The h Bar

Semiclassical

12:05 AM

yeah, sorta lost track of that

ACuriousMind

probably similar to what some students did when trying to solve that :P

Semiclassical

12:31 AM

lol

a calculation in the book is to compute the energy level shift in a Fermi gas, starting without interactions and then adding Coulomb repulsion

in momentum space that gives integrals with $1/k^2$ showing up, b/c huzzah fourier transform

prof wanted them to redo the calculation without momentum dependence, which amounts to a contact interaction

the weird thing is that, if i do the calculation "right" by my lights

then if i include a repulsive interaction, the energy levels go up. but that's the opposite of what i'd expect, and the opposite of what happens for Coulomb repulsion

and it all hinges on whether or not a certain intermediate expression needs a minus sign or not

i think it should, and so do the lecture notes

but then i get a result i can't make sense of

uhoh

12:52 AM

2

Q: What is the maximum amount of carbon atoms that can comprise a fullerene?

As the fullerene sum formula can be defined as Cn, with n being any integer that is not odd, is there any fundamental theoretical limit as to how big "n" can get? If there is no limit to the chemistry but the difficulties in synthesizing large fullerene's, what has been the largest fullerene ever...

+200 bounty just added

Semiclassical

4:15 AM

to put it into different terms: introducing a repulsive contact potential certainly needn't have the same effect on the electron energy levels of a free electron gas as a repulsive Coulomb potential. but how tf can the results have different signs?

Manas Dogra

5:25 AM

How many nearest neighbor pairs can be formed from an integer lattice in n dimensions?
Is there a closed form expression?

John Rennie

Do you mean close packed spheres?

Manas Dogra

No I mean if we take N particles, put them in lattice(integer) points in n dimensional space, then how many pair can we form such that the distance between them is minimal?

John Rennie

But what lattice? Primitive cubic?

Manas Dogra

Integer lattice $\mathbb{Z}^n$

Like in 1 dimension it will be simply N-1, because the required lattice points will be (1,2),(2,3),(3,4),...(N-1,N).

It's just the number of "pair terms" in the n dimensional Ising model.

John Rennie

If you pack the N particles into a hypersphere then it's 2n for each particle in the interior plus, erm, a number less than 2n for each particle on the surface.

Manas Dogra

5:47 AM

But I am not packing into a hypersphere, you can say I am packing into a hypercube.

John Rennie

But you'll get more pairs by packing into a hypersphere than you will by packing into a hypercube.

Manas Dogra

I am not trying to pack anything, it's just a counting problem I thought of..
as I said, it's the number of terms in the summation of the interaction term in Ising model.

Slereah

7:55 AM

morning

Slereah

8:35 AM

twitter.com/JFluidMech/status/1450588172744806401

No preview of twitter posts on the chat

for shame

The diagram

ACuriousMind

8:49 AM

@Slereah Used to work, but apparently they can't figure out how to get it to work again in a secure way

Slereah

Tell me about it

My job at work involves displaying tweets on the local server

Angular really doesn't like me just dumping foreign HTML on the page

ACuriousMind

At least it's HTML :P Any day I don't have to parse half a dozen of antique proprietary formats is a good day

Slereah

One of my job involved some antique Windows database format that nobody has used since the 90's

Access maybe I guess?

although at least I'm not a COBOL programmer

ACuriousMind

I think "nobody has used Access since the 90s" is wishful thinking :P

Slereah

Well obviously they still do

But I would guess mostly for legacy project

One of my biggest job project ever involved me making a new online project management tool for like a huge company that made car parts

This is like a multibillion dollar company, and up until recently, their project management tool was a gigantic, ultraslow Microsoft Excel document from hell, with terrible VBA code

ACuriousMind

8:57 AM

@Slereah the world is positively filled to the brim with "legacy projects"!

Slereah

Code obviously written by someone unfamiliar with both VBA and coding in general

This is why I am skeptical of conspiracy theories

John Rennie

@ACuriousMind Do you have to hand write a parser each time, or are there general purpose parsers that you can make work by writing a definition of the data format?

ACuriousMind

@JohnRennie depends on how nice the format is

most general parsing frameworks require you to be able to write down a nice grammar

Slereah

I'm pretty sure that if there's a sinister conspiracy theory to poison us peasants we can probably skip through the cracks because the management for it is written on some COBOL database

that nobody has known how to work since the 80's

John Rennie

Yes, that's the sort of thing I had in mind, though I guess it may not be easy or even possible to write a grammar for the less organised formats.

Slereah

9:01 AM

it probably depends on the

CHOMSKY GRAMMAR HIERARCHY

Chomsky hierarchy

In formal language theory, computer science and linguistics, the Chomsky hierarchy (occasionally referred to as the Chomsky–Schützenberger hierarchy) is a containment hierarchy of classes of formal grammars. This hierarchy of grammars was described by Noam Chomsky in 1956. It is also named after Marcel-Paul Schützenberger, who played a crucial role in the development of the theory of formal languages. == Formal grammars == A formal grammar of this type consists of a finite set of production rules (left-hand side → right-hand side), where each side consists of a finite sequence of the following...

Apparently JSON is a type-2 grammar

ACuriousMind

the point where it usually starts to get annoying is if they don't reserve their keywords

Slereah

Properly written HTML is also type-2, but apparently general HTML is type-1

Since most HTML parsers are pretty accepting of errors

I guess few formats are type-3

CSV maybe

CSV is pretty easy to parse

ah yes, CSV is type-3

1

A: Wormholes & Time Machines - for *experts* in GR/maths

Let's consider the case of a thin-shell wormhole in Minkowski space $M = \Bbb R^{2,1}$ (We need at least two spatial dimensions for a proper wormhole without getting topology change involved). First we need to define two timelike hypersurfaces $S_1$ and $S_2$. To simplify things, we'll consider...

Probably one of my best answer and yet it languishes at the bottom

For shame

ACuriousMind

that's probably because most people have lost the will to live before they've finished reading the question :P

Slereah

It is a pretty long-ass question

But I was really curious about how to do cut and paste wormholes properly, too

You can indeed do it, turns out

although doing it entirely properly is horribly long, too

You can't even assume the existence of normal neighbourhoods outright

I haven't even yet discovered all the basic stuff for it yet

ACuriousMind

really a terrible example for SE not being suited to back-and-forth between question and answers

Slereah

9:16 AM

like where is the chronological horizon if the wormholes aren't precisely aligned!

yeah kinda

maybe I should hang out more at physics forum, too

it seems nice enough

Slereah

12:33 PM

"It's common knowledge that Mt. Everest is the tallest mountain on Earth, measured from sea level. A somewhat more obscure piece of trivia is that the point on the Earth's surface farthest from its center is the summit of Mt. Chimborazo in Ecuador, due to the fact that the planet bulges out at the equator.
Even more obscure is the question of which point on the Earth's surface moves the fastest as the Earth spins, which is the same as asking which point is farthest from the Earth's axis. The answer isn't Chimborazo or Everest. The fastest point turns out to be the peak of Mt. Cayambe, a vol…

Slereah

12:59 PM

Mad Spaces

2:13 PM

In special relativity why is for when $ds^2 = 0$ then it must be also for any other inertial system that $ds'^2 = 0$ ? i am not really understanding this point and i have checked many pages and books with very vague explanation i cant seem to get my head around it

so if in the none moving intertial frame two points are connected by a beam of light, then ds is zero, alright, but why does it have to be zero in another intertial system? are there always beams of light that connect the two points? how do we prove this?

Obviously it is shown later that $ ds^2 = ds'^2 $ but the above is usually shown before this relationship

Mad Spaces

2:27 PM

i am thinking like this. in any inertial systems, the points $x_1' x_2' $ must be connected by a light beem, since even for lets say the system moves with speed of $ 0.99c$ away from one point, light will eventually reach it and connect both points.
Makes sense?

Also. Are all infinitsmals of same order poprortional to eachother?

Daniel Underwood

Isn't $ds^2 = 0 \implies ds'^2 = 0$ just a different form of saying the speed of light is frame invariant?

ACuriousMind

@MadSpaces the point is that the existence of a beam of light is not frame-dependent!

Mad Spaces

@ACuriousMind Okay. so i am thinking right now that you agree with my small thought process i did?

ACuriousMind

there's nothing to prove, it's just that physics wouldn't work if different frames could disagree on reality, i.e. whether or not a photon emitted at A can reach B

Mad Spaces

I know its not allowed.. but if a system moves with speed greater than c then the photon could never reach, true?

but since the speed is less than c, it will reach, even if it takes different time.

ACuriousMind

2:33 PM

@MadSpaces Uh...rather, I don't agree with the part where you seem to think that "connected by a beam of light" is a frame-dependent characteristic

@MadSpaces I don't understand the question

Mad Spaces

i am refering to this :
i am thinking like this. in any inertial systems, the points x′1x′2 must be connected by a light beem, since even for lets say the system moves with speed of 0.99c away from one point, light will eventually reach it and connect both points.
Makes sense?

ACuriousMind

I think we have very different ideas what "connected by a light beam" means :P

Mad Spaces

Okay. For me, it means light omitted from A reaches B then AB is connected by a beam of light

ACuriousMind

first, the "points" in relativity are not spatial points, they are spacetime points, i.e. "events"

when the spacetime interval $\mathrm{d}s^2$ between two events is zero, that means they are positioned exactly so that (regardless of frame) light emitted at the position and time represented by one of them can reach the position represented by the other at exactly the time represented by the other

(I'm saying this because your "light will eventually reach it and connect both points" makes me think you're thinking about the "points" here in a purely spatial fashion)

Mad Spaces

yes i did i see the error i am trying to take in your statement right now

"light emitted at the position and time represented by one of them can reach the position represented by the other at exactly the time represented by the other" i am having hard time imaigning or understanding this statement

ACuriousMind

2:40 PM

let's take a simple example: I'm standing at my window and fire a laser at the moon. This is the first event: The time and position at which I trigger the laser

On the moon there's a target that will emit some sort of signal when I hit it with the laser - the time and position at which the signal is triggered is the second event

Ryan Unger

is this illegal

shining lasers in the sky

they used to tell us it was

ACuriousMind

the two events are "connected" by the laser beam - their times and places are exactly so that the light from the laser can travel from the first event to the second

@RyanUnger it's illegal if you hit something other than the moon :P

I dunno, really, I'm not up-to-date on laser law

Mad Spaces

Okay. when you mean their times and places are exactly so such that the laser beam can reach, you basically mean, that in reference othe minkowski light cone diagramm, its connectable

ACuriousMind

in the typical diagram they would lie along a 45° line

Mad Spaces

So for example, if a point B lies in the past, we can not connect it from a point A in the present, since the times will not be eligible

And so on for the spatial coordinates.

Are points inside the Cone also connectable ?

ACuriousMind

2:46 PM

@MadSpaces You can always connect two points by a line :P

the question that's relevant for whether or not the two events are at lightlike separation is whether that line is at an 45° angle or not

Mad Spaces

I think i got it

so any points that lie on the lines at 45 degree of the minkowski diagramm are connectable by a beam of light?

bolbteppa

You need to use the two principles of relativity: a) the laws of physics are the same in all inertial frames, b) finite velocity of propagation of interactions (i.e. finite speed of light), to argue this point.

Mad Spaces

I know these postulates i just do not see the connection between them and the fact that we are discussing

bolbteppa

In a frame $K$, b) says the speed of light $c$ is constant, and so if $x_a = (t_1,x_1,y_1,z_1)$ and $x_b$ are connected by a light-beam we have that $(cdt)^2 = d \mathbf{r}^2$ i.e. $ds_K^2 = (cdt)^2 - d \mathbf{r}^2 = 0$ holds between these two points.

In an inertial frame $K'$, a) says that b) is true in K' too, i.e. the speed of light in $K'$, $c'$ is constant. Thus for $x_a$ and $x_b$ as viewed from $K'$, denoted $x_a'$ and $x_b'$, the light beam satisfies $(c'dt')^2 = d \mathbf{r}'^2$ i.e. $ds_K'^2 = (c'dt')^2 - d \mathbf{r}'^2 = 0$. a) actually implies that $c' = c$ also holds, otherwise the motion of a light beam would be different in different frames.

ACuriousMind

@MadSpaces yes, that's the very definition of how the diagram is set up!

light rays go at an 45° angle

how did you define how that diagram works if not by how light rays look in it?

bolbteppa

2:53 PM

Thus if the interval $ds^2$ vanishes in one inertial frame $K$ between two points, it vanihes in all other inertial frames between those two points as viewed in their frames

Mad Spaces

@bolbteppa alright so basically in $K$ we find the result, $ds= 0$ , thus in all other inertial systems, according to the first postulate, it must also apply that $ds'=0$ this is very easy to understand.

bolbteppa

You can't make the argument without using a) and b) because the argument fails otherwise

Mad Spaces

And at this point i would argue, you do not even need b?

ACuriousMind

well, if light didn't have a finite speed, the notion of a "light beam" would be very different :P

Mad Spaces

@ACuriousMind Okay what is confusing me is, for a point on one of the lines, it means that point could be in the future, or the past, how could we from the present connect it with a light beam? (shining lazer on the moon analogy), thus how can you shine a laser on a past event?!

ACuriousMind

2:55 PM

@MadSpaces I'm not sure what you're asking

again, the points are events, not locations

ah, well, there's (usually) only one direction in which one event can reach the other with light, time has a direction in SR, after all

it's the past event that could reach us with light, not we the past event

Mad Spaces

Truly a bruh moment, this is very complicated to fathom.

ACuriousMind

it's just a matter of viewpoint - the "connectivity" doesn't change in my example whether you view it from my point of view or from the point of view of the detector on the moon

in both cases one of the events can emit light that the other can detect

Mad Spaces

Okay this makes more sense. i think the final unclear point is to bolbteppa

bolbteppa

If you don't invoke b) then it one assumes it is infinite as is the case in non-relativistic mechanics which is the whole difference between Galileo and Einstein and the whole point of relativity

ACuriousMind

note that this line of thought is only about the possibility of sending light between the two events - nothing about $\mathrm{d}s^2 = 0$ changes if I don't use the laser at all and just look at the moon in that instant

Mad Spaces

3:01 PM

@bolbteppa I mean, looking up at the discussion we had, that according to postulat 1 observers must agree that $ds= ds'=0$ we did not invoke it anywhere. We can just leave $ c, c'$ in the equations, and it would not change the fact, the $ds =0$ in every inertial system!

Yet again, when i was previously reading books " landau, Nolting, " watching yt videos, everystime this relationship is brought up, they refer to " this is true due to c being const" and move on. so i am guessing there is something more to it

bolbteppa

Postulate a) implies that e.g. the speed of light is the same in all inertial frames. It is an unquestioned assumption for Galileo that this value is $\infty$, which is why we need to impose b) as an additional postulate. Once b) is imposed, a) and b) imply $c' = c$, I just put it at the end to try to make this point clearer

Mad Spaces

I understand why the speed of light needs to be constant and why galielo transformation does not satisfy this. i am argung only about $ds = 0 \Rightarrow ds'=0 $
Question. Do we need postulate (b) to prove\imply this relation? if yes, howcome, if no why do authors say they do.

ACuriousMind

@MadSpaces if the speed of light is not finite and constant, nothing in the entire argument works

you can't even define $\mathrm{d}s^2$ without that

Semiclassical

o/

ACuriousMind

\o

Mad Spaces

3:10 PM

I understand. I believe however the use of the senteance, $ds = 0 \Rightarrow ds'=0$ due to postulate b. is really misleading. a way better senteance owuld have been postulate a AND b imply it

You know, this is very confusing

ACuriousMind

I'd say you have standards for precise formulation not shared by many physicists :P

bolbteppa

Re-read my explanation: on beginning in a frame $K$, the first thing I did was invoke b) in $K$, I used this to show $ds_K^2 = 0$ in $K$ (it's literally just expressing b) as an equation $(cdt)^2 = d \mathbf{r}^2$ and putting it all on one side). I then went to an inertial frame $K'$ and used a) to argue that b) must also hold in $K'$ as well. I then just repeated the beginning argument in $K'$, I invoked b) to argue that $ds_{K'}^2 = 0$ must also hold in $K'$.

If $c = \infty$ you simply can't talk about intervals or $ds_K^2$ it's nonsensical, so there's no way to blindly repeat the SR argument to show $t' \neq t$ in general. In the Galilean case we just directly postulate the Galilean transformation between frames as the next step, and $t' = t$ is part of how it's defined.

Even writing down something as simple as $U(\mathbf{r}_1,\mathbf{r}_2,...)$ assumes an instantaneous interaction which contradicts Einstein, one has to be careful with this stuff

2

A: Lagrangian of a Relativistic Harmonic Oscillator

This is a problem that requires the extension to general relativity. It is not possible to solve this problem remaining in special relativity. Make the relativistic ansatz (in 2d spacetime) $L = - m c \sqrt{-g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}}$ where the dot is with respect to the geodes...

Semiclassical

3:32 PM

so here's a problem i was getting frustrated with yesterday

bolbteppa

If you go back to the first book they frame Galileo in words first i.e. the laws of space and time are the same for a free particle in all inertial frames, and then formulate it in equations with $t' = t$ as one of them

(Coincidentally the fact that's phrased in terms of free particles is part of why it's simply unbelievable to think qm can't describe a free particle except in all these qualified ways, especially since one actually derives the Schrodinger equation by invoking this, but alas)

Semiclassical

Suppose i start with a non-interacting spin-1/2 Fermi gas of $N$ particles (zero temperature). as usual, one can compute the total energy as $\frac35 \epsilon_F N$ where $\epsilon_F$ is the Fermi energy

one can then introduce the (repulsive) Coulomb interaction and perturbatively compute, for instance, the change in the total energy

and this leads to a lowering of the total energy

here's what's been driving me batty: suppose you replace the Coulomb interaction with a repulsive contact potential $V(x,x')=\lambda \delta^3(x-x')$

what i consistently seem to find is that this repulsive interaction makes the total energy go up, not down

i have a sense of what's going on, but i'm not sure how to pin down the detail: if you do the Coulomb interaction calculation naively, you end up with a divergent contribution over long distances because a gas of electrons by itself is unstable

so 'really' you need to include a background of positive ions and show that this (along with screening effects) cancels out the divergence

by contrast, the repulsive contact potential doesn't seem to require such apologies to function: you get a finite positive energy increase

Mad Spaces

@bolbteppa let c be finite but not infinite and then you can do the same argument without b...?

Semiclassical

so i guess that means that the contact potential isn't exactly stable either, but not 'infinitely' unstable

physically, the interesting thing about a contact interaction in a spin-1/2 gas is that it only allows electrons with opposite spins to interact

Mad Spaces

@ACuriousMind Well.. i like to think i have the mind of a mathematican. i am occasionally in disagreement and arguments with my physics professor due to this mental state i posess. i belive however firmly, that the universe expresses itself in rigorous mathematically manner, and we should adjust accordingly.

Semiclassical

3:47 PM

@MadSpaces QFT would like a word with you

Mad Spaces

@Semiclassical if a rigorous proof has not been provided, it means not it does not exist!

Semiclassical

not all of physics has been successfully made mathematically rigorous yet, is my point

but SR and QM certainly have

so this is a bit of a distraction :P

bolbteppa

Assume b) but don't assume b)?

Mad Spaces

I agree. refer to my comment.
i think it is my goal i set for myself, to try to formalize physics in mathematical language. maybe one day i will achieve it, or maybe i will die before i do that.

Semiclassical

i guess the point is that, since not all of physics has been made mathematically rigorous, physicists are fairly comfortable with not being rigorous about everything

Mad Spaces

3:49 PM

And i reallyyyyyy believe in very very rigorus mathematical proofs for physics.. not only on a superficial level.. but also on a very deeper level.

@semi

i understand, i have to fight with this on a daily basis. i can not change my mindset

Semiclassical

to be romantic about it, i'll quote e.e. cummings:

since feeling is first
who pays any attention
to the syntax of things
will never wholly kiss you;

bolbteppa

Rigor is impossible to achieve in physicss, it's basically just a crutch to try to understand the basic claims

Semiclassical

that's how physicists feel about math sometimes i think

Mad Spaces

@bolbteppa if we assume c to be finite but not infinite and not make the assumption that c is the same in all IS we still reach the assumption

Semiclassical

so allow for a finite but frame-dependent speed of light?

Mad Spaces

3:50 PM

@bolbteppa I do not think your statement is correct. "its impossible to achieve..."

bolbteppa

That's exactly what is being done...

Mad Spaces

but the second postulate states the invariance of speed of light in all IS.

Semiclassical

i'm not sure "impossible" is the right word, but it has not yet been achieved

Mad Spaces

Obviously.. maybe one day we can have a book.. called " the language of god " i do not believe in god.. but i think its a great name? we start with the first page " axioms of physics" .. and then proof all laws and relationships in Lemmas.. mathematical senteances.. and such on.. like an Analysis Script :))

Semiclassical

so long as QFT remains mathematically 'in progress', i'd say the case isn't settled as far as whether the dream of rigorous physics can be achieved

though, even if it was, there'd remain the question of whether that rigor is 'useful' for working physicists

newtonian mechanics is perfectly rigorous, but an engineer doesn't use mathematical proof to determine whether a bridge will stay up

Mad Spaces

3:54 PM

It does not seem plausible to me that with disregard of mathematical logic some interaction can be explained, since if it did, then "any" interaction could exist.
Similiar to the argument that for an element in empty set, all properities apply :)

Semiclassical

i do agree that physics does involve reasoning and argumentation. whether those arguments are at a mathematician's level of rigor is a different question

but it's certainly not "anything goes"

Mad Spaces

Let this be then my hypothesis before i die and not write it written here 20.10.2021 17:55 by mad spaces on this forum from Germany:

The mad spaces hypothisis: Every Process of physics must be described throught mathematical rigorous statements.

Prove it and you get a thank you card from Mad spaces himself :)

Semiclassical

i'm also not paying attention to what the actual question was :P

Mad Spaces

I dont think it matters, it is jsut me being stupid and not understanding what they are trying to explain to me haha

ACuriousMind

@MadSpaces I sympathize - I like rigor, too. But what you need to come to terms with is that most existing literature is not written by people that prefer strict rigor, that this approach seems to work well for most people most of the time and so that if you want to understand physics, you'll have to be able to read these non-rigorous texts without getting stopped in your track by every little thing that doesn't conform to your standards of argumentation

Semiclassical

3:57 PM

@MadSpaces then you apply that principle to the first Millenium problem and get yer million bucks

Mad Spaces

@ACuriousMind i understand, i am however a narcisit and look down on these simpletons

Semiclassical

aka showing Yang Mills has a mass gap, which physicists believe but for which there's no rigorous proof

Mad Spaces

@Semiclassical Money is not relevant to me, we will all die someday rich or not. I think the only thing worthwhile is a legacy. Refer to Gilgamesh crying on the walls of Uruk and reliazing that the walls is his legacy and not eternal life )

ACuriousMind

@Semiclassical the problem there isn't so much the gap itself but first setting up the rigorous QFT formalism you could prove that in

Semiclassical

@ACuriousMind yeah, true

ACuriousMind

3:59 PM

@MadSpaces well, let's see how that works out for you, then :P

Mad Spaces

But we can argue legacy is meaningless, since it is only in human minds, which will perish. So we get into nihilisim. i prefer not to depress myself with these thoughts anymore

Semiclassical

the Yang-Mills problem is really a stalking horse for making physicist QFT rigorous

Mad Spaces

@ACuriousMind You sound like my father.. who is an engineer.. i do not like engineers

Semiclassical

currently, i feel like the state of mathematical rigor in QFT is: QFT can be rigorous or it can be useful, but it can't be both

that's maybe a bit overstated but not by much

Balarka Sen

looks like something is happening here

Semiclassical

4:02 PM

kvetching about the state of QFT at the moment

aka the unreasonable effectiveness of unreasonable formalism :P

Mad Spaces

Once more thanks to Bolbteppa and Acuriousminds for helping me with SRT! i will now proceed on my endovours to understands Einsteins mind.. which is proving to be quite freaking difficult :((

Balarka Sen

physicists know things mathematicians dont. the decades of math that grew out of trying to make physicists right which independently proves different problems in math proves them right

the formalism lags 50 years behind the actual guesses/observations physicists make

no question about it

Mad Spaces

These forums have turned out to be quite helpful so far in my studies.. i really appreciate being able to come here and ask questions. i learned alot from here..glad i looked it up :D )

bolbteppa

5:01 PM

@MadSpaces Re-read the first two pages carefully looking for what they say on a) and b) and how they introduce them and when they combine them. I would also advise re-reading the Galilean discussion in the earlier book for comparison, and maybe even try to read the first two sections where they introduce GR from first principles to see where the SR arguments limits are

My point about rigor was more: look how hard this simple SR argument is, rigor is not going to help with understanding it, instead one is more likely to one get distracted by how to set up $ds^2$ 'properly' or talking about quadratic forms or something. There is a simple logic to the argument that's hard enough on it's own, it might take ages to appreciate (it often does) but it's worth sticking with the basic logic instead of thinking one just needs to carefully define the math and it'll follow

@Semiclassical well currently people can't even agree on how qm is supposed to be interpreted :p

fqq

@MadSpaces science.org/doi/10.1126/science.177.4047.393

glS

7:30 PM

@bolbteppa the "hardness" of an argument is really mostly a function of your current understanding of it and the underlying context. What might seem like an incredibly complex argument at first, might be a completely straightforward statement to someone that studied deeply the underlying context and thus understand why things are stated the - possibly mathematically involved- way they are

the way I see it, most of the time, doing things "rigorously" amounts to understanding the precise level of abstraction that makes the statement works (and having a good understanding of the mathematical objects involved). That's not useless at all, because it means to understand which parts of the statements you are making are useful, and which ones are actually "formal baggage"

@bolbteppa that's the nature of what an "interpretation" is though. If there was an objective reason to see a theory one way or another, that wouldn't be an "interpretation" anymore

Semiclassical

8:03 PM

@glS it's a bit of semantics: you'd be interpreting quantum phenomena as "really" some other kind of modified theory.

but i do tend to stick with "interpretation" for cases which don't change the empirical content of QM, but do reify certain aspects of the formalism

e.g., many-worlds reifies wavefunction branching as different universes

while pilot wave reifies the velocity field you get from $\vec{j}/\rho$, i.e., the probability current density normalized by the probability density

whereas stuff like GRW i'd classify as modifications to QM, even if the modifications are "stuff we can't see yet"

(that's a little bit of a fuzzy line, though. with pilot wave you could say that, in any everyday problem, you'll be working in a state of quantum equilibrium and therefore orthodox QM applies. but you could still hold out the thought that, in "some" scenarios (early universe I think?) there'd be opportunities for departures from ordinary QM. so does that count as a modification or not?)

bolbteppa

8:26 PM

It's interesting that Everett didn't use the term many-worlds he talked about 'relative states' and that came over a decade later by DeWitt who was initially skeptical of it

Semiclassical

11:11 PM

@bolbteppa the geneaology of interpretations is interesting to me

for instance, i wouldn't say that pilot wave stuff is what Schrodinger would've understood "wave mechanics", but you can certainly situate within that tradition. by contrast it gives short shrift to matrix mechanics

whereas something like consistent histories, for instance, is matrix mechanics front and center

obviously there's mathematical equivalence between wave and matrix mechanics, but in terms of how interpretations arise there remains space between them

@bolbteppa i'd probably find that language easier to digest tbh. "many-worlds" has always made me gag, in much the same way that "absolute reference frame" does in pilot wave theory

bolbteppa

11:38 PM

Yeah, his short 9 page 'relative state' paper is definitely worth reading/thinking-about, he begins using quantizing gr to motivate it and at least in my reading roughly is just trying to argue for normal qm but applied to everything and that 'collapse' doesn't apply, which I think is right in places but also has real issues but I'm just not 100% sure on his stuff yet

Semiclassical

the funny bit for me is that i do remain sympathetic to pilot wave, but mostly b/c i think people give the wrong objections

if all we had to worry about was non-relativistic QM, i'd have a lot easier time defending it

but that's not exactly an option :P

bolbteppa

I'll definitely go over those Bohm papers again in the future and give it another chance, I'd still say the relativity issue is fixable (of course it has been claimed to be but...)

Semiclassical

i tend to follow Bell's formulation of it more

Bell vs Bohm is interesting in that, if you follow it through, it's the same formalism

but the starting points and emphasis are different

Bohm's starting point to my brain is doing stuff to the Schrodinger equation to make the emergence of classical Newtonian mechanics easier to see

bolbteppa

I was quoting one of his papers in a recent answer where he quotes sections of L&L (which he co-translated) to make his objections as part of this stuff yeah

Semiclassical

hence the whole business of writing $\Psi=Re^{iS}$ and going to real/imaginary parts etc

whereas Bell's story is to show that the mathematical structure of the Schrodinger equation always contains a trajectory story

the problem, of course, is that just because the math gives you trajectories, doesn't mean you must (or should) reify them further

insisting on those trajectories being real does make certain features intuitive, but it does come with a price

for instance, it's hard to avoid the implication that the story, applied to ye olde Bell test, only makes sense if there's some absolute meaning to "Alice measures her particle before Bob measures his"

which, given what we know about relativity of simultaneity, is extremely hard to swallow

the incompatibility of pilot wave theory (as i know it, at least) with relativity of simultaneity is...uncomfortable

especially because the QM formalism, by contrast, does not care a whit about relativity of simultaneity

which makes it hard to avoid the implication that we've taken the Schrodinger equation too seriously somewhere

bolbteppa

11:54 PM

The wave function is defined on all space, one can obviously choose to artificially restrict it to a path/trajectory, so trajectories are not a surprise - that you can use the equation to isolate a trajectory in the space by associating a 'velocity' to it from $v = j/\rho = (\rho v)/\rho = v$ and call that a particle velocity is just an unbelievable leap but you can at least see it isolates a specific notion of velocity

Semiclassical

i mean, that's how one defines velocity fields in the rest of physics: you find a current and divide by the associated density

so i don't think it's wholly unnatural

and in 1D there is a rather cute meaning to the streamlines of that vector field. Suppose you pick points $a,b$ and compute (for a given wavefunction) the probability to find the particle in that interval

now allow those two points to move according to the velocity field. no matter how long you wait, the probability to find the particle between the new points a',b' will be the same as it was originally

The analogy i like to use for this business stems from intro physics labs, though

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