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Leaky Nun

03:20

General Relativity: Top 05 Mishaps [inc INTERSTELLAR]

2

I have nothing against this channel, I'm just using it as an example of a more widespread phenomenon

here they're explaining special relativity by saying that even if I'm travelling at 0.5c and I see the photons travelling at c, you (stationary observer) don't see the photons travelling at 1.5c, etc.

The phenomenon here is that usually popular explanations (i.e. scipop) of special relativity often interpret photons as tiny particles that themselves travel at the "speed of light"

cf. the conversation chat.stackexchange.com/rooms/71/conversation/photon-confusion

Loong

07:37

@LeakyNun Well, if you do the usual coincidence experiment with gamma radiation, it exactly looks like that.

PM 2Ring

07:48

@LeakyNun That's a bit sloppy. They could avoid the problem by talking about light rays, rather than photons. But that kind of sloppy language is fairly common, eg when talking about light clocks, and I exoect people will continue to use it. ;)

There's an old slogan: "light travels like a wave and interacts like a particle". But that can be a bit misleading too. See physics.stackexchange.com/q/473566/123208 & the links ACM posted in the question comments.

And of course you can't actually see light travelling. You only see it when it hits your retina. When we say stuff like "Observer A in frame F sees light travel from point P to point Q over time interval I", we really mean that they have made measurements that are consistent with a light-like path between those events.

Slereah

also light travelling isn't best modelled as photon states

Since photon states aren't localized

user223626865

That's not popsci, it's more like popcornsci :P

2

PM 2Ring

@Slereah Right. Which brings us back to the start of the conversation that LeakyNun linked.

user223626865

08:03

Apr 18 at 17:09, by ACuriousMind

@LeakyNun a highly uncomfortable question, since photons don't actually have "position"

Slereah

If you want to model the travelling of light it is better to consider like coherent states

PM 2Ring

When we try to make an intuitive internal picture of quantum stuff we can't avoid imposing our classical notions onto it, to some extent. The "shut up & calculate" school says, "you're always going to screw up your intuitive internal model of a quantum thing, so just don't do it".

Personally, I think it's fine to have such internal pictures, as long as you never forget that they're just cartoons.

user223626865

Chasing a beam of light: Einstein's most famous thought experiment @LeakyNun

2

imbAF

09:01

Can anyone explain to me how the zeeman effect plays a role in Magneto optic trap?

I am reading about it, but I don't see how it brings confinement of the cooled atoms

PM 2Ring

09:45

@imbAF Does this help? en.wikipedia.org/wiki/Zeeman_slower

> The spatially varying magnetic field is designed to Zeeman shift the resonant frequency to match the decreasing Doppler shift as the atoms are slowed to lower velocities while they propagate through the Zeeman slower allowing [the] pump laser to be continuously resonant and provide a slowing force.

imbAF

This gets more complicated lol

PM 2Ring

And the start of this article explains the general principle of laser cooling: en.wikipedia.org/wiki/Laser_cooling

imbAF

doppler shift?

If we assume that through lazer cooling we slow down the atoms

now we need to entrap them somehow in some region

and this occurs via the Zeeman effect

But the zeeman effect, causes only position dependent splitting of the energy levels

Until this very moment all is good

Magneto-optical trap

A magneto-optical trap (MOT) is an apparatus which uses laser cooling and a spatially-varying magnetic field to create a trap which can produce samples of cold, trapped, neutral atoms. Temperatures achieved in a MOT can be as low as several microkelvin, depending on the atomic species, which is two or three times below the photon recoil limit. However, for atoms with an unresolved hyperfine structure, such as 7 L i {\displaystyle ^{7}\mathrm {Li} } , the temperature...

PM 2Ring

@imbAF By changing the strength of the magnetic field we change the distance between the split lines. That means we're shifting the frequencies that the atoms "want" to absorb.

imbAF

Does it mean

that because of the shifting, while the transition frequency decreases (transition frequency is the frequency difference between two states that differ by +/- 1 in their orbital angular number)

and the laser frequency does not

I would continue by saying

that the momentum kick would be greater

but I have no clue as to how this happens

The atom can only absorb one frequency

the one provided by the laser

the shifting only makes it possible, as how i understand it, for the atom to absorb this at a higher probability

PM 2Ring

09:55

The laser emission frequency is fixed. But the moving atoms don't "see" that frequency, they see a Doppler shifted frequency because they're moving. So as the atoms slow down, we need some way to match the laser to the slower atoms. And so we adjust the atoms' absorption frequency via the Zeeman effect.

imbAF

"we need some way to match the laser to the slower atoms."

what do you mean by match?

aaa, because the atoms slow down, the doppler effect is less and less effective if one can say so, so we make up to it, by doing a spliting of the energy levels

so that we increase the chances for absorbtion

of the photon

which will result

in a stronger kick to the atom. Which sends it toward the center

PM 2Ring

@imbAF The (Doppler shifted) frequency that the atom wants to absorb has to be close to the frequency of the photons.

imbAF

is this somehow correct understanding ?

yes

PM 2Ring

@imbAF I think so. ;) As the atoms slow down towards being at rest in the lab frame, the Doppler effect is reduced. So we use the Zeeman effect to compensate for that.

imbAF

nice

Now I get it

PM 2Ring

10:01

Excellent.

imbAF

One can say that through the zeeman effect we increase the "kick" in the atoms

@PM2Ring one final question

PM 2Ring

ok

imbAF

Let's say that the Magnetic field is prop. to the z axis. So it's positive for z positive and negative for z negative. This is our set up

Now, the circular polarized light, forces certain transitions. Namely, transitions between states, whose orbital quantum number $l$ changes by +/-1 value.

Small in between question: The reason why we need a +/-1 change in the orbital quantum number is so that the dipole matrix element is non-zero, right?

PM 2Ring

I'm not quite sure what you're asking. But the +/-1 change is simply because the spin of a photon is 1, so a single photon can only change the angular momentum of the atom by +/-1.

imbAF

Ok

Let's go to my main questin

question*

PM 2Ring

10:12

Ok. However, I may not have enough QED expertise to answer it. ;)

imbAF

Here it says that if we consider atoms propagating in the positive z direction, and we consider the transition between states with J=0 and J=1 (mj=-1,0,1) than the transition that takes place is the one between (J=0,mj=0) and (J=1,mj=-1). And for atoms propagating in the negative z direction the transition that occurs is between (J=0,mj=0) and (J=1,mj=+1). How is this decided ?

Magneto-optical trap

A magneto-optical trap (MOT) is an apparatus which uses laser cooling and a spatially-varying magnetic field to create a trap which can produce samples of cold, trapped, neutral atoms. Temperatures achieved in a MOT can be as low as several microkelvin, depending on the atomic species, which is two or three times below the photon recoil limit. However, for atoms with an unresolved hyperfine structure, such as 7 L i {\displaystyle ^{7}\mathrm {Li} } , the temperature...

And the reason given there is because

the magn field is negative for z<0

Now, if we consider this splitting under the zeeman effect

This is for the 2nd energy level n=2

for z>0 the magnetic field is positive, and what happens is that the lines split further appart as z increases. We said that the magnetic field is linear proportional to the z axis

And because we said that we need a +/-1 difference in angular momentum quantum nr. In our case we would consider the transition (n=1,J=1/2,mj=0) and (n=2,J=3/2,mj=3/2,1/2,-1/2,-3/2)

We can see that the sub level that corresponds to mj=-3/2 goes closer to the ground level

Now, if we have an atom that propagates in the negative z direction, because the magn. field is negative,

should the lines flip

meaning

the +3/2 sub level changes position with the -3/2 ?

thus allowing the transition (J=1/2,mj=0) and (J=3/2,mj=+3/2)

Which is in alignament to what it's being said in the link about the magneto optic trap

(but there J=0 and J=1)

It's a long description of my question

PM 2Ring

Sorry, I'll have to bow out, and let someone else answer. I don't want to accidentally say something wrong or confusing.

imbAF

but as you can see there is no way around it

Ok

but thanks for your help

PM 2Ring

I think this stuff just comes down to the interplay between charge and parity. When you reverse the field direction, you get the mirror image, and so clockwise spin gets mirrored to anticlockwise spin.

imbAF

spin of which?

cuz we are having at least 3 particles involved here

the nucleus, the electron and the photon

Unfortunately I cannot link a highlighted part

PM 2Ring

10:25

If you have a scenario with a bunch of things that have angular momentum, if you look at that scenario in a mirror, all the spins get reversed.

imbAF

Ok

PM 2Ring

EM preserves CP symmetry, but the weak interaction doesn't. See en.wikipedia.org/wiki/CPT_symmetry & en.wikipedia.org/wiki/CP_violation

imbAF

but in the description in Wikipedia it says that if an atom propagates in the positive Z direction (and in our setup we have a magnetic field in the z direction and linearly proportional to the z and we also look at transition among J=0 and J=1) the transition that takes place is (J=0,mj=0)->(J=1,mj=-1) and if it propagates in the neg z direction the transition that takes place is (J=0,mj=0)->(J=1,mj=+1). And the reason why this transition takes place is because the magn.

field is negative for z<0. My question is simple. How does whether the magn. field is positive or negative effects whether the transition is between (J=0,mj=0)->(J=1,mj=-1) or (J=1,mj=+1)

PM 2Ring

So if you have some valid EM setup, then you create a mirror image duplicate version and invert all the charges, then the duplicate is also valid.

imbAF

EM ?

PM 2Ring

10:46

@imbAF Electromagnetism.

imbAF

Ok

Relativisticcucumber

12:25

hi - i have a question about plateau's problem. in the images below, for the second one, i can see why if we have these two boundaries this cylinder might be the minimal surface to connect the boundaries, however, i do not see why for the first image the paraboloid shape is the minimal surface? why can't it just be the boundary (the circle filled in)

ACuriousMind

13:21

@Relativisticcucumber The circle - or a plane surface in general - is a minimal surface

Relativisticcucumber

so what is this paraboloid business about

ACuriousMind

I think the problem here is the definition of the minimal surface: The claim that the shape in your first picture is a minimal surface is not the claim that it has the least area among all shapes with the given boundary, it's that it's locally a shape of least area, i.e. all infinitesimal variations of it have greater area

It's the usual thing about local extrema: A local minimum or maximum of a functional is not necessarily a global minimum or maximum

Relativisticcucumber

hm i guess i just dont see what that surface is of any particular interest

ACuriousMind

I mean why you care about that surface depends on why you started looking at the problem of minimal surfaces to begin with

usually the physical meaning here is that there is some physical process that means e.g. a soap film should be a minimal surface

Relativisticcucumber

in everything i see about ads cft this is the type of minimal surface shown and i cant figure out why this is the important minimal surface

ACuriousMind

13:30

I don't think it is "the" important minimal surface

it's just that AdS/CFT relates (iirc, I'm not really up to date on AdS/CFT) minimal surfaces to stuff like Wilson loops

so any minimal surface is interesting

Relativisticcucumber

okay okay i see

PM 2Ring

14:18

A new member on Space.SE noticed this interesting coincidence: the cumulative mass of Mercury, Venus, Earth's moon, and Mars total 99% of Earth's mass. That is, the mass of the significant inner Solar System bodies is almost exactly equally partitioned between the Earth and the other 4 bodies. space.stackexchange.com/q/63426/38535

Ryder Rude

15:07

Omg did anyone see Evil dead Rise?

It is extremely good horror. I will recommend it 10/10

It does not hav many jump scares and the suspense is 10/10

Amit

18:40

@PM2Ring Too bad he chose to ask the question only about the "dry" facts, that can be easily verified by anyone... I would like to know if someone can tell whether it's just a fluke or has any deeper significance. OTOH, we all know everyone are constantly trying to avoid "opinion based" questions, maybe that's why he wrote it this way...

Slereah

20:10

I think I need a little GR break

I'm gonna do a dumb article

What if you use Hilbert's axioms to rewrite ancient astronomy

Obviously Euclid's axioms are all terrible

Amit

@Slereah GR break? Where can we find your work though?? :)

Slereah

A good question

some may argue here : samuel-lereah.com

Although literally none of the articles are finished

Amit

Ahhh, so that's you! I visited this before

Slereah

just various sketches of articles

they are slowly filling up

Amit

Cool, thanks

Slereah

20:15

Once in a while I do a dumb one to have a break

samuel-lereah.com/articles/Physics/flat-vs-round-earth

Amit

Is Cohomology important to understand for GR research?

I think I got to the point where I understand most of the math background to GR at least to a certain degree, but when I read something about Cohomology I'm almost completely lost

imbAF

20:33

For incident particles in a potential barrier, the corresponding wave function is in general written in the following form: $\psi(x)=A_1e^{ik_1x}+B_1e^{-ik_1x}$. Why do we leave the time component out?

DIRAC1930

1

A: Why would the "in/out'' states asymptotically approach the free Hamiltonian eigenstates?

In fact, at least for QED, it is not true because there is still interaction with "vacuum" or said differently, the "in" and "out" particles are "dressed". It is easily seen that we cannot scatter from an electron elastically because it emits photons with the probability 1, i.e. the scattering is...

22 hours ago, by DIRAC1930

On second thoughts, perhaps a heuristic argument is that $\hat{S}$ is an operator that is defined on the free Hilbert space. Therefore particles we observe in the theory should be able to be built up from superpositions of free eigenstates Consequently if we have a particle that we have measured with definite momentum, we must have a free plane wave

Has anyone thought about my argument above

ACuriousMind

again, this is a solved problem, Haag-Ruelle scattering theory constructs the asymptotic states in a rigorous manner (but, as usual on rigorous QFT, it is unclear whether this argument holds for realistic theories like the standard model)

If you plug Haag-Ruelle theory into Epstein-Glaser renormalization, perturbative QFT has a completely rigorous foundation. The reason this is not discussed in most standard texts is that it needs a lot of mathematical machinery to arrive at the same practical results (in the sense of what you have to compute to get the actual amplitudes) as the standard non-rigorous QFT arguments, and that we can't actually prove the Wightman axioms in practice for arbitrary field theories

DIRAC1930

Are there any physically motivated reasons?

ACuriousMind

physically motivated reasons for what?

DIRAC1930

20:48

For asymptotic states being free

I'm guessing it's just a lie that works

ACuriousMind

well the physical reason is just that obviously when stuff is far apart it shouldn't matter whether it potentially can interact with other stuff or not

if you're asking whether there's a reason all asymptotic states should be free, then no, obviously not - a stable bound state doesn't magically become free when you let it evolve for an infinite time

DIRAC1930

If I view the S matrix as an operator that acts on the in states, near the pole I will recover the propagator $Z/E-E' + \imath \epsilon$ or something.

Does this have any use?

ACuriousMind

the stipulation that the asymptotic states should be "all states" is known as asymptotic completeness and a rather contentious issue in general

DIRAC1930

Also, if I have a single excitation above the vacuum, I cannot tell if it is a free theory or a dressed theory can I?

ACuriousMind

@DIRAC1930 what does "free" or "dressed" even mean there?

DIRAC1930

20:53

Free would be the free Lagrangian, dressed would be the free lagrangian plus interaction terms

ACuriousMind

the whole point of the asymptotic states is that they behave like free states

so no, of course you can't tell whether it's a "free" or "non-free" state when you look at an asymptotic state because by construction it looks like a free state

DIRAC1930

I am talking about a general state not just asymptotic

general one particle state anyway

ACuriousMind

what does "interacting" even mean when you only have a single particle state?

the interaction part comes in when you compute the n-point function

DIRAC1930

It seems like the only difference between a free theory and an interacting theory is that I have a finite part for the GF along with the singular part

ACuriousMind

the whole point of Haag's theorem is that the field representation fixes the first four Wightman functions, so that you cannot have a unitary equivalence between a free and an interacting field (because unitary equivalences leave the n-point functions invariant)

DIRAC1930

20:56

Sorry I am asking more about the interpretation of the interacting part in general QFT

ACuriousMind

you can generally tell an interacting theory by looking at the mass spectrum

a free theory has some lowest-lying mass $m$, and then the next $p^2$ eigenstate comes at $\geq (2m)^2$

DIRAC1930

I need to look more into the mass spectrum

ACuriousMind

an interacting theory will usually have bound states below $(2m)^2$

DIRAC1930

won't I get multiparticle states for a free theory though as well?

Is it just bound states that are different?

ACuriousMind

the mass gap is crucial: Many of the "simple" rigorous approaches fail completely in the presence of massless particles

because, heuristically, you cannot distinguish a "free" particle from another particle produced with a cloud of low-energy massless particles- this is the "infrared problem"

DIRAC1930

21:00

Also, a field is integrated over all space $\phi^\dagger = \int \mathrm{d}X A^\dagger(X) \psi^*(X)$. Therefore in the interaction terms, even farly seperated particles interact

ACuriousMind

@DIRAC1930 again, this is a solved problem: Haag-Ruelle theory constructs asymptotic states as limits of fully interacting states

DIRAC1930

Okay I will have to look into that

ACuriousMind

there is a lot of literature about the correct formulation of scattering theory, a good starting point is Reed and Simon, vol. 3

DIRAC1930

Also, when people say that farly seperated particles are non-interacting, are they talking about the Yukawa potential dying for large $r$

ACuriousMind

there's essentially a parallel literature where people like Haag, Reed and Simon, Hepp etc. ran along what the QFT pragmatists were doing and tried to make rigorous sense of it

@DIRAC1930 as I just said when I mentioned the infrared problem, the presence of massless particles like in EM makes the construction of the "free" asymptotic states much more difficult

you could probably see this as being related to the Yukawa potential falling off faster than the massless version

but technically it's more due to the absence of a mass gap

DIRAC1930

21:11

QFT is harder than what it is first let on to be

Understanding a consistent picture is difficult

ACuriousMind

depends on how you look at it; these issues are completely irrelevant for most of its practical applications, even if we don't exactly understand why they are irrelevant

I think we're just very spoiled as physics students today that all the other areas of physics have been figured out to a much larger degree: The way QFT looks to you now was how most of physics looked to most of its practitioners in the past!

DIRAC1930

That's true

I still think it will be 500 years till we make progress

Unless it comes from String Theory however I know nothing about that

ACuriousMind

and it's not as if other subfields don't have similar trouble - once you go the "research level" stuff, you'll always run into these situations that don't make quite as much sense as you'd like them to

DIRAC1930

I hope E. Witten writes a book now that he's retired

But looking at that recent course he did, maybe that isn't such a good idea lol

How someone in their 4th year can understand that is beyond me

Imagine if he has multiple books already written exactly like L&L but more updated

imbAF

21:33

Does the electron spin plays a role when talking about the Stark effect?

Mr. Feynman

21:47

@ACuriousMind that is the reason why I disagree with this

Jul 26, 2022 at 17:57, by ACuriousMind

absolutely, yes :P

During the last century the growth of Physics was truly immense

Like an inflationary period :P

DIRAC1930

I mean really it was from around 1915 to late 40s

Also, people forget that the whole of theoretical chemistry is included in that

Since that is reliant on QM

Up to mid 60s was probably tying up loose ends

But it's crazy how much of a paradigm shift QM and GR are

QFT is really an extension of QM

And only radiative corrections was left to sort of which was done in 1949 by Dyson

Actually I suppose stat mech was also a paradigm shift

The fact that it was discovered before QM seems crazy

But it seems like how to progress now is not clear at all

Mr. Feynman

22:12

@DIRAC1930 Well, I would say this is common to Physics in any era :P

DIRAC1930

That's true. People say that all physical phenomenon is explainable through modern theory but I suppose there are massive ones such as dark matter

Which are most likely going to be a complete paradigm shift

And who knows what's going on at really small lengths

bolbteppa

23:35

@DIRAC1930 The fact you are quoting this answer should tell you something

I don't understand what's so difficult about the idea of 'free particle comes in, interacts, then scatters off into a free particle'

DIRAC1930

Because you cannot switch off the interacting part of the Lagrangian. In nature, you never observe a free particle. Free particles do not exist. Dressed particles exist. There is clearly a physical/mathematical reason why this protocol is used.

bolbteppa

This is ridiculous, free particles are the most basic thing in QM, waving your hands and denying a prediction of QM is right in line with quoting answers by alternative researchers denying (or rather, trying to 'reformulate') things like QED, it's just hand-waving

DIRAC1930

The free particles in QM are the dressed particles of QFT

I'm not sure how this limit is achieved but this is a completely different issue

And quite literally, in cond mat qft, you can switch the interaction off. If the particle is outside of the crystal, there is no emergent phonon field etc.

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