The h Bar

0537

00:01

what is "an isotropic component with nonzero trace"

0celo7

component of?

what's the context

0537

it defines it as $\frac{1}{3}\alpha I$

decomposition of velocity gradient.

0celo7

explain symbols...

0537

though it doesn't say what $\alpha$ is.

look here.

it doesn't say why or how.

or maybe i'm dumb...

0celo7

ah

the velocity gradient is some 2-tensor, let's call it $t_{ij}$

then, we have $t_{ij}=t_{(ij)}+t_{[ij]}$ where () is symmetrization and [] is antisymmetrization

(verify this)

furthermore, we can take out the trace of the symmetric part

this is the $\frac{1}{3}\alpha I$ part

0537

00:07

oh.

0celo7

note that this is isotropic because $I$ is rotationally invariant

0537

thanks.

0celo7

@0537 so $\alpha=\operatorname{tr}t$ if you didn't figure it out already

0537

the trace of this?

0celo7

why is there an $\alpha$ in there

0537

00:11

idk

0celo7

no, the trace of $\mathbf{L}$

0537

i see.

0celo7

$\alpha=\operatorname{tr}\mathbf{L}$

0537

i don't understand the values there though.

0celo7

what values where

0537

00:12

in the matrix.

$\mathbf{E}$

0celo7

$\mathbf{E}=\operatorname{sym}\mathbf{L}-\frac{1}{3}\operatorname{tr}(\mathbf{L}‌)\mathbf{I}$

you can verify that this is symmetric and traceless

0537

ok.

0celo7

@0537 alles klar?

0537

00:31

yeah for now.

ty.

0celo7

np, I'm off to bed

0537

good night.

AngusTheMan

@Qmechanic I hope you don't mind me asking you a question, I am just wondering if you can shed a little light on the following problem. I have two equations that requires the following equality to hold $d(p\dot q)=\dot q dp-\dot pdq$ where $d$ is the exterior derivative, $p$ is a conjugate momentum, $q$ is a generalised coordinate and $\dot q$ is a velocity. I believe that they should 100%. However I can't see how (if indeed it does). Any ideas? Many thanks :)

Joshua Lin

01:53

Is Mathjax/Latex supposed to work in chat? Whenever people post math equations into chat like $F = ma$ or something its never nicely formatted

AngusTheMan

you have to download chat jax and put it in your book marks

Mike Miller

03:40

@AngusTheMan I like Floer's original paper (Morse theory for Lagrangian intersections), but the geometry is not easy. If you're worried about wedge products, it might be best to put off thinking about Floer homology much. If you're looking for an expository account of what it's good for I don't know of one (though one surely exists).

Floer homology is somehow the right intersection theory of Lagrangians; there are a number of applications where one has an analogy between some construction in differential topology and a related thing in symplectic geometry, and in the former case intersection numbers appear and in the latter Floer homology does.

Famous applications: the Arnold conjecture; Seidel's work on symplectic mapping class groups (and a lot of later work); topological restrictions on Lagrangians. I'm not much of a symplectic geometer so I couldn't tell you much about the general story here.

ffahim

04:22

hey, i found a problem in chemistry.

@MikeMiller can i tell u?

Mike Miller

Probably not, sorry.

vzn

04:54

@Huy wild, is that a real book?

FenderLesPaul

06:31

@Prahar do you work for Strominger?

Huy

06:45

@vzn Of course.

@0celo7 then don't watch it

0celo7

08:23

The Girl on TV

►

Qmechanic

08:38

0

Q: Could our universe be destroyed within seconds?

I'm not that deep in physics. But some days ago I read, there could be different coexisting universes, which could even share "space" (or what ever it is) where each of them could have absoloutely different laws of physic. Now when I'm thinking about the expanding of our universe is described by...

Primarily opinion-based?

kevin Tah N.

11:35

Several years ago I said something incorrect about Banach and Hilbert spaces. Today, I finally understood that it is most likely the most embarrassing thing I have ever said. I feel soooooo ashamed, and mortified. In fact I want to hide in a my blanket. I feel like I need to find those people, call and apologize for my statement.

Huy

do it

kevin Tah N.

I am looking up the numbers of the people involved lol

also I just realized differential equations is really hard

I have been following some lectures and just discovered the differential equations class I took was a joke

It was at AMS(AIM ?) in Palo Alto near Fry's electronics, I made some very wrong statements about Banach space. I should have just shut up lol. I should definitely send them an apology lol. Gosh, I was soooo dumb back then

Bass

12:38

hi guys, can the components $\omega_{i...j}$ of the exterior derivative $\text{d}\omega$ of a differential form $\omega$ be expressed in terms of the Levi-Civita symbol $\epsilon_{i...j}$? I think it would make sense if it could, but I'm not sure how to use the same $\epsilon$ for the whole de Rham complex..

at least in some dimensions it must be possible, since the Bianchi identity of the EM tensor can be expressed as $\text{d}F=0$ and as $\epsilon^{\mu\nu\lambda\kappa}\partial_\mu F_{\nu\lambda}=0$. I'm pretty sure this can be extended to other dimensions, since the exterior derivative is the "anti-symmetrized sum of partial derivatives".. but I haven't found anything

The Dark Side

@Qmechanic IMO, it probably doesn't qualify as primarily opinion based, especially in light of CuriousOne's answer, which does a reasonably satisfactory job of settling it.

In other news: 997 LQ reviews

physics.stackexchange.com/review/low-quality-posts/stats

3 away from the Steward, but I'm more excited about the El Dorado hat than the badge :)

0celo7

@Bass note that dx^dx=εd^2x in some sense :)

On mobile in Globus so that's the best comment I can give you right now.

@Bass Check chapter X of Zee's Gravity book. Been a year since I've read it but I think he explains the equation I just wrote in there.

Slereah

12:56

a form is a totally antisymmetric tensor

and you can antisymmetrize a tensor using the Levi Civitta tensor

0celo7

Basically ^

@Slereah what do you know about minimal surfaces

Slereah

Nothing

I assume they are surfaces

That are minimal in some way

0celo7

@Slereah what's a top. space for which the closed, bounded sets are not compact

Slereah

I'll go get my counterexample book

Huy

@0celo7: most spaces in FA

Slereah

13:10

can't find an example

Huy

@0celo7: take any infinitely dimensional space

Bass

@0celo7 you can take some sick metric like $m(a,b)=1$ except $m(a,a)=0$. Then you have the discrete topology (every set is open and closed). Every set is also bounded, since the metric does not go higher than $1$. But it's easy to pick covers without finite subcovers, so it's not compact

@0celo7 could not find anything there. I know that "in some sense" $dx\wedge dy=\epsilon d^2x$, but I'm looking for the exact equation..

0celo7

@Huy Doing Riem. Geom. No infinite stuff pls

@Bass Did you check the TFT chapter

@Bass well...that is an exact equation under an integral!

d^2x is the standard measure of the 2-surface spanned by dx and dy

And the epsilon gives an "orientation"

Bass

@0celo7 I mean the exact equation for $(\text{d}\omega)_{i...j}=\epsilon_{?}?\omega_{?}$

0celo7

Oh.

Well...that does not make sense then

What is omega on the RHS

Bass

13:20

some differential form. I'm looking for an equation that expresses the exterior derivative of a k-form in terms of the Levi-Civita symbol

0celo7

Why?

@Bass So on that space the metric topology is discrete?

Slereah

I think you can express it as just like... $T_{[abc]} = \frac{1}{3!}\varepsilon^{\alpha\beta\gamma}_{abc} T_{\alpha\beta\gamma}$?

Or something

0celo7

Could you please explain that, @Bass

Slereah

Oh wait no that doesn't make sense

Hm

Bass

@Slereah where's the exterior derivative?

Slereah

13:24

Exterior derivative is just $\nabla_{[a} T_{bcd]}$

Bass

@0celo7 I think it is. Take for any point $x_0$ the open ball with radius $\frac 12$. This ball contains only $x_0$. So the set $\{x_0\}$ is open. Since this is true for all points, the topology contains all single-point sets.

Now, for a closed and bounded set $A$, take a series of points $x_i$ such that the series of sets $A_i:=\{x_i\}$ is a cover of $A$ (if $A$ has uncountable many points, then define $A_0$ to be $A$ without the points $x_i$).

This cover does not have a finite subcover, so $A$ is not compact.

note that $A_i$ is a cover of open sets

@Slereah what is $\nabla$ here? what is $T$? where's $\epsilon$?

ACuriousMind

14:03

@Slereah Ah, yes, I remember

0celo7

15:11

@ACuriousMind Suppose $(M,g,\nabla)$ is the usual manifold with LC connection and $\phi$ is an isometry. Is there a good formula for $\phi_*(\nabla_YX)$?

The Dark Side

15:58

:: Waits, as seconds pass by like years :: This badge script is slooo....ooow

:: Refresh, refresh ::

ACuriousMind

@TheDarkSide Some scripts only run once per day, you might be waiting some decades, then ;)

The Dark Side

@ACuriousMind A Danuesque shaddap :P

@ACuriousMind Lol. But no, I'm sure it's only a few more minutes.

Secret

Hi guys, I have a scifi self exercise question. I am wondering what other things can be ascribed physical meaning in my working shown here:

Schrodinger equation x scifi: A hypothetical two time physics speculation investigation

Given the schrodinger equation
$$\mathcal{i}\hbar\frac{\partial}{\partial t}\Psi=\left[-\frac{\hbar^2}{2m}\nabla^2+ V\right]\Psi$$

where $\Psi=\Psi(\mathbf{x},t)$

Now consider free particle solution i.e. $V=0,\Psi=Ae^{i(\mathbf{k\cdot x}-\omega t)}$

Suppose the following hypothetical two time scenario where t is defined as

$$\mathbf{t}=\cos (t_1)\hat i-\cos(t_2)\hat{j}$$
and
$$\mathbf{(bold)\omega}=\omega(\hat{i}+\hat{j})$$ with $\omega \in \mathbb{R}$

In particular, what are the conditions I need to verify to ensure t_1 and t_2 behaves like time?

The Dark Side

@Secret Those conditions are Secret :D

Secret

Put it in another way, what are the list of mathematical constraints to ensure some arbitrary defined variable $a$ actually give the same physics and observable outcomes as time does? (what mathematically makes a variable 'time'?)

ACuriousMind

16:15

@Secret That does not make any sense. The Schrödinger equation is essentially just a statement that the Hamiltonian generates time translations. Your "modified Schrödinger equation" has nothing to do with that at all. You just wrote down a strange equation and played around with it, that's not physics.

In particular, calling anything there "energy" is wholly unjustified.

The Dark Side

Ahh, this is taking so long, I am going off for now. I'll get back and brag about my El Dorado in a couple more hours.

privetDruzia

Hello

I have a very short question related to physics

Secret

@ACuriousMind
Page 4 http://arxiv.org/pdf/1308.1065.pdf
Given there are n particles each expressed with their own 4-tuples (x,y,z,t), the energy operator of the schrodinger equation becomes a linear combination of partial derivatives in the time variables $t_j$

I thought I have understood this properly and applied it into my case above (which is basically a 2 time 1 particle system instead of a 2 time 2 particle system described in that article)?

privetDruzia

and bycicles

What causes the frontwheel to turn wehn you lean on your bycicle?

ACuriousMind

@privetDruzia What about the answer to the question your question was closed as duplicate of does not satisfy you?

privetDruzia

16:26

no indeed

it does only partially answer my questions

because the question I am asking now stayed unanswered on the other thread as well

*partially unanswered

Secret

@ACuriousMind and therefore that $\sum_{j=1}^2\frac{\partial}{\partial t_j}$ in the above working is the same thing as the energy operator used in the article (with n=2)? Maybe you can help me to point out where I have misunderstood?

ACuriousMind

@Secret No, apparently you have not understood that. The article is explicitly about n-particle systems with one time for each particle. What makes you think you could somehow make that into more than one time for a single particle? Also, you write $\mathbf{t} = {\cos(t_1)\choose\cos(t_2)}$ and assume the free particle solution then has $\mathrm{e}^{\mathrm{i}\mathbf{\omega}\cdot\mathbf{t}}$ in it - but that doesn't even solve your equation.

Secret

@Acuriousmind
Hmm... it seems that means I cannot just generalise the result naively for time dimensions as in spatial dimension cases...

...

What is the correct thinking process to extend the schrodinger equation to multiple time single particle cases (because that article, which is referred by seller earlier, applies only for n particle n time cases (one time for each particle)). I recall Slereah mentioned that it becomes ultra hyperbolic but I 'd like to see it myself by doing the exercise. But it seems I don't have the correct physical intuition to work out whether some of my manipula…

typo: seller is Slereah, stupid autocorrect

ACuriousMind

@Secret Well, the "correct" generalization of $\mathrm{i}\hbar \partial_t\psi = H\psi$ is obviously $\mathrm{i}\hbar\sum_i\partial_{t_i}\psi = H\psi$. But that's not physics, because the SE is the statement that the Hamiltonian generates time translations, and a single operator cannot generate more than a one-parameter group of evolutions - so you then got an equation without physical meaning.

The reason that paper can do it is because every particle has its own Hamiltonian in the absence of interaction (probably the very reason why you can't have interaction), so it makes sense to say "Let's just evolve particles 2 and 4".

Note that they have $n$ evolution equations which are just the SE for each particle. That's not a generalization of the SE to multiple times, but it is the correct generalization of the physics to their multitimes.

I don't know how you ever got the impression it would be meaningful within standard physics to have more than one time variable for a single particle.

Secret

16:49

Everyone in the physics community often abhors multiple time dimensions because of how it carries causality issues and the PDE are ultra hyperbolic. But I am still felt like I am only being told that this is how it works (and the workings that showed the conclusions) but I want to understand how under different scenarios it generates those problems, (more like how we do exercises, to see how exactly it breaks down and it must break down under all plausible circumstances)

In particular, if my memory serves, to be a dimension of time means the 4-vector is always future pointing. Then an n di…

---
In short, I felt like I don't understand enough on how multiple time breaks things and I want to convince myself by doing some exercise, but I am not sure how to ensure I am on the right track

because if my memory serves, a lot of people often found conceptualising the dynamics of something under multiple time dimension hard. Which is why I am currently relying on the maths to tell me how it breaks down

ACuriousMind

@Secret Honestly, you should learn to think more carefully about what you say first. Why is the 4- in 4-vector? Because you have 3 space dimensions + 1 time dimension. So why do you conclude that n time dimensions have n 4-vectors, and not one (3+n)-vector?

You're always jumping into these things and write a lot of things when even your starting points are not well-founded.

Also, the main reason everyone abhors multiple times in the physics community is that the world doesn't seem to have more than one time.

You don't have to understand anything about equations or causality to understand that there is just no evidence pointing to multiple-time descriptions being anything more than a play of some bored people's imagination.

Secret

right... (I am thinking too much into that article, I should have said one n+3 vector)

I don't know why, but despite spending my entire undergrad on doing physics, it seems my physical intuition is still a mess, and I keep on jumping to points without realising my starting points are unfounded (probably because either I don't know or I don't remember)

*on bold statement*: That's the statement I am trying to understand. One thing I don't really understand about how the conclusions that result form multiple time dimensions linked to that statement is that how would physics would have been d…

ACuriousMind

@Secret There's certainly has a large discussion of "What is time?" that could be had here in order to answer why I say the world doesn't have more than one time. But there's a far easier operational answer: Physics with just one time describes everything thus far perfectly fine (setting "minor" issues like quantized gravity aside).

And, regardless of how you define time, I find it hard to believe that any human could be seriously in doubt of whether we experience one or multiple time directions. Events that happen in our lives are linearly, totally ordered. There's no "second direction".

Secret

17:21

"How would an n time world differ form a single time world?"
I wish n time models that is being worked by some in the past have predicted something that is otherwise not observed thus can provide more support in a physical perspective that we are in single time and not n time

causality violation, ultra hyperbolicity and "we would have not existed" are testable predictions that are too broad to be pinpointed as the consequence of n time (which is why my confusion on how n time physics would have worked, independent of our senses (which aren't always reliable for example we don't encounter e…

David Z

Idea: next chat session we start seriously talking about changing the homework-like close reason to one about calculation requests

3

Secret

typo: "Predicted something that is a more specific consequence because of n time models" instead of "Predicted something"

I sometimes felt really frustrated when I failed to convince myself on the fundamental understanding of something. The n time discussion above is an example, because (as far my memory serves) all known predictions from n time models (which have shown to be false so far) are not specific enough to be said to be a consequence of n time models (or related) alone, thus these predictions shed little light on how the world would have looked different under these models

This is quite different from models like general relativity, where you can test it various ways from gravitational lensing to clocks running differently at different height (and only models that are basically equivalent in terms of predictions will predict the same thing), quantum mechanics, where you can demonstrate an electron behaves like a wave by looking at the interference pattern formed. and even back to classical mechanics, on when will the pendulum be at the next moment in time

etc. etc.

DanielSank

@DavidZ Yes!

Secret

Stand model predicts the higgs boson, and its decay pathways (and we have observed it). Even string theory have some testable predictions (however unable we are to test them right now) that are pretty unique to them

Secret

Why not n time models, rant I hate you (ie. time) (and myself) for being so hard to understand and I cannot analyse you mathematically to understand what you are

DanielSank

17:34

@DavidZ Reference.

David Z

Yeah, it's been on the tip of our collective tongue (to misuse a metaphor in a slightly weird way) for a long time now - a couple years I think - but I think things have started to clear up enough that we can actually do something about it in 2016

DanielSank

@DavidZ Is there a good way for me to participate in the discussion if I can't make the chat session? For example, perhaps there will be a meta post with suggested changes on which everyone can vote?

The Dark Side

Yippie

DanielSank

@TheDarkSide Wow. I'll never get that badge.

The Dark Side

And El Dorado

DanielSank

17:37

I dislike reviewing.

The Dark Side

@DanielSank Lol. You'll get many a gold for your answers.

Secret

http://physics.stackexchange.com/questions/54475/parallel-time-perpendicular-space-on-multiple-time-dimensions

more than one time dimensions leads to CTCs and hence violating causality. But there are many ways to make a CTC, thus it does not really help to understand what really happens in n time models

DanielSank

@TheDarkSide I don't get it.

David Z

@DanielSank yeah, we never make any policy decisions solely based on chat sessions. The sessions are just for brainstorming ideas and getting a sense of where the consensus lies, and then we turn that into a meta post (or use it to guide the discussion on an existing meta post)

DanielSank

@Danu Light chapter plz do

David Z

17:38

Well, maybe if it's something really small, but this is not

DanielSank

@DavidZ Groovy.

The Dark Side

@DanielSank I meant, I haven't got a single answer gold badge, like Great Answer/Great Question, or Populist. My answers on this site, and more generally, my knowledge of Physics is simply not in that league. So, review golds are all I can accomplish.

Ok. Got the badge, tick. Showed it off in chat, tick. Time to sign out. :P

ACuriousMind

17:56

@TheDarkSide Answer gold badges, at least the "great Answer" one on phys.SE are mainly the result of the HNQ effect, anyway :P

And the "Populist" one requires you to give a great answer that doesn't actually answer the specific question asked so it doesn't get accepted, which is quite difficult to do on purpose/just by writing a good answer.

DanielSank

@ACuriousMind Yeah, my closest approach was when I got to talk about Johnson noise. For a while I thought I'd get that sweet 100.

ACuriousMind

Ah, mentioning the question in chat also usually gives a few votes ;)

DanielSank

@ACuriousMind I deny nothing.

Hey guys, I posted my math question about delta functions over on the math site:

1

Q: Integrate $\delta(f(x))$, where $f:\mathbb{R}^N\rightarrow \mathbb{R}$, over one variable

Let $f: \mathbb{R}^N \rightarrow \mathbb{R}$ such that the solutions of the equation $f(x)=0$ form an $N-1$ dimensional manifold $M$. Consider a probability distribution $P: \mathbb{R}^N \rightarrow \mathbb{R}$ which is uniformly distributed over $M$. We could write this as $P(x) = \delta(f(x))$ ...

probability-distributions dirac-delta

Slereah

18:23

sup me droogs

user54412

Question for avid chalk users: Is there an awesome eraser out there? I know of great chalk, and of great slate, but I've never seen an eraser that impressed me.

ACuriousMind

18:39

@ChrisWhite I hear there is this new-fangled thing called a "sponge" ;)

Emilio Pisanty

@ChrisWhite I've seen some pretty reasonable chalkboard erasers.

Nothing too amazing, though.

Has anyone seen this one, by the way?

skyandtelescope.com/astronomy-news/…

user54412

user54412

@EmilioPisanty Yes, very cool. I made sure to check the publication from 9 months ago, and sure enough the predicted location was right where the image appeared.

Emilio Pisanty

@ChrisWhite Completely blew my mind.

Slightly oversold title on the S&T piece, though.

user54412

gravitational lensing is really eerie

2

user54412

18:52

what would we have thought if we saw multiple images of the same galaxy before GR?

Emilio Pisanty

@ChrisWhite Chalk it up to coincidence?

Attempt to cross Maxwellian EM with Newtonian gravity, most like.

Slereah

There is light deviation in Newtonian gravity

Well, there is for light particles

Not EM waves

i guess they would have just added a term to the Lagrangian to take that into effect

no_choice99

20:12

Hi people. I downvoted an answer but didn't lost any point. I thought I would lose 1 point?

ACuriousMind

@no_choice99 Yes, you do lose a point when downvoting an answer. (You don't lose a point when downvoting a question, and you get the point back if the answer is deleted.)

Mike Miller

@DavidZ: I love the phrase "our collective tongue".

no_choice99

20:44

Ok thanks @ACuriousMind

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