The h Bar

Slereah

00:00

there's an exact formula for quartic equations

ACuriousMind

@0celo7 Does substituting $y=x^2$ make it a quadratic one? If no, are any of $-1,0,1$ roots?

If no to both, sucks to be you :P

Slereah

Well it's very simple

Apparently the delta is just

$user image$

0celo7

@ACuriousMind that stubstitution works

wtf now I have to solve a quadratic

ACuriousMind

lol

0celo7

@ACuriousMind HELP

ACuriousMind

00:02

Y'know, the formula for quadratic equations is known in Germany as the Mitternachtsformel - "midnight formula"

Because you are expected to be able to recite it even if woken up in the middle of the night.

0celo7

nie davon gehort

Ich denke du lugst

nur ein Mathe Professor kann das machen

ACuriousMind

Quadratische Gleichung

Eine quadratische Gleichung ist eine Gleichung, die sich in der Form mit schreiben lässt. Hierbei sind Koeffizienten; ist die Unbekannte. Ist , spricht man von einer reinquadratischen Gleichung. Die linke Seite dieser Gleichung ist der Term einer quadratischen Funktion (allgemeiner ausgedrückt: ein Polynom zweiten Grades), ; der Funktionsgraph dieser Funktion im Kartesischen Koordinatensystem ist eine Parabel. Geometrisch beschreibt die quadratische Gleichung die Nullstellen dieser Parabel. == Allgemeine Form und NormalformBearbeiten == Die allgemeine Form der quadratischen Gleichung lautet…

^not lying :P

0celo7

that's still BS

even if I memorize it I still won't be able to use it...

no real solutions

great

according to my calculations no orbits are possible with real initial velocities

ACuriousMind

seems legit

0celo7

indeed!

let's try hyperbolic next

Slereah

00:07

Too bad

I didn't really want to crash in the sun

0celo7

hyperbolic is possible

parabolic is possible

welp

I give up

either I'm wrong or Newton

and sadly I think he was smarter

Slereah

Hm

Apparently Penrose calls a normal convex neighbourhood a "simple region"

0celo7

penrose is a woo peddler

Slereah

He is actually

But mostly about consciousness

0celo7

if physics can't be understood by a dumb German middle schooler

Slereah

00:14

On GR he's fine

0celo7

it ain't physics

it's woo

Slereah

Is Duffield a German middle schooler

0celo7

don't think so

he'd've flunked out by now

@dmckee is "he'd've" real

@HDE226868 hey

astro nerd

I need your help

HDE 226868

@TanMath Yesterday.

@0celo7 Ooh, interesting. Go on.

Ah, you've been doing orbital mechanics.

0celo7

@HDE226868 would you believe it if told you, for $F=-k/r^2$ that $$1-e^2=\frac{2h^2}{rk}-\frac{h^2v^2}{k^2}$$

dmckee

00:19

@0celo7 Casual, but real enough for me. Which is to say that it can be written as $a + i\epsilon$.

0celo7

$e$ is the eccentricity, $h$ is angular momentum per unit mass

my ODE prof loves orbital mechanics

I think his time at the Max Planck gravity place ruined him

HDE 226868

@0celo7 $r$ being the distance, and $k$ an arbitrary constant?

0celo7

@HDE226868 k is the constant in the force

$r$ is the distance, yes.

@dmckee why don't you \mathrm your imaginary unit

dmckee

@0celo7 Lazy. Hadn't thought of it. Distracted. I have any number of excuses.

0celo7

@ACuriousMind for the record, I did know about that substitution, it was an attempt to make light of my failure to study for an exam

@dmckee you do it in general?

TanMath

00:23

@HDE226868 oh.. ok!

dmckee

I don't write things in complex form in LaTeX often enough to have a habit.

HDE 226868

@0celo7 It's making some sense. Working from$$\epsilon=-(1-e^2)\frac{\mu^2}{2h^2}=-\frac{\mu}{2r}$$is the way, I'm assuming? Here, $\epsilon$ is specific orbital energy.

0celo7

@HDE226868 would you also believe me if I told you $h=rv\sin\alpha$, where $\alpha$ is the angle from the radial vector to the velocity

@HDE226868 dunno what $\mu$ is

probably some reduced mass thing

HDE 226868

@0celo7 That sounds about right.

@0celo7 Nope, sum of the standard gravitational parameters ($GM$).

Gotta go to dinner; I'll think it over.

0celo7

@HDE226868 so if I told you $$1-e^2=\frac{2rv^2\sin^2\alpha}{k}-\frac{r^2v^4\sin^2\alpha}{k^2}$$

@HDE226868 NO

that's not my question

do you believe that last equation

which I have so carefully derived

@HDE226868 uh, test body does not contribute, so its mass never comes in

I think

yeah, I'm going everything per unit mass

@Slereah I once programmed by Ti-84 to do that

took about an hour to type it in

was very very difficult

Slereah

00:42

I should make a shortcut for $\mathcal{M}$

I use it a lot

0celo7

\M

HDE 226868

@0celo7 Yeah, I know. Dinner called.

0celo7

@HDE226868 should I get kid diddler or gay hating tonight?

Subway vs. Chick-Fil-A

HDE 226868

@0celo7 Okay, why are there two terms here? If I use$$h=\sqrt{a(1-e^2)G(M+m)}$$then I have$$1-e^2=\frac{h^2}{G(M+m)}=\frac{rv^2\sin^2\alpha}{k}$$

I'm missing a factor of $2$, and the entire second term.

By the way, I hate your notation.

0celo7

lemme see if that gives the correct answer

no need for attacks here

HDE 226868

00:51

@0celo7 . . . Yes. What equation did you start from?

0celo7

that's wrong too

Slereah

guys

I have a confession

I still don't know what the exponential map is in GR, exactly

0celo7

@Slereah me too

@Slereah lol

Slereah

I know it's a mapping from the tangent space to the manifold

But that's about it

0celo7

@Slereah we'll discuss it later

Slereah

00:52

I assume exponentials are involved

Nah, I'll look it up

Gotta do some proofs with convex normal hoods

0celo7

you basically go along a geodesic

for some time

and then you get a point

Slereah

I should also make a shortcut for "convex normal neighbourhood"

\hood

0celo7

and this point is unique in a small nbd

@HDE226868 uh

$G_{\mu\nu}=T_{\mu\nu}$

then I made some simplifications

ACuriousMind

@Slereah For Riemannian geometry, the exponential map $T_x M\to M$ maps a tangent vector $v$ to the point that lies at the end of the geodesic of length $\lvert v \rvert$ in the direction of $v$. For Lorentzian geometry, this is no longer accurate because vectors can have length 0, but the idea is still "Just march in the direction of $v$".

0celo7

got to $\epsilon =v^2/2-k/r$

HDE 226868

00:54

@0celo7 Haha. But I'm being serious. There are loads of expressions for $1-e^2$; you could start with any one of them and try your luck.

@0celo7 Okay, that's right.

Specific orbital energy.

0celo7

deciphering my class notes here

wtf is happening here

HDE 226868

Ohhhh. . . Did you set that equal to$$-\frac{\mu^2}{2h^2}(1-e^2)$$?

Yes. You did. There you are.

0celo7

I don't know what this $\mu$ thing is

HDE 226868

That should give you the right answer.

0celo7

is that not what I had above

HDE 226868

00:56

@0celo7 It's $k$; I just refuse to succumb to your notation.

0celo7

@HDE226868 this is standard notation for classical mechanics in math books

oh

I used the polar requation

solved for $\cos\theta$

HDE 226868

@0celo7 Do the substitution. Start with$$1-e^2=\left(\frac{v^2}{2}-\frac{k}{r}\right)\left(-\frac{2h^2}{k^2}\right)‌$$Then just substitute in everything.

0celo7

took a derivative

then used $\sin^2+\cos^2=1$

then some algebra

Slereah

@ACuriousMind But what is the map itself

0celo7

@HDE226868 that's exactly what I had...

Slereah

00:58

I know that generally there's an exponential map between manifolds and tangent spaces

0celo7

and that's wrong!

ACuriousMind

@Slereah I don't understand the question.

HDE 226868

You'll end up with what you got here. So yes, I believe that step.

Slereah

But I only know it for Lie manifolds

0celo7

MAYBE THE BOOK IS WRONG

ACuriousMind

00:59

@Slereah Ah!

0celo7

screw you book

Slereah

Well for Lie manifolds, the exponential map is like

$e^{\tau_i x^i}$

HDE 226868

@0celo7 I'm getting exactly what your first equation said.

Slereah

Wuth $\tau$ the generators of the group

or algebra

whatever

0celo7

@HDE226868 no, the answer of this exercise

HDE 226868

01:00

@0celo7 Well, what are you supposed to find? I don't think you ever told me that.

You just asked me if the equations made sense.

They do.

0celo7

@HDE226868 given $e=1/2$ and $\vec v_0\bot \vec r$, find $v_0$

Slereah

Anything you can do I can do better.avi

►

ACuriousMind

@Slereah Ah, well, it is the "exponential", because for a unit vector $v$, you have $\partial_t \exp_x(vt)\lvert_{t=0} = v$, which is what the exponential generally has, too, and in the case of Lie groups, it is indeed the actual power series

Slereah

I see

ACuriousMind

But, in general, you can't write the map any different than this "solve for the geodesic, go to the end-point"

Slereah

01:02

Why not

Can't you use like

The generators of the Poincaré group

ACuriousMind

There is no formal expression that magically turns a tangent vector into a point on the manifold

Slereah

Or something

I dunno

ACuriousMind

For Lie groups, both the group and the tangent space are matrices of the same dimensionality, so formal expressions like the power series for the exponential have a chance to turn a tangent object into a point

HDE 226868

@0celo7 Ah, so you need $\theta$, because $\vec{v}_{0_{\bot}}=\vec{v}_0\sin\theta$.

ACuriousMind

But for a general manifold, you don't have such a thing, the elements of the manifold are just...points, there is no further structure

0celo7

01:03

@HDE226868 that's just 90 degrees

Slereah

Too bad

0celo7

because I know what direction it is pointing in

@ACuriousMind yes there is

ACuriousMind

@0celo7 Enlighten me, then!

0celo7

solve the geodesic equation in closed form

easy

ACuriousMind

...

0celo7

01:06

I don't see what the issue is.

@ACuriousMind will you please explain what the "..." means

Is that wrong?

ACuriousMind

@0celo7 I said this:

4 mins ago, by ACuriousMind

But, in general, you can't write the map any different than this "solve for the geodesic, go to the end-point"

0celo7

Or just stupid?

ACuriousMind

And you tell me "no, you're wrong", and then say exactly the same thing

0celo7

@ACuriousMind :D

I didn't see that

I sory

Slereah

0

Q: Daniel Thomas Sank - a psychopatic monster that moderates physics.stackexchange.com

If you see this message save it immediately because Daniel Thomas Sank will delete the topic soon. Do not be discouraged by the erratic comments of the users: Gert, dmckee, ACuriousMind. The real name of the man hiding behind these sock-puppets is Daniel Thomas Sank. He is a known psychopathic ...

classical-mechanics

All true

ACuriousMind

01:09

Seriously, what is it with that guy?

HDE 226868

Who the hell upvoted that???

Slereah

@ACuriousMind : Yes, what is it, DANIEL SANK

Dun dun duuuun

0celo7

GONE

WOW

that was quick

ACuriousMind

Automatic deletion by rude/spam flags.

The system works.

0celo7

I wonder who that Daniel Sank *sshole is, anyway

Seems to be a real tyrant

@HDE226868 I would be forever grateful if you were to solve that problem

and tell me what you got

so I can confirm my sanity

I asked by study buddy from the class and she didn't get it either

but she doesn't know what she got

so your answer would be appreciated

HDE 226868

01:14

@0celo7 Okay, so $h_0=v_{0_{\bot}}r$, which you know. If you knew $\mu$, then you could solve for $v$, by the equations for specific orbital energy.

Slereah

Help

HDE 226868

Can we make any assumptions about $\mu$ whatsoever?

ACuriousMind

@Slereah No, you brought this on yourself :D

Slereah

Shouldn't be too hard, I suppose

i just have to prove it is < 0

0celo7

@HDE226868 Uh I'll give you the full problem statement.

HDE 226868

01:16

$$v=\sqrt{2 \left( -\frac{1}{2} \frac{\mu^2}{v_{\bot}^2r^2} (1-e^2)+\frac{\mu}{r} \right) }$$

0celo7

Slereah

Hm

0celo7

Note they're dumb mathematicians and mixed up pounds and slugs

Slereah

How do you even deal with composition of trig functions

ACuriousMind

@Slereah Reluctantly.

0celo7

01:17

I think they're fun!

Pain is fun!

Slereah

I guess my best bet is going to exponentials

HDE 226868

@0celo7 You didn't mention that the particle's mass is given, nor that $r$ and $F$ are given!

0celo7

I thought that was implied

Slereah

I guess for a start I should plot it, see if it's actually correct :p

HDE 226868

I don't see anything given for $v_{\bot}$, though.

0celo7

01:20

You just know the direction

Slereah

Wait a moment

It's not at all!

0celo7

@HDE226868 did I mention it's number 3

HDE 226868

@0celo7 So that should be enough, right? You can just substitute in the expression for $h$ as a function of $r$, then rearrange and solve.

@0celo7 I figured that one out.

0celo7

@HDE226868 I know it's enough.

I also got an answer.

The issue is that my answer is wrong

HDE 226868

What answer did you get?

And what is the correct answer?

0celo7

01:23

I'm at Chick Fil A

Will tell you when I get home

Slereah

How do you make a non time orientable spacetime

HDE 226868

See, I was gonna vote for Subway. . . Although I have teammates who swear by Chick-Fil-A.

Slereah

Literally no book that mentions them offer an example

the closest that does only gives vectors and their product

Not the metric itself

hm

what to do

I guess take those vectors as basis

Then switch the basis to something more reasonable

HDE 226868

@0celo7 Okay, if I help you solve this, will you answer my question about the use of the $\text{co}$ notation in sets?

dmckee

I would find it very easy to think that the people who ask homework questions here have really bad teachers because they so often frame things in ways that are wrong before they even start.

0celo7

01:27

@HDE226868 you're implying I know what that is

Slereah

Let's see, that would be... $ds^2 = -2 dX_1 dX2$

dmckee

Except that my student do the same thing, even after I have tried to convince them that correct structure will lead to understanding.

Alas, they only see the immediate bother, and can't seem to imagine what the beneficial end result would be.

HDE 226868

2 days ago, by HDE 226868

I came across the equation$$\partial f(x)=\text{co}\left\{ \lim_{i\to+\infty}\mathrm{d}f(x_i)|x_i\mapsto x,x_i\notin S\cup\Omega_f \right\}$$Does anyone have an idea what the $\text{co}$ means?

Slereah

And $X_1 = \cos(x) \partial_x + \sin(x) \partial_y$, $X_2 = -\sin(x) \partial_x + \cos(x) \partial_y$

0celo7

@HDE226868 book: 5.48 ft/sec, me: 2.24 or 3.87 ft/sec

@HDE226868 yeah, I dont know what that is

we can think about it

ACuriousMind

01:29

@Slereah Googling "not time orientable spacetime" gives me a hit in HE where they say "The simplest such identification [of deSitter] is that of antipodal points on the hyperboloid. The resulting space is not time orientable;[...]" so your assertion that no book offers an example is false :P

Slereah

Hm

I should try that I guess

0celo7

I like how my notation is standard here

ACuriousMind

@HDE226868 Where the hell did you find that?

Slereah

What page of HE is it btw

0celo7

is that like the anlysis def of a partial derivative?

ACuriousMind

01:31

@Slereah Google books is buggy and doesn't show the page numbers. Can you follow this link?

If yes, bottom of the page.

Slereah

I'll find it with the diagrams I guess :p

HDE 226868

@ACuriousMind That would be here. Page 8. The same paper that had the statement about the affine Grassmannian of line on $\mathbb{R}^2$.

@0celo7 Aw, they couldn't have gone with SI units?

0celo7

@HDE226868 no

Slereah

(it is page 130)

HDE 226868

@0celo7 Give me ten minutes or so. I'll solve in SI units and then do the conversion.

0celo7

01:35

@HDE226868 why do you need to convert anything

you don't need G

read the problem again

HDE 226868

@0celo7 You need $\mu$, right? For that, you need $G$.

ACuriousMind

@HDE226868 I'm 95% certain that $\mathrm{co}$ denotes the convex hull.

0celo7

@HDE226868 you have the force

it's 100/r^2

HDE 226868

@0celo7 Yeah, but . . . oh, I did it the longer way. Never mind.

@0celo7 Okay, I'm getting 9.36 feet per second, which is clearly wrong.

Oh, wait, did something wrong.

Slereah

Apparently there's a whole paper on it

"Relativistic space forms" by Calabi

you know, I wonder when GR people started going all topological

This 61 paper starts with the usual spiel

"A Lorentz manifold $M^n$ is a differentiable n-manifold bla bla bla"

HDE 226868

01:45

Okay, now I'm getting ~53 feet per second.

0celo7

@HDE226868 lol

order of magnitude difference

Slereah

"A Lorentz manifold Mn is called isochro- nous (time sense conserved), isochirous (reversing time whenever orientation is reversed), orientable, or proper (isochronous and orientable, and hence isochirous), in case the holonomy group belongs to the correspond- ing subgroup of the Lorentz group."

No it's not

You liar

0celo7

what the heck are they talking about

Slereah

Who knows

"DEFINITION. A relativistic space form is a complete Lorentz manifold of constant curvature."

It must be so embarrassing to make up a definition and nobody uses it

HDE 226868

Ooh, now I'm getting 9.92. I'm just completely messing up plugging things in.

0celo7

01:49

@HDE226868 lol

@ACuriousMind what is that

ACuriousMind

@0celo7 The smallest convex set containing the set.

Slereah

"Non-isochronous relativistic spherical space forms"

This is what the nomenclature looked like before Hawking Ellis

People made up whatever they wanted!

0celo7

@ACuriousMind what is a convex set

ACuriousMind

@0celo7 A set that is convex

0celo7

@ACuriousMind I have a c word for you...

Slereah

01:51

Don't you need a topology to define convexness

0celo7

wtf is convex

ACuriousMind

@0celo7 cuddly?

0celo7

@ACuriousMind you're a skeli

Slereah

a skellington?

0celo7

not cuddly

Bohnenstange

HDE 226868

01:52

Okay, so$$\frac{v^2}{2}-\frac{\mu}{r}=-\frac{\mu}{2a}$$But $a=r$, so$$\frac{v^2}{2}=\frac{\mu}{2r}$$so$$v=\sqrt{\frac{\mu}{r}}=17.46\text{ feet/second}$$

0celo7

when did you use $e$

Slereah

I think "Isochronous" is supposed to be time orientable, "isochirous" is space orientable and "orientable" is spacetime orientable

0celo7

also what is that first equation

HDE 226868

Oh, wait. It's not true that $a=r$. Scratch that.

0celo7

@HDE226868 it is not

$r=a(1\pm e)$

HDE 226868

01:54

So$$v=\sqrt{2\left(\frac{\mu}{a(1\pm e)}-\frac{\mu}{2a}\right)}$$

But we know $r$, not $a$. That means that$$a=\frac{r}{1\pm e}$$

0celo7

what is that first equation you keep using

HDE 226868

Let's say it's $+e$. Then we have$$v=\sqrt{2\left(\frac{\mu}{r}-\frac{\mu(1+e)}{2r}\right)}$$

@0celo7 Hold on. I'll get back to that.

Okay, now I have 8.73.

0celo7

you've had five answers :D

HDE 226868

Now 12.35 (that wasn't 8.73) or ~21.

0celo7

lol

HDE 226868

01:59

Specific orbital energy

In the gravitational two-body problem, the specific orbital energy (or vis-viva energy) of two orbiting bodies is the constant sum of their mutual potential energy () and their total kinetic energy (), divided by the reduced mass. According to the orbital energy conservation equation (also referred to as vis-viva equation), it does not vary with time: where is the relative orbital speed; is the orbital distance between the bodies; is the sum of the standard gravitational parameters of the bodies; is the specific relative angular momentum in the sense of relative angular momentum divided by...

0celo7

@ACuriousMind is just shaking his head, thinking "why would anyone use numbers"

Slereah

Hm

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