The h Bar

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

what

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

that looks about right dude

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

Nothing about smoothness

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

Partial fraction decomposition

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

ok please do

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(+si‌n%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

wolframalpha.com/input/?i=plot+cos(x)+and+exp(-sec%5E2(x))

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

Fucking Timaeus will answer

Give me a week or so please

I will figure it out and write a full proof

alright

Koolman

2:04 AM

@Kenshin construct for measuring

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

yeah

Koolman

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

I never really read it

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 $$\frac{\theta}{\pi-2\chi}=\frac{1}{4}$$

is that what you had?

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

wolframalpha.com/input/?i=plot+cos(x)+and+exp(-sec%5E2(x))

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 $\alpha=\theta\pi/4\chi$

idk about the derivatives...

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

wolframalpha.com/input/…

wot

lol

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

Q: Plotting an implicit polar equation

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?

plotting

@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 different—more comprehensible—way to visualize them.

Hmm

I'll download mathematica

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