12:28 AM
@Slereah Are you going to prove Penrose's gluing theorem in detail?
I think it would be nice to have a detailed explanation of the trips stuff

The theorem isn't too hard to understand, but i'm having some troubles understanding the smoothing of the joint

What do you have so far?
I mean, the joints part is the whole content, no?

$$R\cos(\frac{\theta \pi}{\pi - 2\chi}) = \exp(R^2 \sin^2(\frac{\theta \pi}{\pi - 2\chi}) - 1)^{-1}$$

Yeah that

That's the function he's using for the joint

12:30 AM
wtf is that

well he's using polar coordinates on the $(t,x)$ plane for the tangent plane
and $\chi$ is the angle between the joints
But I'm having troubles understanding what that curve looks like

christ
idk

and I guess I need to show that this is always timelike
Wait does it even make sense at $R = 0$

well why not

$$0 = \exp(1)$$

12:35 AM
what's $R$?

Or is the -1 supposed to be outside of the exponential
$R$ is the radial coordinate in $(t,x)$
$t = R \sin \theta$, $x = R \cos \theta$

So...we have a curve in polar coordinates. What's the curve parameter supposed to be?
This isn't making much sense

it's not parametric, just a contraint equation
Oh wait, I think the smoothing just doesn't go through 0
I guess if a curve in $TM$ is itself timelike, its projection on the manifold is timelike?

lemme get Penrose...

If a curve $\gamma$ in $TM$ has timelike tangent vector in $TTM$, is $\exp(\gamma)$ timelike?

12:42 AM

Who knows

Check O'Neill

All things are impermanent
Yeah O'neill prolly has it
How are you supposed to extract the tangent vector from a constraint equation tho, is what I'm wondering
Do you just derive both sides and then... something?

well, look what he says
one must take the region small enough for $\eta$ to be guaranteed timelike
I'd believe that

yeah, but
How to check that assertion
You'd need the tangent vector for that

12:49 AM
Ok, what should I look at first?

what do u mean

I'm done with French math for today, and my next goal is a YCB paper on weighted sobolev spaces
screw that
I'm going to work out this proof

Hm, let's see
Let's pick something simple
Constraint equation of a circle
$(x^2 + y^2) = 1$

@Slereah there is Lemma 5.33 in O'Neill
Maybe that will help

Derivative wrt $x$ is $2x + y^2 = 0$
Oh wait, just $2x = 0$
does that help at all

12:52 AM
what are you trying to do?

trying to figure out the tangent vector from the equation of the curve
I guess it would help if I had the parametric definition
But
no such luck

@Slereah btw
Yvonne has Chern 1944 Ann. Math. 45 747-752 for the Steenrod proof

Oh my
Let's check it out

(I am checking Yvonne for this curve stuff)

I mean I'm sure this is a dumb thing
Probably not in Yvonne but in some basic geometry book

12:55 AM
Huh?

But I can't think of the way to get the tangent curve from that

I'm talkin about making $\eta$ timelike
I have her right here, might as well check
but it's probably hidden in O'Neill

that doesn't seem related to steenrod?

12:57 AM
Yvonne is a smart woman
She gets away with it by taking paths to be piecewise $C^1$

I have a quesiton about potential energy

So perhaps piecewise $C^1$ is the "correct" notion

I just gave it

if a ball is above a table by 1m, then the potential energy of the ball is PE = mgh (where h = 1 here)

12:58 AM

Is PE real, or just a construct for understeanding what would happen if the ball fell

@Slereah oh, I misunderstood her
or I did
nvm
back to trips

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...
Found it

Ok, so he's plotting this implicitly
Can you mathematica it?

I'll do it later
It's 3 AM here
I have odd gaps in my knowledge
Why didn't I know that formula
Let's see, what would that formula be for a circle

1:02 AM
So, is that $1/\exp$ or $\exp$ of the inverse of the stuff inside...
Probably exp of the inverse
because $1/\exp=\exp-$

$$2x_0 (x - x_0) + 2y_0(y - y_0) = 0$$
Let's take $x = 0$ and $y = 1$, the tangent should be $(1,0)$
Well that certainly works out
Seems trustworthy enough

Just checking if that formula works for a simple example
Though maybe I should have just used it for a line really

what is this $R$ supposed to be

Radial coordinate, so $R = \sqrt{t^2 + x^2}$

1:10 AM
well, but look
$t=\tau_0R\sin\theta$
so it's $\tau_0R$ that's the radial coordinate

Oh right, it's scaled

so $R$ is a renormalized radial coordinate

Yeah, $R \leq 1$

So we're in a disk in $R^2$, basically.
Minkowski space even
Ah
The $|\theta|\le \pi/2$ makes sense to me
the range is then $[-\pi/2,\pi/2]$
see this picture

Well it's gonna be in between the two segments, certainly
and the angle is normalized to rescale the interval

1:15 AM
dashed line is the timecone
So really...the proof is obvious...
The only hard part is actually writing down the curve

yeah it's a fairly nasty equation

Now $\chi$ is the angle from the axis so the line $\mu$
This is a strange ass function

the assiest function

Isn't $\theta\pi$ strange
the units don't make sense
mmm, I guess you want to divide for some reason

Well it's $$\theta \frac{\pi}{\pi - 2\chi}$$
So the "units" are fine

1:19 AM
yeah

but yeah since I have the formula for the tangent vector now it shouldn't be too hard to do?
Not trivial because proving that it's timelike sounds like a bit of work

I would really like to know what this damn thing looks like

but doable

I'm trying to evaluate it along points it should connect
it's not working
Like for $R=1$ it should be $\theta=\pm\chi$

Odd

1:23 AM
I get
$$\cos(\theta\frac{\pi}{\pi-2\chi})=\exp[-\sec^2(\theta\frac{\pi}{\pi-2\chi})]$$
there's nothing special about $\theta=\chi$ in there as far as I can tell

yeah seems hard to show anything from that
What about $\theta = \pi - 2\chi$

Maybe Penrose just wrote something and figured no one would ever check

All GR is wrong
Except that would be $\cos = -1$ and i don't think the exponential is gonna be negative

wtf
so I guess $R=1,\theta=\pi-2\chi$ cannot happen

Let's see

1:28 AM
It should be $R=1,\theta=\chi$
Maybe one can solve for $R(\theta)$
this should be a regular old polar thing
I have a picture in my mind of what it's supposed to be

Maybe but seems like a tough one

yeah maybe not
the best I get is $$R^2\sin^2\alpha\ln R-\ln R+R^2\sin^2\alpha\ln \cos \alpha=1+\ln\cos\alpha$$
$\alpha$ being that angle thing
How the hell did penrose do this
he's a god damn genius

If we just pick $\chi = \pi / 4$ that's $$\frac{\pi}{4} \frac{\pi}{\pi/2} = \frac{\pi}{2}$$
$\cos$ is $0$, the exponential will be $-\infty$ so $0$ as well
I guess that checks out

proving it in the general case sounds tough

1:39 AM
I would be happy with a picture here
Because to really check that this is smooth...boy
you have to compute all the partial derivatives
ALL

I think it's basically supposed to be a bump function, except the real line is "bent"?
Well do we need smooth?

@Slereah exactly
@Slereah do you just want $C^1$?

We could just check continuity of the function and first derivatives
The theorem only says a timelike curve

So for continuity we just need to check that it actually connects the two segments

So I guess we need to check continuity, first derivatives, and that it's a timelike curve

1:42 AM
But it's not clear to me why that's actually true

yeah it's a bit tough
Let's try some partial fraction decomposition
$$\frac{\xi \pi}{\pi - 2 \xi} = -\frac{\pi}{2} + \frac{\pi^2}{2\pi - 4 \xi}$$
Hm
Does that help

uhhh what's that

In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is, a fraction such that the numerator and the denominator are both polynomials) is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator. The importance of the partial fraction decomposition lies in the fact that it provides an algorithm for computing the antiderivative of a rational function. In symbols, one can use partial fraction expansion to change a rational fraction in the form ...

well, yeah
but what does that do here

nothing useful, I think
First I think I'm gonna plot both functions because I really want to be sure the ^-1 is correctly interpreted

1:54 AM
i plotted it above
it looked right to me
Apparently for $R=1$, there's no solution of $\theta$ though
hmm, no
there are solutions

yeah neither plot seems to give the same values overall for $R = 1$
The hell was Penrose on about
Let's find his email

can you post the plot

www.wolframalpha.com/input/?i=plot+(cos(+(x+*+pi)+%2F+(pi+-+2*x)+),+exp(1%2F(+si‌​n%5E2(+(x+*+pi)+%2F+(pi+-+2*x)+)-1))+)+from+0+to+1.5

that's broken
you need to put it like this [link](...)

copy paste it

2:00 AM
what is this supposed to be?

Penrose has no email listed, but his university website has a contact form for him

also the plot breaks
lol a fucking contact form

@0celouvskyopoulo7 The plot of both sides at $R = 1$

well there's something wrong here

Ideally they should be equal for $\theta = \chi$

2:01 AM
Wolfram says the $R=1$ solution only has imaginary solutions
but my calculator seems to work just fine

Did you try with ^-1 meaning the inverse of the exponential?
I tried with the plot it's still not working

so inverse of the exponential just flips the sign on the argument
I find it hard to believe that penrose would do that and not just flip the sign

well yeah, but who knows
I think I'll make a PSE post for this

No
Not yet
Give me a week or so please
I will figure it out and write a full proof

alright

2:04 AM
@Kenshin construct for measuring

Worst case scenario, we can throw out that function and find another smoothing, perhaps

we can measure the height from any level
for a system @Kenshin

I think there's an abstract way via variational theory
And my aadvisor once wrote a proof
So, for $R=1$ we have $\cos\alpha=\exp[-\sec^2\alpha]$, agreed?
So that means $\alpha=\pm\pi/4$

yes

2:07 AM
because the left is $0$ and the right is $\exp [-\infty]$
So we have $$\theta\frac{\pi}{\pi-2\chi}=\frac{\pi}{4}$$
So $$\frac{\theta}{\pi-2\chi}=\frac{1}{4}$$

Wait what
They are not required to both be $0$
$$\alpha = -\frac{\pi}{2} + \frac{\pi^2}{2\pi - 4 \chi}$$
That's obviously not gonna be always $0$
Unless $\chi = \pi / 4$

wait what do you think doesn't have to be zero

The left hand side

Oh I meant in $\cos\alpha=\exp(-\sec^2\alpha)$
the only way both sides can be equal is if they're both zero
so if $\alpha=\pi/4$

But that's not true though
Just take $\chi = 0$

2:11 AM
does that not convince you?

Then $\cos \alpha = 1$
Well that's what it should be

@Slereah $\exp -1$ is not $1$...

But it's not what it is
Yes, most certainly
Did Penrose just fuck it up

so what are you on about
why is $\chi=0$ bad?

That $\alpha$ is only gonna be $\pi/4$ in one specific case
If $\chi = 0$ then $\alpha = 0$

2:13 AM
for what $R$

Any $R$

If $\chi=0$ then $\alpha=\theta$...

You mean $\alpha = \frac{\theta \pi}{\pi - 2\chi}$, right?
Oh right, I mean for $R = 1$ here
at the boundary

Then $\theta=\pi/4$

Since we should have $\theta = \chi$

2:15 AM
@Slereah Yeah, we should
But I think he fucked up
I see what you're saying
@Slereah So what should we do
are you gonna try to contact him lol

I am

He won't remember
He's too old and crazy by now.

We can only hope

So how about $\alpha=\theta\pi/4\chi$

I think the trick is that the angle spanned by $\theta \in [-\pi / 2, \pi / 2]$ should be in between the two segments
The two segments span an angle of... $2 \chi$, I think?

2:27 AM
yeah
see my earlier picture

So $\pm \pi/2$ should correspond to $\pm \chi$
Wait, are we 100% sure that at the point, the angle should be $\chi$?
Shouldn't it be $\pi/2$?
Although I guess that's still the same problem as before
$\theta \pi / 4 \chi$ does get rid of that problem

The angle is fixed
we can't change $\chi$ because in the exponential map, the angles are preserved at the origin
so it's whatever angle they intersect at in the manifold

If that is true though
Now the hard part is
Showing the derivatives also join up

Yeah....how on Earth is that supposed to work
that still doesn't make sense ...
Hmmmm
$\chi<\pi/4$
is that true
yeah, $\chi=[0,\pi/4)$
Now I'm confused.
I think I had the wrong picture

doesn't Mathematica have polar plots

2:42 AM
Fuck
$t$ is horizontal
Which idiot decided that was a good idea.
This isn't QFT

oh wait I forgot 1/r^2
no that don't work

Ok that's the right picture.
So let's assume $\chi=\pi/8$ and plot Penrose's picture.
$\pi/8$ is halfway down, right?

It is an angle of 22.5°
So bisecting the light cone

2:50 AM
right
so let's plot Penrose's curve for that.
workin on it...
So $\alpha_\text{Pen}=4\theta/3$
wonder if it will plot
Doesn't look right...

are those sharks
also note that you can restrict your plot

@Slereah Yes. The rare and deadly function-nosed shark.

@Slereah Uh, I'm giving it the implicit function directly
I have no clue how to actually use this
it's not parametric

I mean restrict the range
You can write "for r in [-1, 1]"

Now for $\theta_\text{Ryan}=2\theta$

2:55 AM
it should make things a bit clearer

@Slereah ok
Hmm.
I just realized
This is wrong
It's plotting $\theta$ vs. $R$
NOT polars

well yes
obviously

you could have said that hours ago

you need to type "polar plot"
I didn't know you did not notice

2:58 AM
odd
I'd run it through Mathematica but I don't have it installed right now

20

Mathematica can use ContourPlot to draw implicit Cartesian equations, but doesn't seem to have a similar function to plot an implicit polar equation, for example $\theta ^2=\left(\frac{3 \pi }{4}\right)^2 \cos (r)$ What's the best way to do this?

@dmckee Can you help the cause?

@0celouvskyopoulo7 No experience with mathematica.
You might be able to understand it better in some transformed expression.
What does it look like if you use $\xi = \cos(2 \theta)$, and remove the nasty sine in the denominator with as $1 - \xi^2$?

$$R\xi=\exp\frac{1}{R(1-\xi^2)-1}?$$
@dmckee idk what that's supposed to do...I want $R$ and $\theta$ to be polar coordinates

Yeah. Don't know if that is going any where. It's going to have nastiness here and there.

It's an implicit polar plot that I don't know how to do in Wolfram

3:05 AM
@0celouvskyopoulo7 You want them to be polar in the original space. The game with throwing transformation at them is about finding a differentâ€”more comprehensibleâ€”way to visualize them.

Hmm

@0celouvskyopoulo7 Your school makes it available to you? That would be cool.

They do
I spend a gorillion bucks to go here
at least I get free software
@dmckee Apparently I just go to the wolfram website and give them my school email