@imbAF There is no way to know the actual motion of the centre of mass of the beam along the beam axis. The accelerator physics must still make sense but doing statistical averaging along the beam axis.

@SirCumference this is much less deep than it appears. Microscopic collisions would have localised times and space so that the equality would hold.

@Obliv it must be QM, or else the N in pV=NkT ought to disappear. The derivation is just internally inconsistent and we are really lying to students when considering classical mechanics.

what are the reasons to believe in crossing symmetry?

@Mr.Feynman no no, it's a good book I used it too

how did you guys get good at looking into whether an idea exists or not (e.g. finding a paper in the literature that answers some question about whether someone has tried to do X)

@SillyGoose to me that's difficult. you kinda need to know the litterature and having a supervisor could help

@SillyGoose I google things a lot

@SillyGoose I ask ACM or Slereah

They're more efficient than google

Beware that I am often wrong :p

It was a half joke, blind trust is never good

But my point is that although one can potentially find anything using google (or any other engine, e.g. SE), you also have to know what to search to find the desired results. Talking to people who know more or have already searched that in the past is more efficient

Yesterday I spent hours trying to figure out something about my notes: there was this formula $\omega^\mu=\frac{1}{2}\varepsilon^{\mu\nu\sigma\rho}\partial_\nu u_\sigma u_\rho$. I wasted a lot of time before learning that's called vorticity. After that I found many useful stuff. If I had asked here someone would have recognized immediately

One demerit for google is that you cannot google formulas

Yeah but there is also a notation issue

@Mr.Feynman to add to this often papers have titles which are not really very descriptive of their contents

Like this lol

There must be a strip for that

@ekardnam_ or you search or basic GR stuff and only find the keyword in QG articles

One common issue I have is that a lot of symmetry-related stuff only occurs in QFT papers

Even though they are also classical notions

@Mr.Feynman yeah searching basic stuff almost always end up badly into papers which are more advanced than necessary for me

@SillyGoose You don't need to "believe" in it, it should be obvious from the Feynman diagram formalism: The only difference between the diagrams for the processes related by crossing symmetry is which of the external legs are labeled as "in" and which are labeled as "out"

on a non-diagram level, it's essentially just CPT symmetry

@SillyGoose It's mostly just persistence (don't give up if the first few searches only result in garbage) and being willing to skim papers that look only slightly related (don't be fooled by the abstract not exactly matching what you're looking for). Of course, the best starting point is to ask someone you consider an expert in the area (or at least more knowledgeable than yourself).

Can be tough if the paper is in an entirely unrelated field :p

Like how the Filipov differential inclusion formalism that is sometimes used in GR was originally made for some robotics paper

For stationary spacetimes, is the EH horizon defined by $g^{rr}=0$ or $(g^{tt})^{-1}=0$ (ZAMO Limit)?

Because apparently I could find both definition

And the fact that those are equivalent for Kerr is not quite satisfying

No idea

What surfaces do they define for Kerr?

My intuition would be that you should probably pick $g^{tt}$ because what does $r$ mean in Kerr

If your spacetime is stationary you have a specific families of time coordinates defined, at least

@Slereah $r=r_+$

@Mr.Feynman For both?

Yes, they're equivalent for Kerr

They both boil down to setting a polynomial to zero

If you considered for instance say Kerr expressed in pseudo-Cartesian coordinates, what would it mean to say that $g^{rr} = 0$

It doesn't sound like a coordinate independent notion

If it is stationary though you can say $g(K,K)^{-1} = 0$

"Finally I acknowledge my ultimate support in the knowledge and love of my Lord Jesus Christ, through Him all things are possible!"

Holy thesis aknowledgement

@Slereah it comes from a long ass message I wrote yesterday

This and below

From what I can find, for a stationary surface you have that $g(K,K) = 0$ for some stationary observer with velocity matching the Killing vector

The point being that this surface has a normal vector to it, which is null on this surface, but in the case of the Kerr metric in standard coordinates, you can see that this is identical to $dr$

But that's just an extension of the normal vector to outside the stationary surface

This doesn't need to be unique

It just happens to be fortuitously the gradient of the radial coordinates for Kerr

Presumably you can always define a coordinate patch since you can always extend normal vectors to some tubular neighbourhood around the surface, idk if you can always extend it in some uniquely interesting way

Presumably you can always extend it in a way such that the normal vector and the static timelike vector span it, along with two other vectors

I don't know if there's anything special about it beyond that tho

You just need to show that the Killing field and the normal are never colinear idk

Also to define coordinates you need to be able to integrate out those two vector fields together so that $[\partial_t, \partial_r] = 0$

@Slereah note that this is $g_{tt}=0$, not $1/g^{tt}=0$ (the metric is non diagonal)

I'm calling $t$ the coordinate defined by the timelike Killing

I asked a question on the site to be sure

Man, classical GR is worse than QFT :P

I mean, these are basic topics, aren't they?

@Mr.Feynman Same principle

There might be a deeper reason why $dr$ is the normal vector, but I guess it's just that since it's the event horizon you can just define that vector to be your radial coordinates?

And if you have additional Killing vectors maybe it works out to be indeed the appropriate radial coordinate

Like if your spacetime is spherically symmetric, by symmetry your horizon also has to be a sphere and you can just pick the normal vector to be the generator of some radial coordinates

And probably something similar with Kerr being just axisymmetric

Something like that idk

The one-parameter congruence of null hypersurfaces will basically be your $r$ coordinates

Oh wait that would be only for null coordinates

The time and radial coordinates are only null on the surface

is the reason we study quasistatic processes in thermodynamics (equilibrium thermodynamics, to be precise) simply because if we didn't it would become non equilibrium thermodynamics?

i guess metal expansion can be considered relatively quasistatic....

suppose we hav an n order group and we identify an element with a permutation

this permutation must be made of equal length cycles

there cant be cycles of different lengths

is this correct

the element satisfies $g^n=I$ for some minimum $n$. then for any element $g_1$, $g^n g_1= g_1$

so the cycle that starts at $g_1$ must be of length $n$

so all cycles must be of length $n$

sorry the n for the grp order and the n for the cycle length are different ofc

use m for grp order

@RyderRude I have no idea idea what you're doing here: What does it mean to "identify an element with a permutation" or for a permutation to be "made of equal length cycles"?

it's Cayley's theorem that every grp element of an n-order grp is a sub-grp element of permutatations S_n

and permutations r made of cycles

what do you mean by "made of"?

use precise language

so i take a grp element , extract the permutation element, and then decompose it into cycles

@ACuriousMind for e.g. (123)(45) is a permutation 1-->2, 2-->3 , 3---1, 4--->5, 5--->1

every permutation can be written like this decomposition into cycles

(123) is a cycle and (45) is another

what you mean is that every permutation is a product of cycles

lets say i hav an initial arrangment 12345. and any final arrangment 34512

and in fact one can show that every permutation is a product of 2-cycles (transpositions) so I don't understand the question about the length of the cycles at all

oooh yeah

ok so i meant decomposition into non-common cycles, i.e. cycles that dont share any element

this decomposition is unique

(123)(45) is one such decomposition

(123) and (45) dont share elements

but if u decompose (123) into (12)(23), now they share the element 2

so u can write the same permutation as (123)(45) or (12)(23)(45)

but the im decomposing the grp elements the former way

I know all that, you just didn't clearly state what you were doing (in particular you omitted the "non-common" condition)

yeah. sorry

And yes, then the statement that all the cycles in this representation will have the same length is correct

thankss

i think this proves that in any grp of an odd number of elements, no non-identity element can square to identity. if an element squared to identity, then its cycle rep wud hav a bunch of 2-cycles

which is a convoluted way to state that the cosets under the subgroup $H = \{1,g,g^2,\dots\}$ have the same size

and it wud imply 2k=n . which wud make n even

is this theorem correct?

you really don't need to do any theory of permutations here, this is just aspects of Lagrange's theorem

@ACuriousMind oh

I'm not sure what you're reading, but it's very strange to do this via permutations, all these conclusions can be arrived at straightforwardly in abstract group theory

so this theorem is actually correct??

the source im reading didnt mention this theorem. it only mentioned that, in an even ordered grp, some non identity must square to identity

sure, the order of every element has to divide the order of the group, that's Lagrange's theorem. As a special case there are no elements of even order in groups of odd order

yeah it's so much easier from Lagrange's theorem becuz the 2-cycle forms a sub-group!!

i didnt realise that

i think the book didnt mention it becuz it's a trivial theorem

im reading A Zee's book

why is Lagrange's name there

Group theory didn't exist

It is adorably old French

I don't understand it :P

There does seem to be a trick to turn the foliation by null hypersurfaces into the more proper foliation

Basically by switching from null coordinates to normal coordinates

You just pick your two null vectors and get $t,r = u \pm C_\pm v$, with $C$ some scalar function

I am guessing you are meant to pick so that this extension coincide with $t$ being the killing vector

"Other members of this same school say there are ten principles, which they arrange in two columns of cognates-limit and unlimited, odd and even, one and plurality, right and left, male and female, resting and moving, straight and curved, light and darkness, good and bad, square and oblong."

I think the Pythagoreans started the whole adjoint functor thing

but then again Hegel always claimed that his notions of unity of opposites found its roots in all previous philosophical traditions, so it may be literally related

@Slereah have you read the part of lichnerowicz' book about stationary asymptotically flat vacuum solutions being flat

would be nice if someone translated that!

I did not

how long is it

male and female arent even opposites

Oh they went quite a bit further in those oppositions, saying that they were in fact all the same opposition

Odd numbers are male while even numbers are female

were they 5 years old

lol

"all dogs are daddys and all cats are mommys"

@Slereah but which are square and which are oblong?

that's the opposition I find really strange here

I'm guessing Aristotle lists them in the same order as given by pythagoreans

So men are squares while women are oblong

and by that logic men are good and women are bad

I wish I was surprised :P

I like how everyone just calls the Pythagoreans "The Italians"

@nickbros123 Given external circumstances and constraints, e.g. fixing p,T, or V,S, every system has unique equlibrium, or at most a few possible equilibria. There is no learning path whereby people neglect and abandon the learning of something that can be universally agreed upon to have sensible results, and instead go straight into difficulties.

"his making the other entity besides the One a dyad was due to the belief that the numbers, except those which were prime, could be neatly produced out of the dyad as out of some plastic material."

I'm not entirely sure what the point of Pythagoras is here

out of one many by adding another one

But why wouldn't prime be produced that way

And if it is meant to be interpreted as a multiplication, why would only primes be not included

This may relate to the sentence right after that :

> Yet what happens is the contrary; the theory is not a reasonable one.

@naturallyInconsistent I understand every system has a unique equilibrium under those constraints. its just that i would like some examples where the infinitesimal unbalanced force assumption (basically the quasistatic assumption) hold to a good approximation. my intuition said thermal stretching of metal could be one, but in general it appears a little unreasonable to assume such a thing

btw, GR heads can now get a 200 point bounty if they can explain to me the proper application of BRST to its Hamiltonian formalism

@ACuriousMind that sounds more like deserving a research grant than internet points

But if I think them up I'll be sure to post them

@Slereah I'm still hoping that this already exists somewhere in the literature :P

but if this is the second question of mine that provokes someone into writing an entire paper, I'm not against that, either :D

what was the first?

the one about Majoranas

It only took 3 years to get the answer

I mean, I have time :P

"The present paper is motivated by a Physics StackExchange question asked by member ACuriousMind"

speaking of spinors

oh no

Found someone claim that there's a nice geometric interpretation of spinors in some paper, but said paper is in a conference book once more

inspirehep.net/literature/1624323

oh wait, it is on inspirehep?

inspirehep.net/files/73b3ce09515487a8b803624513cfc852

Didn't even know inspirehep hosted papers

@ACuriousMind did it happen again?

@Mr.Feynman uh, no?

Damn, I had my line ready :(

that was a response to Slereah's comment about a research grant

lol Hoyle wrote in that conference

There's something I've never heard of

apparently the modern explanation of the phenomenon involves the influence of the sun on Earth's volcanism???

Seems weird but I don't know enough vulcanology to judge

@nickbros123 Some examples? I am not sure if you have realised how practically insane any of the thermodynamics processes are, but this is also where Ian Ford shines: He explicitly computes what would be the spontaneous entropy generation if there is, say, a temperature mismatch of so-and-so, is at most quadratic in the mismatch, such that the integral of all the mismatches over time is at most linear in the mismatch, so that if you let the mismatch go to zero, the ideal limit is achieved.

If I have the following function represented via a power series: $f(\rho) = \sum_{q = 0}^{\infty}c_q\rho^q$ and I've found that the asymptotic behavior of the coefficients is $$c_q /c_{q-1} \sim 2b_0 / q$$ and I've renamed $N = q / (2b_0)$. Therefore I reconstructed the function and obtained the following $\sum_{N = 0}^{\infty}\frac{(\rho^{2b_0})^{N}}{N!} = \exp[\rho^{2b_0}]$,

but apparently the solution tells me that the function is of the form $f(rho) \sim e^{2 \rho b_0} $. Could you tell me if you find any mistakes in my procedure?

It's not even that important because, in the end, what matters is that it's still a growing exponential, so I'd have to discard this solution for the wavefunction (normalization condition), but I just wanna make sure that the solution is correct

@ClaudioMenchinelli looks to me like your solution just got the power laws wrong (i.e. mistakenly thought that $\mathrm{e}^{(\rho^{2b_0})} = \mathrm{e}^{2b_0\rho}$)

Ok thanks @ACuriousMind that was in fact my conclusion as well