The h Bar

Phase

00:55

@BalarkaSen @0celo7 @ooolb sadness

0celo7

01:07

@Phase sad indeed

ooolb

02:24

f

is that a math seminar?

like fucking what

Bernardo Meurer

03:29

@Phase Damn

What even

@JohnRennie Audio setup working :)

But the remote volume control on the amp has died. I can hear the motor relay switching, but the motor no longer lives

Rest in peace friend

John Rennie

06:23

@BernardoMeurer Cool! You've got your hi-fi stuff then :-)

Bernardo Meurer

06:38

@JohnRennie Ye :)

Just need furniture now

Also, I'm in love with SQLITE

SQL is such a neat cookie

John Rennie

Meh. You have your laptop and a hifi. Everything else is optional :-)

Bernardo Meurer

Got a new PSU for the laptop today, the old one was murdered

a guy tripped on it in the airport, broke the cable

John Rennie

Did you get a Dell PSU or an aftermarket one?

Bernardo Meurer

Dell, from Staples

The new one looks great, is 90W, and is small

And works for the new Latitudes and XPS' as well

@JohnRennie Did you see the new XPS 13?

I've been thinking of going down to 13 inches...

Balarka Sen

@Phase Jesus

John Rennie

06:44

@BernardoMeurer 10nm? :-)

Bernardo Meurer

Still waiting!

:P

I'll wait until it gets here or this laptop dies

John Rennie

@BalarkaSen that's the Numberphile video on the subject isn't it?

Numberphile are generally pretty good.

Balarka Sen

Yes but the -1/12 video was really bad

It's basically a bunch of physicists saying "hey we could do this false computation and the number comes up as -1/12, and this is what we get in string theory too!!11!"

John Rennie

@BalarkaSen I have watched that one but I confess I remember nothing about it

Bernardo Meurer

@BalarkaSen Tbh I like what they did, even though it's not true, it's a cool party trick

You can get some good things by betting you can prove the sum of all positive integers is negative

skullpatrol

06:48

Indeed.

Balarka Sen

@Bernardo It gets less cool when you know there is this theorem: en.wikipedia.org/wiki/Riemann_series_theorem

So technically I can give you a "trick" computation of 1 + 2 + 3 + 4 + ... which spits out any real number whatsoever

At least I think so

skullpatrol

Yep.

Bernardo Meurer

@BalarkaSen I know about that, we saw it in Chub n' Tuck Analysis

But it's still cool for people who don't know what a riemann is :)

Look, I don't care for correctness here, I'm talking about taking advantage of people

Balarka Sen

lol

skullpatrol

;-)

John Rennie

07:00

@BalarkaSen perhaps this is the mathematician's equivalent of the warm feeling physicists get when they see a pop sci programme depicting the Big Bang as an explosion :-)

Balarka Sen

looool

Bernardo Meurer

Hahahaha

Not sure what the analogous device is for computer scientists

Maybe just banking infrastructure

Anonymous

@BalarkaSen That series isn't conditionally convergent, is it? I'm not sure how the Riemann series theorem applies

Balarka Sen

There should be an analogue of it for divergent serieses

Oh I see, I get it

Look at 1 - 1 + 1 - 1 + 1 - 1 + ...

This is conditionally convergent, right?

You can rearrange this to get 1 - 2 + 3 - 4 + ...

Ah I guess you're right

Anonymous

1, 0, 1, 0, 1, ...

Anonymous

07:14

Yeah, so it isn't conditionally convergent

Balarka Sen

But I think what I want to do is to have a conditionally convergent series, rearrange it to 1 - 2 + 3 - 4 + ...

Which in turn gives a value for 1 + 2 + 3 + 4 + ...

That conditionally convergent series can rearrange to whatever value

So 1 + 2 + 3 + 4 + ... can have whatever value :P

This should be workable

Anonymous

The point is even though you can generate any value with conditionally convergent series according to the theorem you stated, you can't rearrange it to 1-2+3-4...

Anonymous

Since you don't know it's value

Anonymous

1-2+3-4..

Balarka Sen

No you can.

Anonymous

07:16

isn't convergent

Balarka Sen

You can rearrange conditionally convergent series to divergent serieses!!

This is part of the Riemann rearrangement theorem

Look in the wiki page

Secret

(Reading transcript) Somehow reading that philosophical stuff a few hours ago in the transcript and one of 0celo's comment just give me this word in mind: Free range chicken

I guess I must be hungry or something

Anonymous

You can rearrange any conditionally convergent series to 1-2+3-4+...? Or only specific ones (chosen carefully)?

Balarka Sen

Specific ones, carefully chosen.

The point is your "1 - 2 + 3 - 4 + ... isn't convergent" argument doesn't hold because of this

There is no mathematical obstruction to rearranging a conditionally convergent series to a divergent series

As far as I can see

Fixed @skull

Anonymous

So basically you're saying that suppose there exists a certain sequence of $\{1,2,3,4,....\}$ (including positive and negative signs) which is conditionally convergent. Or there may exist a certain conditionally convergent series for which if you re-group the terms you can end up with $1-2+3-4+5-6....$. OK so far.

Anonymous

07:29

Then, similarly if you discover a conditionally convergent sequence which you can re-group to form $1+2+3+4+....$, then that conditionally convergent sequence's value would be the value of $1+2+3+4+...$

Balarka Sen

Yes, but let's try to actually come up with an example. I'm wondering if 1 - 1/2 + 1/3 - 1/4 + ... rearranges to Grandi's series

Anonymous

Now perhaps the natural question is, how do you prove that such a re-groupable conditionally convergent sequence even exists for any such divergent series as 1+2+3+4+....?

Balarka Sen

Like, 1 - 1/2(1 + 1/2 + 1/4 + ...) + ... or something. I'm not sure

I guess it probably doesn't.

Anonymous

I doubt a proof of that exists though... If one is lucky enough they might come up with a conditionally convergent series that can be re-grouped to 1+2+3+4+...

Balarka Sen

A proof in this case is equivalent to coming up with an example...

Anonymous

07:35

@BalarkaSen I'm looking for a general proof which works for any divergent series

Anonymous

This case already has been done I guess:

Anonymous

1 + 2 + 3 + 4 + ⋯

The infinite series whose terms are the natural numbers 1 + 2 + 3 + 4 + ⋯ is a divergent series. The nth partial sum of the series is the triangular number ∑ k = 1 n k = n ( n + 1 ) 2 , {\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},} which increases without...

Anonymous

Have to read it

Anonymous

I have heard of the -1/12 thingy

Anonymous

There might be others

Joshua Lin

07:37

Can't you just take
1 - 1/2 + 1/3 - 1/4 + 1/5 ...
You can rearrange that to get "epsilon close" to 1+2+3+4... (as in each term is epsilon away)
So there is some epsilon big "error sequence", which we can include in the original 1-1/2+1/3-1/4... so all good?

wait no I'm high

Anonymous

@JoshuaLin No, that would diverge as epsilon is finite

Anonymous

*could

Joshua Lin

like the epsilons can get smaller and smaller

Balarka Sen

In this case it doesn't does it

You're looking at $n - 1/n$

That grows as $n \to \infty$

Joshua Lin

like to be formal; for any sequence
a1+a2+a3+a4...
We can rearrange the sequence
1-1/2+1/3-1/4...
Into a sequence
b1+b2+b3+b4...
where
|a_i - b_i| < 2^{-i}
So then we can look at the conditionally convergent sequence
1+(a_1-b_1)-1/2+(a_2-b_2)+1/3+(a_3-b_3)...
Which can be arranged to get
a_1+a_2+a_3+...

nah maybe makes no sense

who knows

Anonymous

07:40

If you subtract a convergent series from a divergent one, you really shouldn't expect a convergent sequence of errors :P

Joshua Lin

its not convergent after rearranging thouhg?

Like we are rearranging it to be epsilon close to the divergent series we want to approximate

Anonymous

Please use MathJax to type

Anonymous

It's a pain to read otherwise

Joshua Lin

Wait maybe I'm not getting the question; are we allowed rearranging?

Anonymous

Yeah, re-ordering of terms is allowed

Anonymous

07:42

But I can't read the wall of text you wrote without MathJax

Anonymous

Just use $$ to enclose the lines

Joshua Lin

We got sequence
$$a_1+a_2+...$$
And have sequence
$$1-\frac{1}{2}+\frac{1}{3}...$$
we can rearrange this to get
$$b_1+b_2+...$$
where
$$|a_i - b_i| < 2^{-i}$$
So that the conditionally convergent sequence
$$1+(a_1-b_1)-\frac{1}{2}+(a_2-b_2)+\frac{1}{3}+(a_3-b_3)-...$$
can be rearranged into
$$a_1+a_2+a_3...$$
why am i typing this

Anonymous

@JoshuaLin The point is $\{a_n\}$ is divergent here.. So you can't do that

Joshua Lin

cant do what?

Anonymous

$|a_i - b_i| < 2^{-i}$

Joshua Lin

07:44

why not

1-1/2+1/3-1/4... is nice enough; seems alright?

Anonymous

Try doing it for $a_i=i$ and $b_i=(-1)^{i+1} \frac{1}{i}$

Joshua Lin

1 + (-1/2+1/3+1/5+1/7+....+1/big number) + (-1/4+1/(big number) +... + 1/(really big number)) + ....

doesnt work?

does
$$\sum \frac{1}{2n}$$
not diverge?

whys there an alternating sign?

if 1/3+1/5+1/7+1/9+... is divergent; then it works right

Balarka Sen

I think $\sum_{k = 1}^\infty 1/(ak + b)$ diverges for any given integers $a, b$

Anonymous

@JoshuaLin Yup, there's no alternating sign. I was wrong

Joshua Lin

all g then?
Wait isnt there something saying
$$\sum 1/p$$
diverges where p ranges over all primes or something?

Anonymous

07:52

What do you do next?

Balarka Sen

@Joshua Yes, that diverges

Joshua Lin

Well then
$1+(a1−b1)−1/2+(a2−b2)+1/3+(a3−b3)−...$
is conditionally convergent; and can be rearranged to
$a_1+a_2+...$

unless im bonkers

Anonymous

You're subtracting from $1+2+3+4+5+...$, $1-(1/2+1/4+1/6+...)+(1/3+1/5+...)$

Anonymous

If I'm getting you right

Joshua Lin

? wait what

Anonymous

07:56

Can you show that difference (which gives you the error terms) is convergent?

Joshua Lin

The error terms satisfy
$|a_i - b_i| < 2^{-i}$
by construction, so they converge

Balarka Sen

You can rearrange to get 1 - 1/2 + 1/3 - 1/4 + ... = 1 - 1/2$a_1$ + 1/3$a_2$ - 1/5$a_3$ - ... where $a_1$ is sum of reciprocal of the first $n_1$ natural numbers, $a_2$ is the sum of reciprocals of the first $n_1$ odd natural numbers, $a_3$ is the sum of reciprocals of the first $n_3$ numbers not divisible by $2$ or $3$, etc.

Anonymous

@JoshuaLin What are your $a_i$ and $b_i$ ?

Joshua Lin

$a_i$ is the divergent (1+2+...), $b_i$ is our rearrangement of $1-1/2+1/3...$ which is 'epsilon close' to $a_i$?

Anonymous

$a_i$ is a term of a series. It can't be a whole series.

Joshua Lin

07:59

$a_i = i$

Balarka Sen

By choosing $n_i$ to be large, as each of those $a_i$'s diverge, you should be able to get a rearranged series of the form $1 - c_1 + c_2 - c_3 + \cdots$ where $c_i$ are fixed real numbers such that $|c_i - c_j| < \varepsilon$ for all $i, j$.

That should approximately be the idea

Anonymous

@JoshuaLin You're telling me that $|i-(-1)^{i+1}\frac{1}{i}|<2^{-i}$ ?

Anonymous

@BalarkaSen Yeah, that sounds good

Balarka Sen

This needs to be written down carefully though.

Joshua Lin

$b_i$ is the rearrangement of the series; like we have
$1-1/2+1/3-1/4+...$
rearrange:
$1+(-1/2+1/3+1/5+...+1/big number)+(-1/4+1/bignumber + ...)...$
$b_1 + b_2 + b_3 + ..$

Balarka Sen

08:01

Rearrangement is dodgy business

Anonymous

$b_i$ isn't a series, nor a rearrangement of a series. It is just a term of a series

Joshua Lin

yeah; $b_i$

Anonymous

So please write down the general term of $b_i$ down carefully

Joshua Lin

ok I guess I've been loose with my language lol

There's like a million choices for $b_i$ though... like its arbitrary

Balarka Sen

Don't. This is a very subtle argument, not a loose one.

Anonymous

08:03

...

Joshua Lin

lol

Balarka Sen

If you can't write it down rigorously, you should be very doubtful about your whole idea.

Anonymous

And without even choosing $b_i$ you claim that $|a_i-b_i|<2^{i}$...gosh

Joshua Lin

normally I don't understand why people get angry online explaining things; but to me the proof is pretty clear

Balarka Sen

It might be, but to us it isn't

Mathematics is about communication and skepticism

Joshua Lin

08:05

1/3+1/5+1/7... diverges
So we can have
$b_i = -1/2i + \sum_{m \in S} 1/(2m+1)$

where $S$ is a set chosen such that

$|a_i - b_i| < 2^{-i}$

Anonymous

No one is angry. But it's annoying if you don't write down things carefully. And that's one of the reasons mathematicians hate physicists :P

Joshua Lin

Such a set $S$ can always be found, since 1/3+1/5+1/7... diverges

No; I'm angry lol just trying to not show it

Maybe its because I've done nothing all day

Anonymous

@JoshuaLin "chosen such that". Choose it first

Joshua Lin

Well in a proof I don't need to choose it; all I need to show is that such a set can be chosen

constructive proofs aren't the only proofs!!

And such a set can always be chosen, since 1/3+1/5+1/7... diverges

Anonymous

@JoshuaLin Proof?

Anonymous

08:09

Convergence is so intricate a topic that if one goes a bit loose, that there can drastic effects on the conclusion

Joshua Lin

oh geez ok I give up on this whole enterprise
Math isnt always about being formal; its about seeing the idea of the proof; and the idea isnt really getting across
Ok maybe I wont give up; I hate myself, I suck
wlog; lets consider a positive number $a$. We can always find an integer $m_0$ such that
$1/(2m_0+1) < a/2, 1/(2m_0+1) > a/4$
Now, inductively, we can choose $m_{n}$ such that
$1/(2m_{n+1}+1) < (a - \sum_{k=1}^{n} 1/(2m_{n}+1)/2, 1/(2m_{n+1}+1) > (a - \sum_{k=1}^{n} 1/(2m_{n}+1)/4$
Now, we can pick a really large $N$, and we have

how far down the rabbit hole do we go

OH WAIT I KNOW HOW TO EXPLAIN

Anonymous

Unfortunately, I'll have to leave now. I'll surely read what you wrote when I get back and ping you. Cya

Joshua Lin

https://en.wikipedia.org/wiki/Riemann_series_theorem
So we can rearrange a conditionally convergent series to get any number right? Cool. So we have divergent series $a_i$, and conditionally convergent series $b_i$, so we can rearrange $b_i$ to converge to $a_1$. Truncate this rearrangement, and this is $\epsilon$ close to $a_1$. Now with the rest of the $b$ sequence, we can rearrange to approximate $a_2$, and so on and so forth..

Well since I'm here; I have a question?
We have $F$ is the collection of all maps $x : \mathbb{R} \to \mathbb{R^n}$, and $d$ is the differential on $\mathbb{R}$, and $\delta$ is the differential on $F$, why do $\delta$ and $d$ anticommute on $F \times \mathbb{R}$? And its not me being loose with language here; this is how its written in these notes :'(

Secret

08:35

because I am going to carpet bomb that place

Secret

08:47

Weirdness level in asf chat restored to pre-critical levels. Ceased carpet bombing issued

Carpet bombing will be resumed when he left the scene

Physics Meta

09:27

0

Q: Duration between two questions which I can ask

How frequently I an ask a question? Stack exchange says we are no longer accepting questions from this account why?

Secret

10:55

reddit.com/r/askscience/comments/1douyn/…

gah, thought its nubmer crunching and parallel searching capabilities will allow more high detailed games

Slereah

11:10

According to Gourgoulhon, Fermi coordinates were discovered by Synge

why is Fermi involved

Secret

12:06

smbc-comics.com/index.php?id=3855

0celo7

12:41

@ACuriousMind please ping me if you’re here I have a doubt

Slereah

13:15

Hm, from the paper on point particle quantization I think the issue is that you can quantize it for free particles due to the $\Bbb Z_2$ part of the gauge group for time reversal, but this symmetry is broken when you include EM fields

Hence the issue for interacting RQM

0celo7

@Slereah do you have PUBG?

Slereah

$$S = g^{-1} \dot x + g m $$ is invariant under $\Bbb Z^2\times \mathcal F_+$ while $$S = g^{-1} \dot x + g m - e\dot x A $$ is only invariant under $\mathcal F_+$

@0celo7 I do not

0celo7

@ACuriousMind Basically I'm having that thing again where physicists don't state clearly what they mean, instead wanting me to apply my intuition, which is something I cannot do, and also refuse to attempt.

so please help

Slereah

what are they talking aboot

0celo7

Why the naive Noether theorem doesn’t work for rotations

Slereah

13:25

It doesn't? :O

0celo7

Something about the “tensorial character of the metric” which is some horseshit

Slereah

which one is the naive one

0celo7

@Slereah no, you need the Belinfante Rosenfeld tensor

Slereah

Oh that shit

Though what do you mean by "not working", exactly

The Noether theorem gives the appropriate angular momentum

But it doesn't give the same result as the GR stress energy tensor yeah

0celo7

You need to use the BR tensor and not the canonical EM tensor

Slereah

13:30

yeah

0celo7

It has to do with the way the fields transform under the rotation but there’s no fucking reason given why you need to account for that

And the way that Christo firststates the noether theorem makes it seem like it should apply perfectly fine in all cases, but then it doesn’t!

Slereah

What theorem is it, exactly?

ACuriousMind

What exactly is the problem - in what way does the Noether theorem "fail"?

Slereah

Is it just the "Symmetry implies conserved current"

0celo7

Hold on bruddah

Gotta die in PUBG now

0celo7

13:48

@Slereah apparently it doesn't work all the time, which is confusing

@ACuriousMind Do you have Weinberg 1 handy?

Slereah

Is it the part in Weinberg where he says that most symmetries also leave the Lagrangian invariant, but not general Lorentz transforms

0celo7

well it's equation (7.4.7)

Slereah

I think the difference is that, from what I can see?

It's the action vs lagrangian being invariant

Which changes the form of the theorem slightly

0celo7

well so I sat down and worked it out yesterday

what I got was i.gyazo.com/83e44235d1b8a219eb47ebe2a96c67e1.png

and as far as I can tell, the result does not change when $u$ is replaced by a vector variable

now just take the flow to be generated by $\epsilon_{ijk}x^j\partial_k$, and then you get a Noether theorem for rotations!

But it turns out that implies $T^i{}_j$ must be symmetric, which is false

so I don't know what the hell is going on

@ACuriousMind help please

Slereah

Well it is symmetric, for the most part

The symmetry is mostly lost when you use fields that don't transform trivially under the rotation

I can't really tell what fields you're using in that proof

ACuriousMind

13:59

@0celo7 Why is it false?

0celo7

@ACuriousMind the canonical energy momentum tensor is rarely symmetric

that's why Weinberg does the whole belinfante rosenfeld thing

but to get that, he needs to transform the fields according to a rep of the Lorentz group

ACuriousMind

@0celo7 And what has that to do with Noether's theorem applied to rotations?

Slereah

"Suppose that $L$ is invariant under the flow of $X$"

0celo7

@ACuriousMind Because if you work out the conservation law according to rotations, it's exactly $T_{ij}=T_{ji}$, but that's not true

Slereah

Is $X$ the generator of the rotation

If so there's your problem

0celo7

14:01

it's a very difficult thing to understand

@Slereah hmm?

Slereah

The Lagrangian isn't invariant under rotation

At least not generally

0celo7

I am supposing that it is

Slereah

Well then no problem

0celo7

but it gives the wrong conclusion!

Slereah

It is invariant for scalar fields for instance, and in this case you do have a symmetric canonical SET

ACuriousMind

14:02

@0celo7 I'm very confused. The stress-energy tensor is the conserved current for translations, not rotations. Can you please spell out in more detail how exactly Noether's theorem "fails"?

0celo7

@ACuriousMind do you see the corollary 2?

ACuriousMind

yes

0celo7

@ACuriousMind it shows that $T^i{}_jX^j$ is the conserved current for any vector field $X$ that generates a flow leaving the Lagrangian invariant

do you agree?

ACuriousMind

What is $u$ in there?

0celo7

@ACuriousMind dependent variable

ACuriousMind

14:06

what

0celo7

I am writing the lagrangian as $L(x,q,v)$ btw so that's how the derivatives are written

@ACuriousMind a Lagrangian theory has an independent variable $x$, a dependent variable $u$, and a velocity $Du$

ACuriousMind

...is it a field?

0celo7

it could be

ACuriousMind

Why would you tell me it's a "dependent variable" instead of just saying "it's a field" or "it's a generalized coordinate"

0celo7

I'm not a physicist so I don't want to restrict myself to fields necessarily

@ACuriousMind Because in calculus of variations (my field) we call it the dependent variable.

blame the greeks

Slereah

14:08

I think the issue is just assuming $L$ to be invariant

The Belifantes tensor is only for weird fields with spin densities

Which doesn't leave the lagrangian invariant under rotation

0celo7

@Slereah such as GR...

ACuriousMind

@0celo7 Okay, so why is that thing there called $T$ - what has it to do with the stress-energy?

Ah, wait, do you write $v$ when differentiating w.r.t. to the generalized velocity?

0celo7

@ACuriousMind Yes. This needs to be edited. Most of those equalities are modulo eom.

ACuriousMind

Then that is the stress-energy alright

0celo7

like I mean $\partial L/\partial v(x,u(x),Du(x))$, where $u$ solves the eom.

@ACuriousMind yes

ACuriousMind

14:11

Sure. So what's this problem I keep hearing about?

0celo7

so do you agree with the corollary as stated?

ACuriousMind

@0celo7 Yup

0celo7

@ACuriousMind ok so now suppose $u:\Omega\to\Bbb R^k$, which we write as $u^a(x)$. Then I claim that with the same hypothesis, $J_X$ is conserved if we write $$T^i{}_j=\frac{\partial u^a}{\partial x^i}\frac{\partial L}{\partial v_i^a}-L\delta^i{}_j.$$ Do you believe that?

ACuriousMind

Sure

0celo7

Now by identifying $\mathrm{Hom}(\Bbb R^n,\Bbb R^n)$ with $\Bbb R^{n^2}$, we can view a Lorentzian metric $g$ as a map $\Omega\to\Bbb R^{n^2}$, agreed?

ACuriousMind

14:17

Yes. Skip all the technical details if you just want to convince me that you'Re allowed to plug $g^{\mu\nu}$ in there as a field :P

0celo7

I wasn't going to bore either of us with details, not even sure what those would be

@ACuriousMind Ok, but this is appretly now an issue

so suppose we have some lagrangian $L$ depending on $g$ and $\partial g$, but not explicitly on $x$

well, then corollary 2 should hold as stated for such a Lagrangian

agreed?

ACuriousMind

Yes.

0celo7

rotations in the alpha-beta plane are generated by $$X_{(\alpha\beta)}=x_\alpha\partial_\beta-x_\beta\partial_\alpha,$$ where $x_\alpha=\eta_{\alpha\beta}x^\beta$

so we should have $$T^\mu{}_\nu X^\nu_{(\alpha\beta)}$$ be a conserved current

but in fact, the divergence of that is $$T^\mu{}_\beta \eta_{\alpha\mu}-T^\mu{}_\alpha\eta_{\beta\mu},$$ which is not zero!

so, what gives?

ACuriousMind

Oh, I think you made a mistake where you identify the field variation $w$ as $u(y,(x,t))$ - you are implicitly disallowing the target space of the field to transform non-trivially under the transformation, there, but for fields of non-zero spin, the target space does transform under rotations.

Slereah

@ACuriousMind that's what I said!

ACuriousMind

14:27

That is, your derivation works for fields of zero spin, and indeed the canonical stress-energy is symmetric if and only if you have no fields of higher spin

0celo7

@ACuriousMind Ignore "spin" I am looking at this from a purely mathematical perspective

$g$ is just a map from $\Omega$ to $\Bbb R^{n^2}$

ACuriousMind

@0celo7 Yes, and from a purely mathematical perspective, you are incorrectly identifying what the variation $w$ is for rotations.

Slereah

But how do you know that the stress energy tensor isn't symmetric if you're looking at it mathematically?

the only clue that it isn't is physics

and only for spinned fields

0celo7

@ACuriousMind Why can't I just do $w^a=u^a(y(x,\tau))$ like usual?

what I need is for (1) to hold

ACuriousMind

Because even from a purely mathematical perspective, a vector field with values in the tangent bundle transforms differently under the map induced by a diffeomorphism than a field with values in some random vector space.

0celo7

14:30

@ACuriousMind Well that's how you eventually "fix" it but I still don't understand why the naive point of view fails

I'm pretending my metric is just a thing with values in a vector space

because in a chart, that's exactly what it is

ACuriousMind

Sure. If it was, you would be correct.

0celo7

in a chart, it is!

are you saying (1) does not hold when $u$ is a metric and the variation is a rotation?

ACuriousMind

That is, in a theory where the metric is not cotangent-valued and therefore does not transform under a diffeomorphism by the Jacobian, your derivation would be correct.

Slereah

Basically what you have is just a theory of $n$ scalar fields

ACuriousMind

@0celo7 I am saying you are incorrectly identifying what the variation $w$ is for the transformation induced by a vector field.

0celo7

14:33

I am using the variation $w$ that gives me (1), and (1) then implies corollary 2

once I have (1), I have corollary 2, which you agreed to

ACuriousMind

I.e. what is lacking is your proof of corollary 2, and that that $J_X$ of corollary 2 needs to include an additional term that accounts for the transformation behaviour of the fields.

0celo7

so somehow my $w$ does not give (1), which makes absolutely no sense

ACuriousMind

@0celo7 I am rescinding my agreement to corollary 2, see above :P

0celo7

@ACuriousMind Change that to theorem 1 then

ACuriousMind

Change what to theorem 1?

0celo7

14:34

corollary 2 in my above sentences

Do I have (1) or not?

ACuriousMind

You have (1) only if you adapt how you define $w$ in that proof.

0celo7

so you're saying (1) does not hold if I define $w_{\mu\nu}(x,\tau)=g_{\mu\nu}(y(x,\tau))$?

ACuriousMind

No, I'm saying that that's not the correct definition if $g$ is tangent/cotangent valued and you claim this is supposed to be "invariance under the flow of a vector field".

That is, (1) may well hold but it does not follow from the claim that the Lagrangian is invariant under the flow.

0celo7

from what claim

ACuriousMind

Basically, your (2). where I know realize you actually should have written something like $q',v'$ on the r.h.s. to allow for $q$ and $v$ to vary, because as written you only account for the variation of $x$ but assume there's no variation of the fields. But a vector field $A$ varies onder a flow $\phi$ more than just by $A\circ \phi$, as I'm sure you'll agree!

0celo7

14:43

Unless you define what "varies" means, I agree to nothing.

(2) is just expressing that $L$ depends only on the orbit of $x$ along the flow

this is trivially true for any vector field when $L$ does not depend on $x$

Slereah

I mean you could have a theory with an identical Lagrangian that transforms trivially, but in that case you'd just have a theory of many scalar fields

The spin would be all wrong

It would not be the same theory as GR

0celo7

spin might be a mental illness

I don't see a reason for it as of right now

ACuriousMind

@0celo7 Aha! Have you checked that, for $u$ a field of non-zero spin, the claim that $m(x) = X^i \frac{\partial u}{\partial x^i}$ holds?

(ignore that I said, spin, just say tensor field)

0celo7

@ACuriousMind Of course it holds, $u^a$ is just a collection of functions.

@ACuriousMind I know you want me to pull back by the flow, and that's what everyone else wants me to do as well, but I can't actually see why that should be necessary.

ACuriousMind

Yes, I realize that

0celo7

14:56

brb breakfast

Slereah

I think Schwarz has a thing about that

Proving that you can't have a vector field have the same form as a KG field

Due to the transformation rules

Tho of course it's a physicist thing

0celo7

@ACuriousMind The way you might convince me is perhaps that $$\int_{\phi_\tau\omega}L=\int_\omega\phi_\tau^*L$$ is important here

but that pullback is kind of useless because it's only in the $x$ variable. or something

idk

and the Lie derivative of $L$ would just be $X(L)$ because it's a scalar function of $x$

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