2:25 AM
@imbAF See the graph in my answer here: physics.stackexchange.com/a/595175/123208 As you can see, the error in using Newtonian physics is still quiite small at 0.01c

Odd question, but is there a special name for functions f, g such that f(g(x)) = g(f(x)) or just "commutation under composition"?

1 hour later…
3:47 AM
I am about to make a small torsion spring, reducing the number of coils in it will make the spring stronger ?

5 hours later…
8:34 AM
Oooh apparently there's a whole gravity theory based on the projective structure
Awful idea but interesting

8:44 AM
They even work out the spinor structure
What is even a projective spinor

What's a projective gravity

Gravity where the gauge group is PGL(n)
Or possibly PO(n,1), not quite sure
It is called Thomas Whitehead gravity if you want a look
Ominous tit'e

@DanielUnderwood I'd just say $f$ and $g$ commute. After all, when matrices "commute", you are saying that the corresponding linear operators commute under composition

8:59 AM
I was expecting nlab Whitehead towers
> 'the field outside a spherically symmetric source is not just given by the Schwarzschild solution but in general contains an additional constant of integration which is in principle measurable'

@bolbteppa a man can work on more than one thing!
He also cowrote the principia mathematica

9:14 AM
Oh yeah, man...

Just had the bad luck on working on a version of GR that didn't work
No prize for second place
I wonder if there's a gravity theory for every GR structure
There is for metric, conformal, affine and projective, at least
Although I think beyond that you get into structures too boring to tell you much
like all the $\mathbb{Z}_2$ structures

9:38 AM
In theoretical physics, Whitehead's theory of gravitation was introduced by the mathematician and philosopher Alfred North Whitehead in 1922. While never broadly accepted, at one time it was a scientifically plausible alternative to general relativity. However, after further experimental and theoretical consideration, the theory is now generally regarded as obsolete. == Principal features == Whitehead developed his theory of gravitation by considering how the world line of a particle is affected by those of nearby particles. He arrived at an expression for what he called the "potential impetus...
> Whitehead developed his theory of gravitation by considering how the world line of a particle is affected by those of nearby particles. He arrived at an expression for what he called the "potential impetus" of one particle due to another, which modified Newton's law of universal gravitation by including a time delay for the propagation of gravitational influences.

Papers on jets of structures use the world "osculating" a lot
I fear I'll have to check out what that is

2 hours later…
11:32 AM
Hey all is this correct composition law for relative velocity?
So I wanted to verify the composition law (Given the relative velocity of $AB$ and $BC$ find relative velocity $AC$):

$v_B = \gamma_{AB} (v_A + v_{BA})$
and
$v_C = \gamma_{AC} (v_A + v_{CA})$
and
$v_C = \gamma_{BC} (v_B + v_{CB})$

Substituting $v_B$ from the first equation:

$v_C = \gamma_{BC} (\gamma_{BA} (v_A + v_{BA}) + v_{BC} )$

but as previously mentioned

$v_C = \gamma_{AC} (v_A + v_{CA})$

Comparing the above 2 for all $v_A$ and $v_C$ we get:

$\gamma_{BC} \gamma_{BA} = \gamma_{AC}$
Sorry for the tedious algebra but I couldn't find this result on google
:/

11:54 AM
Hey actually the fancy definition of the jet space does have a natural relation between jets of functions and curves
wonderful
"Two maps $f, g : M \to N$ are said to determine the same $r$-jet at $x \in M$, if for every curve $\gamma : \mathbb{R} \to M$ with $\gamma(0) = x$ the curves $f \circ \gamma$ and $g \circ \gamma$ have the $r$-th order contact at zero."
I bet this will blossom into a wonderful friendship between the tangent bundle and the frame bundle

@Slereah One day I will know enough GR to make sense of your words :) Have you read visual differential geometry by Tristan btw ?

I have not
a jet is just fancy talk for "the derivatives"
The jet of a function $f$ is just $$j f = (f, f', f'', f''', \ldots)$$

@Slereah Oh you might like it ... Ive read bits of it
@Slereah Is there any advantage to that notation?
I guess now u can write the taylor expansion in matrix form with a jet?

It's basically the coordinate invariant version of a Taylor expansion

12:09 PM
@Slereah what do you take its inner product with?
And Im guessing you use the metric in this inner product?

No inner product defined yet
It is but a building block towards it

I see ... spooky stuff

12:24 PM
@glS I'm sorry if I was annoying yesterday, I was coming down with fever from my booster shot and wasn't exactly on top of my game

@ACuriousMind Meanwhile I got the omnicron virus in India instead :P

I get the impression that one can use $J^k(M, M)$ instead of $J^k(\mathbb{R}^n, M)$, but it's not a hundred percent clear
I'm guessing the atlas gets involved somehow

@MoreAnonymous sorry to hear that, hope you'll be fine

Oh Im chirpy ... Sorry to bother you on this again but I cant find the algebra on the internet for the composition law of relative velocity

https://chat.stackexchange.com/transcript/message/60192450#60192450
Any idea where on the internet to look?

There is apparently some mapping from $\mathrm{Diff}(M)$ to $\mathrm{Inv}J^k(M,M)$
v. nice

12:38 PM
@Slereah Historically physicists have never been that rigorous. Why are starting now ?
I swear rigor has been trending far too much for my liking

There's plenty of rigor
Just gotta know where to find it

I can see the day where they teach us Godel's incompleteness theorem in physics class

It doesn't have much application in physics
Nor is it particularly more rigorous than most math

@Slereah Wasn't hawking moaning about it once?

He probably moans about many topics that aren't physics
Especially now that he's a ghost

12:51 PM
If only I knew hawking while he was alive
Heck if we're wishing I might as well wish I was the person who pushed his wheel chair
He'd be on the driver's seat
But I'd be the one behind the wheels
:P

I almost sent him an email once
I had a question on a theorem of Hawking-Ellis and I wanted to ask him about it
But then turns out that the email of Hawking wasn't publicly available anymore

@Slereah Did he respond?

1:19 PM
Hypothetical question if I flag "not an answer" do I ever get to know what happens to that flag?

He did

@Slereah thats cool ... I've never been gutsy enough to write to a physicist

In my experience you have about a 50% chance of an answer
they're not usually fancy people

@Slereah in my experience it's way higher. But then again, my partner and many of my closest friends are physicists and I mostly write to them :)

@fqq fqq aren't you a theoretical physicist yourself?

1:30 PM
@ACuriousMind you were being more annoying than usual? I hadn't noticed :P. Jokes aside, obviously that was totally fine, some degree of pedantry is great to get a better understanding of these things. Hopefully you're better now. I thought the third booster was usually asymptomatic though? At least it was for me. Guess it depends on which (combination of) kinds vaccines you got

@glS each of the three shots was Biontech and has knocked me out for a day, really not much difference between them in my personal experience

Well the physicists who didn't answer me I think probably just let them slip through the cracks
I don't answer all my emails either

@Slereah I ve noticed :p

@glS it's hard to predict

@ACuriousMind first time I hear pfizer referred to as "Biontech" (assuming that's what you mean)

1:33 PM
Hey I'm not even supposed to be doing physics right now
I'm supposed to work

@glS I think that's a German quirk, everyone calls it that here
(probably not coincidentally, Biontech is a German company :P)

Sometimes I wish google would let me search for latex expressions

@MoreAnonymous there should be. It used to be in your activity tab of the profile at least meta.stackexchange.com/a/229139/276202. Though things might have changes because I can't find it now

@MoreAnonymous yes, I write to myself sometimes but I find it annoying and don't reply :(

@glS it's still there, in the "impact" box under the number of people reached

1:40 PM
ah, right, found it
@MoreAnonymous also more directly, physics.stackexchange.com/users/flag-summary/332703

@Slereah why doesnt someone make that>
?

It would be somewhat challenging

@glS that link goes nowhere, 332703 is not a valid physics.SE user ID

@ACuriousMind Yea was about to say

you used MA's chat ID, but you ned to use the physics.SE ID: 150174

1:43 PM
ah, sorry
@Slereah I think there is some search engine for equations. I saw it on some ad on math.SE some time ago
(I mean, not google, only SE Wiki and some other sources, but still useful)

5

Given a connected smooth manifold $M$ and an invertible jet $\xi \in {\rm inv} J^k_p(M,M)_q$, what are the required conditions for the existence of a diffeomorphism $\phi \in {\rm Diff}(M)$ such that $j^k_p \phi = \xi$? What about if we want $\phi$ to be isotopic to the identity? Given a smooth f...

Closest thing I can find to what I want is this
But there are zero references
Though I did find his thesis :

@fqq Hmm ... I would have written to you too but the response rate seems low :P Are you a condensed matter physicist or cosmologist or particle physicist?

2:19 PM
Anybody seen this question before or know where I can find a solution? I’m giving a supervision where I’m meant to teach the answer but it has me stumped

@JakeRose Use the sudden approximation?

It’s meant to use time dependent perturbation theory
so not necessarily in the spirit of the question

@JakeRose Well to be fair the sudden approximation uses time dependent perturbation theory

I don’t think the question is pushing that route just from my experience if the guy who set the question
I was just going to find a a Textbook solution

Really? Seems a lot like this question to me
3

I am trying to solve a problem where a system (a quantum harmonic oscillator) is suddenly perturbed by a large field of strength $E$. I want to calculate the transition probability for it to go from the ground state of the original Hamiltonian ($\frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$) to the $... 2:30 PM I don’t see it personally It very specifically washes you to use the Dyson eq *wanrs Isnt the sudden approximation derived by that? The adiabatic theorem is a concept in quantum mechanics. Its original form, due to Max Born and Vladimir Fock (1928), was stated as follows: A physical system remains in its instantaneous eigenstate if a given perturbation is acting on it slowly enough and if there is a gap between the eigenvalue and the rest of the Hamiltonian's spectrum.In simpler terms, a quantum mechanical system subjected to gradually changing external conditions adapts its functional form, but when subjected to rapidly varying conditions there is insufficient time for the functional form to adapt, so the spatial probability... sudden/adibatic approximation ? Probably, idk. Cambridge question sheets like you too very much work out stuff like that yourself 3 hours later… 5:53 PM Why is the conformal group$SO(n) \times \mathbb{R}$, but conformal transformations also admit that one transformation that's neither a rotation nor a dilation 6:13 PM @MoreAnonymous I've worked mostly on statistical physics/quantum many-body so "condensed matter" I guess @Slereah it's not the same group What group may it be "conformal group" is one of these annoying terms that has slightly different definitions depending on context so whatever you're reading that says the "conformal group" is that doesn't actually mean the same object as you're thinking of when you say "conformal transformations" Dagnabbit But otoh, the conformal transformation that has "the conformal transformation" has the whole$g \to e^f g$property, which certainly looks like a rotation + dilation locally? I don't know where the Moebius transformation comes from Or is it something that only shows up as a global thing rather than locally?$O(n) \times \mathbb{R}$is the "conformal orthogonal" group, i.e. linear transformations that preserve a quadratic form up to a global scalar Like I could be wrong, but as far as I can tell the conformal symmetry comes from a local gauge group of$CO$6:23 PM what you're probably thinking of is the group of global conformal transformations, which is$\mathrm{SO}(p+1,q+1)$Hm (for$p+q>2$) Are they related in some sense? for$p+q=2$, the global group gets pretty subtle the best mathematical treatment of conformal transformations I know is the beginning of the book on CFT by Schottenloher, downloadable here Will Di Francesco do 6:24 PM no :P It's the one that's right next to me why must everything be hard Got a five word summary of the connection between the two? what two?$CO(p,q)$and$SO(p+1, q+1)$I suppose I have no idea what$\mathrm{CO}(p,q)$is The conformal group 6:26 PM and what do you mean by that?$CO(p,q) \cong \mathbb{R} \times O(p,q)$as I said, "the conformal group" can have different definitions in different contexts Preserve the metric up to a factor @Slereah so that's just the group of linear transformation on a vector space and preserve the inner product up to a factor Ah so it's a global effect? 6:27 PM this is not the group of diffeomorphisms on a (pseudo-)Riemannian manifold that preserve the metric up to a factor, which is what you probably mean by "conformal transformation" Perhaps, but otoh, I think I've seen the "conformal transformation" in a CO context too? But I think that may be on a different space, tho possible Like it acts on the Moebius space I think? but the latter is what one usually means by "conformal transformation" in the context of CFT Involving the point at infinity 6:30 PM what's a "Möbius space" supposed to be? are you talking about the conformal compactification? Yeah Actually I think we may be talking about the same group? But I think$CO$is the stabilizer while$O(p+1, q+1)$is the full group And the Moebius space is the quotient of the two I guess maybe it's just that the Moebius transformation isn't a "rotation"? stabilizer of what? the stabilizer subgroup of$O(p+1, q+1)$a stabilizer subgroup is the subgroup that stabilizes some point or subspace in a space being acted upon by a group there's no "the stabilizer" there's only e.g. "the stabilizer of the point at infinity" or "the stabilizer of the origin" Hm Trying to find its description as a Klein geometry Doesn't help that there appears to be multiple conformal geometries apparently depending on what you stabilize indeed "There exist different representations of the groups G and H." 😡 I think the difference between the two seems to be that one acts on a vector space and the other on the sphere 6:56 PM According to this both$CO(p,q)$and$O(p+1, q+1)$are involved in the sordid affair But I'll have to read it for a clearer picture "Since the corresponding group$CO(n)$of conformal geometry is not compact, one might expect some compact manifolds to have noncompact conformal groups. There is a simple example: the Euclidean sphere has conformal group$SO(1, n + 1)$." It would help if they did not call everything a conformal group is "special conformal transformation" a term math people use or is it only physicists 6 First I have the definition of a conformal map. Let$(M,g)$and$(M',g')$be two pseudo-Riemannian manifolds of same dimension. Let$U\subset M$and$V\subset M'$, we say that a smooth map of maximal rank$\Phi : U\to V$is a conformal map if there is some smooth$\Omega : U\to [0,+\infty)$such ... Looks like it's a sham It's only the conformal group for compactified Minkowski space 7:26 PM In special relativity, when one system K' moves with velocity$v$in relation to a system K, we find out that$\Delta t= \gamma \Delta t' > \Delta t'$which is time dilation. In this example it is not implied that any system is accelerating, and I assume is not the case. But then I was reading about the proper time and I find this part odd: "An accelerated clock will measure a smaller elapsed time between two events than that measured by a non-accelerated (inertial) clock between the same two events." But in the example I gave earlier, the time dilation happens without the need of one of the system's to accelerate. Hm, in the 2D case I guess that would be$\mathrm{CSO}(2) = \mathbb{R}^+ \times \mathrm{SO}(2)$, but for the sphere,$\mathrm{SO}(3, 1)$which is... I guess$\approx \mathrm{SO}(3) \times \mathbb{R}^3$???$\mathrm{SO}(3)$is the appropriate motion group for the sphere but why$\mathbb{R}^3$I could understand$\mathbb{R}$but that's a lot more than needed Is this a reply to my question? Because if that is the case,I am not familiar with the notations you are using It is not. I guess the SCT does take$n$parameters so that's exactly where that goes It's just weird that the same concept has an entirely different number of dimensions depending on the topology @imbAF People use "time dilation" for two different effects - on the one hand for the effect due to relative velocity you wrote down first ("kinematic time dilation") and on the other hand for the effect due to acceleration(/gravity in GR) 7:42 PM and in both cases an observer who is, I don't know if it is correct to say, in the reference frame K, measures in both cases a bigger value for the time between two events? I don't really understand the question but both cases involve observers disagreeing in how much time they measure, sure that's why it's called "time dilation" in both cases, after all :P which observer counts more time yes i know it should be the observer who does not move with relativistic speed @imbAF kinematic time dilation is symmetric what you mean by that? there's no one observer who objectively measures a "greater" time than the other 7:44 PM doesn't that depend on where the clock is located? ok thx @ACuriousMind one more thing, related to relativity but not this particular topic. If we have two events E1 and E2 and i.e E2 is not in the light cone of E1, in our lecture we said (and not explained): " LT is possible, so that$t'=0$" What it is meant with this? that's probably a confusing way to state that for space-like separated events there's always a frame where both events happen at the same time yes that's the end result that is derived from that expression I wrote. I just can't see the connection they're just saying that there exists a Lorentz transformation that reduces the time difference between the events to 0 The bad news keep on coming : the SCT comes from a second order part of the coordinate transformation this is exactly the same as saying what I said, there's nothing to derive 7:50 PM Meaning that it's a lousy 2-frame bundle issue ok Why does the sphere have the SCT but not the plane even though the algebra exists for both, is it one of those obstruction issue There's no global section of the whatever again, read Schottenloher's first few chapters Ah well, let's try I will proceed to purchase it legally he's very careful about what transformations are defined globally where, and it's really good compared to the "infinitesimal speak" in most physics texts @Slereah I already gave a download link from his own webpage above it's out of print so he offers it for free, if in a somewhat annoying format 7:53 PM No can do, I can only steal books No free books for me Is it also possible to experience time dilation because both phenomenons happen at the same time ? sure You do it all the time If you move on earth It's in fact something to account for for really precise relativity experiments I see good to know But why is einstein the face of relativity and not lorentz, who If i am not wrong, did he or did he not wrote about his transformation before of einsteins theory of relativity ? You will have to ask historians on that one kinda odd 8:02 PM Oh the SCT has no global flow is the problem? 8:21 PM Hm Actually I suspect something sinister is afoot I think that as a G-structure, the conformal group is$CO(n)$But the SCT actually acts on the second order frame bundle "Donde$\mathfrak{co}(n)_1$es la primera prolongaciòn del algebra de Lie$\mathfrak{co}(n)_1$de$CO(n)$" Mama mia when we find the four velocity or four force, why do we consider the proper time? And not the time that the observer counts when he observes the event? 8:40 PM Four-velocity is with respect to the invariant$ds$:$u^{\mu} = \frac{dx^{\mu}}{ds}$. Using$ds = c d \tau$it has a natural expression in terms of proper time but where does this come from ?$u^{\mu} = \frac{dx^{\mu}}{ds}$in our lecture we took$u^{\mu} = \frac{dx^{\mu}}{d\tau}$which it makes sense fundamentally, since velocity is the derivative of position while in what you wrote, we are deriving 4 position with ds, which i don't know what to name$ds$is an invariant, so a velocity with respect to an invariant parameter will transform as$dx^{\mu}$does 9:01 PM ^there is some prolongation shenanigans going on between$O(n+1, 1)$and$CO(n)$9:18 PM Decomposing as$\mathbb{R}^n \oplus\mathfrak{co}(n) \oplus\mathbb{R}^n$What's$CO(n)$? hey all Conformal orthogonal group does the SCT do anything to the frames? 9:52 PM Apparently the Jacobian is ~ the rotation times a scale factor SCT doesn't enter into it If this is a gauge trasformation$\vec A(\vec r,t)'=\vec A(\vec r,t)+ \nabla g(\vec r,t)$the this is the lorenz gauge$\frac 1{c^2} \frac{\partial \phi}{\partial t}+\nabla \vec A(\vec r,t)=0$. Does this imply that$\frac 1{c^2} \frac{\partial \phi'}{\partial t}+\nabla \vec A(\vec r,t)'=0\$ must also satisfy this condition?

10:29 PM
I guess looking at it, all the angles are preserved?