The h Bar - 2024-03-18

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naturallyInconsistent

2:34 AM

@SirCrackpot I'm much more with Luca on this. We should call things like the charge and fine structure constant the constants that are measured by low-energy, on-mass-shell, photons i.e. $q\to0$ limit of photons. This way, they are constants, and much less confusing. Also, if you scrutinise the various connecting bits, you should get the conclusion that things like $e$ and $c$ and so on are all irrelevant, and it is $\alpha(q)$ that is possibly changing i.e. only the coupling constant is moving.

This is particularly clear if you rewrite the standard equations so as to extract all the stupid constants out, and that the only way that the quantum electrodynamics version of Maxwell's equations and the minimal coupling term of Dirac equation interacts to show the charge behaviour, is actually solely in the energy content of E & B fields (one coupling constant there) and in the equivalent of the thing that leads to Abraham-Lorentz force law.

Everything else is just that we are using stupid units (ok, the pioneers were amazing for being able to see this much from so little that they knew). Natural units for the win, please.

@Slereah This is utterly saddening.

@lucabtz That's definitely mostly because it is the only approximately tolerable thing that maths of undergrad allows profs to cover. And it is wonderfully successful at being a decent first approximation too.

1 hour later…

3:49 AM

@naturallyInconsistent as long as one establishes a convention for that, it's fine

The important thing is not to get tricked by names

naturallyInconsistent

@SirCrackpot but it is simply the case that a tremendous proportion of students are confused by this. The naming is just horrible.

4:09 AM

Not that I care, I'm quite the evil guy

(I agree, horrible naming but not so much to be a danger imho :P)

Meanwhile, Huang QFT is confirming what I said yesterday. Now I need to compare it to Wilson Kogut 1974

naturallyInconsistent

@SirCrackpot which particular thing? Yall said a lot!

@naturallyInconsistent Basically that after one step of the Wilsonian RG the rescaled cutoff $\Lambda/b$ is the running scale $\mu$ which appears in Callan-Symanzik equation

So the bare couplings $g_0(\Lambda)$ are mapped into the running renormalized couplings $g_R(\mu)$

Not that I'm convinced though. There also exists a RG equation for the running of bare couplings

4:45 AM

@SirCrackpot archive.org/details/space-and-time-in-the-microworld/page/94/… chapter VI has an explanation

1 hour later…

5:49 AM

Just to be sure. Type $IIB$ string theory has $(1,1)$ worldsheet SUSY but $(2,0)$ spacetime SUSY, right? (Wiki says it has $(2,0)$ SUSY (vague? WS or spacetime?) but Polchinski discuss Type II theory in a chapter that promises to discuss $(1,1)$ SUSY and also talk of $(0,2)$ SUSY in the chapter on heterotic string theory)

@DIRAC1930 an explanation of what?

6:17 AM

Sorry I meant for the thing I was worried about

naturallyInconsistent

6:49 AM

@SirCrackpot way beyond myow pay grade.

7:07 AM

@naturallyInconsistent considering that I'm not paid, also beyond minyan

naturallyInconsistent

lol

7:48 AM

@naturallyInconsistent yeah I agree, also it got right quite some results

1 hour later…

8:56 AM

> "Let me begin," I replied, "with the subject of Natural Philosophy, which is Epicurus’ particular boast. Here, in the first place, he is entirely second-hand. His doctrines are those of Democritus, with a very few modifications. And as for the latter, where he attempts to improve upon his original, in my opinion he only succeeds in making things worse."

lol

> Democritus believes in certain things which he terms ‘atoms,’ that is, bodies so solid as to be indivisible, moving about in a vacuum of infinite extent, which has neither top, bottom nor middle, neither center nor circumference.

The OG space

> He believes that these same indivisible solid bodies are borne by their own weight perpendicularly downward, which he holds is the natural motion of all bodies; but thereupon this clever fellow, being met with the difficulty that if they all traveled downwards in a straight line, and, as I said, perpendicularly, no one atom would ever be able to overtake any other atom, accordingly introduced an idea of his own invention: he said that the atom makes a very tiny swerve—the smallest divergence possible; and thus produces entanglements and combinations and cohesion of atoms with atoms, which…

"A tiny swerve"

He does not give much to work with as far as mechanics goes

> this riotous hurly-burly of atoms could not possibly result in the ordered beauty of the world we know

9:24 AM

@JohnRennie would you like to explain it now?

@MRVOREX Hi :-)

Although I only showed the force acting at one point it acts all round the circumference of the drop. So assuming the drop is radially symmetric the force is not moving the drop. It either expands or contracts the (circular) contact patch with the substrate.

i.e. it pulls or pushes on the edge of the drop all round its circumference.

That means you are taking the liquid particles as your object right?

The object on which you are balancing the forces

Not really. As an analogy, imagine you have a spherical balloon and you pull outwards round the circumference of the balloon. The result would be to deform the balloon into an oblate spheroid.

The deformation is determined by the elastic stresses in the balloon skin and the pressure of the air inside the balloon. Yes?

Yupp

In this case you have a drop of some liquid being pulled outwards (or pushed inwards) at the edge due to the interaction of the liquid with the substrate.

And the surface of the drop has elastic energy due to the surface tension at the air water interface.

So you're treating the drop as a deformable object and calculating how it deforms.

But it's simpler than a balloon because surface tension is constant. Unlike the rubber balloon it doesn't depend on how much the surface has been stretched.

9:31 AM

I do get this...and when I was first presented with the young's equation my first idea was virtual work as it results in a way simpler analogy

OK ...

I am having trouble understanding this part- " if we show the forces then why should we show the force due to solid vapor interface

To me it seems wrong

Surface tension can be interpreted both as a force and as a surface energy. They are exactly the same thing.

As the forces of interaction due to solid vapor interface will act along the solid surafce

That is, the air-liquid interface has some energy per unit area, and to create new surface you have to do work and that means applying a force.

The force associated with the surface energy is just the surface tension.

9:37 AM

So ...the force is just a representation of the energy and we have to show the force due to the solid vapor interface as it is a part of the energy

Have you read Wang ?

If we expand the drop outwards it replaces some of the air-solid interface by liquid-solid interface, and it expands the air-liquid interface.

So to get the total energy change we subtract the lost air-solid energy, then add the increased liquid-solid energy and the increased air-liquid energy.

@JohnRennie yes that is exactly what is done in virtual work method

And then the change has to be equal to the work done by the force acting at the edge of the drop as the drop expands outwards.

But ..I am just curious...have you ever read wang?

@MRVOREX Yes, so the point is we don't need to care exactly how the force acts.

We know there must be a force since the energy changes as the drop expands, and we just set it to zero to find the equilibrium state.

It is effectively the same as using virtual work.

9:40 AM

@JohnRennie makes sense

@MRVOREX No I have not read Wang.

@MRVOREX So when you and Gaurang Agrawal ask me about details of how exactly the force acts I can only say I don't know and I don't care. The details don't matter.

@JohnRennie I didn't have this doubt until I read it.. it has presented a erroneous proof to young's equation and it has mentioned the reasons to why the proof is wrong

If you could just go through it..then I think it would be better

Do you have a link to the book?

But then again when wang itself presented its own proof to young's equation it used a force diagram without explaining where the force acts so it becomes confusing

@JohnRennie let me see

openmaktaba.com/read-online

You can read it from here..or just download it

Just post a link to the book on Amazon, or the publisher's site or some similar site. I'm not going to register on some dodgy download site.

9:50 AM

amazon.in/Competitive-Physics-Mechanics-General-Aspects-ebook/…

There you go

What page?

486

in the end i didnt find a standard way to convert bessel I functions to the J function in mathematica and had to do this `ruleBesselIToJ =
BesselI[\[Alpha]_, x_] :> I^(-\[Alpha]) BesselJ[\[Alpha], I x];`

@MRVOREX I've read through it and I can see his point, but I am not very impressed. He dismisses the idea of a force, but as I explained we can view the force as just a convenient way to describe the changes in surface energy.

10:23 AM

Hmmm

I guess too much pondering is just a waste of time after all

@MRVOREX I think it depends on what you are trying to achieve. Most of us don't care that much about the details of the derivation so a plausibility argument is just fine.

I was under the impression that I was missing something in my understanding of surface tension

So I stopped going through the subject and looked for some answers here and there

No, you're fine :-)

I hope so

Btw..how old are you?..and from where?

10:38 AM

I am a retired physicist, so I'm ancient :-)

For real?

physics.stackexchange.com/users/1325/john-rennie?tab=profile

I am incredibly younger than you then

About 50 years younger, which isn't really incredible :-)

Maybe it does look incredible from where you're standing :-)

That's amazing actually..a PhD scholar just helped me out

10:41 AM

This chat room is stuffed with PhD scholars. We even have a string theorist!

Seems like I am the most stupid guy here😂

"inexperienced" yes, "stupid" no!

The question you asked about the Young's equation derivation is a perfectly good one.

Thanks...I will be back with more questions some other day

1 hour later…

12:04 PM

hi

im having a hard time really understanding why tangent vector components transform according to the jacobian of $x^{\mu'}(x^{\mu})$. it can be easily proved using the chain rule, but still..

how to visualise this

12:25 PM

Draw the two coordinate systems on the manifold and draw their tangents at a point

2 hours later…

2:07 PM

thanks i got it

Could someone help me understand the solution discussed in ashcroft solid state? Chapter 8 problems, problem 1. I don't understand this solution . Why are they taking scattering states as the solution in the region -a/2 <x <a/2 ? Could someone help me understand the idea?

1 hour later…

3:16 PM

Let's say we have a light detector that can detect light falling on it from any direction, and whenever it senses one or more light pulses at a time it will increment some internal counter by 1. This detector is placed at the center of a trolley, with two lasers at the front and back of the trolley facing the detector at equal distances to it.

Now the trolley is moving relative to the ground, and both the lasers are triggered simultaneously (relative to the trolley frame), sending a light pulse towards the detector. The detector will sense both the light pulses simultaneously and increments the counter once. Now from the ground frame, it will look like the pulse from the laser at the front of the trolley reached the detector first followed by the other pulse.

Yet the counter will say 1, making it look like the sensors did not detect the second pulse from the ground frame. How can this be? What am I doing wrong here?

naturallyInconsistent

@shahrOZe That is not what Ashcroft and Mermin are saying. The shape of the solution in -a/2 < x < a/2 is not being constrained to be anything.

@naturallyInconsistent So the solution in -a/2 <x< a/2 is just a general solution and may not contain sines and cosines? But how have the solution parts of psi_L and psi_R? They found solutions outside this region. Since the potential is greater than the energy in the region -a/2 <x< a/2, shouldn't the solution consist of exponential parts? I am confused how they are trying to solve it. If you could help me out from the basics that would be really helpful. Thanks

naturallyInconsistent

@shahrOZe psi_L and psi_R are general functions in -a/2 < x < a/2 but they are specified in x <= -a/2 and x >= a/2; that they are specified at the points -a/2 and +a/2 is crucial to the solution, and the derivative is also specified there. That is enough to follow the problem.

@naturallyInconsistent so at the end we are just imposing conditions of continuity at +- a/2?

@naturallyInconsistent also is it somewhat like the Kronig-Penney but the coefficients that is the r,t and i are undetermined here?

naturallyInconsistent

3:38 PM

@shahrOZe yes. do it

@shahrOZe it will end up becoming determined by Bloch's theorem

3:52 PM

@0xVikas hm I think this is wrong, It should look like the laser at the back was triggered first to the observer on the ground, followed by the laser at the front, so that the observer on the ground will also see both the pulses reach the detector at the same time. Is this correct?

4:05 PM

@naturallyInconsistent thanks

4:52 PM

can i say that a diffeomorphism maps the tangent vectors at p to tangent vectors at the point p' that p gets mapped to

suppose i hav a curve passing thru p. then diffeomorphism maps this to a curve passing thru p'

therefore, it maps a tangent vector at p to a tangent vector at p'

Yes

If you're doing it "actively" your curve simply gets transported to another point

and therefore so does its jet to another jet

really cool

the active view is good sometimes as opposed to co ordinate changes

suppose i hav a function f(p) on a manifold, and we do a diffeomorphism, does the value of the function also get transported to the new point?

Yes

If your function $f(p)$ has some value and you have some diffeomorphism $\phi(p) = q$, then you have a few function $f' = f \circ \phi$ which is the image of that function under the diffeomorphism

That's why you have the formula that for $f'(q)$, you have the equality $f'(q) = f(\phi^{-1}(q))$, as you will see in books

5:10 PM

kind of trippy

right. we just compose

Or should it be $f'(p) = f(\phi^{-1}(p))$

Something like that

Same deal as long as it's a diffeomorphism I guess

Basically you should just have the inverse annihilate the original diffeomorphism

It's almost equinox time again.

user image

ssd.jpl.nasa.gov/api/…

That graph should look straight, but we get artifacts with such a tiny time step.

5:33 PM

Hey all!

3 hours later…

8:22 PM

@JohnRennie It must have been quite an experience to have done your PhD in Cambridge. Seems like such a magical place

8:41 PM

@ACuriousMind I am really confused about what is meant by charge and how this fits into charge quantization. It seems like all that is happening is that Coloumbs law gets modified and reaches the asymptotic form at large distances which we define to be the fundamental constant $e$. However is this really $e(r)$ changing (here $r$ is the radius from the electron to observation point) or just Coloumbs law changing with $e$ always being constant?

If $e$ really was changing with 'energy scale' which is a confusing sentence to begin with, wouldn't this mean that the quantization of charges by $e,2e,3e,\dots$ becomes inconsistent

I don't know what you mean by $e(r)$ or Coulomb's law "getting modified" or what either of these has to do with "charge quantization"

the running coupling is running with the energy scale, not with the distance between the electron and some "observation point"

Chapter IV here archive.org/details/space-and-time-in-the-microworld/page/94/…

I'm not reading your weird book

Higher energy just means getting closer to the electron i.e. energy scale going up

no, that's not really what the running coupling is about in the mainstream interpretation

it's just the value of $e$ you'd have to choose so that at that particular energy level (which in itself is a notion dependent on the renormalization scheme) the fully summed (or summed to whatever order in perturbation theory we're working at) interaction vertex is equal to what you'd compute with that $e(\Lambda)$ at tree-level

heuristically we say that the energy level is related to the "spatial resolution" of the process, but it's not about a distance between an electron and some observer, it's about what size of structures can play a meaningful role in the interaction

8:50 PM

Okay so say if I integrate out higher modes and I'm left with an effective action, will all interaction processes in this range have the same value of $e$?

I don't know what that question means

the running coupling $e(\Lambda)$ is a function of a particular renormalization scheme, not a physical property of a process

it's well-known that beyond tree-order (and at the very least at third order in perturbation theory), the running depends on the scheme

How do i understand this using normal renormalization and not RG stuff

I didn't use any RG notions here

Okay, when we renormalize we set things based on physical conditions

yes, but those conditions are already not unique and part of what defines the scheme

8:54 PM

So the condition depending on $e$ seems fo be to set it such that $D\rightarrow 4 \pi/k^2$ when $k^2\rightarrow 0$

e.g. you might define the scale $\Lambda$ of a diagram by $s^2 = \Lambda, t^2 = u^2 = 0$ or $s^2 = t^2 = u^2 = \Lambda$ (I don't recall all the possible conditions exactly off the top of my head), and then setting $e(\Lambda = 0) = e_0$ (the usual elementary charge) is your "physical condition", but these two notions of $\Lambda$ might define different running couplings

there's a reason why when you look up e.g. running couplings in the PDG data, they're usually careful to specify things like "in the MS (=minimal subtraction) scheme"

@DIRAC1930 I really don't understand why you are so stubborn in avoiding all standard texts on the subject

Oh so there are two things here I am confusing. One, the fundamental charge, and the other is an effective coupling

the "fundamental charge" is just a coupling

without renormalizations, you'd like it to be the "bare" coupling you write into the Lagrangian

alas, renormalization messes everything up; that's how it goes

But then why is this said to be a constant en.wikipedia.org/wiki/Elementary_charge

8:59 PM

it's the value $e(\Lambda = 0)$

@DIRAC1930 again that is defined at a specific scale by a specific measurement process

So it's the asyptotic form of Coloumbs law

I don't know what you mean by "asymptotic"

@DIRAC1930 what does this mean

there's no asymptote here, you can just put $\Lambda = 0$

9:01 PM

@lucabtz archive.org/details/space-and-time-in-the-microworld/page/94/…

A proper book

@ACuriousMind technically he said asyptotic

@DIRAC1930 if you want to talk to other physicists about physics, you have to learn the standard ways in which we talk about things

by insisting on idiosyncratic approaches and refusing to explain them from the mainstream viewpoint, you're just putting yourself outside the mainstream

you're not gaining anything else

@DIRAC1930 I never saw that before

@ACuriousMind I fully understand what you are saying but I am genuinely more interested in how Coulombs law gets modified at short distances than how it is generally done. I am just following my interests

Which I can only find from books like these

I'm the first one to admit that there's a lot left to be desired with the standard approaches to QFT, but you need to understand the standard approach in order to communicate effectively about its shortcomings

9:04 PM

If you want a book that's very physical you should read Zee

@DIRAC1930 the uehling potential is pretty standard

For sure there is an explanation in Schwartz which is a standard book

@DIRAC1930 The modification of Coulomb's law by QFT effects is relatively straightforward, if you understand the derivation of the normal Coulomb potential from the tree-order diagram you can easily extend it to higher orders (thus including running couplings, in a sense).

I outline the idea here, and this should be - maybe in different words - part of most standard treatments of QFT

the potential is the Fourier transform of the propagator, if you compute higher-order corrections to the propagator, you compute higher-order corrections to the potential

Yes but what is changing, $e$ is constant in all these formula en.wikipedia.org/wiki/….. My question is what quantitiy does the electron actually carry.

I don't think that's a well-defined question

the value $e(\Lambda = 0)$ is purely mathematically a constant, a renormalization parameter to be determined by experiment

So here en.wikipedia.org/wiki/Elementary_charge it states the electron carries 1e which is defined to be $e$ = 1.602176634×10−19 coulombs,

in the low-energy limit that yields most of classical physics and non-rel quantum mechanics, this constant is the only value you need, since higher-order corrections are intrinsically suppressed by orders of $\hbar$

and actually, if we're being really careful, $e$ is meaningless anyway since it's a dimensionful constant

the thing that runs is the fine-structure constant $\alpha$, and it's meaningless to attribute its change to a change in "the elementary charge" and not one of the other constants

9:11 PM

By energy scale, do you mean momenta between $0$-$p1$ and $p_1$-$p_2$ and not momenta between $0$-$p_1$ and $0$-$p_2$

Jan 27 at 14:19, by ACuriousMind

see this Q&A by Emilio Pisanty on how (not) to talk about changes in dimensionful constants

@DIRAC1930 I mean whatever your renormalization scheme defined as "the energy scale" - again, this is a scheme-dependent notion

usually you want it to correspond to some intuitive idea about the "total energy" of the inputs of the scattering process

But then I will have a scattering experiment at low energy with one effective charge and another at a higher energy with a different effective charge and then the charges $e_1$ and $e_2$ won't add up to $2 e_1$

so what?

Then there isn't any charge quantization

no, it just means you haven't understood what the "charge" is that is quantized :P

$e$ is just a number

the discretization of charge happens on a level of representations, where representations of the electromagnetic U(1) are labeled by integers $n$, and in the low-energy limit you see particles from those with charge $e_0 n$

no one claims that the "running charge" is conserved; it's entirely unclear what that would even mean

9:17 PM

So this $U(1)$ is $e^{\imath e \theta(x)}$

you just need to stop imbuing functional expressions that depend on the renormalization scheme with this kind of immediate physical importance; precisely because they depend on the scheme they cannot have such importance

I cant rememebr if there is an e up there

$e(\Lambda)$ isn't a "charge", it's a running coupling, a technical thing that appears in the perturbative renormalization treatment of QFT scattering

Okay so what is the correct definition of charge

This $U(1)$ stuff

that's at least the thing that's independent of any scattering theory or renormalization

we can say that the electron is charged three times as much as a quark with charge 1/3 entirely without having to specify some unit $e$ for this charge

9:19 PM

Okay so wont this $U(1)$ transformation be different depending on your effective theory i.e. $e^{\imath e(\Lambda) \theta(x)}$

that's it: The $e$ is just a unit

@DIRAC1930 I don't know what you mean

all these things are not so straightforwardly related to each other

How fermions transform under a gauge transformation. If we have an effective action, we set $e$ by the coupling constant then everything is fine

you can't just take $e(\Lambda)$ from some perturbative renormalization scheme and plug it into parts of the theory that have nothing to do with renormalization and expect everything to work out

all these things are just much more complicated than you're willing to give them credit for; there is no short few-sentence explanation of how exactly renormalization works - I'm not even sure there's an explanation that fits into a single book

(also the "how does a fermion transform under this" shows you still have not internalized that it's really forbidden to talk about definite particle states like that anywhere but in the asymptotically free parts of the theory)

What is the normal renormalzation scheme done historically

So this MS stuff was found in 1973

@DIRAC1930 physical quantities are scheme independent

It doesn't matter which scheme you pick, the physical intuition is scheme independent

I think you should really just pick up a standard book and read it, stuff will be way easier

9:30 PM

I dunno, I might just give up with this all

You're just too stubborn in avoiding the standard treatment. You will find it's way easier if you start with a book and continue consistently with it

2 hours later…

11:24 PM

I think I'll just do a full QED calculation

Here is my confusion. Using the renormalization conditions that the pole of the propagator is the mass, we just have to define $m_B^2$ and $\delta m^2$ such that $m^2$ is the pole position

through $m^2=m_B^2+\delta m ^2$

Lets just say $m_B^2$ is just fixed since this is what is in the Lagrangian, so we only have the freedom to set $\delta m^2$ such that $m$ is at the location of the pole

I don't see what freedom we have

Plus why doesn't Weinberg mention anything else

@ACuriousMind Oh so it seems that I've only heard about the On shell renormalization scheme en.wikipedia.org/wiki/On_shell_renormalization_scheme

I wont worry about other renormalization schemes unless I learn QCD which I doubt at this point

So probably it's best for me to understand the physical consequences of vacuum polarization

So this on shell (physical) renormalization scheme is completely different and you can interpret intermediate quantities indico.ibs.re.kr/event/153/contributions/250/attachments/200/… page 6

This modification of Coloumbs law makes complete sense and even thinking about other renormalization schemes completely overcomplicates this

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