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
0

The 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?

Sure

Thanks @bolbteppa !

12:43 PM

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
0

Consider 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 -2 The 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 IF IT DOESN'T 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 thx 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$$It has ALL THE SYMMETRIES 5:38 PM I like that name @Slereah Well, you likely should restrict your action functionals to those that actually have equations of motions :P 0 = 0 is a fine equation of motion Of course the theory that predicts nothing is symmetric under everything! Although \phi has some (all) gauge freedom @Slereah Let me say it like this: You should require your action functionals to have isolated extremal points. 5:39 PM I'm trying to come up with really dumb actions with weird symmetries but it's hard to think of ones that are continuous @ACuriousMind Could be worse Henneaux had$$S[q] = \int q(t) dt$$With EOM 1 = 0 why is basic physics so complicated :V @Slereah Because it needs to support a world as complicated as ours :P Even zero dimensional QFT is bad Why is it complicated it's just a point Although fortunately zero dimensional classical mechanics is alright It's just the stationary point x = 0 6:23 PM Heyyy I've got a dumb idea$$S = \int \ddot{x}^2 dt$$hm isn't zero-dimensional QFT = stat mech? no I'm not sure there's any stranger symmetries here nvm (up to wick rotation etc) @Semiclassical Zero dimensional QFT is just the Lebesgue integral mmkay 6:25 PM @Slereah No, bad Slereah! (physics.stackexchange.com/a/44183/50583) Since functions from \{0\} to \mathbb{R} are \approx \mathbb{R} @ACuriousMind I know it's bad! I just wonder about its symmetries Just call it symmetries of a functional, if you prefer :p what are you talking about @ACuriousMind this answer is wrong, there are many natural higher order actions say the Willmore energy @RyanUnger math @RyanUnger Maybe you missed that Qmechanic is talking about quantum theories? why would he be doing that 6:39 PM ...because he's not Cmechanic? :P 4 Yo @Semiclassical I'd like to have a crack at the 2D quantum harmonic oscillator, but with funky stuff in the middle what sorta funky stuff? i.e. I don't care about irregular behaviour at the origin, I'll fix that later 6:44 PM gotcha I just want the solutions to the radial equation, for arbitrary energy, that are regular at infinity (i.e. \mathrm{poly}(r) \exp(-r^2/2)) for zero angular momentum I've pinned them down to the parabolic cylinder functions hmm. remind me what the radial part of the SE equation looks like? do you know off the top of your head what class of special functions I should be looking at? nope the relevant Schrödinger equation has the form$$-\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$$the fact that you're only interested in good behavior at r=\infty translates into behavior at u\gg 0 6:59 PM I mean, sure but this has to have been done before, right? yeah i'm just trying to find a change of variables which makes it look familiar I'm hoping I just need to identify the ODE and that'll lead me to some suitable Authoritative Resource oh. should've been e^{2u} E \psi after the change of variables I tried setting \psi(r) = r^\ell e^{-r^2/2} \phi(r), which then has one of its solutions as a bare confluent hypergeometric but it's not regular, and the other solution is r^{-2\ell} times a different confluent hypergeometric ODE comes to$$ -r\phi''(r) + (1-2r^2-2\ell)\phi'(r) + 2r(1-E+\ell) \phi(r) = 0 $$in case that tells you anything looking at the solutions I get when \ell=0 in Wolfram, I see why you resist my change of variables it doesn't make things any nicer i think your ode is in the form of the generalized hypergeometric ODE, but pFq gives me headaches oh, wait. ignore that. i was thinking of the wrong thing 7:19 PM Bam, got it hypergeometric U thanks, guys =) oh, the tricomi one? (had to look it up) neat 1 hour later… 8:27 PM My heart, your heart, tide up like two ship. Drifting, weightless, waves try to break it. Why is it so hard to say it. I don't wanna wait till you're gone. Cuz you make me strong... I'm sorry if I say I need ya, but I don't care, I'm not scared of love. Hm How to prove that the action is only invariant under diffeomorphism of the field if it's a Galilean transformation x(t) \to f(x(t)), with the time derivative being$$\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 $f(x) = \pm x(t) + a$
Heyyyy
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
but
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

o.O

@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
Sure
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

oh
so it's...geodesically complete but some "attainable" curves end in singularities
lol
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

Yeah

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

right

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

hmmm

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

26

In 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

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?

Sure

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

en.m.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor

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
right...so...
eh...

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