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12:00 AM
@Slereah is time orientable $\Leftrightarrow$ nonvanishing timelike vector
or is it only one way
@ACuriousMind yeah, what's up with that
 
All spacetimes have a non-vanishing timelike vector
 
He goes on the main site regularly, but never comes here :/
 
That's lorentzian metric equivalent to nonvanishing timelike vector
But the trick is
All spacetime manifolds are time orientable
Not all metrics are time oriented, though
 
what does that even mean
the metric is what tells you what is timelike
how can the spacetime been time oriented
 
Given a spacetime topology, you can always define a metric on it that is time oriented
 
12:03 AM
what?
where is the proof
 
@Slereah Are metrics which generate the same topology physically equivalent?
 
Theorem 2.4.
For any connected smooth manifold, the following properties are equivalent:
(1) M admits a Lorentz metric.
(2) M admits a time-orientable Lorentz metric.
(3) M admits a non-vanishing vector field X.
(4) Either M is non-compact or its Euler characteristic is 0.
Well metrics don't generate topologies
 
Lorentz metrics don't
 
"(1) ⇒ (2) (The converse is trivial.) The time orientable double covering ( ̃M, ̃g), satisfies (3) and hence (4). So, the latter is satisfied obviously by M."
 
12:06 AM
Wait a moment
 
ah, I misunderstood what you meant by "spacetime topology"
 
How the heck are the open sets in spacetime even defined :o
Preimage of open sets of R^4 under a chart?
 
Well yes
It's a manifold
The spacetime metric isn't the same notion as the metric space that the manifold is
 
I must be stupid or something
 
@Slereah How is the manifold a metric space?
That's only true in the Riemannian case
 
12:08 AM
but there's also the "Alexandrov topology"
 
Eh I dunno
I'm not a math guy
 
Ok I must be really dumb
How are open sets on the manifold defined
D:
 
What you said.
Images of open sets in $R^4$
 
@ACuriousMind is this true
 
Well it's true in 4 dimensions
 
12:11 AM
ah, the charts are homeomorphisms
ok got it
derpy moment
 
Evening guys. A classical harmonic oscillator doesn\t have time translational symmetry does it?
 
It does
 
Of course it does
 
does it...
 
The Hamiltonian doesn't have time dependence
 
12:12 AM
$L = m \dot x^2 + \omega x^2$
No time involved
 
@0celo7 Well, the formal answer is, they aren't "defined" at all: A manifold is by definition a topological space that fulfills some extra conditions. But it is true that the topology is generated by the (pre)images of open sets in $\mathbb{R}^n$ under some chart.
 
should that be $H=$
I get the signs mixed up with the SHO
 
w/e
 
@ACuriousMind Exactly.
Forgive my moment of weakness there.
 
well I agree with you guys but look at Qmechanics answer here .. what am I missing?
 
12:14 AM
...where?
 
11
Q: Do an action and its Euler-Lagrange equations have the same symmetries?

DilatonAssume a certain action $S$ with certain symmetries, from which according to the Lagrangian formalism, the equations of motion (EOM) of the system are the corresponding Euler-Lagrange equations. Can it happen that the equations of motion derived by this procedure have different kinds and/or numb...

 
@ACuriousMind ^
 
example 3
haha
 
He talks about symmetry in the solution, not the dynamic itself
That is a different thing
 
@AngusTheMan That's time-scaling, not time translation.
 
12:15 AM
embarrassed emoji face
:p
cheers :)
 
@AngusTheMan don't worry, it happens ;)
 
flipping through Lee
he wrote a 700 page diff geo book without talking about connections
Like a personal trainer I taught coke how to stretch. - Jay Z
 
How
I mean
Technically you don't need connections for a metric space
Still feels a bit of an omission
 
get it "legally" and don't look
he only does Riem geom for a few pages
enough to introduct geodesics so he can talk about the exponential map
 
You can't do geodesics without a connection >:|
 
12:20 AM
To be fair the actual book is 600 pages, the rest is review and index
@Slereah lol
no shit...
lemma skim
 
@0celo7 Look at it. It is about smooth manifolds not pseudo-Riemannian menifolds.
And a generic smooth manifold just doesn't come with any kind of connection
 
@ACuriousMind Look at it! I own it!
Not gonna read the whole thing.
ah
I'm really dumb
It's the exponential map of Lie algebras, not geodesics
I just assumed embarrassed emoji face
I wonder what the Lie derivative on twistors looks like
vacuum $\to$ scalar $\to$ vector $\to$ tensor $\to$ spinor $\to$ tensor valued spinor $\to$ twistor $\to$ Xor
solve for X
also why isn't it a scalor
 
I still don't know what a twistor is
Also where is the schwingor
We decided that spin 3/2 fields were schwingors
 
I got a book on twistor
but I returned it to the library
 
@Slereah Not raritars?
 
12:26 AM
is anyone else getting results for that?
 
"Twistor theory is unique to 4D Minkowski space and the (2,2) signature, and does not generalize to other dimensions or signatures."
Oh, so it's boring
Oct 1 at 13:43, by 0celo7
Rarita-Schwingors
Oct 1 at 13:44, by Slereah
Schwingors
 
dude it gives me "zero results"
 
mb u r just bad at it
 
@0celo7 It's case-sensitive
 
@ACuriousMind wow
that's pretty stupid
 
12:28 AM
Yep
 
lies
 
Hm, perhaps it lowercases uppercase but does not uppercase lowercase?
 
Hm, no
Ah!
You typed schwingor
But slereah said schwingors
It's the s
 
ha!
 
12:31 AM
Huh?
lol
 
What?
 
1:01 AM
@Slereah I don't need a connection. A geodesic solves $\delta\int \langle v,v\rangle\,\mathrm{d}t=0$
@ACuriousMind why do you hate $\oint$
Even in CFT?
What about all of those loop integrals :o
 
That's not a geodesic >:|
That is just the shortest path!
 
@Slereah Aren't you asking multiple questions in your question
 
Eh
 
Geodesic := shortest path
 
Not if there's torsion >:|
 
1:04 AM
"is this the right action"
"what is the infinite sum formulation"
@Slereah I have yet to read that torsion document you sent me
 
Hm
I recall reading that torsion may have geodesics that are not the shortest path
But I'm not sure
In the geodesic equation IIRC the torsion term vanishes
 
@Slereah it's true
you can calculate it
 
Since it's $T^a_{bc} \dot x^c \dot x^b$
 
well vary that integral
 
And the tensor is antisymmetric in b and c
 
1:06 AM
see what you get
then add the torsion in by hand
and you get a torsion equation
because the connection is different with torsion I think
then you get the shortest path with torsion
the geodesic equation $\nabla_vv=0$ has no torsion in it
right, @ACuriousMind ?
 
It does
It's right in there
$\nabla$
 
no
damn, what the hell is torsion, again?
 
Torsion is the antisymmetric part of the connection
 
ok ok
torsion doesn't appear explicitly in the geodesic equation
actually shouldn't it fall out completely?
 
Yeah that's what I wonder
Since it only appears in a symmetrical term
 
1:10 AM
because you get $\Gamma^i{}_{jk}v^jv^k$
 
"The antisymmetric part of the Christoffel symbols will not contribute here in any case, but the torsion will change the symmetric part (and thus invalidate the conclusion that extremal length paths are geodesics) if and only if $T_{cab} \neq T_{[cab]}$."
Hm
 
torsion changes the symmetric part o.o
where are you reading this
 
@ACuriousMind why isn't the plural of "album" "alba" and not "albums"
which idiot put that minus there
he should be shot
 
Geodesic for torsion might be a bit more delicate to derive I suppose
I dunno
Hm
$u^a \nabla_a u^b = u^a (\partial _a u^b + {\Gamma^a}_{bc} u^b) = u^a \partial _a u^b + {\Gamma^a}_{bc} u^b u^a = u^a \partial _a u^b + {\Gamma^a}_{\{bc\}} u^b u^a$
 
1:19 AM
f that
 
No it should still be the symmetrical part
 
oh, that's not bad
yes it's the symmetric part
 
Wait
No
I fucked up my indices
 
I've never seen braces on indices
yeah you did
 
a should be on the bottom
(heheheh bottom)
 
1:20 AM
I thought you were some kind of GR guy
can't even indices
 
Braces indicate symmetrization
 
ok so what is the symmetric part of the connection when there's torsion
 
it's the regular old connection
 
no
 
Maybe I dunno
 
1:22 AM
there's something we don't understand going on
 
There's a few different tensors
Let's check
 
> The other is the source of the name
(as I understand it): a “shortest possible path” between any two of its points
(or more generally, an “extremal length path” between them). In the presence
of torsion, these two concepts need no longer be equivalent
that's what we need
 
Oh wait
I might have confused the torsion tensor with the contortion tensor
Maybe
I don't know
 
contorsion tensor?
 
Contorsion tensor is the difference between the Levi Civitta tensor and the general tensor
 
1:24 AM
what?
 
The contorsion tensor in differential geometry expresses the difference between a metric-compatible affine connection with Christoffel symbol and the unique torsion-free Levi-Civita connection for the same metric. The contortion tensor is defined in terms of the torsion tensor as where the indices are being raised and lowered with respect to the metric: . The reason for the non-obvious sum in the definition is that the contortion tensor, being the difference between two metric-compatible Christoffel symbols, must be antisymmetric in the last two indices, whilst the torsion tensor itself is...
 
ah
that last bit is the contorsion tensor
 
yeah the contorsion tensor is a bit different from the torsion tensor
Not the same symmetries
 
ah that's in BBS
you need torsion in string theory
torsion is BS
no wonder Einstein said fuck it
wait why is there no torsion in GR
I always forget it
 
Because nobody cares about torsion in classical mechanics
No classical field carries torsion
 
1:28 AM
why
I still haven't read that torsion thing you gave me
no clue where I put it
 
The torsion equation is roughly $T = \Theta$
With $\Theta$ being the spin tensor
It is 0 for scalar fields
I think it's also 0 for EM fields
Let me check
 
spin tensor
dude gimme a link to that shiz niz
arXiv paper
 
The spin tensor is the variation of the lagrangian with respect to the torsion tensor
(or contorsion, whatever)
 
o.o
 
Scalar fields and EM fields don't use the connection at all
 
1:31 AM
just give me the thing
 
So their variation is 0.
 
thanks
sigh
 
So there really isn't much point of doing torsion for GR since basically you'll never use it
 
sometimes GR sucks balls
speaking of balls
Big dog; big nuts. - Lamar Davis
 
"Only the symmetric part Γi(kl) of the connection enters the autoparallel equation [...] geodesics do not depend on torsion"
>:\
 
1:46 AM
lelelele
 
Which one is it!
Maybe when they say "In the presence of torsion, these two concepts need no longer be equivalent", they don't actually mean that it's not
 
OK, then. Someone who doesn't understand first semester basics about the centripetal force has been bothering us with endless questions about his efforts to overthrow the current understanding of large scale cosmology. Nice.
 
He has Einstein and the evidence backing him up
Well 3 AM
Best go to bed
 
night
@dmckee Not me!
 
@0celo7 No, someone with much less self awareness than you. Shocking, I know.
 
2:02 AM
@dmckee damn, way to rek me
what did I do to you :'(
@dmckee who is this person
 
That's for you to figure out.
 
@dmckee so I'm not self aware
in what sense
 
@0celo7 In the sense that your buddies, even casual ones, pass casual insults just for the hell of it. I don't know you well enough to hazard a guess about your inner self.
 
> In the sense that your buddies, even casual ones, pass casual insults just for the hell of it.
 
If you want an appology, you can have it. But then I'll think you thin skinned. (And so the "society" of lads is propagated.)
 
2:07 AM
I don't know what that means.
I don't want an apology, I want to know why you think it.
 
Why ask why?
 
So I can "look inward" and decide if it's something I want to correct.
@dmckee Donald?
 
What does "self awareness" mean to you?
 
Not sure, that's why I'm asking.
 
Are you asking for a definition?
 
2:16 AM
Sure + why I lack it
 
Each person defines it differently for themselves, because of the "self" aspect.
 
ok, so @dmckee has to define what he means by it
 
No, you have to decide what you think of yourself.
That^ is self awareness.
:-)
 
can @dmckee just explain?
 
He has not interrupted, has he?
 
2:23 AM
@0celo7 I did explain. It was a pointless dig of no importance. Basic ladismo at work.
 
>ladismo
What?
 
In al seriousness, I'll lay off that sort of nonsense if you want, but I had interpreted some of your actions here as being of the same kind.
 
"same kind"
I have no idea what you're talking about
 
Take "lad" in the slightly pejorative sense the British use and mash the ending of machismo on it...
@0celo7 In that case I might have been right. But that was purely by chance. Think nothing of it.
 
32 mins ago, by dmckee
OK, then. Someone who doesn't understand first semester basics about the centripetal force has been bothering us with endless questions about his efforts to overthrow the current understanding of large scale cosmology. Nice.
The reason I said "not me" is because I thought the first few words were about me or something
then I read on and figured out it wasn't
 
2:25 AM
@0celo7 Donald has been trying to overthrow big bang cosmology for the last few months. But it seems he would have struggled on the test I gave last week.
Of course, my class struggles, too. So he'd have been right at home.
 
But I'm certainly not challenging any theory!
So I don't know what "I had interpreted some of your actions here as being of the same kind" means
What actions, specifically?
 
Several times when you engaged in seemingly pointless rounds of minor insults with other users. Or should I have taken them seriously and move in with my moderator nightstick?
 
They must have been minor enough for me to not remember them at all.
You mind saying which users?
And how does that make me lack "self awareness", whatever that means?
 
3:00 AM
Sorry couldn't get past the laugh.
 
@skullpetrol What?
 
The laugh was too annoying.
 
I didn't notice it.
 
You are a troll.
Put that^ on your "self awareness" list :P
 
I wish @dmckee responded
 
 
6 hours later…
9:07 AM
anybody here???
This is my problem: The bungee jumper jumps to the water from the bridge that is 24 metres high... First 12 metres of the fall the cord is not extended, it is only the free fall... The next 12 metres the cord is extended and the total fall is 24 metres, so the jumper falls in a way that he almost touches the water... The weight of the jumper is 60 kg. What is the maximum speed that the jumper attains during the fall?
 
9:31 AM
The maxim speed is reached after falling more than 12 metres because the speed is growing as long as the downward force of gravity is greater than than the upward force of extension of the spring. F = m*a , m*g - k*Δx = m*a . The speed stops growing when a=0, when those two forces are equal. At this moment the jumper attains the maximum speed.

The thing is that I don't know how to calculate this maximum speed.
 
 
3 hours later…
12:48 PM
@bolbteppa : I know plenty of maths. IMHO what's important is to understand what the terms really mean. If you've ever seen any of my stuff you can see how I've looked hard at things like E and m and c and C. As for rotation and reflection, please provide a link and I'll comment.
 
1:25 PM
@JANORTS : see this for something that might help.
 
I see 0celo on a downward spiral :\ linear algebra is a tough subject, would drive anybody down that road, even moreso if they had been studying string theory 2 months before
 
His problem is that he is studying math
He should study physics!
 
linear algebra is ok when compared to real analysis
 
@JohnDuffield if you know "plenty of maths" explain to me right here, without any links why can you never divide by zero?
 
1:36 PM
Well I did a math course so that the time would come when I could learn physics at the best level without math holding me back, so it's a good idea if you are willing to wait 2 years
 
RpfnoR 7 (this mathematical structure has some binary operator, and it is its own inverse, it is closed under the operation and)
but it lacks an identity)

I tend to rely on maths to gain insights on the physics problem
 
But I did not go off studying string theory books without real analysis or linear algebra, which is like the biggest sin a student could commit in my eyes
@Secret I find it's very helpful at times, but I'm always conscious of that Feynman warning where he says math might ruin everything haha
 
Yeah, when you are too relied on maths, you have to be careful that some solutions to the equations can corespond to something unphysical,

Also, ASSUMPTIONS, (ugh I suck at these!)
 
@JohnDuffield your profile here science20.com/profile/john_duffield (not working for me at the moment for some reason) says you don't know much math, even the stand up physicist guy was admonishing you on that site, personally I would take it as a sign I may be doing something wrong hehe
 
Maybe the fact that all physicists do would be a warning sign
 
1:40 PM
typo: Red should be +-odd number
 
@JohnDuffield But you say you've tried hard to understand energy $E$, do you know how to write down the energy in special relativity from a Lagrangian?
I'm not trying to catch you out, merely trying to convince you to study math more thoroughly, and better physics books than the ones you've read
"Something of an analyst and logician, well read on physics, but maths relatively weak and technically an amateur."
 
From the above diagram I am suspecting the set {odd, even,+,-} forms some kind of rng
 
I imagine now Duffield frenetically looking up on wikipedia the formula to copypaste it
 
although odds are low
He will probably just go "The math isn't important!", as usual
 
1:44 PM
@Secret I don't get the diagram, looks like you're just permuting $1$ and $2$ and giving a sign to the permutation as odd or even?
 
I am trying to illustrate the relationship when you add an odd number to an odd number, and even number to an even number and an odd umber to an even number
It's just something that pops up when I am procratinating because of being frustrated too much by some matlab code
we know that
odd+odd=even
even + even=even
odd + even=odd
I am just jotting these out some some kind of map or graph or something
because I am curious on its structure
 
Maybe you can write it out with delta functions as something like $(2n + \delta^i_j ) + (2m + \delta ^k_l)$ haha
So you want the structure of the quotient group of the additive integer group mod 2
 
probably, I am just curious on what operations lead to where

Let's use a london subway map as an analogy, what I am doing above is trying to construct something like a guide so that I can see where and how I get to places
Becuase I have a tendency to like to make "roadmaps" of mathematical objects, and then start doing some extra things to it
 
Nooooo
@Secret : This is your mistake :
 
uh...
 
1:54 PM
haha
$E = RF$
 
I didn't say I have a clever analogy, I am just saying how I think about it.
I don't force people to use them, and I am actually pretty dumb


or am I still making That Mistake?
I could have jsut say I want to make a map for mathemtical objects for myself to explore it, but I cannot say that because maths peopel here will think I am talkign about maps as defined in mathematics

Sometimes I just don't have enough volcabulary to express what I am trying to say
hence all that confusing analogy speak
 
One of my personal hobby is to watch a maths object breaks down when you start messing with it and see what pops out
 

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