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5:19 AM
Sunrise in Chester. It is almost worth starting work at 5 a.m. when the view from my desk includes sunrises like this.
Almost ...
6:20 AM
@EmilioPisanty next round: meta.stackexchange.com/q/332974/271002
7:00 AM
@Loong thanks for the heads up
@ACuriousMind Achtung
7:28 AM
Q: Translation invariance of point particles as a field theory

SlereahThe case of point particles, relativistic or not, can be treated as a field theory in general, ie for the $(1+1)$-dimensional case this is the theory of a field theory on the vector bundle $$\pi : \mathbb{R}^2 \to \mathbb{R} $$ with $x(t)$ a section of that bundle. I thought therefore to look a...

plz halp
Actually thinking about it, why don't we talk about symmetries of field theories for scalar fields?
$\phi \to \phi + \alpha$ is also an entirely fine field symmetry for massless fields
Although I guess that may simply be the conformal symmetry
Although... not quite, no
same problem as before
7:45 AM
Are we supposed to use an affine bundle
8:05 AM
"An affine space or affine linear space is a vector space that has forgotten its origin."
@KyleKanos Typically well-educated people can be more motivated by secrets. The urge to know and to learn things, is more strong in them.
Well there's $A_{\mu} \to A_{\mu} + \partial_{\mu} \alpha$
8:56 AM
@bolbteppa True
Is the tangent space of the $U(1)$ bundle an affine bundle
What's a bundle
I'm sure if the answer is somewhere it's in Naber
Naber is literally two whole books on EM from the bundle perspective
The tangent space to a $U(1)$ principal bundle is going to be (part of) the $u(1)$ Lie algebra right
Hopefully yes
It does seem to be affine
Does this derive from some fundamental idea or is it just decided by fiat
We need the bundle to do this therefore it is an affine bundle
I guess for generality it is best to treat the fields as affine bundles
9:18 AM
Skimming it they seem to be saying the connection bundle is affine (like in GR) but not the $U(1)$ starting point
I don't think there's any general issues with that
Well $U(1)$ isn't even a vector space
it's not gonna be affine
Though the $U(1)$ action on it is basically the equivalent of an affine action I s'ppose
2 hours later…
11:17 AM
@JohnRennie, Hi. Could you please answer my query - Do you feel Halliday, Resnick and Jearl Walker is a good book for physics. I am interested in physics and know till High School level.
@Intellex I can't help I'm afraid. I'm not familiar with the Halliday book.
@JohnRennie, Or could you please suggest a good book for Physics?
@Intellex yes it is, the calculus-based one, and GRE's use it
@Intellex I can't help with books I'm afraid as I'm out of touch with what books are good these days.
@JohnRennie most good books are before all of our time :p
11:39 AM
@bolbteppa Thank you for your reply. I am not preparing for GRE but I wish to read since I am interested in Physics.
@JohnRennie No problem. Thank you for spending your time helping me.
@bolbteppa Yes. I agree.
What are your views on Feynman Lectures?
You will learn how to do physics better from Halliday
What do you mean by "do physics" ? Do you mean numericals?
Thanks @bolbteppa !
12:43 PM
@bolbteppa I always read your name as Bob Teppa
You're supposed to read it as
Professor Vito Volterra (, Italian: [ˈviːto volˈtɛrra]; 3 May 1860 – 11 October 1940) was an Italian mathematician and physicist, known for his contributions to mathematical biology and integral equations, being one of the founders of functional analysis. == Biography == Born in Ancona, then part of the Papal States, into a very poor Jewish family: his father was Abramo Volterra and mother, Angelica Almagia.Volterra showed early promise in mathematics before attending the University of Pisa, where he fell under the influence of Enrico Betti, and where he became professor of rational mechanics...
adjusting my sleep schedule to normal hours is proving difficult
@RyanUnger it's impossible to do
I really want to know the whole integral equation stuff properly
1:32 PM
Q: Weak Solutions to the Einstein Equation across a Junction

Naveen BalajiConsider the principle part, i.e., the part which contains the highest derivatives of the metric (which is the $2^{nd}$ derivative) is $$\mathcal{P}\{R_{ab}\}=\frac{1}{2}g^{cd}\left(\partial_{a}\partial_{c}g_{db}+\partial_{b}\partial_{c}g_{da}-\partial_{a}\partial_{b}g_{cd}-\partial_{c}\partial_{...

For those interested in GR
1:47 PM
@NaveenBalaji There's the treatment of stress energy tensor at junctions in Visser if you want
2:25 PM
@Slereah Yeah, in fact, that was what I referred when I learned the subject (and Poisson too). This question addresses the shell problem in specific with only the principle parts of the Einstein tensor considered.
3 hours later…
5:02 PM
Q: If the string is inelastic (conserves Hamiltonian engergy) then by Gauss's theorema egregium there is only one equation of motion solution, right?

Dr. Terence B AllenThe idea the string equation of motion has many solutions results because the differential wave equation cannot define a Riemann integral on mass-and-spring elements. As the atoms get smaller under large numbers they do not converge on a limit while the spring constant is inexplicably conserved. ...

"Since the tangent-cotangent product space of the Hamiltonian is fiber-co-fiber space, the manifold is time and space averaged."
5:21 PM
@Slereah I'm not sure what you're really asking - since you haven't given any reference that would claim that you need to restrict yourself to vector bundles, nothing is stopping you from considering the tangent bundle as an affine bundle. So what's your problem?
@ACuriousMind yeah I suppose that is the solution
Although now I'm wondering
Do I even have to consider affine bundles at all?
Are there any physical reasons to not consider non-linear transformations?
Not really
But generic fiber bundles are a rather weak structure, so there's a mathematical reason to not want to relax the notions that far if one doesn't need to :P
Are there any reasonable system that has nonlinear symmetries?
Sine Gordon has a bunch of symmetries IIRC, mb that one
let's see
@Slereah Gauge symmetries are realized non-linearly in spontaneously broken phases with Goldstone bosons.
yeah so I guess it's best to consider them
Although I guess it's probably useful to sometimes restrict them to affine or linear transformations
Are all Lie symmetries linear btw?
I guess all the ones generated by a principal bundle are
5:29 PM
@Slereah Gauge symmetry groups are usually Lie groups, so the SSB statement above applies to them.
Dang it :V
Why is nothing ever simple
I wonder if it's doable to show that the point particle has no nonlinear symmetries
dun dun
@Slereah However, you can get away with only considering the linear representation of a subalgebra and inducing the non-linear realization of the full group as in en.wikipedia.org/wiki/Nonlinear_realization
I should check Henneaux too
i'm sure there's some weird ass symmetries in there
Alas, in full generality, gauge algebras are not even required to close, i.e. there are "gauge theories" in the broad sense of constrained Hamiltonian systems which do not really have a gauge group and cannot be described in the principal bundle language at all
Henneaux/Teitelboim also has some examples for the failure of the Dirac conjecture, which provides another class of theories we can't really treat with standard methods.
yeah that is what I suspected
I mean
just consider this dumbass action :
$$S[\phi] = \int_\Omega 1 d^nx$$
5:38 PM
I like that name
@Slereah Well, you likely should restrict your action functionals to those that actually have equations of motions :P
$0 = 0$ is a fine equation of motion
Of course the theory that predicts nothing is symmetric under everything!
Although $\phi$ has some (all) gauge freedom
@Slereah Let me say it like this: You should require your action functionals to have isolated extremal points.
5:39 PM
I'm trying to come up with really dumb actions with weird symmetries but it's hard to think of ones that are continuous
@ACuriousMind Could be worse
Henneaux had $$S[q] = \int q(t) dt$$
With EOM $1 = 0$
why is basic physics so complicated :V
@Slereah Because it needs to support a world as complicated as ours :P
Even zero dimensional QFT is bad
Why is it complicated
it's just a point
Although fortunately zero dimensional classical mechanics is alright
It's just the stationary point $x = 0$
6:23 PM
I've got a dumb idea
$$S = \int \ddot{x}^2 dt$$
isn't zero-dimensional QFT = stat mech?
no I'm not sure there's any stranger symmetries here
(up to wick rotation etc)
@Semiclassical Zero dimensional QFT is just the Lebesgue integral
6:25 PM
@Slereah No, bad Slereah! (physics.stackexchange.com/a/44183/50583)
Since functions from $\{0\}$ to $\mathbb{R}$ are $\approx \mathbb{R}$
@ACuriousMind I know it's bad!
I just wonder about its symmetries
Just call it symmetries of a functional, if you prefer :p
what are you talking about
@ACuriousMind this answer is wrong, there are many natural higher order actions
say the Willmore energy
@RyanUnger math
@RyanUnger Maybe you missed that Qmechanic is talking about quantum theories?
why would he be doing that
6:39 PM
...because he's not Cmechanic? :P
Yo @Semiclassical
I'd like to have a crack at the 2D quantum harmonic oscillator, but with funky stuff in the middle
what sorta funky stuff?
i.e. I don't care about irregular behaviour at the origin, I'll fix that later
6:44 PM
I just want the solutions to the radial equation, for arbitrary energy, that are regular at infinity (i.e. $\mathrm{poly}(r) \exp(-r^2/2)$)
for zero angular momentum I've pinned them down to the parabolic cylinder functions
hmm. remind me what the radial part of the SE equation looks like?
do you know off the top of your head what class of special functions I should be looking at?
the relevant Schrödinger equation has the form $$-\frac12 \frac{\partial^2\psi}{\partial r^2} -\frac{1}{2r} \frac{\partial \psi}{\partial r} + \frac{ \ell^2}{ 2r^2}\psi(r) + \frac12 r^2 \psi(r) = E \psi(r)$$
6:47 PM
hmm, yeah.
what are the special functions if you do assume it's bounded at the origin
if I feed it in that form to Mathematica it returns a confluent hypergeometric and a Laguerre polynomial, times a bunch of prefactors and an exploding exponential, and neither is regular at infinity except for the usual energy eigenvalues
@RyanUnger laguerres times gaussians
the change of variables I'd be inclined to consider is $r=e^u$, so that $\partial_u = r \partial_r$
and $\partial_u(\partial_u-1)=r\partial_r( r\partial_r-1)=r^2\partial_r^2$
@Semiclassical wooot
that looks insane
meh. it's the usual change of variables to turn a Cauchy-Euler ode into one with constant coefficients.
it turns the ode into $$-\frac12 \partial_u^2 \psi +\frac{\ell^2}{2}\psi+\frac12 e^{4u}\psi=E\psi$$
the fact that you're only interested in good behavior at $r=\infty$ translates into behavior at $u\gg 0
6:59 PM
I mean, sure
but this has to have been done before, right?
i'm just trying to find a change of variables which makes it look familiar
I'm hoping I just need to identify the ODE and that'll lead me to some suitable Authoritative Resource
oh. should've been $e^{2u} E \psi$ after the change of variables
I tried setting $\psi(r) = r^\ell e^{-r^2/2} \phi(r)$, which then has one of its solutions as a bare confluent hypergeometric
but it's not regular, and the other solution is $r^{-2\ell}$ times a different confluent hypergeometric
ODE comes to $$ -r\phi''(r) + (1-2r^2-2\ell)\phi'(r) + 2r(1-E+\ell) \phi(r) = 0 $$
in case that tells you anything
looking at the solutions I get when $\ell=0$ in Wolfram, I see why you resist my change of variables
it doesn't make things any nicer
i think your ode is in the form of the generalized hypergeometric ODE, but pFq gives me headaches
oh, wait. ignore that. i was thinking of the wrong thing
7:19 PM
Bam, got it
hypergeometric $U$
thanks, guys =)
oh, the tricomi one? (had to look it up) neat
1 hour later…
8:27 PM
My heart, your heart, tide up like two ship.
Drifting, weightless, waves try to break it.
Why is it so hard to say it.
I don't wanna wait till you're gone. Cuz you make me strong...
I'm sorry if I say I need ya,
but I don't care, I'm not scared of love.
How to prove that the action is only invariant under diffeomorphism of the field if it's a Galilean transformation
$x(t) \to f(x(t))$, with the time derivative being $$\frac{d}{dt}f(x(t)) = f'(x(t)) \dot{x}(t)$$
anybody here have visual studio 2019?
@enumaris Yes, but I've never used it :P
Can I say that $$\int f(x) g(x) dx = \int f(x) dx$$ for all $f$ iff $g(x) = 1$
Is that a theorem that is a thing
Do you know if this functionality is in it: blogs.msdn.microsoft.com/onecode/p/devassistant
8:36 PM
Oh wait
@enumaris I'm pro
Shuffling it it's the fundamental theorem of calculus of variation
Specifically the part where you can search a "how do I" question and it pops up example code snippets
So indeed all transformations are $f'(x)^2 = 1$
So $f'(x) = \pm 1$ or $f(x) = \pm x(t) + a$
That works
now to prove all the symmetries of the MSSM
I think $n$-dimensional point particles should be about as complicated
I don't have access to visual studio to see if the functionality is already fully integrated or if they just discontinued that functionality
8:39 PM
just $(\vec{f}')^2 = 1$ instead
@enumaris Dunno. I could check tomorrow
All the links to that plug in give me 404 error now
I know they integrated that functionality into Bing itself and you can try it
I wonder what happened to it in the IDE lol
@ACuriousMind that would be very helpful :D
8:51 PM
@NovaliumCompany if you could check if this functionality exists in VS2019 that'd be helpful :D
@Slereah great find
10:02 PM
@Slereah "fundamental lemma" not "theorem" ;)
In mathematics, specifically in the calculus of variations, a variation δf of a function f can be concentrated on an arbitrarily small interval, but not a single point. Accordingly, the necessary condition of extremum (functional derivative equal zero) appears in a weak formulation (variational form) integrated with an arbitrary function δf. The fundamental lemma of the calculus of variations is typically used to transform this weak formulation into the strong formulation (differential equation), free of the integration with arbitrary function. The proof usually exploits the possibility to choose...
what the difference is
Going by the etymology 'lemma' means 'something assumed' which means it's basically something not important enough to be a theorem but something you absolutely need to assume to do more important things
Lemmas are theorems
"Lemma" just means you think it's not important enough!
But I think it's plenty important
In that case a "fundamental lemma" might be an oxymoron...
but lemmas should be proved...unlike axioms
10:20 PM
Not a whole lot of axioms in analysis tho
They tend to be lower down on the totem pole
I just make every statement I make an axiom
this way I can never be wrong
It certainly makes proofs easier
Although you may end up with an inconsistent axiom set
It is an oxymoron until you look at the lemma, it's like 'if $\int f(x) g(x) dx = 0$ for all $g$ then $f$ must be zero', it's completely obvious, but it is used all the time to invoke the EL equations, and it can fail for some monster examples no sane person would think of, so it's trivial and the most fundamental thing ever in CoV at the same time
I would call it an oxygenius
I don't trust any obvious statement in mathematics
10:29 PM
Physics takes all the good from math and thankfully leaves all these boring counterexamples that fail (usually for bizarre real functions it took 200 years to think up)
intuitive but very hard to prove
@bolbteppa oh sure
Until they pop back up!
All the paths in path integrals are continuous nowhere differentiable!
is that all paths or almost all paths
The other paths are of measure 0
so almost all paths
10:33 PM
My favorite convoluted physics example is probably the spacetime constructed to be geodesically complete but with a singularity
It's the worst spacetime ever concocted, and I've seen a few
what definition of singularity is that one using
cus isn't geodesic incompleteness the usual definition
in which case you'd have a contradiction
like it's geodesically complete but some curvature scalar is not bound?
There are non geodesic curves of bounded acceleration which are of finite proper time
It is an extremely convoluted example
so it's...geodesically complete but some "attainable" curves end in singularities
yeah good luck with that...
like some singularity you'd have to accelerate into to hit and can't hit if you just moved naturally
who had enough free time to come up with that?
10:38 PM
It has to be of bounded acceleration because of course, if unbounded, even Minkowski space is singular
It was Robert Geroch
Basically it's an infinite stack of spacetimes designed to trap geodesics
So that only an accelerated curve can emerge
I wonder how much free time he had to concoct it lol
Hell he's not the only one to have done it
There are two such examples
fun times...
does it violate the energy conditions?
10:44 PM
Well it's based on AdS and also has a lot of conformally minkoswki stuff
So in places yeah
Although I don't know if it's mandatory for such a spacetime
Nobody cares that much about the topic
I don't think there's a proof for it
I mean there's a bunch of space times with CTCs...but we can sort of just exclude those...
like the Godel spacetime
GR is just full of pathological cases we just sort of hope real hard aren't physical
Earman wrote an entire book on the topic
It's entirely possible that our universe could at some point just naturally evolve with a naked singularity for no reason
Just take the natural spacetime development of our universe, remove an open set from it
This is still an entirely acceptable spacetime
"Thus, the lesson to learn is that Energy is only conserved if there's translational time symmetry in the problem.

Which brings us to General Relativity: in several interesting cases in GR, it's simply impossible to properly define a "time" direction! Technically speaking, this would imply a certain global property (called "global hyperbolicity") which not all 4-dimensional spacetimes have. So, in general, Energy is not conserved in GR."
10:50 PM
It's kind of assumes any spacetime will be inextendible but really there's nothing against it
Q: Is energy really conserved?

CasebashIn high school I was taught energy was conserved. Then I learned that nuclear reactions allow energy to be converted into mass. Then I also heard that apparently energy can spontaneously appear in quantum mechanics. So, are there any other caveats with the conservation of energy?

What's the deal with energy conservation in GR
Define energy
Although really
Things are even worse than you'd assume
It's not just the lack of time translation
Time translation charge
Take Minkowski space with the origin removed
That's a singularity at every time t
From that singularity, an arbitrary EM radiation can emerge or get absorbed
You could start with zero energy in flat soace and end up with arbitrary energy
I think this is one of those "we don't define the action properly at the boundary" sort of issue
10:56 PM
We just always assume that there's onlynboundary conditions on the hypersurfaces and everything works out wrt the action
Basically this is a mess
Oh yes
It is
Although you can of course do the split $g = \eta + h$
And then it's just an awful field theory that is time invariant
Once you start fucking with the topology and geometry of spacetime you have to throw out a lot of notions
"We will not delve into definitions of energy in general relativity such as the hamiltonian (amusingly, the energy of a closed universe always works out to be zero according to this definition),..."
from my understanding you can define energy conservation in different ways in GR. Locally, you can describe it as the covariant derivative of the stress-energy tensor has to vanish. But globally you would require some time-like Killing field to exist. But not all space-times have a time-like Killing field. Even non-pathological ones. For example...the FLRW universe...
That's why a lot of times people just assume a very mild spacetime
11:01 PM
this is due to a problem in GR that global quantities are often ill defined
but my understanding could very well be flawed :)
At the end of the day, when you remove a gravitational force from the equation and equate geometry with gravitation, you can no longer easily define gravitational potential energy (except when you have a time-like Killing field).
And so...things like gravitational red-shift become a purely geometric effect...but now you have no good way to describe where the lost energy is going
Well it's not that bad
but IF you have a time-like Killing field like in the Schwarzschild solution then it's all restored :)
You can define a gravitational potential energy
Einstein whatever pseudotensor
in spacetimes without time-like Killing fields?
11:04 PM
I think there's other ways you can try, but it's always not quite satisfactory...
but when you do have a time-like Killing field it's pretty satisfactory
Some people don't like it because it's not a tensor
But I'm not prejudiced
right pseudotensors are eh
I remember something in Wald about it
but it seemed like you need to have some particular set of coordinates to make it work or something?
Well it's a pseudotensor
Ergo not covariant
11:08 PM
But that is fine
like in the sense that the dog trapped in a burning building is fine?
"One way to get round this is to pick one coordinate system, and transport vectors so their components stay constant. Partial derivatives replace covariant derivatives, and Gauss's theorem is restored. The energy pseudo-tensors take this approach (at least some of them do)"
"Indeed, the issue of energy in general relativity has a lot to do with the notorious "problem of time" in quantum gravity... but that's another can of worms." oh good
seems legit

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