The h Bar

Naveen V

18:00

speaking of symmetries , does symmetry ever physically exist ?? Like even if its perfect in macroscopic scale , sometimes can particles be generated like electron positron pair during the measurement and break the symmetry ??

Amit

All of particle physics is about symmetry I think

Bosons are all the same and stuff

With respect to certain things at least

Naveen V

is it :o

Amit

Yeah if you ask people what math you need for Particle physics they always say group theory right? And what is group theory? Study of symmetries

But it's not necessarily the symmetry you know about today. I don't know what you are currently calling symmetry

Naveen V

but those symmetries are sometimes abstract right , which we ourselves term it ?

Amit

In math it's roughly, anything you do to a thing, and leaves at least some aspect of "thing" the same as it was

Not necessarily, some of these symmetries correspond to physical processes

Naveen V

18:04

Im finding it hard to differentiate physical and abstract here in terms of symmetry

Amit

Like exchanging two particles

Because we study both types, some symmetries we know what they mean physically, and others less so

Naveen V

yeah

my mind I is saying that the abstractness aspect comes from electron = exist , positron = must exist which was decided by us so would you call this physical or abstract @Amit

Amit

I'm not sure what you mean, the positron apparently physically exists yeah

Whether Dirac said that because of some symmetry, maybe, I don't know enough to tell you

Probably it had something to do with it

Naveen V

@Amit I got it , I was confusing myself with symmetries in algebra and stuff like if we have a circle and we cut in half its symmetric but electron positron are not same stuff right !

Amit

Different types of symmetries. It's just that "Symmetry" is such a wide term. If you see how general the meaning is you'll understand why it applies to both

Silly Goose

18:12

well group theoretic symmetry is motivated sometimes by colloquial physical symmetry

you can define the symmetric group $S_n$ as the set of all rigid rotations and reflections of an n-gon

Amit

Yeah but physical symmetry in the sense of symmetry you deduce from experiments, is not necessarily geometric symmetry :) Which is more like the example with the circle

Naveen V

@Amit what would be the general meaning here ?

Amit

The general meaning is do something to a "system" that preserves some property of the "system".

Exchange two halves of your circles and you leave the circle the same

Naveen V

ahhh

Amit

Exchange the LHS and the RHS of an equation and you leave the equation the same (silly example but correct lol)

Naveen V

18:16

exchange two proton antiproton you get boom and nothing from which it created from !

Amit

Are you giving a counterexample? :)

Silly Goose

i tried to minimize a function of one million parameters before sleeping XD it was not complete by the time i woke up

Naveen V

@Amit counter example ? , nothing --> proton antiproton ---> nothing symmetry right ?

Amit

Hmm no, I think the proton and antiproton while we say they "annihilate", this process still has to conserve energy somehow, probably as photons

Silly Goose

i think you'd like to denote what is symmetric with respect to what? or no

Amit

18:20

I don't think pair production happens out of "true" nothingness

Silly Goose

like if i am interested in conservation of angular momentum w.r.t. time, then i'd like to explicitly specify that my hamiltonian is invariant under spatial rotations, which would be my "symmetry"

Naveen V

@Amit hmm , from gravitational potential energy we get proton antiproton pair and boom energy conserved ( hawking radiation) but here the two particles escape the black hole to get a good boom

Amit

@SillyGoose Definitely, we need to know both what is the transformation and what is the conserved quantity to define the symmetry

@NaveenV Did you get that from a pop science book?

I'm asking mainly about the "good boom" part

Silly Goose

i finally learn about limits and sequences thank god lol

Naveen V

@Amit well erm , I dont remember but I think its like a video I was watching which indicated some particles escape but others are not so fortunate and one pair is out and other will be consumed by black hole

in that escaping particles , it collides with themselves and hence "good boom"

@Amit Please correct me , if I am wrong because Im 99% sure from this that I am wrong

Amit

18:28

Well I'm quite sure there's no meaning in saying that particles are or are not fortunate

Naveen V

yeah , but I think its great to give emotion to particles :D

Silly Goose

does anyone know if there is any import or work on finding an equivalence class of quantum circuits just with respect to anything

Amit

If your question is about Hawking radiation... I know very little about that, but I'm quite sure you won't get a very accurate picture from watching a pop-sci program on that :)

Naveen V

not necessarily about hawking radiation , but the creation of particles and annihilation of them , and I do agree on pop-sci part , for the topics that I have seen in pop-sci stuff its very different from original thing I have learnt

Amit

@SillyGoose Sure, for quantum programming right? You may be able to find the most efficient implementation of a program that way, maybe

@NaveenV So why do you need the black hole? You don't need it to discuss pair production

Naveen V

18:34

@Amit for conservation of energy of pair production

Amit

Energy is conserved anyway, what's the black hole got to do with it

Naveen V

@Amit I mean , the first thing that comes to my mind , for nothing --> pair production and you describing to conserve energy was the gravitational potential energy of black hole made use to create this pair

Amit

who said it's nothing first of all?

PM 2Ring

@NaveenV That doesn't work. You can't make a proton + antiproton from nothing, and when they annihilate they certainly don't leave nothing.

Naveen V

ah then I must be wrong

Amit

18:39

look up pair production anywhere, you'll see the process

PM 2Ring

29

A: What happens to the quantum information of a particle and an antiparticle when they annihilate?

Particle + antiparticle annihilation preserves quantum information. It is often said that annihilation creates a pair of photons, but that's a big simplification. It only applies to the electron + positron, and even in that case it may lead to more than 2 photons. If the electron's and positron's...

Naveen V

thank you very much !

PM 2Ring

Also, that stuff about black holes and virtual particles is just a heuristic, it's not really a good way to think about Hawking radiation.

42

A: An explanation of Hawking Radiation

To answer this we need to talk a bit about how particles are described in quantum field theory. For every type of particle there is an associated quantum field. So for the electron there is an electron field, for the photon there is a photon field, and so on. These quantum fields occupy all of s...

Besides, most Hawking radiation is in the form of photons, which are their own antiparticle. You don't start to get heavier stuff (like electrons & positrons) being emitted until the BH is tiny, with a Schwarzschild radius around the size of a proton. vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator

Physics Meta

21:25

0

Q: Why was this considered a 'bad question'?

Yesterday, I asked a question about capacitors. It was closed under the "homework-and-exercises" tag and my submit-to-review appeal was rejected. I'm not coming here to complain, I'm just genuinely curious what the actual issue is. See, I am used to the culture of close votes over on MSE (mathema...

Nihar Karve

21:52

@bolbteppa re this, could you get it from the spin-j term ($\propto P_j(\cos\theta)$) in the partial wave decomposition of the t-channel amplitude (equiv to s-channel via crossing symmetry)? Then high energy = high s, so $P_j(\cos\theta)\sim (\cos\theta)^j \sim (-s)^j$.

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