@Amit In the end, we can directly see neither matter nor spacetime. We r only aware of our mental states. Matter, EM field and Metric field r required only in our model of the universe. I think it wud b arbitrary to say that Matter and EM "exist" while the metric is just a "tool" for predictions.

The metric can also carry energy like in the ADM formalism and can hav gravitons which carry momentum and energy. So it is a very real thing.

I mean it is required 4 energy conservation in ADM

@Amit no, i wudnt call it the medium. We cud begin our theory by postulating a bunch of fields attached 2 a common topology. The metric field is just one of those fields. one could have metric-less theories too like topological field thoeies. This also shows that spacetime need not b emergent from other things that exist

But the thing is, by "spacetime" we can mean both the metric and the manifold. I think, more often, we mean the manifold

The manifold can be seen as a side-effect of the fields that exist, I think. There is no such theory as the manifold existing on its own. Again, this is just a matter of semantics about which mathematical tool you name to "exist". The manifold is a real tool to sew the fields together

@JackRod Yes, the details are in this paper.

I don't know the details, but the experiment studied the dark matter present when the CMB radiation was generated about 380 thousand years after the Big Bang. The dark matter distribution was measured by detecting how it gravitationally lensed the CMB.

In the Rutherford scattering why the solid angle is as $d\Omega=sin^2\theta d\phi d\theta$? Where does the 2nd $sin\theta$ comes from? When I try to calculate the solid angle for the unit sphere I get $d\Omega=sin\theta d\phi d\theta$

I have a soft question about science. I remember reading about a principle of science that every new theory must be able to reproduce the results of the old theory in the zones the old theory is valid in. Relativity for example reduces to Newtonian mechanics in the regime of small m and v << c. What is the name of this principle? I vaguely remembered it as "principle of complementarity", but I can't find results for that in Google. Is asking about this appropriate on the SE?

@imbAF Looks to me like the first formula (with an extra sin(theta)) is extraneous. Where'd you get the formula from?

@Allure Sounds to me like a general question about philosophy of science rather than a question specifically about physics, so I'd first try at Philosophy

@Allure correspondence principle?

Thanks, I will ask on Philosophy SE

Or nevermind because Nihar Karve answered it :D

I don't think I understand atomic orbitals, so lemme ask a stupid question

if we ever only observe an electron at a definite position, and wavefunction is "not physical" (inb4 sabine), how can atomic orbitals exist?

Hello. Is anyone here into jewellery?

Is it worth it? Especially diamond jewellery? I think it looks great but u cud use the money on smthing meaningful

Plus it'll attract thieves

I dont see any point

then why are you asking

Just asking if smone has a dffrnt Perspective

value is subjective

so only you can decide its worth to you

Yeah.. I guess if it looks pretty it can b worth it depending on the person

But it wud b stupid to buy it if u havnt got much money

@LeakyNun wavefunction is a real object used to describe ur probbailistic knowledge of a physical system

The predictions made from the wavefunction r testable

well that goes against copenhagen

Y

copenhagen says that the wavefunction collapse is not physical

I dunno. People r very confused on what copenhagen means. I think Copenhagen talks about a quantum-classical divide. The classical apparatus objectively collapses the wavefunction of the quantum system

So it speaks of a physical collapse

Unlike Relational QM which speaks of relative collapse

@RyderRude well I guess I can modify Copenhagen to be consistent with this, by saying that the collapse of a decohered wavefunction is not physical

idk i guess this doesn't make much sense either

@LeakyNun Again we have to be careful with our words - what do you mean by the "existence" of an atomic orbital

What do u mean by collapse of a decohered wavefunction? By "decohered wavefunction", do u mean density matrix with classical probabilities in it? Decoherence already assumes the usual collapse to arrive at that conclusion

@ACuriousMind I thought the point was that if the electron was at a definite position then it would slowly fall into the nucleus

@RyderRude i don't think the last statement is true

@LeakyNun I don't see how that's a response to my question :P

@ACuriousMind that the electron really doesn't have a definite position

No, Decoherence applies the Born rule to arrive at classical-ish probabilities

It already assumes collapse

@LeakyNun well, an electron in an orbital doesn't; an electron whose position you just measured obviously does have a definition position

I'm not sure where exactly the problem here is - the orbitals are energy eigenstates, not position eigenstates

there are like geometrical shapes that represent the 95% probability right

you mean pictures like this?

yeah

I think I see where the problem is

the pictures are pretty but the specific shape of the wavefunction is really the least interesting thing about orbitals :P

well those allow you to have pi-bonds etc

we're not interested in orbitals because we have some sort of electron-catcher that can position-measure electrons in flight

you make MO geometrically

we're interested in them because they're the energy eigenstates, i.e. these are the time-stable states an electron can be in

Decoherence just means that u can pretend that the probabilities hav become classical and u wud pretty much make the same predictions as if u had used the usual collapse of the wavefunction. It holds upto a good accuracy

but sure, these shapes represent the probability to detect an electron in those volumes if you were to make a position measurement on it

what exactly is your problem with their "existence"?

@ACuriousMind it's unobservable

@LeakyNun so? that's not specific to orbitals, any wavefunction is "unobservable" by a single measurement

@ACuriousMind tell that to the chemists

that's just how probabilities in general work, isn't it? If you claim that something follows a specific probability distribution, then that is only a frequentist claim about the distribution of measurement outcomes in the limit where you repeat the measurement infinitely often on an identically prepared system

@ACuriousMind Also isn't the law of large numbers only for specific distributions

Like if you don't have independence it's gonna be tough

I think Leaky Nun means whether we can ontologically attribute this orbital to the electron or not. Everyone agrees that the wavefunction "exists" at least as a tool to make verifiable predictions

@Slereah I didn't make any statements about expectation values!

The ontological problm with quantum mechanics is unsettled. So the only use of these orbitals is as a tool for making predictions relative to an observer

and the law of large numbers holds for any distribution that has an expectation value

the existence of distributions whose expectation value diverges is indeed somewhat annoying but irrelevant for the claim I made

Probbailitiea r always a relative object, even classicaly. They hav no place in the ontology. This is y im also skeptical of Many worlds

I used 2 love Many worlds but it has many problms like a preferred time-direction

Relational QM deals away with simultaneity of collapse issues

I generally don't consider it useful to start arguing about interpretations when someone starts the conversation with "I don't understand atomic orbitals"

@RyderRude classically it is deterministic

before we can start arguing about interpretations we first must agree on what the interpretation-independent facts are

it is with QM that you start having intrinsically indeterministic stuff (at least accoridng to copenhagen)

@ACuriousMind yeah... but Bell's theorem kinda has to be involved with interpretations right

what does Bell's theorem have to do with your confusion about orbitals?

I wud say these orbitals exist in the sense that they make testable predictions? What is the sense in which u mean for these orbitals to exist? @LeakyNun Do u want the electron's objective ontology to b these orbitals/wavefunction?

or probabilities in general

I'm still not really sure what the problem even is

Ooh i got ur problem. Different observers will not use these orbitals to describe the electron. It's a relative object, indeed

@ACuriousMind an electron always has a definite position; an electron with definite position cannot orbit a nucleus stably

@LeakyNun "an electron always has a definite position" what do you mean by this?

the core idea of quantum mechanics is that this statement is simply not true

the canonical demonstration is shooting electrons at a double slit

right ok, when we observe an electron it always has a definite position

we have never observed an electron that is 50% at A and 50% at B

@LeakyNun again, double slit

all you're saying is that the outcome of a position measurement is always a definite position; that's true

but the outcome of any measurement is always a definite value

@LeakyNun this is only tru when we measure the position of an electron. And delta functions r not physical anyway. So u never observe a sharp position even in that case

that this does not imply that all objects have definite values of all quantities when not being measured is the crucial idea behind QM

hmm

I guess that makes sense

but then you can only ever indirectly (e.g. through double slit) "observe" that phenomenon

Yes.

That is as "direct" as it gets

@LeakyNun Also, the claim that an electron with definite position cannot orbit a nucleus stably is too vague. What is true that an electron with definite position obeying classical electrodynamics cannot do so.

it's not as if just saying "the electron can't have a definite position" somehow determines the physical theory that gives us the orbitals

I can always spot a post by @Qmechanic without even reading it

LOL

and i shouldn't ask the question "what is the electron really doing?" i suppose

I mean you can ask that question but I wouldn't expect an uncontroversial answer to that

the pragmatic fact is that it doesn't matter - you don't need to know what the wavefunction "is" to use it to correctly predict what happens

@LeakyNun technically, u cant even speak of an electron's wavefunction on its own. Nothing is isolated from interaction. As soon as u account for interactions, u cant speak of wavefunctions of particles individually

U instead hav to speak of the tensor product wavefunction

This is different from classical mechanics where u can speak of mathematical objects corresponding to individual particles (points)

Given a "point" exists :P

So not only can u not speak of "what trajectory an electron is following", u cant even say " What is the wavefunction of the electron right now? " This "individual wavefunction" stuff is an approximation

But it is a great approximation 4 experiments

I mean you also can't observe a trajectory, you only observe the end result

and assume that the electron didn't magically teleport to the endpoint in the process

That's why we postulate axioms.

the emergence of classical trajectories can be explained from the quantum behaviour, see rob's answer on Mott's classic derivation

@ACuriousMind that's brilliant

+1 👍

🙏

Speaking of orbitals, Wikipedia has some nice anims based on dropping a dimension, so instead of looking at spherical vibrations we're looking at the vibrations of a disc. See en.wikipedia.org/wiki/… The shapes of atomic orbitals can be qualitatively understood by considering the analogous case of standing waves on a circular drum.

Also see Greg Egan's Quantum Oscillator app that he created 25 years ago (originally in Java, but he translated all his apps to JavaScript a few years ago). gregegan.net/APPLETS/07/07.html

Trying to write that article on my site about proving that the earth is round and boy it's not fun

Having to do actual astronomy

On the topic of classical trajectories emerging from quantum rules, Feynman's classic popular book QED: The Strange Theory of Light and Matter has a great explanation of how the classical rules of the reflection of a light beam are consistent with a quantum picture.

There's a lot of papers on tracking the position of the sun in the sky

iopscience.iop.org/article/10.1088/0031-9120/45/6/010

This paper was written by druids

Well, tracking the Sun is kind of important when you use mean solar time.

also to predict the next sacrifice

Cutting edge tech

Part of what makes it hard is that I don't know the standard textbooks

they seem to recommend "Astronomical Algorithms"

Mama always told me not to look into the eyes of the sun, but Mama that's where the fun is.

@Slereah As a Discworld Druid said, just when you finish building your 32 megalith processor, it's obsolete, and you have to start building a 64 megalith system.

@Slereah Do you have to? What's wrong with the more elementary demonstration of curvature by looking at long flat lakes?

Honestly stone age astronomy isn't nearly as fun as renaissance astronomy

or is the problem that that only demonstrates "local" curvature not the full shape

they had all kinds of wacky machines

@ACuriousMind Gotta look at all the angles

the horizon is one indeed, although you have to consider REFRACTION

the horror

@Slereah I can with 90% accuracy as other answers with numbered lists, numbered equations and "OP" often fool me

ugh why is it so hard to find a good libgen mirror

The ones I usually detect more easily are Moretti's posts

@Slereah I would try to help but I don't think it's allowed here

wouldn't even break the law to help a starving man

Did you learn nothing from Les Misérables

There is a big 19th century book proving that the earth is flat, I should give it a look

like this is the standard "good" book on the topic

Are the things you write on your blog not found on books?

Because in that case I have a question :P

Most of the stuff on my site is known stuff

Although sometimes a bit obscure

The book is "One hundred proofs that the earth is not a globe"

By OG modern flat earther William Carpenter

gutenberg.org/files/55387/55387-h/55387-h.htm

@Slereah The book by Jean Meeus? Yes, it's quite popular, and the algorithms are pretty accurate. I have Sage / Python code comparing Meeus's Sun positions to JPL Horizons values. They're generally pretty close. It's not online, but here's a similar comparison I did a while ago, based on a combination of Meuss algorithms & those of Hughes et al. astronomy.stackexchange.com/a/49546/16685

@Slereah Oh ok, because otherwise I would have asked how you could be sure everything is right

I'm often scared of writing original stuff, even about known theories :P

I mean proving that the earth is round is not a new one certainly

I do have new-ish ideas to try to maybe checkout but we'll get there when we get there

I have Suspiscions about the scale bundle

I think it is secretly much cooler than given credit for

@Slereah It's amusing at first, but it soon becomes tedious and annoying. I think I managed to read about 30 or so of those "proofs" before I got sick of it.

A few of them are a little tricky

@Slereah Yeah, I didn't mean that proving the Earth is round is something new. Coming up with a new method is, though :P

It's unfortunate I don't have a garden because doing actual experiment may get a little tricky

I don't even own a gnomon

I do have a garden but the Pauli effect would blow it up if I tried

Probably the hardest part wrt flat earth disproving is showing that the hydrostatic limit for planets is reasonable

Planetologists seem kind of vague about why this applies

The count of the possible realities might be aleph_1 or alpeh_2 in the QFT. One of them is selected as "here and now". Or, as we could call it, "reality". Question is, what did it. Who did it.

Mathematician / artist Jos Leys did some anti Flat Earth stuff: josleys.com/show_gallery.php?galid=384

There is some notion of the potato radius for the limit at which an object becomes fluid, but the proofs for it are a bit handwavey

how can we assume the highlighted equation

@Slereah because of "rocks"?

@Mr.Feynman Well yes, rocks are not fluids

They usually handwave it as "gravity >>> rupture stress" or something but idk

it feels like a complicated matter*

???

The fracture point thing

@PrateekMourya you assume it and you then check if you can find a solution that satisfies such condition

ok

@PM2Ring Imagine an Universe with a little bit uncommon spatial dimensions: 20000km height, 20000km width and 20000km length. It would be cyclic on the X and Y axis, but it wold not be on the Z axis. There would be also some curvature resulting a constant gravity into the positiv Z axis. I think it would be like a flat earth Universe.

It's not a priori obvious that there is one

yess

Also like even if rocks break under gravity, you still have to prove that granular flow behaves like that

@PM2Ring Question is, what type of matter distribution could cause such a spacetime configuration in the GR. Probably there would be no impulse conservation on the Z axis on Noether theorem.

@Slereah Rocks are a bit more fluid when they're sufficiently hot. And large planetesimals can get pretty warm with the heat of formation. OTOH, none of the Main Belt asteroids apart from Ceres are spherical, and even Ceres "cheats" by containing a lot of ice.

@PM2Ring I mean many planets are cold now!

Yet still in hydrostatic equilibrium

I think they are metastable. If you dig there a well, it will remain.

as I said there is a concept of such a radius where planets are big enough that they will be at hydrostatic equilibrium

But it's hard to find a lot of derivations for it

arxiv.org/abs/1004.1091

If you dig a 10 km deep well in the Earth, without some really strong support, it will collapse. In this sense, the Earth is in hidrostatic equilibrium, even its land area. Doing the same with a 10m deep well, it will likely remain

@Slereah Sure, but planets are huge compared to asteroids, so the gravitational energy can easily dominate the chemical energy. But even so, I bet the cores of the gas & ice giants are fairly warm.

"To our knowledge, the published derivation of something most analogous to the potato radius is in Chap. 1.2.1 of Stevenson (2009)."

Hm

But even if rocks didn't break, wouldn't they behave completely differently from a fluid?

I have no idea

Wrt e.g. pressure

Maybe this is one of those things I should go ask the planetary science stack exchange people

or the astronomy SE

"Focusing on energy (not force or pressure) the electronic bond energy of ~ 1 eV is set equal to the gravitational energy ~ GMµ/R where µ is the mass of the particle in question and R is the radius of the object. The result is a radius of a few thousand kilometers for a rocky body and ~ 1000 km for an icy body."

I guess it's a good enough ballpark estimate

"This suggests the potato radius is inversely proportional to the square root of density and inversely proportional to the square root of particle mass. This obviously is not the case; the observed potato radius of rocky objects is about 50% more than that of ice objects, even though rocky objects have the greater density and are made of more massive molecules."

Potato science is hard

@peterh Ok, but typical Flat Earth theories aren't that sophisticated. But I kind of like the idea of a universe with that topology. :) Yes, it would have gravity (and I think the field would be uniform), but it still needs a big chunk of mass for the gravity to be significant.

Earth is conformally flat

"the potato radius depends on the density ρ and the yield strength σ y of the material "

@Slereah David Hammen (who has worked for NASA) has a few posts on the potato radius: astronomy.stackexchange.com/search?q=user%3A2618+potato

It's kind of weird how recent most of that stuff seems to be

all the potato radius stuff seems to be post 2000

But maybe it's just because I don't know the field well enough

how did maxwell know that c wasn't infinite?

The speed of light had been previously measured

@LeakyNun $c$ appears in Maxwell equations as a combination of parameters that were known i.e. the dielectric constant and the permittivity of vacuum

Also on the EM front, the value of $\mu$ and $\varepsilon$ were known

Also really whatever value they had, the speed of propagation of EM waves would be finite

We don't have a lot of good data about what's really going on inside solar system bodies. We have some rough ideas by looking at their surfaces, and at meteorites. It would be great if we could get a few core samples from asteroids...

But it was found to be very close to the speed of light as measured by previous experiments

which one of μ and ε would be affected if c = inf?

So they were like Ooooooooh

@LeakyNun It depends!

@LeakyNun either of them should be zero

Usually people will say that there would be only electric fields then

but there are multiple limits of electromagnetism in that limit

@LeakyNun that's an ill-formed question; $c\to\infty$ does not (alone) well-define a specific limit of electromagnetism

what's your favourite model in that case i guess

@LeakyNun c = inf means inf is a real number

I'm not even sure it makes sense as the $c\to\infty$ limit is the galilean limit and electrodynamics is naturally relativistic

@Mr.Feynman I mean you can do it

Via the Inonu-Wigner contraction and all

The "classic" limit is just the universe without magnetism

And where the electric field is just the Coulomb field

Even gravity surveys would be useful. When the Juno mission started analysing the gravity field of Jupiter we learned that it appears to have a large diffuse core. That was quite surprising, and we're only beginning to come up with models which could account for such a core.

I was just about to mention that the only time I've ever really thought about $c\to\infty$ is in the context of the IW contraction

@Slereah is that the group contraction of Poincaré group?

But there are a few other possibilities IIRC

@Mr.Feynman Yes

Schrodinger-like fields r also non relativisitc but they never appear in newtonian mechanics

I should read about it to see how nR spinor fields appear

I guess a massful non relativistic classical field wud b schrodinger-like and newtonian

For massless non relativistic fields, we get the Poisson eqn in the non relativistic regime like Coulomb's law or Newton's gravity law

is it possible to have only magentism and no electric field?

@LeakyNun Yes!

That is another of the limits

Is it true that a mass-ful non-rel complex classical field wud obey the schrodigern eqn?

Btw, a more down to earth way would be to take the galilean limit of the electromagnetic field of a moving charge, and only the electric Coulomb field would survive as Slereah said

We use this field to obtain non rel QFT

also are these hypothetical universes devoid of atoms?

hi

hi

@LeakyNun Why would they be

not sure how like electrons would "orbit" around protons

the topic is called galilean electromagnetism

if you want to learn more

link.springer.com/article/10.1140/epjp/i2013-13081-5

many such things

Isn't Galilean EM Coulomb interaction only?

@Mr.Feynman See above

There are multiple limits

galileo is just sitting in his grave being like "i ain't never come up with any of these"

Think about poor Euclid

the limits are apparently $E \gg c B$ and $B \gg c E$

$v\ll c$ (low velocity limit) is not the same as instantaneous interaction $c\to\infty$ (which I call galilean limit)

s3.cern.ch/inspire-prod-files-1/…

I once read something about it and the "Carroll transformations" (not Sean)

this paper does it with Inonu-Wigner it seems

Carroll transformations are a different topic

This is a contraction of the Poincaré group that isn't Galilean

Basically the limit $c \to 0$

Why does a dressed particle of QFT (which is a free particle in QM) behave like a free particle in QM?

is the non-instantenous interaction observable "in a school lab setting"?

Where the light cone closes to a line

@DIRAC1930 classically, it looks like this: the original particle has mass negative infinity. Dressing raises its mass to normal. The combined system behaves like one free particle of a normal mass

But in QFT, it is much more complicated

Haag solved this stuff

@Slereah interesting, thanks

@DIRAC1930 Have you looked at the Haag-Ruelle construction of scattering states that I pointed you to the last time you asked this question?

Apparently the two limits of the EM field relate to the two inequivalent vector representations of the Galilean group

I don't understand why you keep coming back here to ask the same question and somehow expect different answers :P

Haag-Ruelle theory replaced spinors

Except there are 9 irreps of the Galilean group

So you can have a lot of weird versions of classical EM if you really want to

@Slereah A different topic I couldn't find much about I wanted to know more about the difference between the instantaneous interaction limit vs low velocity limit in that regard

@LeakyNun if you count a telescope as a "school lab tool" :P

@Mr.Feynman Low velocity limit is presumably the velocity of charges, I guess?

@ACuriousMind ok, how do you use a telescope to observe that?

@Slereah yes, the $v\ll c$

@LeakyNun see en.wikipedia.org/wiki/…

Not sure I have the wig to perform such an experiment

I haven't asked this question yet

The Galilean group is much nastier than the Poincaré group unfortunately

Working out its irreps is even worse than usual

A sensible difference is that e.g. in the low velocity limit we neglect $\mathcal{O}(\beta^2)$ the radiation is still there because this doesn't kill the $1/c^2$ terms

Haag-Ruelle is a bit above my mathematical level so I just accepted someone worked it out properly

While the infinite $c$ limit kills it

they are apparently indexed by three numbers

@DIRAC1930 It sounds identical to all the other questions about asymptotic states to me; this means I haven't understood what you're actually asking. So what do you mean by a "dressed particle", and why is it mysterious that you can treat it as "free" if you accept that the asymptotic states of QFT are indeed free

@Mr.Feynman By low velocity limit, do u mean $v\rightarrow 0$

Do all interference effects observed with coherent light also appear with single photons? I've only ever heard of single-photon double slit experiments, but there's a lot of cool single-slit effects too.

But nothing moves in that limit

I am in my low velocity limit

Ok i think u mean u r keeping upto first order in v

idk if galilean EM is still a gauge theory tho

$v\rightarrow 0$ and $c\rightarrow \infty$ shud b equivalent

Does it maintain the gauge invariance???

@RyderRude $v/c\to0$ with constant $c$, so yes

I meant that in QM problems, we have a particle obeying the Schrodinger equation without a potential $V$ i.e. it is free. In QFT, we have the interaction energy operator $\hat{V}$ being a function of the field operators. So somehow this dressed particle reduces to one that is free in the non-rel limit

But they are not strictly the same as $v\to0$ doesn't kill $1/c$ terms

@DIRAC1930 who claims that just taking the non-relativistic limit on any such dressed state is possible?

or, rather, who claims that this yields a free QM particle?

I'm not sure

Probably noone

In my world, the free particles of QM are all asymptotic states of the underlying QFT

What about in the KL spectral rep. We have a pole at a one-particle state with energy $E$. In the vicinity of $E$ the only contribution to the 2 point function will be the pole. If I have small $p$ (non-rel), expanding around the pole, I will have a GF that obeys the free EOM

yes, so?

I'm not sure what we're trying to show here

Dressed particles in the non-rel limit of QFT are free particles of QM

but you already know that the asymptotic states are free!

I don't get what there is left to show

Yes but asymptotic states, are not states in the real world at finite $t$ like we are experiencing now

@RyderRude That's fine, we use our mental activity to become aware of measurements too. But time is still different. Take a rock, I first perceive it, then decide to measure its length. With time, I never perceive "it", I perceive various processes (rising and setting of the sun or whatever) and then define time to quantify them

@DIRAC1930 sure but the statement "asymptotic states are free" is that there is a construction of time-dependent states $\psi(t)$ such that $\lim_{t\to\infty}\psi(t)$ is free. By the nature of limits, this means that for finite but large $t$, it is approximately free!

combine that with the scattering amplitudes at low energy all being strongly suppressed or zero, and you get that you don't need a particularly large $t$ to have very good approximation to free states

@Amit u cud say that processes r the only things that exist, as there's no such thing as a theory of just a manifold with no fields. The manifold is always occupied by fields. Fields r the real things that exist. Metric field is one of those fields