The h Bar

Ramanujan

12:00

@BalarkaSen, why is it unanswerable?

Balarka Sen

Not enough information. Read the above discussion.

Yashas Samaga

@Ramanujan take 50kg balloon and a 10kg rock for example

@Ramanujan Now consider a 10kg balloon and a 50kg rock

The answers are different in the two cases.

Your question does not provide enough information to answer the question.

Balarka Sen

You can take a very flat rock, and a perfectly spherical rock.

Ramanujan

@YashasSamaga, So is it answerable or not?

Yashas Samaga

.

@Ramanujan It is not answerable without making assumptions.

If the two objects are of the same shape and size, the heavier one will hit the ground first.

The air resistance will be nearly equal if they are of the same size and shape.

Why nearly? because the air resistance is a function of the object's velocity.

Ramanujan

12:03

But the question says shape and sizes are not defined

Yashas Samaga

$F_{air-resistance} = c_1v + c_2v^2 + c_3v^3 + \ldots$

ACuriousMind

@Ramanujan ...that's what we're telling you: You have to assume something about the shapes and sizes for the question to have a well-defined answer. Otherwise the question is ill-posed. That's really all there is to say.

Balarka Sen

I think you can explicitly write down the air resistance as a function of velocity using Stokes' law, can't you?

Yashas Samaga

@BalarkaSen Stoke's law works for spheres only?

ACuriousMind

^

Ramanujan

12:04

@YashasSamaga, @ACuriousMind, @BalarkaSen, what would be the answer if the same question appears in a board examination?

Yashas Samaga

You can write that the question is unanswerable. You'll get a zero.

The best answer if it appeared in the board exam would be what I've told.

Balarka Sen

@Yashas Ah, fair enough.

ACuriousMind

@BalarkaSen In general, the laws for drag forces involves coefficients that are empirical properties of the shape of the object.

Balarka Sen

Makes sense, got it

Yashas Samaga

@Ramanujan I'm telling from my experience. Assume that the objects are of similar shape and size. Explain using inertia.

ACuriousMind

12:06

@Ramanujan I do not understand why the answer would change depending on the context the question appears in. It doesn't become less ill-posed just because someone is demanding an answer.

Yashas Samaga

@ACuriousMind He has to make some assumption to answer it. You can't fight after the exam either. The paper setters will complain that we had to use common sense.

This has happened to me before.

ACuriousMind

From a well-posed exam question, I would expect "Make reasonable assumptions" as either a specific addendum to the question or as a general rule.

Yashas Samaga

I can guarantee you that 30% of the conceptual/MCQ questions in the exam are not well defined.

ACuriousMind

@YashasSamaga Doesn't change the that the answer to the question is ill-defined, and I find guessing what exactly the exam authors thought one should assume profundly uninteresting.

Slereah

@ACuriousMind that is why I say informally :p

Since most of it isn't really formal arguments

ACuriousMind

12:10

@Slereah It's just a quantum theory with an infinite-dimensional algebra of observables.

Slereah

I mean more in a standard model sort of context

Mostafa

@YashasSamaga Really true! This is actually the key to get a higher percentage than 90~ %95...without guessing what was in test author's mind you can never get higher than that...

ACuriousMind

That's my informal definition :P

Slereah

Is it a particularly helpful one though

Not sure it's that good for say solving some phi4 scattering

Yashas Samaga

@Mostafa It has been this way here for years.

India is a weird place.

ACuriousMind

12:12

@Slereah Why would the informal definition need to be good to "solve" a specific theory?

Mostafa

@YashasSamaga I really LOVE India!

Slereah

To have the proper tools to solve it

Yashas Samaga

@Mostafa Occasionally, even JEE questions have such ambiguities, but the question is usually deleted after the test.

ACuriousMind

Also, $\phi^4$ is a bad example because it is probably rigorously trivial :D

Slereah

Well yes, but

Yashas Samaga

12:14

@Mostafa There are 15 year old students who are doing research and there are 25 year old people who are trying to pass the 12th grade exam. This is India. Nobody is in the average.

Slereah

If I use some shitty non rigorous perturbation theory

It's all good

Hence the emphasis on informal

ACuriousMind

@Slereah How do you know it's good? Are there experimental matches to $\phi^4$ perturbation theory?

Slereah

Well the Higgs field is phi4

So I guess it's good enough

ACuriousMind

Yeah, but it's within the SM.

Slereah

Well it's hard to find something not within the standard model

Since that's what it's for

ACuriousMind

12:15

You don't calculate $\phi^4$-theory there, you calculate the Higgs field coupled to a whole bunch of other stuff that has, among others, a $\phi^4$ interaction

The reasons why $\phi^4$-theory is believed to be trivial don't work there

Slereah

If the phi4 theory is coupled to some gauge fields and such, is it likely to be non trivial?

ACuriousMind

You have the same thing with the Landau pole of QED and then embedding it into the SM, really

There's little reason to believe most of the toy theories are actually well-defined

So it is not clear to me that the "proper" definition (formal or informal) of a QFT should include that vast space of what we currently call QFT

Mostafa

@YashasSamaga I'll never forget that the very first math question in the university entrance exam was wrong! and it took me 2 to 3 times more than the max time of a single question, and had a bad effect on my performance for the remaining part of the test...and they deleted the question afterwards!

John Rennie

@ACuriousMind I'm pretty certain the accelerating charge paradox was worked out a few years back, though the paper on it was so impenetrable I gave up at the abstract. If you want I can grub around for a copy.

Yashas Samaga

@Mostafa The hard part is that you cannot even guess that the question is wrong because all the questions are weird questions. I would always hunt for the special thing in the question. As a general rule, everybody in India uses the 2+ pass method to answer the tests.

ACuriousMind

12:19

@JohnRennie Yes, the Wiki article gives the resolution. But the resolution sounds even stranger to me than the paradox :D

Mostafa

Nikola Tesla fell in love with a pigeon!

Yashas Samaga

I have around 20 cats (around? I really don't have a count)

Slereah

But what is a "toy model" really

Are there any non-toy model qft really

Yashas Samaga

What do you do if user X posts an excellent answer and user Y posts another answer which is just part of the excellent answer reworded?

Slereah

Also has there been any attempts to show phi4 nontrivial in some gauge context?

ACuriousMind

12:24

@Slereah In that sentence, I meant "toy theory" in the sense of "theory that is not actually expected to match experiment".

Slereah

Like through numerical simulations or such

ACuriousMind

I'm reasonably sure there are lattice simulations of $\phi^4$ coupled to a gauge field

Since that'd be the simplest implementation of a theory with Higgs mechanism

John Doe

Hi all

John Rennie

Ha, the HNQ strikes again

John Doe

Is a stimulated transition rate in QM defined as the probability of an electron entering a transition state under the influence of an external electromagnetic field?

Slereah

12:30

I wonder how many proofs of QFT are like "well technically we can't but let's do it anyway and see what happens"

Mostafa

@YashasSamaga Is that legal?

Yashas Samaga

@Mostafa There are people with 1000 cats.

Slereah

Like doing fourier transforms in curved spacetime

Even though that is verbotten

Yashas Samaga

@Mostafa Haven't you seen my cats? You were checking my imgur profile yesterday, right?

Koolman

Is the wavelength in both these equation means same . E=hc/$\lambda$ and $\lambda = h/mv$

Yashas Samaga

12:31

@Koolman For a photon, it is the same.

Mostafa

@YashasSamaga I just saw the username VeryCuriousMind :)

Yashas Samaga

@Koolman For other particles, it isn't.

@Mostafa imgur.com/a/8bpXk

Slereah

Also why is it cold instead of not cold

Koolman

@YashasSamaga when we use required equations

Yashas Samaga

?

If you are dealing with photons, you can mix the two equations.

The $lambda$ represents the same quantity for photons.

ACuriousMind

12:33

@Mostafa lol, we also have ConfusinglyCuriousTheThird here

Koolman

Other particles means

Yashas Samaga

$E=h\nu$ isn't valid for particles other than photons.

but the De Broglie equation is valid for all particles

ACuriousMind

I think that is a joke from when CuriousOne was still active, so that questions would often have comments from me, CuriousOne...and ConfusinglyCuriousTheThird :P

Mostafa

@ACuriousMind There was a reason I included you in my PhysicsSE timeline ;)

Yashas Samaga

@Mostafa How are my cats? ^_^

Koolman

12:35

@YashasSamaga we cannot also use it for electrons

Yashas Samaga

Yes.

Koolman

Ohk thanks @YashasSamaga

Yashas Samaga

@Koolman physics.stackexchange.com/questions/304859/…

@Koolman That is what happens if you use E=hf with electrons.

Mostafa

@YashasSamaga Wow! cute! But I can't reconcile that VeryCuriousMind username at the top with those cat photos in my mind!

But how can you handle so many of them? male and female....

(the growth should get exponential after a while....)

Yashas Samaga

The females stay indoores 24x7

The males dissappear for weeks.

Jim

12:39

@ACuriousMind Thanks for the link, but I found my solution. I was using the classical assumption that a charged particle can't interact with its own em fields. If you assume it can, you find that a charged particle undergoing uniform acceleration does not radiate in any frame, which solves the problem GR presents

Yashas Samaga

@Mostafa I sterilize every female after it gets a chance to be a mother.

Mostafa

@YashasSamaga So you don't care about the total population of the cats outside...

Yashas Samaga

?

Mostafa

males dissappear for weeks... :)

Yashas Samaga

And yea, when I said 20, I meant the ones which stay with me regurlarly

I honestly have lost the count

If I count all the kittens which were born inside my home, it might hit 30.

ACuriousMind

12:42

@Jim Hm. Do we know whether uniformly accelerated charges radiate or not?

The Wiki article mentions that at the end...and says we haven't measured it

So...it's not clear what the correct resolution is, it appears

Slereah

I think Feynman has a chapter in his GR book about radiation in GR

I forget what his resolution is

I don't think a coordinate change to Rindler for a static field would be EM waves tho

Jim

@ACuriousMind never been an experiment that could directly observe it

Ramanujan

,@YashasSamaga, When a feather and coin is dropped towards the surface of eartg, do they reach the ground together? If the experiment is done in moon, what is the difference? Write with reason

Slereah

Not on earth, no

Coin will drop forst

Jim

but the wiki article also points out the classical assumption's solution contradicts the bremsstrahlung effect, which is a definite measurement

Slereah

12:49

At least if you have an atmosphere

Yashas Samaga

@Ramanujan On earth, the coin reaches faster. On moon, they reach at the same time.

There is no atmosphere on the moon

Slereah

The earth typically has an atmosphere

Yashas Samaga

typically? lol

Jim

Feynman's solution agrees with bremsstrahlung and says a free falling charge should fall at the same rate as uncharged and NOT radiate. Classical says it radiates. Where is the extra energy coming from in classical?

Mostafa

Brian Cox visits the world's biggest vacuum chamber - Human Universe: Episode 4 Preview - BBC Two

►

Yashas Samaga

12:51

@Ramanujan As a matter of fact, the Apollo astronauts have dropped a hammer and a feather on the moon. They landed at the same time.

Ramanujan

@YashasSamaga , thanks.

Yashas Samaga

Feather & Hammer Drop on Moon

►

Jim

Plus classical has accelerated observers having event horizons and radiation from particles in the same frame entering inaccessible regions of spacetime. That sounds like a load of garbage that's just used to make it sound like it works

Slereah

Accelerated observers do have a horizon

the Rindler horizon

Ramanujan

@YashasSamaga , how does the anomalous expansion of water help aquatic aninales?

animals??

Yashas Samaga

12:53

Interesting question.

Ah!

It does not let the water freeze

The top layer freezes and forms an ice sheet. The water below it does not get frozen.

Slereah

I'm an animal and I am not helped at all by the anomalous expansion of water

Yashas Samaga

@Slereah You aren't an aquatic animal, are you? :P

Slereah

I've been known to go in water

Yashas Samaga

lol

Ramanujan

@YashasSamaga , why do the top layer freezes/

Yashas Samaga

12:55

Ice is at 0 degrees or lower.

Ice is less dense than water.

From 4 to 0, the density of water decreases instead of increasing.

Suppose you are a fish inside a lake.

It is 10 C outside.

Temperature is cooling down.

Once you reach 4 C, the cold water at the top sinks.

the atmosphere is cold, so the water at the top becomes cooler first

after the entire lake has reached 4C, the lake starts to cool further

as the temperature approaches zero

the water at the top freezes

it freezes at 0

and does not sink to the bottom

so it prevents the water below from freezing

Ramanujan

@YashasSamaga oo

Yashas Samaga

The fishes won't get frozen.

This is so amazing.

I am going to keep thinking for days now.

Nature is amazing.

Jim

@Slereah we're talking about an event horizon that would prevent you from seeing radiation from a particle in your frame that could literally be at any position relative to you

Slereah

Well yes?

Is there a problem with that notion

Jim

.... little bit

Yashas Samaga

13:00

@Koolman "According classical mechanics we have other types of energy also " What do you mean by that?

Slereah

Horizons aren't coordinate invariants

Even black hole horizons disappear in some coordinate systems

Ramanujan

@YashasSamaga , why do specific heat capacity of objects differ?

Yashas Samaga

-_-

That question is as good as asking why different people have different weight.

Jim

true, but in my frame on Earth's surface or in free fall over it, I'm fairly sure I don't have an event horizon that prevents me from seeing particles right next to me

I'm not THAT overweight

Ramanujan

@YashasSamaga ,???

Yashas Samaga

13:02

@Ramanujan different objects have different specific heat capacities

How can someone answer why?

they are different things, they have different proeprties

Jim

@YashasSamaga more like why different materials have different textures

Yashas Samaga

@Ramanujan Please don't tell me that you found that question in some question paper.

Jim

no

Ramanujan

@YashasSamaga, Yeah

Slereah

the rindler horizon basically only means that there are things that are not able to "catch up" with the observer

Yashas Samaga

13:03

@Ramanujan Where did you find that question?

Slereah

since it is constantly accelerating

Ramanujan

@YashasSamaga, and theres a hint too.

Jim

@Slereah light can catch up with me no matter how much I accelerate

Slereah

Sure

Yashas Samaga

@Ramanujan huh, there is a hint? -,-

What's the hint?

Jim

13:04

@Slereah The classical solution says there's an event horizon for accelerating observers that makes them unable to see radiation from any charged particle in the same frame.

I call shenanigans

Slereah

Well me I mostly see it this way

Take the static solution

Change it to accelerated coordinates

You're not gonna have radiations out of that

pretty simple to show

Ramanujan

@YashasSamaga, The properties of molecules of different materials are different and different materilas absorb heat energy in different ways. When the same amount of heat is supplied, the intensity of vibration and motion of molecules are different,so..

Jim

yeah, the math says you won't see it, I get it. It also says you will see it in a static frame. So they's clearly a contradiction

Secret

Hey guys, a rather random question. How many microstates are there for a system that is some classical particle accelerating in a straight line?

Yashas Samaga

@Ramanujan That's what I said: different objects have different properties.

Slereah

13:07

Self interaction in classical EM isn't trivial to solve

especially for point particles

Jim

@Slereah Feynman's non-classical solution says charges simply don't radiate under uniform acceleration. That works in all frames, no contradiction whatsoever

Slereah

yeah

sounds pretty reasonable

ACuriousMind

@Secret For a single classical particle, the concept of "microstate" doesn't make sense. It has states, to be precise states in $\mathbb{R}^{6n}$.

Ramanujan

@YashasSamaga , Tje coil of transformer is coated by enamel,why/

Yashas Samaga

It is an insulation.

Ramanujan

13:09

@YashasSamaga, means??

Jim

@Slereah so much more than "Well, we have event horizons that make it so the radiation visible to other frames enters a region of spacetime inaccessible to us"

Yashas Samaga

@Ramanujan It covers the coil wire from the outside so that the internal conducting part does not come into contact with another piece of wire.

Secret

@ACuriousMind Hmm, so I can only meaningfully talk about the entropy related to the kinetic energy of the system only if the system has an ensemble of particles rather than just a single particle present?

Jim

I also like the whole "The classical solution contradicts the observed bremsstrahlung effect whereas Feynman's does not"

Slereah

I have not read that argument so I don't know

Ramanujan

13:10

@YashasSamaga, what does it help for?

Yashas Samaga

:|

If you don't insulate the wire, it will short circuit.

The transformer won't be a transformer.

It will function as a insulator or as an open circuit or allows just a tiny tiny tiny bit of current to flow.

ACuriousMind

@Secret Entropy needs a definition of macrostate and of microstate. Without those, the concept doesn't exist.

Just going to an ensemble doesn't really help you, you still have to declare what the relevant variables that define a macrostate are.

(What a microstate of $N$ particles is is clear - a point in $\mathbb{R}^{6N}$, encoding all their positions and momenta)

Yashas Samaga

@Ramanujan en.wikipedia.org/wiki/Magnet_wire

Ramanujan

@YashasSamaga, why do the satellites not fall while revolving around the earth?

ACuriousMind

@Ramanujan See physics.stackexchange.com/q/9049/50583

Yashas Samaga

13:13

@Ramanujan physics.stackexchange.com/questions/312492/…

@Ramanujan It falls to the earth but the earth curves away enough to prevent it from hitting the earth.

Slereah

@Ramanujan strings

They are attached to the celestial sphere

Yashas Samaga

...

Mostafa

*Young man, in mathematics you don't understand things. You just get used to them.* (Von Neumann)
The better you become at math, the more you get used to it. The more you get used to something, the harder it would become for you to embrace new things (= to innovate).

Secret

Hmm, I am not very sure what the macrostate for a moving object will be. For example, if I have N particles all moving with some momentum p, then clearly they are all in the same momentum state. This configuration will then be a microstate. Going from that, if we define a macrostate to be the number n of particles that move with some specified momentum p, then if n=N, there is only one possible configuration in phase space that can satisfy this macrostate,
hence I counted only one microstate and get the entropy for this macrostate to be zero

Not sure if that's the correct way to think about it

0celo7

@ACuriousMind ...no response?

Jim

13:17

@Mostafa tell that to Newton or Hamilton or Dirac or any of the great mathematicians that became very used to mathematics before innovating and developing new mathematics

wait, you can't.... they're dead

Secret

No sorry, typo:

I mean suppose we have a system of N classical particles

Each of these particles has states given by the ordered pair (x,p), position and momentum

0celo7

Implying newton was a mathematician

Jim

yes

Secret

Suppose a macrostate of this system is defined to be the number n of particles that has some momentum p0

0celo7

No rigor @Jim

Jim

13:20

he wrote a book on it and helped develop calculus

Yashas Samaga

3

Q: Calculating force through ICOR (instantaneous centre of rotation)

A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure). If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ re...

is the answer correct?

radius is treated as displacement, right?

it shouldn't change over change of reference frames?

radius of motion of an object is independent of the frame of reference?

Mostafa

@Jim Does that mean you don't agree with Neumann's quote? (the remaining are just logical implications of it)

ACuriousMind

@0celo7 It looked fine to me; what sort of response do you want :P

Jim

@Mostafa What? Describe, in detail (this is for marks), how the remaining is at all the logical implications of it

Koolman

@YashasSamaga potential energy

Secret

13:24

Then since the phase space of n particles is $\mathbb{R}^{6n}$, where half of these will be momentum. Then for the system to be in the macrostate (x,p0), the number of possible microstates that the n particles can take is continuumly many (because we only need to fix p=p0 for each particle to satisfy this macrostate, leaving x free to take any real value for each of the 3 xyz compoennts).
Therefore I get an cunountably infinite number of microstates that can give this macrostate, and thus my entropy blows up.

Yashas Samaga

Wow! Someone experienced time dilation while driving a car. LOL

0

A: Would things experiencing time-dilation seem to be moving in slow motion to an observer?

I believe I actually experienced a time dilation. I would think it was crazy but I was with a friend when it happened and he experienced the same phenomenon. We were in my car (1980 Honda civic, small 4 cylinder with not modifications and very high miles) we were on the expressway. We were moving...

0celo7

@ACuriousMind "looks fine" "good" "have my babies"

All are acceptable

Koolman

@YashasSamaga rotational energy , vibration ,intermolecular , electronic , relativistic

0celo7

Wait

"I want to have your babies"

There we go

Koolman

@YashasSamaga I don't have enough knowledge , you would be knowing much more about it . So its just a guess

Jim

13:27

@Mostafa I'd argue that the only people who can develop new math or get used to new math are people who are good at it. Nobody who is poor at math is going to develop new legitimate forms of mathematics. I would also argue that if you see a lack of innovation as the only logical implication of the quote, there may be a separate conclusion you failed to account for: you may not have understood the message of the quote to the degree you believe yourself to have done.

Secret

@ACuriousMind That is, if I have calculated this correctly, the number of microstates for n particles that give a macrostate where they all have the same momentum, is the uncountable set $\mathbb{R}^{3n}$. How should I interpret this so that my entropy does not blow up?

ACuriousMind

@Secret you should go back to statistical mechanics and look at how entropy is defined in terms of microstates there - the exact same problem arises. The solution is coarse gaining.

Jim

thus, there would be at least 2 logical avenues of thought. Meaning yours isn't the only conclusion/implication. And, since I only requested that you describe how yours was the ONLY logical conclusion, showing there is at least one other is all I need to do to say "My hair is a bird. Your argument is invalid".

ACuriousMind

However, I would also urge you to think more about what you're doing: why are you looking at "all have the same momentum" as the macrostate? How is that physically relevant? think about how thermodynamic macrostates are characterized by easily accesible and relevant quantities: pressure, temperature, volume

Emilio Pisanty

so, question:

abcabc123

1 6

dirac-equation klein-gordon-equation particle-physics quantum-chromodynamics quantum-electrodynamics quantum-spin

why does that user have reputation 1?

Yashas Samaga

13:34

interesting

he has 4 upvotes

Emilio Pisanty

s/he ought to have 79

physics.stackexchange.com/users/121362/abcabc123?tab=reputation

Secret

Well, the motivation of this question is when I look at a system of classical particles in 1D travelling with constant momentum p, and recalling how entropy is the ln of the number of microstates that give the same macrostates, I then get the aforementioned weird result as any real number for the position x can give the same momentum thus using the definition of entropy, I then end up with a continuum number of
microstates for this system and then got stuck at how to interpret the result correctly.

Mostafa

@Jim I never said that's the ONLY conclusion.

@Jim There are many individuals who are good at math but never developed any truly original math. I think this is enough to get convinced that my argument is to a great extent true.

Emilio Pisanty

ditto with this user: physics.stackexchange.com/users/31119/…

Mostafa

The question is then "how could those very few math practitioners come up with their brilliant ideas"?

Secret

13:37

In short, the physically relevant background of this problem is since we knew entropy and energy is related by temperature (which in statistical mechanics, is the derivative of energy wrt entropy for a system in equlibiurm), and since kinetic energy is a type of energy, it seems logical that entropy is deifned for that, but when I tried to figure out the microstates in the naive way mentioned above, I get continuum many of them

Yashas Samaga

@Koolman I am not good with relativity. I only had a little crash course at NSC.

Koolman

And I had no course

Its just a guess

0celo7

@ACuriousMind what's the DL on von Neumann algebras?

ACuriousMind

@0celo7 What's a DL?

Also, why do you think I know anything about von Neumann algebras? :P

@EmilioPisanty 100 points rep penalty for a post deleted by rude/abusive/spam flags.

Emilio Pisanty

@ACuriousMind ah

fair enough

Secret

13:49

This user explains much clearer in his question on my motivation on why I actually consider the macrostate of "all particles moving with a constant momentum"

4

A: Can a single classical particle have any entropy?

The concept of entropy is very difficult because of the following day-to-day fact: when we have a macroscopic mechanical system, we can look at the system all the time, and know exactly where everything is. In such a situation, we know what each particle is doing at all times, the evolution is de...

Emilio Pisanty

also, wth is "JEE"?

Secret

Joint Entrance Examination

Joint Entrance Examination (JEE) is an all India common engineering entrance examination conducted for admission to various engineering colleges and courses all over the country. The test comprises of two stages - JEE Main and the JEE Advanced. The exams are of the objective pattern. JEE Advanced is regarded internationally as one of the most challenging engineering admission tests. In 2012, the government-run Central Board of Secondary Education (CBSE) that earlier conducted the AIEEE, announced this common examination that replaced the AIEEE and IIT-JEE. JEE consists of two parts, JEE Main and...

Jim

@Mostafa so, because many people familiar with math exist that don't innovate, that means that becoming familiar with math causes lack of innovation? Hmmmm..... Well, I know that any correlation must imply a causation. So I have no choice but to accept your argument as true

0celo7

@ACuriousMind it has algebra in the name, dog

"Down low"

Slereah

"Following Callaway [23], consider the effect of coupling a gauge field to the scalar field of (55). It is demonstrated in [23] that for the combined theory to be non trivial, the renormalized quartic coupling λ_r must not be too strong."

Jim

13:56

@0celo7 got your back, brah

Slereah

The ride never ends

"The breakdown of total screening (entering in nontrivial regime) occurs when the quartic coupling constant λ_r is less than the effective quartic coupling generated by the gauge field interaction"

ACuriousMind

@Secret No, he doesn't. I don't see how "I know the exact momentum but not the position of these particles" is a physically interesting situation in classical physics. Sure, you can consider it - but why? The answer just says that essentially every kind of constraint/knowledge about a system can be used to define a notion of entropy, but this doesn't mean choosing random constraints and looking at the entropy assoicated with them is useful.

Danu

o/

I found a web of dualities, @ACuriousMind

ACuriousMind

@Danu Between what?

Slereah

So apparently the SM may not be trivial

Danu

13:59

@ACuriousMind Almost complex structures on a $G_2$-homogeneous space :P

Slereah

Depending on constraints on the various coupling constants

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