The h Bar - 2021-03-24

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2:37 AM

@Chris re this answer: you say "there's no reason that electric charge, in particular, has to be conserved." This stuff's been decades for me, but wouldn't Noether's thrm + gauge invariance of electric potential determine charge conservation? I may just be badly misremembering things. But if I'm right, I suggest that statement either be differently explained or scaled back.

3:06 AM

*correction ~~awhile~~ a while

Welcome @nitsua60 it has been a while :-)

since you've been in this this chatroom

1 hour later…

4:23 AM

@nitsua60 Yes, gauge invariance implies charge conservation. Or the other way around, that gauge symmetry only exists because electric charge is conserved. There are many different charges in the standard model, and not all of them are conserved.

5 hours later…

8:58 AM

@Chris cc @nitsua60 Noether's first theorem does not apply to gauge symmetries but to global symmetries, so it is not gauge symmetry that leads to (meaningful, on-shell) charge conservation via Noether's theorem, but the global part of the symmetry of electromagnetism.

However the relationship between the gauge symmetry and charge conservation is a bit subtle and the subject of many other questions, see e.g. Qmechanic's answers here and here

9:10 AM

@Chris In particular, you might say that the global part of the symmetry exists because charge is conserved, but the gauge symmetry exists because you cannot have massless vector bosons (=photons, in this case) without having a gauge theory.

9:33 AM

@ACuriousMind Thanks. It's been a little while since I've got too in depth on this stuff ;)

It's a subtle topic that's surprisingly not discussed with the depth it might deserve in most places. The gauge/global distinction in particular is a bit difficult to make precise, we just recently got another question about that.

2 hours later…

11:35 AM

@ACuriousMind how exactly does $\partial_{\mu} A^{\mu} = 0$ allowing harmonic functions $\partial^2 \lambda = 0$ relate to brst

I'm not sure what you mean

After you gauge fix using the first (partial gauge) condition in the path integral, the residual symmetries $\partial^2 \lambda = 0$ are still there for harmonic functions no

sure, but I'm not sure what you mean by this relating to BRST

Is there any relation of that residual symmetry to brst

BRST translates the gauge symmetry to the rigid BRST symmetry, you can't do "partial gauge fixing" in that formalism

12:32 PM

@Mithrandir24601 thanks for saying that :-)

is the roll of toilet paper in your new avatar a Covid-19 shopper's item reference? @rob

1:05 PM

new? that's been his avatar for almost a year now

i guess, we've been out of sync since pandemic started

or I'm finally regaining some of my powers of observation

user image

1:51 PM

@user85795 Yes, I made that painting in March and April of 2020.

2:27 PM

There's no inherent advantage to doing qft in the Heisenberg picture other than manifest covariance is there?

2:43 PM

Is Heisenberg manifestly covariant?

The covariantest of QFTs is the path integral one

Are the field operators not manifestly covariant in the heis. picture? I feel like in the schrodinger picture the covariance isn't obvious bc your objects depend on space not spacetime

I mean in both pictures, you're basing your theory on time evolution

which breaks covariance because you have to select a time variable

3:32 PM

But in quantities like $\langle \Omega|\phi_1(x_n)...\phi_n(x_n)|\Omega\rangle$ which are the physical quantities, you're dealing with entirely covariant objects aren't you? I thought that was the point of manifest covariance

I mean sure

Hilbert space products are just probabilities, they have to be covariant

@Charlie That quantity is not specific to a picture, really

That too

You can compute it in just about any QFT formalism

"picture" refers to how you apply time evolution, there's no time evolution in your example, and in fact you will probably use the interaction picture to evaluate it if you use LSZ

My question mostly stemmed from the fact that (at least in P&S) some effort is made to do qft in the Heisenberg picture where the field operators carry the entire time dependence

As in we define the interaction picture because in it the field operators are the operators of the free theory, but we then define the interacting time evolution operator to take us from the interaction picture to the heisenberg picture, as if there is some preference for it

3:39 PM

well, you have to choose some way to define it, don't you?

there's never a deep reason to choose any of the pictures over the other, it's always just computational convenience

yeah sure you have to pick one eventually, and that's when I wondered if the reason for picking the H picture is manifest covariance, since like you say they are all physically equivalent

Sometimes there are non-computational reasons

But here is my advice

If they give you a non-computational reason, run

that person may be insane

Presumably in the interaction picture manifest covariance is lost since the fields themselves are not longer covariant in expressions like n-point functions, even if the overall quantity is

I guess if the answer is just computational convenience that's probably the most satisfying answer, since I would have assumed beforehand that they are equivalent

Every theory is eventually covariant because in the end you have measurements which are independent of the coordinates

it's the middle steps you have to watch out for

In any operator-based theory, the Hamiltonian is gonna break the covariance since it is defined to be orthogonal to a choice of foliation

how does that work? orthogonal wrt the hilbert space inner product or the spacetime one?

3:52 PM

Orthogonal to the spacelike hypersurface

Although slightly more complicated because the hamiltonian is a form on the phase space or something

but the foliation still applies

is this in phase space quantisation or something?

i've heard of that but not really taken the time to read about it

@Charlie it's not the Hamiltonian, it's the time direction that's orthogonal

the two are related 'course since the Hamiltonian generates time evolution

sure, but I think Charlie was troubled by the notion of a function (the Hamiltonian) being orthogonal to a surface

2 hours later…

6:08 PM

Well the Hamiltonian is a form, technically

6:35 PM

0

Q: Why can't I vote two times to reopen a question?

This question received hot attention initially. Then it was closed. I and others didn't agree and voted to reopen. And so it was. Then, again, the question was closed. I wanted to reopen it but couldn't. I think the ones who closed it could close it again. But why can't I vote to reopen again? Is...

discussion down-votes moderators up-votes

SN11 has launched. It's starting the descent now ...

@JohnRennie How can you focus on that when the great chicken debate ensues?

And it's landed! No bang (yet :-)

No dammit, they're just reshowing the SN10 landing. SN11 hasn't taken off yet.

@BioPhysicist I love Andrew's answer :-)

Ah, the SN11 launch has been cancelled :-(

6:53 PM

@BioPhysicist ::sigh::

really not sure if that question is better or worse than the endless quantum interpretation questions :P

7:13 PM

"removed meta-commentary"

@ACuriousMind Haha. It's ok, there is an answer there applying QM to the chickens. So we can combine the two problems into one.

For some reason the word chickenfunction is really funny

lol

"It can be shown an isomorphism exists between quantum interpretations and the surface area of a chicken"

4

"What does the MWI say about wavechicken collapse?"

7:15 PM

I'm surprised no one pointed out that clothes/armor never perfectly cover the surface area of the wearer. You don't usually cover everything, and there usually isn't a completely tight fit

depends on the clothes :P

but sure, armor doesn't really have to care about surface area

7:33 PM

If playing as a female character in rpg's has taught me anything, it's that surface area is always a secondary consideration when it comes to armour

I think the chick(en)s in chainmail bikinis era is fortunately over

8:04 PM

If that question gets reopened I'm done lol

1 hour later…

Mithrandir24601

9:06 PM

@Charlie I agree

-1

Q: "State of research" questions

Are questions asking for what the current research topics in a certain field allowed? Under what circumstance would they be? In case they are allowed, after how much time would it be allowed to ask the same question again, perhaps after a big discovery? Regarding this last point, while it's true ...

9:41 PM

@ACuriousMind Chicken chicken chicken: chicken chicken

10:05 PM

Chicken. Chicken chicken chicken chicken.

*Heh. I forgot about that.

classic

1 hour later…

Mithrandir24601

11:28 PM

Physics of chicken:

I Cooked a Chicken by Slapping It

When you think about it, the ability to fan oneself with our hand is a pretty unique cooling method for a mammal

Like a dog would probably just have to endure the heat

@SirCumference that's why dogs pant - the evaporation of saliva from the mouth/tongue cools them

huh, interesting

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