The h Bar

12:24 AM

hey guys

just saying hello, today will be my quasi free day, so might just hang out and lurk a bit :D

2:22 AM

@Blue One thing that I really like about Abraham Marsden is that is its very self contained unlike Arnold, which needs a good bit of familiarity of differential geometry notation(I feel notation is a big thing to get a hold of differential geometry)

Slereah

7:09 AM

morning

Balarka Sen

@Slereah u didn't respond to my challenge

Slereah

which

Balarka Sen

this

Slereah

Is it the whole QFT as a worldline thing

Balarka Sen

something like that

but on $\infty$-stacks

Slereah

7:14 AM

What is a $\infty$-stack

Is that the famous stack

to exchange

on stackexchange

Balarka Sen

It's a sheaf with values in $\infty$-groupoids

Slereah

Alas I'm not gr8 on category theory

Do u know a thing, btw

Does the worldline formalism actually reproduce QFT

I wonder because there's that whole thing about how the perturbation theory thing doesn't run over every possible state

Balarka Sen

Tangential fact: Prior to stackexchange.com there was a coding-related Q&A site called expertsexchange.com

which people used to call expert sexchange.com

alas they hyphenated it at some point

Slereah

Like perturbation theory does the sin of exchanting the sum and integral

Balarka Sen

@Slereah I know staggeringly little about QFTs if that wasn't clear from my challenge

Slereah

7:17 AM

BUSTED

Balarka Sen

indeed

Slereah

Speaking of worldline formalism

I am learning aboot point particle BRST quantization

(STILL)

It is not an easy task

what with the general covariance

Balarka Sen

I have heard of BRST cohomology

Slereah

Apparently the big difference between the Goto-Nambu action and the Polyakov one is that the mass-shell constraint is first class and second class in them or something

I am v. bad at constraints

Albas

8:47 AM

Yo @BalarkaSen

Balarka Sen

Hey

Albas

What's up

Balarka Sen

Not much; I fly to Bangalore tomorrow

Albas

Ahhh nice nice. All the best at ISI

Balarka Sen

Thanks

How's symplectic geometry going, @Albas?

Albas

8:57 AM

Doing lie brackets right now, will slowly reach there

I was thinking about something. When we talk about vector bundles, we say that locally the preimage of a point should have the vector space structure. It might not yield anything much but can having the structure be an infinite dimensional vector space lead to anything interesting

Balarka Sen

You can do that

Albas

Though I do understand how a symplectic group structure comes from Hamiltons equations of motion and canonical transforms

Balarka Sen

That is precisely what I don't know :)

Albas

9:17 AM

Sorry @BalarkaSen but I am not well versed with it and hence I wont be able to take an attempt to explain it

Balarka Sen

No worries, tell me when you understand it better

Albas

The basic idea I think is that the jacobian is a symplectic matrix

We use that to obtain a symplectic group structure from the canonical transform

Balarka Sen

Aha

Jacobian of a canonical transform?

Albas

Yea

Slereah

You can have fibers that are infinite dimensional

The tensor bundle is an example

Also whatever fiber continuous spin representations have

or the Hilbert bundle

Balarka Sen

9:57 AM

I found an English translation of Lemaitre's paper on "Quaternions and Elliptic Spaces" which I am reading now because google led me down a rabbithole

John Rennie

11:52 AM

I have just succeeded in upgrading the CPU in my desktop PC to one considerably faster than Dell officially supports, and (so far) it hasn't burst into flames and exploded. A good morning's work I feel :-)

Anonymous

@bolbteppa Did you go through Kubiznak's lectures on the Perimeter site? Are they good (in terms of content)? I was watching the first lecture from 2015-2016's PSI and he seems to be following L&L and Arnold. Seems nice so far

user929304

12:46 PM

@JohnRennie Hi, I hope my message finds you well. If you're around today, I wanted to ask a question about one of your posts, thanks a lot in advance.

John Rennie

2:53 PM

@user929304 Hi, I'm around now for a couple of hours.

user929304

@JohnRennie Great, thanks for getting back to me. First I had a somewhat dumb question to sanity check if I ve correctly understood how motion along time axis and spatial axes get mixed when spacetime is no longer flat

@JohnRennie So is it correct that, even if we may not be accelerating, and simply having a constant spatial velocity (say 1/3 of speed of light), then time in our frame evolves more slowly compared to if we were at 0 speed (spatially)? In other words, it doesn't have to be accelerated motion for motion along time axis to be slower than speed of light.

I guess im making a mistake in that if the spacetime is curved, we must be in acceleration by definition, right? Meaning if ever there's a discussion of time "slowing down" or ticking faster, it goes without saying that the object of interest is in a gravitational field and as such experiencing accelerated motion.

John Rennie

@user929304 have a look at this question. Although it might not appear to be directly relevant, in the answer I give a geometrical description of why different observers disagree about the time axis:

76

Q: How can time dilation be symmetric?

Suppose we have two twins travelling away from each other, each twin moving at some speed $v$: Twin $A$ observes twin $B$’s time to be dilated so his clock runs faster than twin $B$’s clock. But twin $B$ observes twin $A$’s time to be dilated so his clock runs faster than twin $A$’s clock. Eac...

The mixing of the time and space axes occurs even in flat spacetime. It doesn't require any curvature.

user929304

@JohnRennie Oh! Exactly what I was trying to get, so I always understood "flat" as time axis being orthogonal to spatial ones, consequently, motion along spatial axes will not influence motion along time axis. But I guess what you mean is that, because the four velocity is conserved and always equals speed of light, if we are on flat spacetime, that is, e.g. when experiencing constant velocity motion in space, time our clock measures is slower than it would be if we were at 0 velocity spatially?

John Rennie

3:09 PM

Your clock, i.e. the clock you are holding, always runs at the same speed of one second per second. But when you observe clocks moving relative to you in your coordinates they will be running slower than one second per second.

That's because what you consider the time axis is not what the moving clocks consider their time axis to be.

user929304

@JohnRennie I just read the linked post, clarified a lot for me already. Why do the chosen directions of time is different between the three? if they are all moving along the x-axis, then they are constantly moving in a direction orthogonal to time axis, what is here making the definition of actual time axis different for each party?

@JohnRennie Aha! so in my frame im at rest and by definition my clock ticks at 1 sec per sec, but what if Im in accelerated motion? such as falling from a mountain, then do I still measure on my clock 1 second per second?

John Rennie

@user929304 I don't know of a really intuitive explanation for why the time axes differ. You can show very easily that they must differ by crunching through a bit of maths. If you haven't already seen it I do this in:

56

Q: What is time dilation really?

Please will someone explain what time dilation really is and how it occurs. There are lots of questions and answers going into how to calculate time dilation, but none that give an intuitive feel for how it happens.

relativity time-dilation

@user929304 in your frame your clock always ticks at one second per second, regardless of whether your frame is inertial or accelerated.

user929304

@JohnRennie then how can the four-velocity I measure in my frame be conserved? is it because by saying "in my frame" it is assumed that I am at rest spatially?

John Rennie

Your frame is, by definition, the frame in which you are at rest (usually at the origin). So your 4-velocity in your frame is $(c, 0, 0, 0)$.

user929304

3:24 PM

@JohnRennie oh I see, makes sense. Then to follow the reasoning, when we send a shuttle with people on it to space, say the shuttle goes at a constant speed +v relative to us (so no acceleration, to simply things), then in our reference frame on earth, we measure their spatial speed to be +v and consequently their velocity in time (according to us) must be slower for the four-velocity to be conserved, and we call this observation: time is dilated for them. Is this about right?

John Rennie

Yes, that's basically it

user929304

@JohnRennie great! but they in their frame (i.e. the shuttle s frame) measure in turn our velocity in time on earth to be slower, so from their point of view, they are at rest and the earth system is experiencing dilated time. But why do we say then that when they come back they ll have aged less...?

John Rennie

@user929304 See this question:

70

Q: What is the proper way to explain the twin paradox?

The paradox in the twin paradox is that the situation appears symmetrical so each twin should think the other has aged less, which is of course impossible. There are a thousand explanations out there for why this doesn't happen, but they all end up saying something vague like it's because one tw...

special-relativity time reference-frames relativity

user929304

I feel i am missing a big point... :(

zounds

why is length contracted and not expanded?

geogebra.org/graphing/s34v3zpg this is my spacetime diagram, which i think is right

John Rennie

3:29 PM

The twin paradox necessarily involves accelerated motion and that changes the way the time dilation is calculated.

user929304

@JohnRennie ohhh! maybe that s what I was missing then! So if there s no accelarated motion and there s only constant relative motion at play between the two frames, we do measure relative time dilations but neither ages faster or slower, right? I guess acceleration somehow introduces an asymmetry.

zounds

each sees the other as younger

they are allowed to disagree because they are far away from each other

John Rennie

@zounds Lorentz contraction isn't a contraction! See:

12

A: "Reality" of length contraction in SR

Lorentz contraction is easy to understand once you realise that it is not a contraction at all. Instead it is a rotation and the length of the object, or more precisely its proper length, doesn't change at all. To see this take the usual example of a rod of length $2a$ aligned along the $x$ axis...

@user929304 correct

zounds

@JohnRennie my question, i guess is about the assymetry of time and space here

John Rennie

@zounds It is?

zounds

3:33 PM

if you look at the diagram i linked,

user929304

Great, so as long as constant relative speed is at hand, we have the scenario
as you described in your post with different definitions of north. Last bit of pestering if I may, is there any intuition for how "accelerated motion" introduces an asymmetry?

zounds

we take the "relativistic time" as being the y-value of a point, but the "relativistic length" as being where the time axis intersects our space axis

John Rennie

@user929304 if you observe my time then what that means is you measure my time in your coordinate system. The time dilation changes the mapping between our two coordinate systems, which is we we disagree about the time.

What acceleration does is change your coordinate system so that your coordinates become curved.

So even if you and I are (momentarily) stationary relative to each other, if you are accelerating then we will disagree about the time because your acceleration has caused the coordinates you are using to measure my time to be curved.

The trouble is that this is complicated, and while it seems intuitive to me that's only because I've been studying it so long. I'm sure it isn't intuitive to you.

@user929304 if you look at that last question I linked I go through the maths to show this, but I have to admit it's a bit short on intuition.

user929304

@JohnRennie indeed it is very hard to get a grip about :( So I guess, only if there's acceleration involved, then one sees the other s frame as curved (though in my frame Im still at rest and clock ticks 1 sec per sec), and experience time evolution differently

@JohnRennie Thanks a lot, I promise to go through it. Is it fair to say that if we sent out shuttles to space but only at constant speed relative to us on earth, then there would be no twin paradox once they return, because we'll have aged with equal pace?

John Rennie

Mathematically an accelerated frame is very like the situation where you are hovering near a black hole i.e. to really understand it requires some of the concepts used in general relativity.

@user929304 the shuttles can't return if they travel at constant speed. To return they must have decelerated to a stop then accelerated back towards us.

A return trip necessarily involved acceleration

user929304

3:43 PM

@JohnRennie ahhhhhhhhhhhhhhh! im so silly, I see what you mean!!! Neat!

@JohnRennie so with constant relative motion, time dilation (disagreement between clocks) is relative, but with accelerated motion the disagreement is "real" because one will have actually aged differently?

John Rennie

Yes. And that's due to the asymmetry in the acceleration. Only one of the twins has accelerated.

user929304

@JohnRennie great! I admit, I feel like I have both understood and still very puzzled at the same time :)

John Rennie

@user929304 it is basically hard to understand.

The funny thing is that after you've understood it then it seems obvious :-)

user929304

@JohnRennie I can only wish that day comes for me... :(

John Rennie

It's like riding a bicycle in a way, once you get the knack for it you wonder how you could ever have found it difficult.

@user929304 are you at school or university?

user929304

3:48 PM

@JohnRennie I am at university but I dont study physics, I read physics just for myself, hence all these naive questions sadly :/

John Rennie

Ah, OK. The problem is that we physicists are very familar with the maths involved, but if you don't have that background the maths is a real obstacle.

Really you need to learn the maths and get to the point where you easily calculate time dilations, and then you'll find suddenly you understand them!

user929304

indeed! I try to catch up but it is very slow.. I also get easily lost by the terminology at play often, proper time, coordinate time, time-like, ... etc.

Balarka Sen

@JohnRennie "The problem is that we physicists are very familar with the maths involved" Hmmmmmmm 🤔

John Rennie

@user929304 yes, you tend to get swamped by the detail. This happens to all of us and the only solution is to keep bashing your head against it :-)

user929304

@JohnRennie Speaking of four-velocity being conserved, applying this to photons, given they move through space at speed of light, can we say that a photon is by definition at rest in time? :)

@JohnRennie haha I promise to keep at it, the struggle is enjoyable!

John Rennie

3:52 PM

@BalarkaSen out, damned ~~spot~~ mathematician!

@user929304 photons do not have a rest frame. You'll find that if you attempt a coordinate transformation into the rest frame of a photon (or anyhting travelling at the speed of light) the transformation becomes singular.

This crops up over and over again in questions on the site. It is meaningless to ask about the time experienced by a photon.

enumaris

the good old twin paradox

user929304

@JohnRennie Right, oh so if there s no definable rest frame, we cannot speak of time evolution meaningfully?

Slereah

what if they're siamese twins

enumaris

it would be very painful

...for you

Slereah

you're a big guy

enumaris

3:56 PM

:D

Slereah

The twin paradox is really pretty easy to solve with GR

Hell for globally hyperbolic spacetimes you can prove it for all spacetimes and observers!

since timelike geodesics are minimal then

enumaris

not sure how much utility you're gonna get out of trying to explain the Twin Paradox to a beginner by invoking globally hyperbolic spacetimes lol

John Rennie

It's not a paradox! If you want a paradox ask a set theorist. They have lots to spare! :-)

Slereah

Well if you're a beginner and you still don't get the twin paradox after all that's on the internet, take it on faith maybe :p

enumaris

"What's a Globally Hyperbolic Spacetime?" "Well you see little Timmy, when you have a spacetime that can be foliated by Cauchy surfaces, you have a globally hyperbolic spacetime"

Slereah

4:00 PM

Globally hyperbolic spacetimes are the boring spacetimes

enumaris

they admit a Hamiltonian formulation of GR so maybe they're the fun spacetimes

don't discriminate bro

Slereah

You know

I still don't know if that's true

I mean

It is true

But I don't know if the converse is true

Hamiltonian GR is a complicated topic

enumaris

but you don't know if they're the only spacetimes that can be formulated that way?

Slereah

yeah

John Rennie

@user929304 time evolution means some process that is mapped onto a time axis. Your time axis is a perfectly good time axis for measuring or calculating time evolution, so there's no problem calculating time evolution of photons. What you can't do is try and work in the rest frame of the photon.

enumaris

4:02 PM

yeah, I'm not sure either, but the treatments of Hamiltonian GR invariably start off by assuming Global Hyperbolicity

Slereah

I do know that at least if it's globally hyperbolic the development is unique

Which might be the issue

But I have seen some people use Hamiltonian GR for non-globally hyperbolic spacetimes

user929304

@JohnRennie Makes sense thanks! Last question for time being if I may, can we say that an object in a gravitational field, always experiences accelerated motion? (i.e. irrespective of the chosen reference frame of the observer)

enumaris

perhaps within some region there's a local Hamiltonian formulation

@user929304 in GR a freely falling body experiences no acceleration, this includes freely falling bodies under the influence of gravity

John Rennie

@user929304 Sadly not. The only unambiguous measurement of acceleration that can be assigned to an observer is their four-acceleration, sometimes called proper acceleration.

Slereah

@enumaris well every spacetime has a globally hyperbolic neighbourhood around every point

John Rennie

4:06 PM

If you are falling freely in a gravitational field then your four-acceleration is zero. The astronauts in the International Space Station are falling freely in the Earth's gravitational field and that's why they are weightless i.e. their four-acceleration is zero.

enumaris

@Slereah yeah, that's just the statement that spacetimes are manifolds right

Slereah

Pretty much

enumaris

so if you just go to a specific coordinate system around some point you can probably do some Hamiltonian stuff

John Rennie

But you and I, sitting stationary on the surface of the Earth, have non-zero acceleration and that's why we aren't weightless.

Slereah

Well, I think the metric has to be $C^2$ or something

So that you can define a convex normal neighbourhood

enumaris

4:08 PM

hmmm

Slereah

$C^0$ metric spacetimes act pretty weirdly

user929304

@JohnRennie exactly, but can we ever have a zero four acceleration if we are in a graviational field?

John Rennie

@user929304 yes, just jump off a cliff. While you're falling freely your four-acceleration is zero.

... though not for long :-)

enumaris

I worry not about $C^0$ metrics

Everything is analytic to me

as all things should be

Slereah

Everything analytic isn't great either

You can't have metrics defined on a compact region then

enumaris

4:11 PM

no big deal

user929304

@JohnRennie aha :) So if my four acceleration is zero (i.e. free fall) I experience a flat spacetime?

Slereah

Having metrics Minkowski except in a compact region is a nice thing to have for proofs

enumaris

eh

John Rennie

@user929304 correct, and in fact that's a very important principle in GR. If you are falling freely then your local area of spacetime appears to you to be flat.

enumaris

then let them be $C^\infty$

Slereah

4:12 PM

Smooth is fine

ALTHOUGH

Thin shell

is nice

enumaris

let them eat cake

danielunderwood

$C^0$ physics anything sounds scary to me

Slereah

$C^0$ isn't too spooky if it's well behaved enough, but it could be like

Non-differentiable anywhere

enumaris

then you would have undefined parallel transport

Slereah

you can get fairly weird spacetimes behaviours

enumaris

4:16 PM

and undefined curvature

user929304

@JohnRennie Many thanks again! You are very kind and patient. I will mull over all this and hopefully get back to you with less silly questions :-) Thanks again.

enumaris

you need at least $C^2$ if you want to have well defined curvature...

Slereah

You can still talk about the causal structure

But it gets weird

Nah, you don't need $C^2$

enumaris

The metric needs to be at least twice differentiable, no?

John Rennie

@user929304 you're welcome. I just hope I haven't just confused you further!

Slereah

4:17 PM

You just need to have the Geroch-Traschen condition

but then they are defined in a distributional sense

Semiclassical

What does Brownian motion count as?

in terms of smoothness, I mean

enumaris

nowhere differentiable

but everywhere continuous

so $C^0$ I think

Semiclassical

sounds right

Slereah

jdegrant.files.wordpress.com/2012/05/continuouscausality.pdf

The horror

Balarka Sen

In fact Brownian motion if Holder continuous

s

is

floopdoodle

Slereah

4:20 PM

Holder continuous $2$, I think?

Balarka Sen

Sounds right

enumaris

took a course on that once a long time ago

danielunderwood

So is doing GR with a $C^0$ metric or similar something physicists do is it just of mathematical interest?

Slereah

Or at least the set of holder continuous $2$ paths isn't of measure $0$

Well $C^0$ metrics might pop up in quantum gravity

For instance in the path integral formulation

enumaris

Feels like you'd have trouble defining a Curvature lol

but if there's some fancy workaround

then so be it

Slereah

4:23 PM

Things get Complicated with path integrals

you don't use the usual definitions

enumaris

paths in path integrals can also be nowhere differentiable amiright

user929304

@JohnRennie I have one more question suddenly if you re still here :)

enumaris

John is always here

Semiclassical

in other news: sunburns aren't fun. I knew that. what I did not know about was sunburn itch

enumaris

he is the silent knight

Semiclassical

4:24 PM

which sounds innocuous enough but

enumaris

Anytime you have an open wound, it'll itch

John Rennie

@user929304 yes?

Slereah

@enumaris Paths that are differentiable anywhere are of measure 0

enumaris

so if your skin peels...itchyness for days

@Slereah sounds about right

John Rennie

@enumaris do we have to discuss skin pathologies while I'm eating lunch? :-)

Emilio Pisanty

4:26 PM

@Slereah give pathological cases an inch and they'll take all of the infinite-dimensional space you had around you that you hadn't even noticed was there

enumaris

@JohnRennie well...fried Chicken skin is pretty good munchies

Semiclassical

see, this is where i like bohmian trajectory stuff. the trajectories there are weird, but the space of such paths is pretty simple. (trajectories are smooth and don't cross in configuration space, weeee)

user929304

@JohnRennie to correctly understand the muon example in SR: one observer is on earth stationary, the other is the muon. Relative to us on earth, the muon is travelling at constant but very high speed (relative to speed of light), so we on earth measure the muon s clock to be ticking very slowly and that s how we explain why it can reach earth before its lifetime is reached (its time is going slowly in our frame so it can reach earth before lifetime ends). Whereas in the muons frame,

enumaris

fried pork skin also not bad

user929304

@JohnRennie the clock is ticking 1 second per second, but it see the length to travel to have contracted, so it needs less time to travel before its lifetime reaches because it sees a lesser distance to travel. And there s no acceleration involved. Is it a sound picture?

John Rennie

4:28 PM

@user929304 yes, and in fact I'm sure I explained exactly this in an answer somewhere on the main site.

danielunderwood

Since you GR guys are here, is Kaluza-Klein producing EM unique? It seems to me like you may be able to come up with an extra dimension in a way that you could add whatever interaction you wanted.

user929304

@JohnRennie And if the muon wanted to measure our time on earth, but in its own frame, it think our time is dilated on earth, right?

user351417

@EmilioPisanty You know that I really hate you? I mathjaxificated 2 equations in this when I saw that while I was doing that stuff somebody else had edited the post.

John Rennie

7

A: The real meaning of time dilation

Take the two events when the muon is created, and when it decays. We'll choose the origins of the muon rest frame and the lab rest frame so their origins coincide at the point and time the muon is created i.e. the muon is created at (0, 0) in both frames. Both frames will agree about their relati...

Slereah

Apparently $C^0$ metrics have that weird thing where $\partial I^+ \neq \partial J^+$

Emilio Pisanty

4:30 PM

@Chair I think there's a meta.PSE question about that

John Rennie

@user929304 there, that's a full treatment of the muon travel done using the Lorentz transformations.

Emilio Pisanty

> what do I do if I'm too slow with my edit

or sommn like that

user351417

@EmilioPisanty Lol really?

Emilio Pisanty

@Chair no, I shouldn't think so

=P

Slereah

@danielunderwood Kaluza-Klein is related to the fiber bundle formulation of EM

John Rennie

4:30 PM

@user929304 yes

Slereah

Other interactions can also be formulated in such a way, but I don't know if that would work, because EM is a pretty simple case

user351417

I've been doing mathjax stuff just for the practice of it though... I'll need to do some stuff with LaTeX later this semester so I thought it'd be a useful skill to pick up

Slereah

ie the gauge group is abelian

user351417

So I guess it's not a huge loss

Semiclassical

@EmilioPisanty oh, there's been further action on the question you bountied. (WimC gave quite a bit more details on what Numerical Recipes was doing)

Slereah

4:31 PM

I don't think you could do the weak or strong interaction

user929304

@JohnRennie great! so I guess the moral is that, in either choice of frame, we consistently end up concluding that it can reach earth before it decays, right?

Emilio Pisanty

@Semiclassical yeah, I saw that

John Rennie

@user929304 Correct! :-)

user351417

@Semiclassical Also someone wrote a pretty lame answer for the 3 mirrors vs 2 one.

Slereah

The trick for KK is that $\mathbb{R}^5$ is locally similar to $\mathbb{R}^4\times U(1)$

Emilio Pisanty

4:32 PM

@Semiclassical is the edit enough for your purposes?

user929304

@JohnRennie Physics is beautiful!

Emilio Pisanty

@Chair yeah, it seems Rod Vance just isn't that interested in answering =(

Slereah

I mean $\mathbb{R}^7$ is locally similar to $\mathbb{R}^4 \times SU(2)$, but the group structure isn't there

Emilio Pisanty

@Chair two lame answers, one now deleted

Semiclassical

yeah, it definitely clarified things a lot. I'd definitely rate it as deserving the bounty. (especially by comparison with the other answer, which while qualitatively helpful was not actually addressing the question being asked)

user351417

4:33 PM

@EmilioPisanty I guess when you have 70k+ rep 200 really won't mean too much?

user351417

@EmilioPisanty Oh, yeah, The ultra-short one got deleted too? Why? It wasn't VLQ standard, though it wasn't 200 bounty-level either.

Semiclassical

(the fact that I'm the only person who has upvoted that answer sorta bugs me, tbh. i guess people just don't look at such questions that much?)

user351417

Never mind... one now deleted

user351417

I saw something a while back on a question with 500 bounty (the incident occurred some years ago, don't ask me how I saw it)... Someone put a placeholder answer which contained just 2 sentences about how it's a placeholder, and that answer won the bounty. Later, when someone put up a good answer, the bounty was re-initiated by the owner of the placeholder, if that made sense.

danielunderwood

@Slereah Well at least I half understood most of that. I don't know that much about physics from the group side of things and combining spacetime with a group like $\mathbb{R}^4 \times U(1)$ seems a bit foreign to me. Guess it's time to learn!

Emilio Pisanty

4:38 PM

@Chair data.stackexchange.com/physics/query/344526/… physics.meta.stackexchange.com/questions/10713/…

danielunderwood

My original thought was that you could arbitrarily add a dimension to GR to add a scalar field. There's obviously something wrong with that picture since EM requires a vector field, but is there any validity in it?

Emilio Pisanty

@Semiclassical cool

I'll read in more detail when the bounty is closer to the end

danielunderwood

And are fiber bundles worth learning? I've seen a couple sources of "this is what a tangent bundle is, which is an example of fiber bundles that we don't need here"

Semiclassical

kk

enumaris

@danielunderwood when you add a dimension to 4 dimensions, and you look at rank 2 tensors like the Stress-Energy Tensor, you are actually adding 9 elements so you should have enough freedom to do some stuff there

Slereah

4:41 PM

fiber bundles are a very common thing in theoretical physics

For gauge theory

danielunderwood

Actually fiber bundles look a lot simpler than tangent bundles

Slereah

Well the idea of a fiber bundle is simple

but there are many things to it

enumaris

Tangent bundle is just a subclass of fiber bundles

where the fibers are tangent spaces

so I dunno how they can be more complicated o.O

Slereah

A fiber bundle is just attaching a space at every point

ie at every point of space, you attack $R^4$, for the tangent bundle

Or $U(1)$, for the EM gauge bundle

enumaris

With some specific rules on how the attachment goes

ACuriousMind

4:42 PM

@enumaris In some sense, objects with less structure are often more complicated because you don't know as much as about them.

Compare, for instance, the theory of fields with that of arbitrary rings.

Slereah

ring ring ring ring ring

enumaris

fair enough I guess

Slereah

bananaphone

danielunderwood

Ahh that sounds fairly simple. I still have trouble thinking in the math way where everything starts with a definition

enumaris

the math way is

just making simple things

way complicated

John Rennie

4:44 PM

5

enumaris

because they have to be "rigorous"

John Rennie

From Facebook :-)

danielunderwood

My real problem is when there are 4 different definitions for the same thing

I understand that they're useful for different things, but it sometimes seems like the definitions talk about different objects

ACuriousMind

@enumaris I often find a bit of rigor much preferable to "the physics way" where you can never really know what you're allowed to do with some symbol or object because people don't care for proper definitions.

danielunderwood

@Slereah So are you saying that $\mathbb{R}^4 \times U(1)$ is an example of a fiber bundle?

Slereah

4:46 PM

Yes.

ACuriousMind

Too much rigor can suffocate, but too little of it leaves you afloat in an ocean of vagueness

Slereah

All products of manifolds are fiber bundles

trivial bundles, they are called

enumaris

@ACuriousMind just do what you need to do

John Rennie

Damn, ninja'd AGAIN!

enumaris

let the mathematicians worry about it later

Just do it™

Emilio Pisanty

4:47 PM

@enumaris "ask for forgiveness, not permission"

enumaris

Indeed

Emilio Pisanty

if it turns out to be justified, then you're going to need to do the calculation anyways

Semiclassical

@ACuriousMind to ape a line of Kant, a bird can't fly in vacuum

Slereah

$™$

Emilio Pisanty

and if you do the calculation and it turns out to not actually be useful, what was the use of worrying about whether it was justified?

Slereah

4:48 PM

@Semiclassical Can an ape?

enumaris

There we go

Emilio Pisanty

^ and I got that argument from a mathematician, I should point out

Semiclassical

@Slereah yes. huzzah for rocket ships :P

Emilio Pisanty

Analysis II, probably close to the most rigorous I got while in undergrad

Slereah

ACuriousMind

4:49 PM

@Slereah See monkey do

enumaris

aaand blocked

Semiclassical

i suspect the counter argument to that is that you can have a calculation that looks right but actually isn't

Slereah

You know I wonder how much that monkey freaked out once in orbit

Semiclassical

and thus one can labor under the delusion that it's useful when it's actually not

ACuriousMind

@Semiclassical Cf. all the questions we have about "paradoxes" arising from pretending too hard that position eigenstates are actual states.

Emilio Pisanty

4:50 PM

@Slereah why would orbit be the most freak-out-inducing part of the trip?

Semiclassical

yeah

danielunderwood

@Slereah So even $\mathbb{R}^n, n > 1$ itself would be a fiber bundle? Assuming that my thought $\mathbb{R}^n = \mathbb{R} \times ... \times \mathbb{R}$ is actually correct, which I'm not entirely certain of

Emilio Pisanty

I would think launch would be by far the most terrifying bit

Slereah

@danielunderwood You can define it as such, yes

You can even do $\mathbb{R}$ as a fiber bundle, with a zero-dimensional fiber

But there's not much point to it

The interesting part is more that you add structures to the manifold

ACuriousMind

@EmilioPisanty The way up may be more directly stressful, but I think the complete absence of gravity may be more fundamentally confusing than just "more" of it.

Emilio Pisanty

4:52 PM

@ACuriousMind meh

if you're strapped in and you can't unbuckle the straps, you probably won't notice it that much

Slereah

"Ham's lever-pushing performance in space was only a fraction of a second slower than on Earth, demonstrating that tasks could be performed in space."

That chimp is more professional than some humans

danielunderwood

That comic made me think of "sea monkey do"

Emilio Pisanty

@Slereah which reminds me, I have a copy of First Man sitting at home waiting for me to be not-busy enough to read it

hopefully I'll get through it before the movie hits theaters