The h Bar

skill patrol

4:00 AM

Negative time is just measured in the opposite direction as positive time.

barrycarter

That's a very Zen-like statement.

skill patrol

The same would go for displacement.

barrycarter

Yes, but we generally accept the "arrow of time" points forward.

We can't go backwards in time, and the fact that simultaneity lets us is just "creative accounting" as it were.

skill patrol

True, but recall the equations work for negative time as well, right?

barrycarter

@skillpatrol That's what I'm saying. We invent simultaneity so that the equations work.

@skillpatrol I believe there's a better way, perhaps involving proper time or breaking the Lorentz transform into two parts: a simple v * t part, and another part for displacement.

skill patrol

4:04 AM

What I'm trying to say is the observer perceives simultaneous events.

barrycarter

Really, though? You could argue that what the observer perceives is what the observer sees, not what the observer computes.

skill patrol

The classic two flashes of lightning along a railway embankment :)

barrycarter

You mean the before-and-after issue?

Observer 1 and Observer 2 assign different ordering to two events?

skill patrol

Nope, he measures out equal distances in opposite directions and stands in the middle and perceives two flash at the same moment.

barrycarter

OK, and?

(I'm not familiar with this example, apparently)

skill patrol

4:10 AM

Yes, then he stipulates the time intervals are equal.

barrycarter

In other words, he assigns the same time to both events?

0celo7

@barrycarter annoys

skill patrol

Yes, for him they happened at the same time.

barrycarter

Ocelot, good to see you're up to "noun verb" constructions.

@skillpatrol OK, so what's the point/paradox?

Ocelot, Dog barks.

0celo7

@barrycarter what

verb means I am doing verb

like yawns

means I'm yawning

barrycarter

4:12 AM

Ocelot, I was helping you with your two word sentences. Cat purrs. Dog barks. Mouse squeaks.

0celo7

I'm annoying you @barrycarter

@barrycarter It was a one word sentence

barrycarter

Ocelot, that would be "annoys barrycarter", not "barrycarter annoys"

0celo7

That's what they teach us in engineering

@barrycarter I don't consider the @name a part of the sentence

barrycarter

Ocelot, so your sentence was literally the word "annoys"?

0celo7

Yes.

barrycarter

4:14 AM

Ocelot, I don't consider verbs part of sentences :P

Ocelot, but unless it's an imperative sentence, I think you need more than one word.

Ocelot, annoy!

0celo7

I can annoy you

$\mathrm{e}^x=1+x$

barrycarter

Ocelot, true, but experience has shown that I annoy you more.

skill patrol

Does the first observer have the same experience with "time" as a second observer in a train moving towards one of the flashes?

barrycarter

@skillpatrol A train that happens to also be between both flashes?

skill patrol

Yes

barrycarter

4:17 AM

@skillpatrol Well, yes, right? The speed of light is constant.

Ocelot, why would I find that annoying?

skill patrol

Yes. But does he perceive them at the same "time" as the embankment observer?

barrycarter

@skillpatrol I would argue yes, but I sense you're leading up to something.

skill patrol

What about his motion?

barrycarter

@skillpatrol Irrelevant if we're talking about light(ning) flashes.

skill patrol

Sure the effect is small.

barrycarter

4:20 AM

@skillpatrol I think I'm missing your point. Are we talking about a super-fast train or a real life situation?

skill patrol

Super fast, thought experiment

Just logic.

barrycarter

@skillpatrol OK. Well, since he's at the same position as the embankment observer AND the two light flashes are equidistant, I would say his experience is identical.

At the very least, he would also say the two flashes are simultaneous, albeit potentially red/blue shifted.

skill patrol

Ah, so there is a potential difference :-)

barrycarter

@skillpatrol Well, of course. Relativity only says the speed of light is constant, not its frequency.

skill patrol

Due to

5 mins ago, by skill patrol

What about his motion?

barrycarter

4:25 AM

@skillpatrol Yes... but what are you driving at? I thought we were talking about time.

skill patrol

Yes the time it takes for light to travel to the observer.

barrycarter

And you're saying it gets to the moving observer faster because of Lorentz contraction?

skill patrol

Well, he is moving towards one of the flashes and away from the other, right?

barrycarter

@skillpatrol True, but that doesn't have an effect because light regards the rest of the universe as stationary.

skill patrol

But he is not stationary, the embankment observer is stationary in this thought experiment

barrycarter

4:29 AM

@skillpatrol With regard to the source of the two light(ning) flashes, right?

skill patrol

Exactly.

barrycarter

But the relative speed of a light source should be irrelevant, that's the point of relativity.

skill patrol

We are just building up to that.

barrycarter

OK

skill patrol

44 mins ago, by skill patrol

how else could you describe it to a beginner?

barrycarter

4:32 AM

Once a beam of light is released, its speed is constant for all observers?

skill patrol

yes it is a constant. So the observer on the train must perceive the flashes as simultaneous at a different moment in "time." Got it? :-)

barrycarter

Not really, no.

skill patrol

:-/

barrycarter

@skillpatrol For convenience, let's have the two observers agree on a t=0. That would be the first step, no?

skill patrol

Yes, at the very moment the one in the train passes the one on the embankment t'=0 and t=0

barrycarter

4:37 AM

Yes, which is also the exact instant they both see the light flashes.

skill patrol

Nope. The light still needs time to travel, right?

barrycarter

OK, wait, you're saying the light flashes are "released" when the pass each other?

skill patrol

Yup

barrycarter

(thinking...)

OK, so both observers still say the flashes occur at the same time, but "stationary observer" says two different times for moving observer?

skill patrol

I have to go...read this

Sorry :(

0celo7

4:44 AM

@barrycarter I can talk relativity with you from a mathematical PoV

barrycarter

Skimmed it, but we can talk more later.

Ocelot, OK, what do you think about our discussion?

0celo7

tl;dr

barrycarter

I claim simultaneity is merely bookkeeping

0celo7

what's your notion of simultaneity?

barrycarter

The time you must assign to a non-local event in order to get the Lorentz transform to work correctly.

0celo7

4:47 AM

@barrycarter There is a book on special relativity for mathy people

I think their notion of simultaneous is that there's a spacelike hypersurface containing the two events

barrycarter

Remember, I think I already know special relativity. I'm just now considering the physical interpretation of it.

0celo7

Why would you do that to yourself

barrycarter

Yes, this seems to be the general consensus: use the formulas, don't try to understand it.

0celo7

I'm not sure what you're trying to understand

have you drawn a picture?

barrycarter

In the above, the two observers must assign different times to Earth.

Bernard Meurer

4:55 AM

@DanielSank robertbowenart.com/wp-content/gallery/art/github-octocat.jpg

0celo7

alright @barrycarter

let's see what's going on here

@barrycarter different times?

barrycarter

Ocelot, yes, the Black observer must assign Earth a time of 2008.

0celo7

the distance is 10 light years in what frame?

barrycarter

Ocelot, proper distance in the Red/Earth frame.

0celo7

@barrycarter can I think about this tomorrow? I have to be awake in 7 hours

I need my beauty sleep

@BernardMeurer can confirm I need it

barrycarter

4:59 AM

Ocelot, sure. I didn't even ask you to think about it in the first place :)

Sleep, dream of birds.

0celo7

I gotta look good for my one-eyed toothless alabaman gf

who plays banjo, of course

barrycarter

What happened to her eye?

0celo7

@barrycarter Well I kinda do relativity in almost all of my spare time

@barrycarter battle with a fucking pit bull

barrycarter

Ocelot, when you're not proving those physics formulas you use?

Ocelot, yes, they can be quite horny.

0celo7

@barrycarter Did you check my blog?

barrycarter

5:01 AM

Ocelot, I did at one point, yes. You do prove some formulas.

But every single formula you use in physics? Yikes.

0celo7

it's not really "physics"

barrycarter

It's derivations, yes.

0celo7

...for Lorentzian geometry

physics has to be done with physics in mind

barrycarter

I understand the matrix of transform. I'm complaining that it's stupid.

I believe there is another way.

0celo7

@barrycarter ok, I'm going to sleep 4 realsies this time

barrycarter

5:02 AM

Rest in peace.

Slereah

7:38 AM

hey hey

skill patrol

8:47 AM

ho ho

Slereah

Is this a valid flag reason

If so I have a lot of flagging to do

skill patrol

nah, it can always get worser :P

Slereah

Also is this a correct answer to a homework question

skill patrol

lol

Git Gud while da gitten's gud

ACuriousMind

11:28 AM

@DanielSank The equilibrium thermal state at inverse temperature $\beta$ is $\rho_\beta = \sum_i \mathrm{e}^{-\beta E_i}\lvert \psi_i\rangle\langle \psi_i \rvert$ for $\lvert \psi_i\rangle$ the energy eigenstates of the system, right? And that's certainly not a pure state.

Oh, wait

nvm

I mixed up being mixed with being entangled, that's embarrassing.

I don't see an immediate argument that the thermal state is not entangled, though.

Slereah

is there a measure of entanglement

Like there is of purity

ACuriousMind

I think it's pretty hard to find a measure that only detects the correlations due to entanglement and not those of the mixedness.

Slereah

What's the difference between mixed and entangled, anyway

What's a state that is mixed but not entangled and vice versa

ACuriousMind

The mixture is statistical, it represents incomplete knowledge about which quantum state the system is in. The entanglement is intrinsic, it's a property of a single quantum state.

Slereah

Oh

Isn't the entangled state not just $\text{Tr}(\rho) \neq 1$

ACuriousMind

11:40 AM

Uh, no?

A density matrix must have $\mathrm{Tr}(\rho)=1$, else it's not a density matrix.

Slereah

Hm

I am pretty bad at density matrix formalism

ACuriousMind

For pure states, you can use the von Neumann entropy to detect entanglement, but it also detects mixedness, so it's useless to decide for mixed states.

You also have to consider that entanglement is always formulated w.r.t. subsystems

Slereah

Is it?

ACuriousMind

It cannot be a mathematical property of the full density matrix alone.

Slereah

Wouldn't $\vert 0 \rangle \vert 1 \rangle + \vert 1 \rangle \vert 0 \rangle $ be entangled, even if it was the whole system

ACuriousMind

11:43 AM

@Slereah Of course, a state of a system is entangled by definition if it is not separable into the tensor product of states of its subsystems.

Slereah

Oh, you mean that

ACuriousMind

Yeah, I mean that searching for a property of the state of the whole system alone that gives a measure for entanglement can't work, you have to know the spaces of the subsystems - while the concept of mixedness makes perfect sense without subsystems.

Slereah

Well here I assume that we know the full theory

What operator can we apply on a perfectly known state to just spit out if it's entangled or not

ACuriousMind

If you know the full theory, it's probably easier to just compute the subset of non-entangled states explicitly and just check if it's in it :P

Slereah

Well then do that to the thermal state mister fancy pants

ACuriousMind

11:48 AM

The problem is that the non-entangled states don't form a vector space, you can't expect there to be a nice operator that tells you whether a state is entangled or not

@Slereah Well, but in the thermal case I don't know any subsystems! It probably depends on the thermal system considered!

Slereah

Well take a particular example then

Gee

Take the thermal state of a free scalar field

ACuriousMind

...why QFT? :P

Slereah

QFT is best

ACuriousMind

I'd just have coupled a finite bunch of oscillators together.

Slereah

Then just pick free particles

in a box

ACuriousMind

11:50 AM

(I don't think my method is actually feasible for Hilbert spaces with more than one-digit dimensions :P)

Slereah

for shame

" At T > 0 mixing of states leads to entanglement"

From random bits of papers, I think thermal states are somewhat entangled

Probably not overly surprising?

It's pretty nonlinear

ACuriousMind

No, I'm not surprised. The higher-dimensional the spaces get, the "smaller" the set of non-entangled states gets, and I don't see anything forcing a thermal state of an interacting system to be non-entangled.

I'm not so sure about the free quantum gas, though.

0celo7

Good morning boys, Bajoran.

Slereah

Mornin

ACuriousMind

Since the Hamiltonian is just a sum of free ones $H = \sum_i H_i$, the thermal state $\mathrm{e}^{-\beta H}$ neatly factorizes into the product $\prod_i \mathrm{e}^{-\beta H_i}$, so it's separable.

0celo7

11:58 AM

@Slereah Kobayashi-Nomizu is coming...eventually. Had to get them through a third party seller.

Slereah

Hm

Could be

Well

At very low temp, it's gonna be like

$\sum_n 1 - \beta H_n$

So...

Fuck I dunno

I haven't read much on thermal states

My biggest ref on thermal states is Birrell, and that's not at all the topics

ACuriousMind

12:13 PM

Oh, no, it's not non-entangled, I wrote the wrong product!

That it factorises into a product of operators is irrelevant, it has to factorize into a tensor product of operators.

Hmm, now I'm very confused.

Slereah

Doing ladder operators is hard on the latex

$\hat a^\dagger _{-\vec p}$

So many extra bits

Physics Meta

0

Q: Can we post question answer in a form of article?

Some days ago I gave answer to my own question " what is dark matter and dark energy ", it was actually a copy of an old question asked months before. Now my concern is that I wanted to share some viewpoints related to the concerned problem and wanted to show it to every member here. But as you ...

Slereah

$FT(\sqrt{2\omega} \hat \varphi) = \hat a_p + \hat a^\dagger_{-p}$

$FT(i \sqrt \frac{2}{\omega}\hat \pi) = \hat a_p - \hat a^\dagger_{-p}$

Sounds about right

So... $\hat{a}_p = \frac{1}{\sqrt{2 \omega}} FT(\omega \varphi + i \pi)$

Hm wait

Is it correct

The document i had has $e^{+ikx}$

Which is an inverse transform

Maybe they use different conventions from Peskin

ah yes, they switch the signs on creation and annihilation

So that is indeed correct

But then that means I still don't know if the commutator is correct!

$[\hat a_p, \hat a_k^\dagger] = \frac{1}{(2\pi)^3 \sqrt{\omega_p \omega_k}} \int d^3x\ e^{i(p-k)x}(\omega_p + \omega_k)$

The form of the function seems correct enough but I don't know what to do with the sum of frequencies there

Is $\frac{\omega_p + \omega_k}{\sqrt{\omega_p \omega_k}} = 1$???

Or did I fuck up

Let's see

0celo7

12:34 PM

You always fuck up :/

ACuriousMind

@Slereah No, what the heck are you doing?

0celo7

@ACuriousMind quantum mechanics

@Slereah What are you doing?

Slereah

$[\omega_p \varphi + i \pi, \omega_k \varphi - i \pi] = i (\omega_k [\pi, \varphi] - \omega_p [ \varphi, \pi])$

Is this correct

Just doing the commutator of ladder operators

$[\omega_p \varphi + i \pi, \omega_k \varphi - i \pi] = \omega_p \delta(x-y)-\omega_k \delta(y - x)$

After integrations on the deltas I get chat.stackexchange.com/transcript/message/28858510#28858510

ACuriousMind

You have to start with imposing $[\varphi(\vec x),\pi(\vec y)] = (2\pi)^3\delta(\vec x -\vec y)$. Then deduce the Fourier transformed version $[\tilde{\varphi}(p),\tilde{\pi}(q)]$ and $[\tilde{\varphi}(p)^\dagger,\tilde{\pi}(q)]$. Then write the ladder operators in terms of those and compute the commutator of ladder operator without doing a Fourier transform again.

Slereah

What is wrong with what I did, though

What particular step is false

ACuriousMind

12:44 PM

Oh. Your expression for $a_p$ is wrong. It's $a_p = \frac{1}{2}\left(\sqrt{2\omega_p}\tilde{\varphi}(p) + \mathrm{i}\sqrt{\frac{2}{\omega_p}}\tilde{\pi}(p)\right)$.

Slereah

(That's what I wrote)

ACuriousMind

Oh, I missed the $\frac{1}{\sqrt{2\omega}}$ in front, sorry

Why did you write it so weird! ;P

Slereah

Less square roots to write

ACuriousMind

Well, then I can't tell what "step" is wrong because you didn't show any steps towards $[a_p,a_k^\dagger]$.

Slereah

Well I didn't write the full thing

It was basically

0celo7

12:48 PM

When life sucks I just enjoy the head -- Lil Wayne

^genius.

@Slereah You're telling me this is in NO QFT books?

Slereah

$[a_p, a^\dagger_k] = \frac{1}{(2\pi)^3\sqrt{\omega_p \omega_k}}[\int d^3x e^{-ipx}(\omega_p \varphi(x) + i \pi(x)), \int d^3y e^{iky}(\omega_k \varphi(y) + i \pi(y))]$

ACuriousMind

@0celo7 I think he wants to do it on his own :P

Slereah

$$[a_p, a^\dagger_k] = \frac{1}{(2\pi)^3\sqrt{\omega_p \omega_k}}\int \int d^3y d^3x e^{-ipx} e^{iky} [ \omega_p \varphi(x) + i \pi(x),\omega_k \varphi(y) + i \pi(y)]$$

And so forth

I could not find it in a QFT book, no

Usually they just say "The commutator is this, check that it is this"

ACuriousMind

@Slereah You didn't conjugate the $\mathrm{i}$ in front of the $\pi$.

Slereah

$$[a_p, a^\dagger_k] = \frac{1}{(2\pi)^3\sqrt{\omega_p \omega_k}}\int \int d^3y d^3x e^{-ipx} e^{iky} (\omega_p \delta(x-y) - \omega_k \delta (y-x))$$

I did in my notes, tho

No worries

Then I integrate out the deltas, and the result is fine for commutators of ladder operators, except

I have omegas I don't know what to do with

0celo7

1:00 PM

@ACuriousMind I don't understand this concept

Slereah

That's why you'll always be an engineer

ACuriousMind

lol

(not to you 0celo7)

I found my notes

Slereah

My professors tended to be quite unkind towards engineers

0celo7

@Slereah yeah, who's better at math than you

ACuriousMind

From my QFT course where we did that

And I did that exercise

Slereah

1:01 PM

Well now I'm wondering if you are

Are you good

Or did you find it in a book

ACuriousMind

I have the same $\omega$s. They are there in one line, they disappear in the next

0celo7

Lol

ACuriousMind

No indication as to why

0celo7

HACKER

Slereah

wot

0celo7

1:02 PM

Won't there be a momentum delta function

So you set $p=k$

Slereah

Do you mean you have $\frac{\omega_p + \omega_k}{\sqrt{\omega_p\omega_k}}$

ACuriousMind

Slereah

and then it is $= 1$

0celo7

Jesus

The momentum delta sets $p=q$, people

ACuriousMind

lol

You're right

Slereah

1:03 PM

It's not a momentum integral you numbnuts

Wait

ACuriousMind

@Slereah No, he is right. That's why I didn't write anything, it's silly.

0celo7

@ACuriousMind who is

ACuriousMind

@0celo7 You

0celo7

Oh no, I'm the engineer

You physicists are of course right

Keep on doing what you were doing

ACuriousMind

$f(p,q)\delta(p-q) = f(p,p)\delta(p-q)$ for all intents and purposes.

0celo7

1:04 PM

For "physics" purposes

Slereah

Isn't it $\delta(x-y)$ tho

0celo7

@Slereah There's two integrals

ACuriousMind

...what?

Slereah

Well yes, but one is over $x$ and the other over $y$

Neither is over momentum

ACuriousMind

What are you talking about?

Slereah

1:05 PM

Isn't $a_p = \int d^3x\ e^{-ipx} (\omega_p \varphi(x) + i \pi(x))$

ACuriousMind

It is.

Why does it matter?

Slereah

Oh wait

This is outside of all integrals, right?

I see

ACuriousMind

Yes

You're "anticipating" the action of the $\delta$ by using it already on the function in front of it

Slereah

a tad odd but understandable

So I get $\frac{\omega_p + \omega_k}{2\sqrt{\omega_p \omega_k}} \delta(p-k) \approx \frac{\omega_p + \omega_p}{2\sqrt{\omega_p \omega_p}} \delta(p-k) $

Welp, onto the next bit

Let's prove it the other way I s'ppose

what are good QFT books btw

I got Peskin and Weinberg

And Grimm and Jaffar

Apparently Zee is good

and Schwartz

I wonder how many physicists try to have a book title that actually stands out

Just call it Bantum Bield Beory I dunno

Slereah

1:51 PM

ugh