of people eating the family pets.

user228700

Ah...

In Eurasia the chicken filled that role.

In fact they occupy a similar niche to regular pigs i.e. they run around and eat scraps and grow quickly. So they don't need much attention.

Just like chickens, as David says.

0celouvskyopoulo7

4:14 PM

@BernardoMeurer I somehow wrote code that completely breaks Mathematica so badly one has to reinstall.

Bernardo Meurer

Lol

Let me try it on my installation ...

Bernardo Meurer

I'm proud

user228700

@JohnRennie Hmm, I see.

0celouvskyopoulo7

@JohnRennie First try:

Manipulate[
ContourPlot[
Evaluate@With[{r = Sqrt[x^2 + y^2], \[Theta] = ArcTan[x, y]},
{r Cos[Pi \[Theta]/(Pi - 2 chi)] - Exp[1/(r^2 Sin[Pi \[Theta]/(Pi - 2 chi)]^2- 1)] == 0,
r^2 == 1}], {x, -2, 2}, {y, -2, 2}],
{chi, 0, Pi/4}]

4:15 PM

Hey, it's started hailing in Chester. It's supposed to be spring!!!

0celouvskyopoulo7

Then try:

Manipulate[
ContourPlot[
Evaluate@With[{r = Sqrt[x^2 + y^2], \[Theta] = ArcTan[x, y]},
{r Cos[Pi \[Theta]/(Pi - 2 chi)] - Exp[1/(r^2 Sin[Pi \[Theta]/(Pi - 2 chi)]^2- 1)] == 0,
r^2 == 1},y==Tan(chi) x], {x, -2, 2}, {y, -2, 2}],
{chi, 0, Pi/4}]

user228700

@JohnRennie Ah, hail. Wonder what that's like...

user image

Bernardo Meurer

@Kaumudi.H It sucks

Who is raising spurious flags?

Bernardo and Jaime are friends

0celouvskyopoulo7

4:18 PM

@JohnRennie Yep

Next one?

There is no stalking going on here.

I get an error on the next one. Possibly because I'm using MM 10 not 11.

user image

@Kaumudi.H hail hurts if you're caught out in it. It makes your ears sting!

user228700

Ears sting? That's weird.

Hailstones are hard and cold and moving fast. If they hit your ears they sting!

0celouvskyopoulo7

@JohnRennie if you open a new notebook does it still work?

@JohnRennie and if they're big enough they can give you a massive headache and concussions

4:22 PM

@0celouvskyopoulo7 I'll try now.

I mean it's pretty rare for them to be that big, but yeah

Bernardo Meurer

Shoo

user228700

@JohnRennie Haha, that I understand.

I think in the US hailstones can get very big. In the UK large hailstones are exceedingly rare.

0celouvskyopoulo7

@JohnRennie Someone wrote an answer on my question

It's not an answer

4:23 PM

Now where's that Mathematica code ...

0celouvskyopoulo7

it's a sign of solidarity

@BernardoMeurer please stop telling me to go away. It's not funny.

@BernardoMeurer Can you take a look at that LaTeX?

Bernardo Meurer

@JaimeGallego Which LaTeX?

→ 1 message moved to Trash

4:24 PM

@BernardoMeurer ...or anything to the same effect. Please be respectful.

@BernardoMeurer dangerous ground ...

→ 2 messages moved to Trashcan

@BernardoMeurer you can't tell people to leave the chat

and doing that will not only incite ONE moderator's wrath, but several

so let's keep to the SE chat global rule of "Be Nice"

Bernardo Meurer

@JohnRennie Why not?

4:26 PM

This one chat.stackexchange.com/transcript/message/37390795#37390795

Everyone has equal right to use a chatroom.

Bernardo Meurer

@ThomasWard I don't give a damn who's wrath I incite

Aint your job to be wrathful

@0celouvskyopoulo7 the first code still works fine. The second generates an error so I can't test it.

Bernardo Meurer

@ArtOfCode Sure, I'm not making you leave

I'm not infringing on your right, I'm just respectfully voicing my opinion

@BernardoMeurer it's not your chatroom. And even it it was you still can't tell people to leave.

4:27 PM

@BernardoMeurer telling people to go away isn't Nice. That's what I'm trying to say. If you don't like me, feel free to use the ignore button on my user card.

Let's all be nice, now. No need to form opinions into the shape of "order"-like sentences.

Relevant:

477

Q: The NEW new "Be Nice" Policy ("Code of Conduct") — Updated with your feedback

Update: Thanks for all the additional feedback below. We incorporated a lot of your suggestions, and this is going live (as https://meta.stackexchange.com/help/be-nice). We're also looking at ways to get this in front of more new users when they sign up, to help them start off on the right foo...

discussion etiquette help-center behavior

Bernardo Meurer

@JohnRennie ::grunts::

@ArtOfCode Yeah yeah whatever

Also relevant: the SE Be Nice Policy Itself

@BernardoMeurer there are no shortage of people who I wouldn't mind if they enjoyed an appointment with the now legendary red hot knitting needle. But I keep these views to myself.

4:29 PM

we have red hot knitting needles now? I thought it was a kitchen knife...

Bernardo Meurer

@JohnRennie Why'd you gotta burn people? Just beat them up

or a nickel ball

@ArtOfCode was it a kitchen knife or a high powered laser?

@ArtOfCode it was a joke from a student magazine in about 1979 ...

@JohnRennie do you remember the swirly coffee thing you posted about a week or so ago ? You then gave instructions on how to recreate such swirlishness. Could you repost the instructions ? It's an awful long scroll up.

4:30 PM

@ThomasWard search red hot knife or red hot nickel ball

Maybe two weeks

Bernardo Meurer

@HsMjstyMstdn You can search the chat history. Top right corner

Oh okay, thanks

@JohnRennie well, you're not wrong.

Bernardo Meurer

@JohnRennie lol

4:31 PM

It ... kind of makes me laugh at the outrageousness of it :-)

NO DON'T STAR THAT - I'LL GET BANNED!!!!

9

user228700

@JohnRennie You've referenced this a lot over the years!: -)

@JohnRennie fzzt Huh, i guess there are uses for industrial laser technology here in chat. :)

stars everything

user image

@HsMjstyMstdn it's really easy. Put milk in the mug and heat in the microwave until it's nearly but not quite boiling. Sprinkle on instant coffee and it should spread out and form a film on top of the milk. Pour on boiling water. It will froth up and form that swirly pattern.

4:34 PM

@JohnRennie thanks, but I don't have instant coffee... trying to recreate with cocoa powder

Do you expect me to use -+++
No I expect you to die Mr @AccidentalFourierTransform

user228700

@JohnRennie Ohhh, you do it with a microwave!

@Kaumudi.H because I can't be bothered to clean a pan. It will work just as well if you heat the milk in a pan.

user228700

Yep, I can confirm that.

I need motivational help to crack open this philosophy book

4:38 PM

you can do eet

Oh well, time to relax into my armchair and read this week's New Scientist.

5:01 PM

So apparently

The theorem that links Grassman bundles to the euler number isn't directly related to vector fields

It's yet another theorem

Although the proof in Steenrod seems to only be done for the sphere

"If $2k \leq n$, then $S^n$ admits a continuous field of tangent k-planes if and only if it admits a continuous k-field"

In trying to answer this question, I've confused the hell out of myself, guys.

If we have two events which are spacelike separated, then an observer in a reference frame moving w.r.t. the rest frame should **still** measure the distance between those events as smaller than their distance measured from within the rest frame, right?

That is, length contraction should still apply, shouldn't it?

And if yes (I'm pretty sure yes), then I apparently have no idea what the spacetime interval is: s^2 = - cdt^2 + dx^2

I'm still not sure why a section of the Grassman bundle of lines is equivalent to $\chi = 0$

because i would think that the spacelike interval between the two events is s^2 = dx^2

and since it's invariant, the spacelike interval between the two events from the moving reference frame is s^2 = dx'^2 - cdt'^2

but that means dx'^2 = dx^2 + cdt'^2

which is like... length... dilation :l

5:21 PM

halp

0celouvskyopoulo7

5:37 PM

user image

@Slereah ...

Working on it

something seems slightly off

0celouvskyopoulo7

user image

Boom.

Penrose was right.

Seems like it

but now, how to prove it

0celouvskyopoulo7

Manipulate[
ContourPlot[{Evaluate@
With[{r = Sqrt[x^2 + y^2], \[Theta] =
ArcTan[x, y]}, {r Cos[Pi \[Theta]/(Pi - 2 chi)] -
Exp[1/(r^2 Sin[Pi \[Theta]/(Pi - 2 chi)]^2 - 1)] == 0,
r^2 == 1}], x == Tan[chi]* y, x == -Tan[chi]*y}, {x, -1.5,
1.5}, {y, -1.5, 1.5}], {chi, 0, Pi/4}]

@JohnRennie My working code ^

It's the best GR code

I guess it should be doable the same way as the bump function, maybe

5:40 PM

Anyone have any idea? I feel like it's a quick misunderstanding, but I"ve been struggling with it for over an hour and have no clue what's going on

0celouvskyopoulo7

@Slereah Hmm, do you see those bits on the other side there?

I wonder if those are the weird angles we were getting

It seems Penrose was right with his angles?

I do not know

0

Q: Resubmitting question

If my question was marked as unclear and I have reworked it what needs to be done to either: Get it approved Or Resubmit it.

support closed-questions

0celouvskyopoulo7

@Slereah I made a discovery though

One wants $x=\tan(\chi)y$, not the other way around

that means the angle $\chi$ is to the $y$-axis

But the polars are measured wrt. the $x$-axis

Well yes

It's the angle from the time axis

That way its range spans all causal vectors

0celouvskyopoulo7

5:47 PM

So if you know all of this then why have you been encouraging my wrong thoughts?

Bernardo Meurer

@0celouvskyopoulo7 Did you make Mathematica work?

0celouvskyopoulo7

Yeah, I actually made a really good plot.

Bernardo Meurer

@0celouvskyopoulo7 To take over the world?

0celouvskyopoulo7

It describes the smoothing at a corner between two timelike geodesics

Bernardo Meurer

I'll show you the smoothening of my corners

0celouvskyopoulo7

5:49 PM

...why?

The X-Files - Main Theme (Extended)

Kate Bush - Wow

0celouvskyopoulo7

@Slereah Dude

wot

0celouvskyopoulo7

$\theta_c=\pi/2-\chi$

That's the intersection in polar coordinates

5:51 PM

Hm, what does that give

0celouvskyopoulo7

But $\alpha(R=1)=\pi/4$ so $$\frac{\theta\pi}{\pi-2\chi}=\frac{\pi}{4}\implies \theta=\pi/2-\chi$$

That's exactly right.

So the curve does indeed connect the two endpoints.

Well that is good

0celouvskyopoulo7

Now to prove it is smooth...

I think the derivative in $R$ should be easy enough to do

First derivative will just be $\cos \alpha = - 2 R \sec \alpha \exp ...$, which is also 0

And then higher ones are just powers of $\sec \alpha$ and zero on the LHS

0celouvskyopoulo7

Wait what are you doing

5:56 PM

Well the tangent vector is $(\partial_R f, \partial_\theta f)$, with $f = R\cos \alpha - \exp (-R^2 \sec^2 \alpha)$

Wait, should it be $0$ for $R$

I'm pretty sure it will be for $\theta$

0celouvskyopoulo7

It should be zero for $\theta$ because its transitioning to a line...

But I don't know what the polar equation for a line through the origin is

You might have issues computing the derivative

I think it might just be $1$ here

0celouvskyopoulo7

How do you plan to check if it's timelike lol

it's clear from the picture

i guess compute the tangent and find its norm?

0celouvskyopoulo7

Christ....

6:02 PM

Well

i don't need its norm

Just a bound on it

Showing that it's $<0$ should be doable

@AmagicalFishy the problem is that for the moving observer one event is rotated into the future while the other is rotated into the past. While the length is indeed contracted it turns out to be surprisingly complicated to explain why.

For a detailed, and necessarily somewhat tedious explanation of exactly what happens see this answer of mine:

2

A: "Reality" of length contraction in SR

Lorentz contraction is easy to understand once you realise that it is not a contraction at all. Instead it is a rotation and the length of the object, or more precisely its proper length, doesn't change at all. To see this take the usual example of a rod of length $2a$ aligned along the $x$ axis...

0celouvskyopoulo7

@Slereah There's something we don't understand about that $\pi/(\pi-2\chi)$...

It somehow folds up the bump function

it's a scaling yeah

user image

user image

Guys, is the choice of the author to pick this "negative" (downward) direction for $da_2$ arbitrary?
Or is it "normal" to do this? as in, is this convention?
In the case of the sphere, I can see that our perpendicular unit vector has to be $\hat r$. However, in the case of the $xy$ plane, I'm wondering if there is some "right-hand" rule (or something like that) which explains the direction of $da_2$

0celouvskyopoulo7

@Slereah eh, not really a scaling

@ShaVuklia no

6:10 PM

Instead of the "line" of the bump function having angle $\pi$, it's mapped to an angle of $\pi - 2\chi$

@0celouvskyopoulo7 so it's arbitrary?

0celouvskyopoulo7

I'm just saying no to you

oh.. you mean i can't ask questions?

0celouvskyopoulo7

not until I understand this

k

0celouvskyopoulo7

6:12 PM

@Slereah I find it hard to believe we can't write it in terms of $x,y$

I really need to diagram the situation by hand to really grasp it

0celouvskyopoulo7

it sure doesn't seem that strange

@0celouvskyopoulo7 well you could

0celouvskyopoulo7

@Slereah You have my mathematica code

I'm not sure it would be that good looking, though

0celouvskyopoulo7

6:12 PM

You can slide chi to get different angles

@ShaVuklia the direction is arbitrary and a matter of convention.

Question is

It doesn't matter what direction you choose for your area element as long as you are consistent.

Is there a good bump function outside of the truncated gaussian one?

0celouvskyopoulo7

huh? Gaussian isn't bump.

6:14 PM

@John i see, thanks

@ShaVuklia see:

1

Q: What is the Direction of area vector?

In case of closed surfaces the area vector is directed outwards the surface. But what is the direction of the area vector in case of an open surface e.g. A thin lamina type of surface. Does it depend upon the curvature of the surface i.e. concave or convex side.

vectors conventions geometry

I guess not

0celouvskyopoulo7

community.wolfram.com/groups/-/m/t/1093186

@John ah right, that makes sense. I understand your dual answer now

@ShaVuklia a top tip is to search the Physics SE using Google not the builtin search. Google does a far better job - it found that question immediately.

6:19 PM

yea true. i'd forgotten to google it

0celouvskyopoulo7

@Slereah the $R$ derivative of $$e^{\frac{1}{R^2s^2-1}}$$ is awful

I'll bet

But

I'm sure it goes to the appropriate number

0celouvskyopoulo7

$$\left(\frac{s}{2(rs+1)^2}-\frac{s}{2(rs-1)^2}\right)e^{\frac{1}{2(rs-1)}}$$

It should go to... 1 or -1, I think?

since the derivative of the cosine will be $0$ at the boundary

0celouvskyopoulo7

where $s:=\sin\alpha$

ok so we want $\alpha=\pi/4$

6:22 PM

Is that $s = 1$ at the boundary?

0celouvskyopoulo7

Wait, why $\pi/4$

Where did we get that from anyway

I thought it was $\pi/2$

Wait, if $s = 1$ the second term is divergent

Unless it's moderated by the exponential

0celouvskyopoulo7

Hmm

Meaning it's gonna be Complicated

Although probably not that much

It's basically the limit of $f^2 e^f$

0celouvskyopoulo7

Yeah so use Hospital

Although does that exponential go to $0$?

oh shit

6:26 PM

that's what we've been using for continuity

0celouvskyopoulo7

it's actually $$e^{\frac{1}{2(rs-1)}-\frac{1}{2(rs+1)}}$$

Maybe my computer did some partial fractions actually

end my life

0celouvskyopoulo7

Maybe I should just do the derivative by hand. One sec...

How do derivatives work

$\partial e^f=e^f\partial f$?

yes

0celouvskyopoulo7

Ok, it's $$\frac{-2Rs^2}{(R^2s^2-1)^2}e^{\frac{1}{R^2s^2-1}}$$

6:29 PM

Chain rule

0celouvskyopoulo7

So we want $R=1$ at the boundary

and then $s=1$, no?

I think so

0celouvskyopoulo7

So that's coming from the negative side

so $R^2s^2-1\uparrow 0$

and the exponential $\to 0$

that's going to beat the exploding denominator

But then the derivative will just be $0$

Which doesn't sound right

Are we sure it's not $\sin^2(\pi / 4)$

In which case it would be... 2?

0celouvskyopoulo7

that's 1

6:32 PM

Oh

Doesn't change much the problem

True

I have a question that's probably contentious since @0celouvskyopoulo7 is in the room. Is it really ok to just treat differentials as fractions?

it feels wrong

In what formalism

0celouvskyopoulo7

It is very wrong

Well, just like standard separation of variables

@0celouvskyopoulo7 could you tell me the correct way to arrive at the same result?

What result

0celouvskyopoulo7

6:34 PM

@Phase The answer is that it works, and you can plug the result back in to verify.

There is no "correct" way to do it, separation of variables is a mnemonic.

So there isn't a way of just writing it so that each step isn't an abuse of notation?

Behold:

0celouvskyopoulo7

@Phase Sigh, there is. See Theorem 6.9 in Jost, Postmodern Analysis.

user image

0celouvskyopoulo7

Flagging for obscenity

6:39 PM

@0celouvskyopoulo7 what does that sigh mean

it fills me with dread

@JaimeGallego Russian?

0celouvskyopoulo7

Because you won't leave me alone

oh

noted

0celouvskyopoulo7

@Slereah Where did you find that thing for the derivative of an implicit function

@JohnRennie Yes

24 hours ago, by Jaime Gallego

My parents have a display case full of Soviet pins in the back of a wardrobe

0celouvskyopoulo7

6:40 PM

@Phase I am in a very bad mood today

It's 2:40, Jesus Christ

hope things get better

@Phase it's wrong but everyone does it and it (mostly) works fine.

Treating differentials like fractions makes us all feel unclean at first, but you get used to that :-)

just another dead part on the inside

I'll chuck it on the pile

@0celouvskyopoulo7 the wikipedia

@Phase your (mathematical) conscience will just get in the way - discard it immediately :-)

6:44 PM

In mathematics an implicit curve is a plane curve which is defined by an equation F ( x , y ) = 0. {\displaystyle F(x,y)=0.} Hence an implicit curve can be considered as the set of zeros of a function of two variables. Implicit means that the equation is not solved for either x or y. If F ( x , y ) {\displaystyle F(x,y)} is a polynomial in two variables, the corresponding curve is called an algebraic curve, and specific...

tangent vector is just the derivative of the implicit function

0celouvskyopoulo7

@Slereah I'm not even sure what we need for this

@Slereah those are in Cartesian coordinates...

would that not translate well in arbitrary coordinates

0celouvskyopoulo7

@Slereah I think I would argue that all you're doing is a coordinate transformation $\theta\mapsto \alpha= \theta\frac{\pi}{\pi-2\chi}$

We know it's smooth for $\theta=\alpha$

So because that coordinate transformation is linear (hence $C^\infty$), it won't change the differentiability properties of the curve

Might be alright yeah

0celouvskyopoulo7

Good?

6:51 PM

That would probably fly

0celouvskyopoulo7

Now, arguing that it's timelike seems terribly hard. I vote to just use the picture.

If you really have to compute $dy/dx$, you can.

You'd have to transform the curve to cartesian coordinates (a priori not impossible)

You need $dy/dx$ because the metrix is $-dy^2+dx^2$

You could do it in geodesic polars, perhaps

But I think the formula on Wiki is only for Cartesian

hm, what would be the metric of Minkowski in polar

I guess it would be similar to the hyperboloid metric

0celouvskyopoulo7

Well, he says to use the Cartesian cone for reference

But just look at the picture...

I'm not sure this is worth spending more time on

Well I'll leave it as is for now

it's probably more effort than any other GR book already

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