The h Bar

Danu

15:00

the "backward derivative" is negative the forward one

Huy

what, why?

the minus signs cancel?

or am I looking at a different derivative

Danu

Hmm right

idk

Huy

:D

Danu

I'm not even sure how this would generalize to this situation anyways

probably reasonably well because of charts

Huy

I mean if differentiable, in the real case, the differential quotient gives same limits by definition

0celo7

15:02

The radius never matters

Huy

because it's only differentiable if the limit exists and the limit exists if it's the same in both directions

0celo7

It's all about the angles

Huy

@0celo7: yes

0celo7

In class

So figure out the optimal angle placement

Maybe?

Alpha is fixed

ACuriousMind

@Danu I have no idea why you would expect $-t$ there.

Danu

15:09

@ACuriousMind Because pulling back :P

but I guess @Huy's remark resolves it :P

0celo7

@Huy take the partial of the log of that mess wrt gamma

See what happens!

ACuriousMind

And, yeah, if the thing is differentiable, then the left- and right-derivatives are equal, anyway.

Danu

lack of analysis knowledge intensifies

:D

Huy

@0celo7: I get $-b/(bc+c^2)$ without Greek letters

0celo7

Eww

Huy

15:16

@0celo7: I'm a few steps behind, I just had to verify that $\alpha$ is fixed

so now basically we want to find $\beta, \gamma$ such that $d$ is minimal

0celo7

Yes

Alpha is fixed because P is fixed

Huy

wait

I think something's wrong

if I take a different point on the curve

and the circle through that point and $P$

that circle will have an entirely different midpoint

so a different $\alpha$ will result

no?

0celo7

Draw it

Wait

You're right

Ok this is nuts

Huy

ikr

but

effectively it only depends on the midpoint of the half circle

which lies on the $x$-axis

so it's just one parameter

so I should be able to derive and find the minimum

no?

0celo7

15:36

Yes

@Huy can we put $g$ through the origin and just shift everything

Huy

@0celo7: that's what I'm doing right now

@0celo7: I'm trying to express $d$ only using $x_M$

0celo7

the hell is $x_M$ :O

Huy

x coordinate of the midpoint of the half circle

0celo7

I see

so make $x_M=0$ the optimal case

Huy

yes

0celo7

15:49

so can you write $\alpha(x_M)$, etc.?

Huy

the $\gamma$ is the hard part

0celo7

getting out paper

Huy

because it depends on the intersection point, and the intersection point of a line and a circle isn't very nice

0celo7

yeah

there has to be a better solution

Huy

idk

I'd hope so but I don't see an easier way

I mean conceptually this way isn't really hard but it's really cumbersome atm

0celo7

15:53

$\beta$ is hard too, no?

lol $\pi-\alpha-\gamma$

stoopid

Huy

^^

ok I have all the formulas

it's not very nice

I hope wolframalpha can derive it

0celo7

$$\alpha(x)=\tan\frac{P_y}{P_x+x_M}$$

Huy

yea

0celo7

differentiate

Huy

why, we don't want to minimize or maximize alpha or gamma

we want $d$

so we have to plug that in

into the log thingy

0celo7

15:56

the word for $d/dx$ is differentiate

not derive

Huy

i mean $d(P,Q)$

also the tan should be on the other side

so you get an arctan

0celo7

tan-1, yeah

Huy

so in the formula for the distance you'll get a product of arctans in the denominator

doesn't look so nice

0celo7

jesus how does $\gamma(x)$ work

the equation for the circle is $(x-x_M)^2+y^2=e(P,x_M)^2$

yes?

Huy

I have the equation

0celo7

15:58

is that the correct equation for the circle

Huy

y

@0celo7: imgur.com/JC5fYzg sorry my writing isn't very nice

0celo7

how do you find the intersection of a circle and a line

sub the line equation in for the circle?

Huy

@0celo7: you plug in and then get a quadratic equation

0celo7

or other way around?

Huy

yea

0celo7

16:00

yeah fuck this

Huy

:D

0celo7

what coordinates did you get for the intersection

Huy

@0celo7: look at the bottom

$Q$ is my intersection point

0celo7

sweet jesus

Huy

yea it's not very nice

and then you have to plug that into the other thing

and the $y_Q$ is similar, just multiplied by $a$

0celo7

16:01

yes

Huy

setting $x_M = 0$ before differentiating doesn't really show it's an obvious minimum

so I guess I'll have to differentiate that thing

0celo7

plot it!

Huy

that's cheating

:D

btw are you sure about your signs in the denominator of your angles?

because you have $x_P + x_M$ and I have minus

0celo7

I agree with that equation

erm

yes, minus

uh crap

we have $x_M$ on opposite sides of zero

we need an absolute value in the denominator

then we agree

$|x_M-P_x|$

don't want that being negative

@Huy are you plotting this

this is a monster

@Huy I'm gonna pick $P=(1,1)$ and $y=x$

Huy

@0celo7: imgur.com/BkdqeFV

why do you think this is a monster

aren't all formulas in physics like this?

0celo7

16:12

wtf you crazy man

HAHA

Huy

:P

0celo7

you had to rewrite log

Huy

xD

I'll tell the freshmen who asked me about this "it's a simple calculation"

0celo7

wait why only $\pi-$ one thing

what exactly is that

pls write using $\alpha\beta\gamma$

Huy

@0celo7: you hat the $\log (a+b)(b+c)/(ac)$

and you know $\pi = a+b+c$

so I use $b = a+c-\pi$

uh

other sign

$b = \pi - a - c$

and so you get $\log (\pi - c)(\pi - a)/(ac)$

0celo7

16:14

my simplification made it more complicated smh

my first equation was right

Huy

:D

0celo7

so yeah, graph that!

Huy

-.-

0celo7

@ACuriousMind yo I heard you like tedious calculations

how long do the freshmen have for this :O

Huy

@0celo7: a week

@0celo7: it's by far the hardest exercise and I have no idea how they would come up let alone try these calculations in their first week

0celo7

16:18

I came up with it

I take credit for the idea, you can take credit for fixing my mistakes :)

Huy

pf

0celo7

pf?

Huy

pf

lets see if mathematica can do this

0celo7

What does that mean

Huy

or more like if I know how to properly use mathematica

0celo7

16:20

What if x is not the minimum

What if we discover new mathematics

Huy

what do you mean by that

0celo7

X=0 I mean

Huy

you mean if $x_M = '

ye

I call it 0celoean geometry

0celo7

Now how the fuck do we derive d(P,Q)

Huy

I already did

see my last formula

0celo7

16:22

From the metric?

Huy

or what do you mean

0celo7

No the General formula

The one with the ks

Huy

the formula they give for $d(P,Q)$ you can obtain quickly from the Riemannian metric

0celo7

Really

Huy

the problem is that we don't know any Riemannian geometry in the first year

well the first one should be obvious

if the points have same $x$ axis

and you take the integral along the segment between them

0celo7

16:24

I've done that one

Never done the General one

John Duffield

@0celo7 : yes, I can see posts from people who ignore me. Mind you, there's nothing much to see. Just ramblings, no physics.

obe

Hello.

0celo7

hello

@Huy D: what do I do for the general case

do I need to find a circle connecting the two points

obe

I have to go to the doctor... then once I'm back I'll be working on the pset until midnight.

0celo7

@Huy pls

Huy

16:34

@0celo7: I think you can just use the cross-ratio and that Mobius transformations are isometries

0celo7

for?

Huy

@0celo7: then you can apply the formula for the easy case to the harder case

and get the formula for the harder case immediately

(cross ratio is Mobius invariant)

0celo7

cross ratio?

Huy

@0celo7: $(z_1, z_2; z_3, z_4) = \frac{(z_1-z_2)(z_3-z_4)}{(z_2-z_3)(z_4-z_1)}$

you can check that this is invariant under Mobius

0celo7

seen that in string theory

Huy

16:36

and you can also prove a bunch of things with this

e.g. it is real iff the points lie on a line

(hyperbolic line. in general, it is real iff they lie on a circle or a line)

0celo7

what what are those numbers supposed to represent here

Huy

points in the complex plane

0celo7

certainly they're not insertion points for conformal fields on the worldsheet

complex plane?

where is a complex plane

Huy

$\mathbb{C} \cong \mathbb{R}^2$?

0celo7

oic

well I'm trying to get it from the metric

failing because I'm not sure what integral to do

Huy

16:39

probably works too but then you'll have to parametrize the half circle

0celo7

ok that's what I thought

Huy

I've always done it this way because we've done Mobius transformations/cross ratio etc several times in geometry and complex analysis

0celo7

so how do you get it from the isometries

ok here's how I do it

wlog we put the origin between the two points

Huy

@0celo7: if the two points don't have the same $x$-axis, you can take the two points and the endpoints of the half-circle connecting them and map it to a ray

0celo7

then we can use polar coordinates for the circle

@Huy equations pls

Huy

16:41

@0celo7: do you know Mobius transformations?

0celo7

in theory

PSL(2,C)

Huy

no, PSL

0celo7

yes

Huy

you can explicitly find a Mobius transformati9on mapping such a half circle to the imaginary axis

0celo7

example?

Slereah

16:43

Hello chums

Are we talking spin groups

0celo7

no

Slereah

o

0celo7

gdi this is hard using the line element

I have to coordinate transform the metric

oh joy

what's the tensor transformation law lol

Slereah

for metrics it's dx'/dx, if I recall correctly

Huy

@0celo7: well you know that inversion maps circles & lines to circles & lines if you look at $\mathbb{C} \cup \{\infty\}$, right?

Slereah

16:48

Oh, it is actually dx/dx'

dang it

0celo7

@Huy in class, can we resume later?

Huy

maybe

@0celo7: en.wikipedia.org/wiki/…

0celo7

$$g=\frac{\mathrm{d}r^2}{r^2\sin^2\theta}+\frac{\mathrm{d}y^2}{\sin^2\theta}$$

Huy

18:12

@0celo7: I tried it with an example but I must have messed up something.

@0celo7: bit.ly/1FB41mF

Hippalectryon

Hello :D

Slereah

Gotta go fast

Hippalectryon

Sanic sanic forever

I'm reposting this from the chem chatroom since I believe I might find a good answer here :D

in The Periodic Table, 23 mins ago, by Hippalectryon

Anyone got an idea of a cool experiment/subject I could use ? Or somewhere I could find one ? It's supposed to be a personal research project on any good enough subject, it should have some "theoretical" part and some "experimental" part. Last year I studied the deformation of ferrofluid under the action of a magnet and caracterized the surface in function of the different parameters using fluid mechanics and whatnot, but I'm out of ideas this year :(

in The Periodic Table, 23 mins ago, by Hippalectryon

It shouldn't be too easy (the research needs to have some value) nor too hard (I should understand it well)

in The Periodic Table, 20 mins ago, by Hippalectryon

I'm willing to spend a bit of money in material/... (up to €50) if necessary

in The Periodic Table, 18 mins ago, by Hippalectryon

It's better if it's a bit 'original' or 'homemade' (for instance 'How fast should I walk in order to keep my tea from spilling on the floor' is better than 'How can I maximize a solar panel's efficiency')

Any idea is welcome :-)

0celo7

18:34

@Huy Suppose we have two points $P,Q$ in the hyperbolic space connected by a geodesic $\gamma$. We know that this geodesic is a semicircle with end points on the $x$-axis. Set up a coordinate system with the midpoint of the circle on the origin and polar coordinates so that $P=(r,\theta_P)$ and $Q=(r,\theta_Q)$ where $r$ is the Euclidean distance from $P$ and $Q$ to the origin.

The metric of this space is $$g=\frac{\mathrm{d}x^2+\mathrm{d}y^2}{y^2}=\frac{\mathrm{d}r^2+r^2\mathrm{d} \theta^2}{r^2\sin^2\theta}$$

Along $\gamma$ $r=\mathrm{const.}$ so we can ignore $\mathrm{d}r$.

We parameterize the circle by $\theta$, and the distance between $P$ and $Q$ along $\gamma$ is $$d(P,Q)=\int_\gamma\mathrm{d}s=\int_{\theta_P}^{\theta_Q}\frac{\mathrm{d}\theta‌}{\sin\theta}=-\log|\csc\theta+\cot\theta|_{\theta_P}^{\theta_Q}$$

@Huy What next?

@Huy Jesus.

@ACuriousMind You're welcome to help as well.

ACuriousMind

I don't even know what the question was.

0celo7

5 hours ago, by Huy

Let's look at the upper half plane model of hyperbolic space only knowing that points are $(x,y) \in \mathbb{R}^2$ with $y > 0$ and lines are either half-circles with midpoint on the $x$-axis or rays parallel to the $y$-axis. Define the distance of two points $P, Q$ as follows:

5 hours ago, by Huy

If they have the same $x$-coordinate, define $d(P,Q) = |\log \frac{Q_y}{P_y}|$. Otherwise, look at the line (half-circle) through $P$ and $Q$ and its endpoints $A, B$ on the $x$-axis ($A$ on the left of $B$). Then, define $d(P,Q) = |\log \frac{k(A,Q) k(B,P)}{k(A,P) k(B,Q)}|$ where $k(A,Q)$ is the Euclidean length of the arc $AQ$.

5 hours ago, by Huy

Now take a Euclidean line $g: y = mx + b$ and a point $P$ that does not lie on $g$. I want to compute the distance from $P$ to $g$. Apparently, if I take the intersection of $g$ with the $x$-axis as the midpoint of a circle running through $P$, the arc on that circle between $P$ and $g$ realizes the minimal distance and therefore is the distance between $P$ and $g$. How do I see that it is this arc and if I don't know it beforehand, how can I find out?

@ACuriousMind How to derive $d(P,Q)$ from the Poincare half-plane metric?

0celo7

19:17

no one wants to help :'(

Hippalectryon

I wish I could :P

0celo7

ACM is not even playing PoE/Kerbals/whatever

Hippalectryon

You don't have an idea for my experiments thingy above by any chance ? ;-)

0celo7

no

projects are pretty dumb unless there's millions of dollars involved

Hippalectryon

:c

There's millions of dogecoins involved

0celo7

19:21

@ACuriousMind huh?

ACuriousMind

...what

I see it too, though

500 late answers just appeared in the queue out of nowhere

0celo7

so you answer that but not my geometry question :O

ACuriousMind

@dmckee @Qmechanic @DavidZ @ManishEarth This is not normal, is it?

@0celo7 I don't know the answer to your geometry question

0celo7

hmmmmmmmm

Qmechanic

@ACuriousMind : A glitch in the matrix? SE perhaps adjusted the criteria for being in the queue?

ACuriousMind

19:25

@Qmechanic Probably...the first answer it asks me to review is from Oct 20 '13

That's definitely not "new user's answer to an old question"

Hippalectryon

in The Periodic Table, 26 secs ago, by Loong

@Jan many sites are affected. The mods are looking into it.

Loong

24

Q: Can we raise the bar for reputation for late answers to enter the review queue?

I would like to raise the bar for late answers to enter the Late Answers Review Queue to 50 rep, which is the threshold after which users gain the ability to comment. Here are some answers that were posted by users with between 10 to 50 rep (yes, there's a little selection bias to be sure. I susp...

0celo7

1

Q: Ricci tensor as relativistic Hamiltonian

I am little bit dissapointment with action integral in General relativity. The action integral is: $$ \int Rd^{4}x=\int R_{ij}g^{ij}d^{4}x\tag{1} $$ Where $$ R_{ij}=\frac{\partial\Gamma^{l}_{ij}}{\partial x^{l}}-\frac{\partial\Gamma^{l}_{li}}{\partial x^{j}}+\Gamma^{l}_{ij}\Gamma^{m}_{lm}-\Gamm...

general-relativity lagrangian-formalism hamiltonian-formalism curvature stress-energy-tensor

> I am little bit dissapointment with action integral in General relativity.

ACuriousMind

@Loong So, they implemented that and it's retroactive?

0celo7

I'm not going through 500 late answers

gl queue warriors

@obe how's the problem set(s) going

Loong

19:33

@ACuriousMind yes

Slereah

Are you never disappointed with actions, @0celo7

I think the Standard Model is very disappointing

I mean seriously, neutrino mixing matrix and quark mixing?

v. amateurish

Pies

19:55

Howdy

skill patrol

Hello

0celo7

@Slereah indeed

what is the SM lagrangian in ST

Pies

I got a joke. Why did the photon go on a diet?

Because it wanted to stay light!

hahaha....wait, photons don't have mass...

so nevermind i guess

0celo7

yes they do

$E=mc^2$ :^)

Pies

No

E^2 = M^2C^4 + P^2C^2

that's the full equation

ACuriousMind

19:59

@0celo7 ...that's not a meaningful question, I think

0celo7

@Pies says who

Pies

it only makes a difference when dealing with massless particles, so it's usually shortened

Mr. Einstein

0celo7

almost

you almost got it

Pies

?

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