The h Bar

ACuriousMind

14:00

@0celo7 Ah, not if you use the Wiki idea of it only being in Euclidean space, okay

Still, it's not a metric, it's a map on tangent spaces that induces a metric

0celo7

@ACuriousMind I can read!

You don't get to do that

Only my profs do :P

@ACuriousMind What's the difference?

Secret

so the minkowski metric is not actually a metric but a FFF because it is indefinite (has signature -+++) ?

ACuriousMind

@0celo7 Well, how is a map on tangent spaces supposed to be a metric for the manifold?! (This is well-known, but was the math student's point)

0celo7

It's a metric in the physics sense.

And a metric in the sense of e.g. Dr. Freire, who did a postdoc at a gravitational math place.

@ACuriousMind Integrate!

ACuriousMind

@Secret The term "metric" is simply overloaded to mean two different things. No point in saying "That's not a metric".

0celo7

14:05

Maybe!

@ACuriousMind Do you not just integrate the inner product of tangent vectors?

Or am I being silly

ACuriousMind

@0celo7 Not finished yet. This particular guy wanted to have a metric on the manifold, which is by definition a map $M\times M \to \mathbb{R}$.

0celo7

beat me then

ACuriousMind

Well, what would your "integrating" have given you?

Slereah

@ACuriousMind : Well how do you call the symmetric bilinear form $TM \times TM \rightarrow \mathbb{R}$

Huh???

Mister smarty pant

ACuriousMind

@Slereah 1. I'm playing advocatus diaboli here, I was not the one saying it's not a metric. 2. Yes, that is a metric, but not on the manifold

Slereah

14:10

But isn't the tensor bundle ALSO A MANIFOLD

0celo7

@ACuriousMind The length of a curve going between two points

You'd probably take the inf of that integral.

ACuriousMind

Exactly

Slereah

@0celo7 : but it is not positive definite!!!

It is a pseudo-metric at best

ACuriousMind

@Slereah Ah - but on the bundle - instead of the individual spaces - it's not a metric since it is degenerate - it sends the zero vector at $x$ and the zero vector at $y$ both to zero. also, it's not defined everywhere, you can't get the "distance" between two vectors at different points.

Slereah

@0celo7 : Also wait

What if there is no path between two points

What if the manifold isn't connected :O

0celo7

14:16

They're connected by definition.

Slereah

Hm

I should make a SE thread for the Helmoltz equation of the torus

0celo7

@ACuriousMind What?

I'm still not sure what you want from me.

ACuriousMind

@0celo7 Huh? I agreed.

You take the infimum over all paths connecting two points to get the distance between them.

0celo7

Yes, so?

I knew that

So why were you making a fuss

ACuriousMind

I did never say you didn't know that. You were the one who wanted to know why someone I talked to thought one shouldn't call the fundamental form a metric.

So I explained

0celo7

14:25

ok

I've been very suspicious of you lately

Must be my spidey sense

ACuriousMind

@0celo7 No, you've always been paranoid that I'm insulting you/thought you were dumb/whatever. Not a new phenomenon :P

0celo7

@ACuriousMind hmm

what is your ulterior motive here

Secret

https://www.youtube.com/watch?v=IOcrHOc23N4

I have a Newton age question

While I understand how the maths told us the correlis force and the centripetal force are the consequence of the rotating frame thus not a real force, I still don't quite understand why relative linear motion does not result in fictious force while rotation does , given that rotation is basically an infintesimal number of linear motion arranged into a circle?

Or perhaps another way to ask is that, given that we can create rotation by doing linear motions in a certain way, why it makes such a big difference that rota…

sorry, I posted the wrong link, it shoudl be corrected now

Slereah

A force is defined as $\frac{dp}{dt}$

A boost will be $p \rightarrow p + a$

0celo7

Newton

Slereah

14:37

And $\frac{da}{dt} = 0$

0celo7

Newton was woo

Slereah

Obviously the speed is proportional to the force

cf Aristotle and the evidence

0celo7

yes, thank you

Slereah

I remember a nice science history SE post about the approximation in which that is true

0celo7

freshman lab :p

I have no reason to think F=ma is true

maybe within 35%

Slereah

14:42

5

A: Why did Aristotle make mistakes in his laws of motion?

Air or more generally medium resistance was not yet treated as a separate effect in Aristotle's time. Nor was there a clear idea of motion in a vacuum, in fact most ancient Greek philosophers, including Aristotle, did not believe that vacuum exists. So he had to explain phenomena as they are obse...

0celo7

@ACuriousMind You said "never ask that"

ACuriousMind

@0celo7 ?

Never ask what?

0celo7

the thing you deleted

ACuriousMind

What thing I deleted?

0celo7

-.-

and you wonder why I'm suspicious

ACuriousMind

14:45

Ah, you mean the comic

Yes, it was "Never ask that" with a link to the PhD comic Chris posted

Related to you, well, asking Chris about his thesis

Slereah

I am sad that abstruse goose stopped updating

Secret

@Slereah So you mean the linear analogue of fictious force is the "suddenly felt being pushed backwards due to inertia" when you are in a frame that is linearly accelerating?

0celo7

@ACuriousMind I also asked you about yours :P

you have cleverly refrained from answering

Slereah

Yes

In classical mechanics, there are basically 4 fictitious forces

ACuriousMind

@Slereah Me too :(

Slereah

14:47

Linear acceleration, the centrifugal force, the coriolis force and the Euler force

0celo7

@Slereah love, hope, greed, centrifugal

@Slereah linear acceleration?

Slereah

John, Paul, Ringo, Centrifugal

0celo7

Huh?

what's the Euler force

Slereah

@0celo7 : $x \rightarrow x + at^2$

Or whatever

0celo7

how is that not Galileo

Slereah

14:49

because of the square

0celo7

then where is the 1/2

Slereah

It's an integration constant

You can sweep it inside $a$

0celo7

that's evil :o

Slereah

Well renormalization does it with infinities

So I can do it with a half

0celo7

that's evil too!

@Slereah that's like saying he can murder ppl so I should be able to steal >:|

Slereah

14:54

Who murders people

0celo7

@ACuriousMind I'm sure you'll be pleased that I'm directing boring questions at profs and TAs instead of you

y'know, using the things I'm paying for

@ACuriousMind you can thank me whenever :)

Secret

0celo7

christ...

Secret

so they are basically the consequence of inertia?

the 4 fictious forces

?

0celo7

wtf is happening there

Slereah

15:06

That square sure can boogy

Fictitious forces are the result of the derivative of the momentum not being invariant under such coordinate changes

0celo7

fictitious forces are due to dancing squares

Secret

so if we use the above results and the results of relativity, is the relativistic analogue of these manifest itself by having energy also becoming a component of momentum hence will slightly differ for each accelerating and rotating reference frame?

0celo7

woo!

Secret

I only recall the equivalence principle which equates gravity with a (linear?) accelerated frame

and that rotating massive objects caused frame dragging (a term which I don't quite understand in detail)

But is there more?

0celo7

Secret

15:13

what are the relativitic analogue of the coelis force and euler force?

if there are any

Slereah

In relativity fictitious forces are due to the change of the connection when changing coordinates

In flat space the geodesic equation is $\ddot x = 0$

0celo7

^^

See Weinberg's book.

Slereah

If you change coordinates, it will be $\ddot x + \Gamma \dot x ^2 = 0$

0celo7

If you force a coordinate change in $\ddot x$ you get the geodesic eq.

Slereah

That is the fictitious force

Secret

15:15

The $\Gamma \dot{x}$ term?

Slereah

yes

0celo7

@Slereah pls it's $\ddot x+\dot x^T\Gamma\dot x=0$

Slereah

I'll write as lazy as I want

0celo7

use proper notation >:|

Slereah

$a + \Gamma v^2 = 0$

0celo7

15:16

ahh

you monster

Secret

@0celo7 the way the fiticious force is written reminds of quadratic forms. Is it really a quadratic form (or some sort of similarity relation (c.f. $x^T A x$ in matrices) or is it just happens to look like that?

0celo7

> similarity relation

no, $x$ is a vector, not a matrix

Secret

I see

0celo7

similarity is three matrices

Slereah

you're three matrices

0celo7

15:18

@Secret I wish I had a legal copy of Weinberg in this state

I'd find the derivation

basically, just change coordinates in $\ddot x$

so $x\mapsto y(x)$

then you find $\ddot y+\dot y^T\Gamma \dot y=0$

there's some boring details

Secret

ok

0celo7

and $\Gamma\sim \partial^2 y/\partial x\partial x$

something like that

Secret

I see

0celo7

actually I do have a (really) legal copy of Weinberg on my iPad

which is in my dorm

hmm, I think zee does it, but omitting the technical details...one moment

ok so it's the geodesic equation in $x$

you can show that $\Gamma$ here is the LC connection of flat space in an accelerated frame

Secret

Slereah, I think i will prefer 0celo7's i.e. the more mathematically rigorous way, because my exploratory nature to maths means I might want to explore something more general, i.e. some kind of non commutative spacetimes (if such term makes sense, or at least I think you guys will know what I am trying to say) where you cannot simply exchange mixed derivatives, hence the $\dot{x}$

@0celo7 Reading the excerpt now

0celo7

15:25

> non commutative spacetimes

What?

ohhh in Zee's thing $y$ are unaccelerated?

yes, $y$ are unaccelerated, $x$ are the general ones

obtaining a legal copy of Weinberg.

Secret

I don't know, I am wondering if there's something more general such that the terms in the geodesic equation are not commutative (hence once must use $\dot{x}^T \Gamma \dot{x}$ instead of $\Gamma \dot{x}^2$

But perhaps it might not be physically relevant

0celo7

dude

$\Gamma$ is a matrix

Slereah

$\Gamma \dot x ^2$ isn't the equation

I just wrote it because I am lazy

Secret

o i see...

0celo7

this is standard linear algebra notation

the proper way to write $\Gamma_{ij}\dot x^i \dot x^j$ as a matrix is $\dot x^T\Gamma \dot x$

they didn't cover this in your GR course?

now that I think about it, it's not in Wald, Straumann or HE

Slereah

15:33

Hm

0celo7

Hm?

Slereah

I wonder if you can write QFT in matrix notation

So terrible

0celo7

Yes!

You can!

Weinberg does it, partially.

Slereah

gross

Secret

Well, our GR course taught to use the gammas in index notation, and then compute each component of the gammas to form a matrix, which is then used to compute the riemann tensor

But we never illustrate it in the linear algebra notation

0celo7

15:34

@Secret not the matrix notation

Slereah

It is a vector of matrices in the matrix notation

0celo7

the fact that accelerated frames in flat space look curved

@Slereah indeed

and the Riemann tensor is a matrix of matrices :D

Secret

Umm, we cover the bending of light in a free falling rocket and gravitational redshift, does that count?

Slereah

It should be $\ddot {\vec x} + \vec x^T \vec \Gamma \vec x$

0celo7

I say again: Einstein in Matrix Form is the greatest mathematical achievement of the age

amazon.com/Einstein-Matrix-Form-Derivation-Relativity/dp/…

Ah, good. I now have a legal copy of Weinberg on hand.

Slereah

15:38

Of course you like that book

You're an engineer

Secret

excerpt of the lecture ntoes used in our course

0celo7

do you knew that

I think

you're just forgetful

Secret

probably, given that at that time, most of the GR stuff is very confusing to me

as that was the first time I took GR

Slereah

THIS WAS A SECRET

NOW EVERYONE WILL KNOW

Secret

so they introduce it by first defining the covarient derivative, which is different from the approach used by weinberg that you show

I think the weinberg is clearer

0celo7

15:40

I showed you Zee

which is the same, really

Weinberg has more boring details

Slereah

Generally speaking it's kind of a GR gauge thing

Secret

is it because it is a coordiante transformation thus this matrix must be invertible?

Slereah

You can define the metric and connection in GR as a way to make things invariant under diffeomorphism and $SO(3,1)$ gauge

0celo7

@Secret yes

Secret

i see

0celo7

15:42

diffs are invertible

I wish ppl stopped calling me an engineer. I'm a student. I'm not an engineer until I pass the PE exam.

Slereah

It's okay

People call me a murderer and I've never been caught

0celo7

I call you a ~~junkie~~ avid drug user

Secret

People said I am a theoretician with the tools of an experimentalist, that's why I get stumped by problems easily

Slereah

Well you're certainly not an artist!

0celo7

REKT

No wonder serious physicists don't visit this chat. We're a bunch of hooligans. ASTRONOMERRRRRRRRR

Slereah

15:47

More like wooligans

0celo7

kek

Slereah

(I mean we believe in woo, not that we are Liverpudlian sheep hooligans)

0celo7

what

Slereah

wool-igan

You can earn a general relativity badge? :O

Hot diggity damn!

0celo7

WAIT A MOMENT

AREN'T YOU AN ENGINEER

Slereah

15:50

No

I am a cat

3

ACuriousMind

@0celo7 You've a habit of using "we" where you mean "I" :P

0celo7

@Slereah Isotropy $\implies$ homogeneity

Confirmed by like 5 people

Slereah

How can someone be a bunch of hooligans

@0celo7 : Yes

(Well isotropy at every point)

0celo7

@Slereah Ofc

ACuriousMind

@Slereah Multiple personalities

0celo7

15:52

@ACuriousMind Not really.

Slereah

I think vice versa also works?

0celo7

@Slereah NO

Slereah

Hm

Oh wait

Godel is homogeneous and not isotropic, right?

0celo7

Indeed.

Slereah

Damn Godel

0celo7

15:54

We star too much, this is the mark of a squash of hooligans

Slereah

I think the monochromatic wave spacetime may also be like that

0celo7

monochromatic wave spacetime?

Slereah

Ah, not quite, no

@0celo7 : Metric generated by a monochromatic EM wave

0celo7

link?

never seen that ne

Slereah

Monochromatic electromagnetic plane wave

In general relativity, the monochromatic electromagnetic plane wave spacetime is the analog of the monochromatic plane waves known from Maxwell's theory. The precise definition of the solution is quite complicated, but very instructive. Any exact solution of the Einstein field equation which models an electromagnetic field must take into account all gravitational effects of the energy of the electromagnetic field itself. If there is no matter and no non-gravitational fields present other than the electromagnetic field, this means that we must simultaneously solve the Einstein field equation and...

It's in MWT

0celo7

15:57

Monday Wednesday Thursday?

Slereah

Malicious Woo Theory

0celo7

haha

MTW is really expensive

I might get it when thesis time comes rolling around

Slereah

It's a big book

0celo7

I think a math thesis on Lorentzian geometry and PDEs would be cool

Slereah

Show that electrons are photons on a moebius strip

Transcript for