The h Bar

DanielSank

05:00

A system being linear tells you a lot.

Bernard Meurer

If I say two things happen simultaneously and they have a synchronicity (as in they didn't just happen at the same time for luck) can I infer that that is no sort of signal that connects the two entities to 'coordinate' the action? Because if it's simultaneous, the speed of that signal would tend towards infinity right?

0celo7

@DanielSank tensors, not voltages

or are voltages tensors?

DanielSank

@0celo7 But capacitance is a tensor!

So is scattering, and stress/strain, and dielectric response, and...

curvature...

I don't need to think about components to understand linearity!

Twice as much charge --> twice as much voltage! Physics!

0celo7

what is the capacitance tensor

DanielSank

Capacitance

Capacitance is the ability of a body to store an electrical charge. A material with a large capacitance holds more electric charge at a given voltage, than one with low capacitance. Any object that can be electrically charged exhibits capacitance, however the concept is particularly important for understanding the operations of the capacitor, one of the three fundamental electronic components (along with resistors and inductors). The SI unit of capacitance is the farad (symbol: F), named after the English physicist Michael Faraday. A 1 farad capacitor, when charged with 1 coulomb of electrical...

That's annoying, it didn't like the part I wanted.

0celo7

05:02

if it's that partial derivative, it's not a tensor :P

DanielSank

Direct link

0celo7

at least not on interesting spaces

both links lead to the same place

DanielSank

@BernardMeurer I'm not sure what you mean.

0celo7

@DanielSank as much as I love math, I fail to see the purpose of shoehorning the formal definition of tensor into places

the tensor transformation law is exactly what coordinate invariance is

DanielSank

@0celo7 Huh?

@0celo7 Yeah, I know that.

I just don't think it's useful to tell students that a tensor is a group of numbers which transform like a certain way.

It obscures the simpler notion of a linear thing.

0celo7

05:08

the problem I have with this is the following: if you do something, do it properly

do you agree?

Bernard Meurer

@DanielSank I was just imagining this. lets say you have two entities in space $x$ and $y$ which are apart by a distance $d$. Now suppose $x$ and $y$ are subject to some form of entanglement typed effect, I could automatically rule out the existence of a signal of some sorta travelling from $x$ to $y$ couldn't I? Since the entanglement would mean the changes would happen simultaneously (at the same time) there can be no signal that would 'travel' through $d$ instantly, right?

dmckee

@BernardMeurer Nothing fundamental prevents them from both being triggered by the same past event, so while neither causes the other they may still be connected.

0celo7

@DanielSank specifically, if one opens up the student to the gorey definition of tensor as a multilinear map, one should present all the details, right?

DanielSank

@0celo7 Uh... no?

The first time you heard about energy did your teacher talk about time translation invariance and Noether's theorem?

0celo7

Yes.

The first physics I ever did was Zee's GR book.

DanielSank

05:11

Also, tensors do not need manifolds, coordinates, charts, or any of that stuff.

Bernard Meurer

@dmckee I see what you mean.

DanielSank

They're just functions of vectors and covectors.

0celo7

I'm not talking about that

DanielSank

@0celo7 My god...

0celo7

@DanielSank Let me ask you a question

what is the contraction of a tensor

DanielSank

05:12

@0celo7 It's a partially applied function.

Bam.

0celo7

what does that mean

DanielSank

It means $f_a(x) \equiv f(a, x)$.

See?

0celo7

huh?

Bernard Meurer

The more time I spend here the more I'm convinced this is all just magic

DanielSank

$f_a$ takes one variable. The result is the same as calling the two-variable function $f$ but with $a$ as the first variable.

0celo7

05:13

I'm talking about $T^a{}_b\to T^a{}_a$

@DanielSank I have no clue wtf you're talking about

DanielSank

@0celo7 Let me do a slightly easier to talk about case.

0celo7

@BernardMeurer well @DanielSank is on drugs right now

DanielSank

$T_{a,b}$ is a function that eats two vectors (or is it covectors, I always forget), right?

0celo7

@DanielSank what?

@DanielSank yes

DanielSank

Ok, now suppose I have a vector $v^c$.

0celo7

05:15

if you're using abstract index notation, "vector $v^c$" is the math equivalent of "ATM machine" :P

DanielSank

Then $v^a T_{a,b} \equiv S_b$ is a tensor that eats a single vector. It's action is defined like this: $S_b w^b = v^a w^b T_{a,b}$.

@0celo7 Oh, sorry.

0celo7

@DanielSank just being overly pedantic...I'm sure everyone does it lol

DanielSank

That's what contraction means: it's just partially applied function.

0celo7

um, nope

DanielSank

@0celo7 Don't be an ass.

0celo7

05:16

no, you're wrong.

I'm not being an ass, that's not what contraction is.

DanielSank

@0celo7 Yes, it is.

0celo7

I'm talking about contraction on the same tensor. Not two different tensors.

DanielSank

@0celo7 I know, I wanted to do the two-tensor case first because it's easier to think about.

0celo7

But that's the trivial case because you can do it by definition of $T$.

Try to contract $T_{ab}$ with itself now.

DanielSank

@0celo7 thinking...

0celo7

05:20

:)

thinking is good

DanielSank

I know that for $T_a^b$ the contraction is the same thing as the trace. I'm trying to think of how I would think about this in a more general situation. I understood this once because I was thinking about density matrices etc. etc. and this came up.

0celo7

oh I meant $T^a{}_b$

DanielSank

What's the difference?

0celo7

dies

DanielSank

Did I typeset something wrong?

0celo7

05:21

no, I did

well

you messed something up and I went along with it, or something

DanielSank

Oh oh you had two down indices before.

Yeah two down indices can't contract in vanilla tensor land. You need another assumption to do that.

Right?

0celo7

yes. some canonical isomorphism between $V$ (vector space) and $V^*$

DanielSank

Right.

Ok @0celo7 can you give me a simple no unnecessary math description of what the contraction means?

I can describe it in words, actually.

0celo7

go for it

you won't believe my definition

DanielSank

typing...

Bernard Meurer

05:25

I'm off to bed now, night folks!

DanielSank

Given a tensor $T$, each down index represents a map from the set of vectors to the set of tensors of rank one less than $T$. This map is linear.

@BernardMeurer ciao

This map is actually itself isomorphic to the set of covectors. So in a very real sense each down index represents a covector.

Similarly, up indices correspond to vectors.

Contraction maps a tensor to another tensor of rank two less. This new tensor is equivalent the the old one but where the vector and covector corresponding to the contracted indices have eaten each other.

@0celo7 good enough?

I realize I could have said that better.

Interestingly, I could describe this with Scala's type system!

Oh actually no I probably couldn't...

0celo7

@DanielSank that's the idea

do you want the true definition?

DanielSank

@0celo7 Will it improve my understanding?

0celo7

"eaten each other" means?

DanielSank

@0celo7 Function application.

Remember, each vector and covector is a function.

It's just function application.

0celo7

05:30

so if I have $f(x,y)$, how do I apply the "y slot" to the "x slot"

say $f(x,y)=xy$

DanielSank

That's not linear. It doesn't work.

This only works for linear functions.

Multilinear, i.e. linear in every argument.

Try again.

Thanks a lot, now my comments make no sense.

0celo7

:)

DanielSank

@0celo7 This is a great question. Thinking...

I am surprisingly confused.

:D

Enlighten me.

0celo7

on?

what you said is nonsense, so I can't help there :P

DanielSank

Your function is definitely linear in both arguments.

It must correspond to a tensor.

Right?

0celo7

05:35

well.

if I take $V=\mathbb{R}$, then maybe

DanielSank

yes, that's the assumption.

0celo7

hmm.

the proper definition is nonsense and won't help us to calculate the thing

so the only thing left to do is...calculate it

I shall now define the tensor contraction in the physicsy way for you

DanielSank

wait wait

NO no no

I wanna understand $f$

0celo7

oh, actually

$f$ is not of the correct type for this

I guess we could just say $\mathbb{R}^*=\mathbb{R}$

@DanielSank $f$ is just the natural pairing of $\mathbb{R}$ and its dual.

If $y\in\mathbb{R}^*,x\in\mathbb{R}$, then $y(x)=xy$

DanielSank

@0celo7 Yes, I was assuming that.

@0celo7 Yes.

So what is $f$?

0celo7

05:39

the map $f:(y,x)\mapsto y(x)$

DanielSank

@0celo7 So what's the contraction in this case? That's where I'm puzzled.

The contraction should result in a single number, shouldn't it?

0celo7

Well this is a terrible example :P

the contraction ends up being $1$

DanielSank

@0celo7 Could you explain that please?

This would be helpful to me.

0celo7

well I was going to

so let's define tensor contraction

DanielSank

yeah

I think I see where this is going now.

0celo7

05:41

suppose we have some (1,1) tensor $T$ over a vector space $V$

let $V$ have a basis $e_i$ and $V^*$ a basis $\theta^i$ s.t. $\theta^i(e_j)=\delta^i{}_j$

$1\le i\le n$, n=dim V

then the contraction of $T$, denoted $CT$, is $$CT:=\sum_{i=1}^nT(e_i,\theta^i)$$

DanielSank

@0celo7 Yeah.... I was really trying to not say anything about bases.

I guess this is ok since we can prove that this definition is basis independent.

0celo7

I was going to get to that

there is a way to do this without a basis

DanielSank

Ah.

0celo7

that's the definition I said you won't believe

DanielSank

lay it on me

0celo7

05:46

We first need this commutative diagram.

Let $h$ be any bilinear map

$V,W,Z$ are all vector spaces

Then there is a $\varphi$ such that $\tilde h$ is unique and the diagram commutes

@DanielSank So we let $c$ be the canonical pairing. Then $\tilde c$ is the contraction.

Rajesh Dachiraju

I am interested in applying this new Fourier contruction formula in Fourier-spectral methods/Gelerkin methods, for shockwave capturing, and I am new to pde, anyone interested to help/collaborate with me (who knows nemerical pde's)

DanielSank

@0celo7 I'm not following.

Sorry.

Wanna give me a reference?

I'm trying to do some lab stuff so I'm not paying as much attention as I would be otherwise.

0celo7

Tensor product

In mathematics, the tensor product, denoted by ⊗, may be applied in different contexts to vectors, matrices, tensors, vector spaces, algebras, topological vector spaces, and modules, among many other structures or objects. In each case the significance of the symbol is the same: the freest bilinear operation. In some contexts, this product is also referred to as outer product. The general concept of a "tensor product" is captured by monoidal categories; that is, the class of all things that have a tensor product is a monoidal category. The variant of ⊗ is used in control theory. == Tenso...

Rajesh Dachiraju

This is the link :

7

Q: Eliminating Gibbs phenomenon, and approximating with jumping functions in Fourier Analysis : An attempt and a question in this regard

This problem seems like a nightmare to me. I tried to expand $K_{\omega}^f(t)$, but I am clueless of getting some kind of a closed form or some kernel like structure. If I try to take derivative, its even worse. I desperately need some clues on line of attack I should do. Please help m...

DanielSank

@0celo7 Thanks.

Mike Miller

05:51

@DanielSank: A lot of words that just mean $\sum_i v_i \otimes \varphi_i \mapsto \sum_i \varphi_i(v_i)$.

2

Rajesh Dachiraju

rajeshd007.wordpress.com/2016/01/19/…

Mike Miller

The lot of words are just needed to prove that this doesn't depend on how I decide to write down the element on the left, since tensors do not have a "unique" representation as such a sum (...without choosing a basis).

Rajesh Dachiraju

Hi @DanielSank

user54412

@HDE226868 Yes

user54412

@HDE226868 Use methanol (just be careful with proper sealing and/or ventilation). Isopropyl's problem is that it's too heavy compared to air. It falls too quickly for there to be much of a saturation region. (It works, just methanol is better.)

user54412

06:05

Dry ice is easy but not sustainable. On the bright side, it's fun to play with the leftovers. We had a really good perpetual chamber driven by a compressor in undergrad.

Rajesh Dachiraju

Hi @ChrisWhite

user54412

hi

Rajesh Dachiraju

Do you solve shockwave pde's?

numerically

user54412

yes, though not in the same context as most people who do such things

Rajesh Dachiraju

Astrophysical?

user54412

06:07

as in, everything I do is relativistic

Rajesh Dachiraju

Ok, I am interseted in the math only though (no concern for the physical quantitythey represent

@ChrisWhite : I have a new type of Fourier reconstruction alternative to Fourier partial sum and I wonder I can use it for Spectral/Galerkin method to see if any good thing in case of discontinuous solutions

I am totally new to pde

user54412

@HDE226868 One thing no one thinks about is lighting. Poor lighting makes the trails harder to see and appreciate. Also, cosmic rays and spontaneous decays from trace elements in your container might well outshine a smoke detector. The alphas from americium don't go very far in air, and I had less luck with that than I had hoped.

Rajesh Dachiraju

Looking for someone who can help me with this

Swapnil Das

Hello :)

user54412

I don't have much experience with spectral methods myself.

user54412

06:11

I wouldn't be surprised though if a wide variety of basis functions could be used, with varying degrees of success for different problems.

Rajesh Dachiraju

@ChrisWhite : I only need Fourier basis methods

I am not using anything other than Fourier basis but the reconstruction I do is different

Swapnil Das

I have a doubt, can anyone help?

Rajesh Dachiraju

I see in solving Burger's equation using Fourier-Galerkin, they substitute u_N instead of u and convert pde into a set of ODE, then later on how they solve I dont know

I don't have a good book which gives me step by step process

Swapnil Das

How fast an object with mass one Pentagram need to be going to have the same 'mass' as an object with rest mass 5.5 Pg?

Should I use E^2 = m^2c^4 + p^2c^2 ?

DanielSank

@RajeshDachiraju Hi.

Rajesh Dachiraju

06:17

@D

@DanielSank : Do you know numerical pde?

user54412

@SwapnilDas I normally don't recommend just looking for any equation that might work, but in relativity there are only really two equations, and you picked the one that doesn't have velocity in it ;)

DanielSank

@MikeMiller yes I believe I understand this.

@RajeshDachiraju No.

user54412

@RajeshDachiraju I actually don't know of any books on the subject :( That's not to say they don't exist, however.

Rajesh Dachiraju

ok np

open source projects?

Swapnil Das

@ChrisWhite Thanks for advice. It is quite confusing, still interesting to me as a ninth grader :P

user54412

06:23

@SwapnilDas I'll give you a hint: there is a simple relation between (1) rest mass, (2) "effective"/"relativistic"/"inertial" mass, and (3) velocity/Lorentz factor. In fact, it's the definition of relativistic mass.

user54412

By the way, relativistic mass is an outdated concept, so no one even writes it in formulas anymore. Instead, they just use the definition in terms of velocity and rest mass.

0celo7

that can't be true, I see tons of PSE questions/answers that use it

maybe you're an outdated concept

Swapnil Das

m/sqrt(1-v^2/c^2) ??

@ChrisWhite

user54412

that looks good

Swapnil Das

Thanks, glad that a physics genius helped me :)

user54412

06:27

also, don't just look to authority to tell you you're right or wrong -- it's a good skill to develop to check your own ideas and see if they convince you

user54412

later in life, if you can't convince yourself, you won't convince anyone else either

0celo7

@ChrisWhite can I haz life lessons too pls

Swapnil Das

True, are you a researcher in physics? You sound like that..

user54412

@0celo7 2/10. Your position as resident troll is in jeopardy.

user54412

@SwapnilDas yes

0celo7

06:30

good

maybe that's because I'm not a troll

Swapnil Das

Wow, then allow me ask a question on a widespread rumour on Mechanics. Is lagrangian mechanics a simplification of Newtonian one?

user54412

Lagrangian mechanics is perfectly equivalent to Newtonian in any case where both apply. The Lagrangian formalism is definitely more complicated, but it makes solving some systems easier.

Swapnil Das

Oh, thanks for clarification :)

0celo7

@SwapnilDas note that as soon as you add friction, Lagrangian becomes a hell of a lot more complicated

user54412

At some level, I don't even acknowledge Lagrangian mechanics for anything but nondissipative, holonomic systems.

DanielSank

06:47

@ChrisWhite tesselate any spheres today?

@SwapnilDas For systems with constraints it is considerably simpler.

@0celo7 Meeeehhhhhhhhhh, I dunno...

It's really not that bad.

It's reallllly not that hard to figure out how to add terms to a Lagrangian to get common forms of dissipation.

TanMath

09:38

Hey all!

TanMath

11:05

@DanielSank here is the gist of the two level system:

gist.github.com/TanMath/63572c48ef02d6f16dad

alarge

@RajeshDachiraju Well Im' guessing that the mixed term, the convection bit, you'd multiply in real space, and everything else in Fourier space, right?

David Z

13:39

0

Q: What is the experimental evidence that light is an electromagnetic wave?

Do we have any experimental evidence to confirm this? How can Maxwell's equation confirm light as an EM wave simply based on a similarity in speed?

electromagnetism visible-light electromagnetic-radiation maxwell-equations

I think that might be worth reviewing for reopening

user116211

14:15

@David Z: I was about to ask this here; but you've done that already. I've voted it to re-open but it would be fine if Kyle or John might tell why they thought it's broad.

user116211

@David Z: Also, how could a user answer this much later after you closed? physics.stackexchange.com/a/231299/36790; you closed it approximately 43 min ago & the user seems to answer hat after you closed it; the time shows he answered 6 min ago ; but by then the question was closed already:/

user116211

@David Z: Could you tell me about this or I'm mistaking?

David Z

Apparently there is a 4-hour grace period for answering a closed question: if you start writing an answer before the question is closed, you have 4 hours to submit it and the submission will be accepted even if the question is closed.

user116211

@David Z: Oh! Got it; never knew about this. O.o

David Z

I didn't know either, for a long time

user116211

14:29

@David Z: Well! I wonder how much things I still need to know yet:/ Thanks, BTW.

0celo7

14:47

holy moley

Some of those are actually good ideas if you remove the "Black" restriction :/

They can't even decide if it's "Africa" or "Afrika"

Bernard Meurer

@0celo7 this that black lives matter movement?

I think that's the name I saw on the news a while ago

0celo7

I think this is even nuttier

Bernard Meurer

I was about to say that

Hell I can't demand a brazillian in every officer

nor a brazillian-only safe space on campus

0celo7

@FenderLesPaul On page 11 they say "Black woman". I thought "woman" is ill-defined in the SJW mind.

Lol what if they got a white man for that position and he just said "I identify as a black woman"

Bernard Meurer

15:05

@0celo7 I think it would short circuit them

much like that capacitor of mine

0celo7

Did you investigate it yet?

yuggib

15:31

@0celo7 Just finished the correction of the Scheinklausur 2 of the analysis 1 course...it was really easy; nevertheless the marks were a tragedy T__T

0celo7

15:42

Because analysis is PhD level math

yuggib

nah

it's because you young bloods are lazy

and don't study enough

0celo7

No, it's fucking hard.

Maybe it isn't for you, but we can't all be smart.

It's snowing...crap

yuggib

I mean "$f:(a,b)\to\mathbb{R}$, mit $(a,b)\subset \mathbb{R}$ ein beschränktes offenes Intervall. Ist $f$ stetig und beschränkt, so existiert $\lim_{x\to b}f(x)$" True or false?

0celo7

Ich kenne keine mathematische Begriffe auf Deutsch

yuggib

I think that more than a half of the students had that wrong...

I don't understand a word of Deutsch ;-P

0celo7

15:56

$f:(a,b)\to\mathbb{R}$ with $(a,b)$ an open, bounded interval

yuggib

yes

0celo7

if $f$ is monotone and bounded

yuggib

continuous and bounded

0celo7

then does the limit exist

ah

what's monotone?

monoton?

yuggib

yes

0celo7

15:57

I think the limit does exist...

yuggib

T__T

think a little bit harder

0celo7

oh

to $b$

of course not

dude they misread the question :P

wait a moment

are you saying it's continuous on the interval?

or over $\mathbb{R}$

yuggib

on the interval

0celo7

poorly worded question

yuggib

it is defined only on the interval

0celo7

15:59

I fucking hate google images

dr282zn36sxxg.cloudfront.net/datastreams/…

DUDE

just set $b=1$ and there's your counterexample

yuggib

well, that's technically not the counterexample

:P

0celo7

well then what is the counterexample

we haven't done functional limits yet

ACuriousMind

@0celo7 There $\lim_{x\to 1-} f(x)$ (taking the limit from the left) exists, it's just not equal to $f(1)$

yuggib

$\sin[(x-b)^{-1}]$

0celo7

ugh, what does that look like?

yuggib

16:02

it oscillates very very fast going towards $b$

0celo7

@ACuriousMind I know

ACuriousMind

Infinite oscillation at b

0celo7

oh right

yuggib

if a one-sided limit do not exist, the rule of thumb is that you're getting a very fast oscillation :-P

0celo7

well I did not know that!

jeez

I'm not a mathematician

have we not been over the fact that I can't math?

yuggib

16:04

"rule of thumb" are words seldom used by a mathematician

0celo7

hmm

I don't quite buy that

I have so much homework to do

yuggib

I have worked also saturday this week

so from tonight to monday morning is complete nothing

0celo7

I need to schlepp so much crap to the library

@ACuriousMind how do I survive the snow

I did not learn this skill in Germany as I should have

@ACuriousMind rip America snow is evil

8 dead already

goblin

Is there any formal proof of the continuity of space/time or is it something we've taken for granted?

0celo7

@goblin There are no formal proofs in physics, period.

goblin

16:16

Right. Still it's something that bothers me.

ACuriousMind

@goblin "formal proof"...for there to have a formal proof, physics would need formal axioms.

And if you look at the partial axiomatizations of some fields, you almost always get "Spacetime is a manifold" (and hence its continuity) as an axiom. Why does that bother you?

goblin

It just suddenly occurred to me that I have no idea or an explanation for why space is assumed to be continuous.

And I don't really know where to look.

ACuriousMind

Because there is no evidence it is discrete?

goblin

Surely that's a sort of a circular argument.

Besides, it makes more intuitive sense if it is, indeed, discrete.

0celo7

I, for one, have no intuition for what spacetime should/shouldn't be.

I've never seen spacetime. (Not mocking you-know-who here.)

ACuriousMind

16:20

@goblin Well, that's of course not the whole argument. That's the argument why you shouldn't believe it's discrete. You should believe it is continuous because the theories assuming that it is continuous give well-tested empirical predictions.

@goblin I don't know what "intuitive sense" means

goblin

That's not true though. They give a well-tested approximations of the actual results, if we assume the discreteness is something of the order of plank time.

ACuriousMind

...what?

In the currently accepted physical theories, Planck units are not the scale of some discretizations. They are the scale where quantum gravity effects are expected to become significant, and nothing more.

goblin

Yes, but what if time was divided in portions of

0celo7

@goblin Are you referencing LQG/spinfoam type theories?

goblin

oops, my comment was cut off

I meant to say, what if time was divided in portions of roughly that size.

I am not referencing anything, I am just entertaining a weird idea I got on Saturday afternoon.

0celo7

16:25

If it's a weird idea, how do you know that they give well-tested approximations?

I'm sure if you just replace all derivatives, etc. with difference quotients, etc. you could get some reasonable numbers.

But that's hardly proof.

goblin

If time was really discrete on such a small scale, wouldn't that make calculus (dealing with infinitesimals) misleadingly correct? I am not very good at maths, sorry. :)

0celo7

Yeah. That's how computers do integrals.

They do the Riemann sum with very small increments, and the result can be as good as the result by hand to many decimals.

goblin

Right. So there's no actual proof that the results obtained via calculus weren't in fact supporting the theory that space/time is discrete.

And by that I mean that it's not correct to rely on results obtained via calculus to dismiss this idea.

0celo7

If you want spacetime to be discrete you have to overhaul everything. It's one thing to obtain a discrete version of a continuous theory, but building the theory from the ground up on discrete spacetime is much harder.

But this is way out of my element, maybe @ACuriousMind can say something more definitive.

goblin

I think I should properly research the matter, certainly somebody else has already made at least some effort in that direction.

ACuriousMind

16:30

@goblin You can always "put your physical theory on a lattice", that's what simulation people do after all. But lattice approaches do have their issues when the equations are unsuited to standard solution methods, and they have other issues as well (e.g. they can't cope well with fermions).

0celo7

Loop quantum gravity

Loop quantum gravity (LQG) is a theory that attempts to describe the quantum properties of the universe and gravity. It is also a theory of quantum spacetime because, according to general relativity, gravity is a manifestation of the geometry of spacetime. LQG is an attempt to merge quantum mechanics and general relativity. The main output of the theory is a physical picture of space where space is granular. The granularity is a direct consequence of the quantization. It has the same nature as the granularity of the photons in the quantum theory of electromagnetism and the discrete levels of the...

@goblin Take a look at that article, but remember that none of that is proven or standard.

ACuriousMind

@goblin Lattice field theory, where you suppose that spacetime is a discrete lattice, is an alive area of study.

goblin

Nice, thank for the input, you two.

0celo7

@ACuriousMind I thought that was for computational convenience, not a fundamental principle.

goblin

Although I think LQG is a bit outside my abilities. :)

ACuriousMind

16:32

However, as I said, there is no evidence to believe spacetime actually is a lattice - but by the very nature of "continuous" things, they are well-approximated by discrete things.

@0celo7 Well, I don't think many of the lattice people actually think that spacetime is a lattice, but that doesn't make much difference to the theory, does it?

0celo7

@ACuriousMind No, but @goblin was asking about what spacetime actually is. So it makes a difference to him.

ACuriousMind

@0celo7 And you then sent him a link to a theory with no experimental confirmation, congrats ;)

3

0celo7

@ACuriousMind At least I didn't link Lumo's blog ;)

ACuriousMind

Something to be thankful for, I guess

0celo7

I used to read it every day.

It's...interesting.

Mike Miller

16:43

I dunno, a theory where the universe is discrete is interesting. I have no reason to believe it but it'd be cool.

0celo7

It would suck for all the GR books out there

I wonder how causality/spacetime topology works in such a universe

FenderLesPaul

17:38

Afternoon folks

looks like a missed an interesting discussion

user54412

Just learned that a former student of mine (one of the brightest and most enthusiastic I've had) was hit by a car and killed this past Christmas.

user54412

(in case this place needed something less upbeat)

user54412

Is this something that happens to older academics? Finding out about the misfortunes of former students who seemed to have the best part of their lives still ahead?

dmckee

17:58

I've yet to learn of a student loss abruptly, but I've lost a couple of colleagues to mischance. You have my sympathy.

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