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12:46 AM
0
Q: Are future physics discoveries properly categorized as non-standard?

wave"Non-standard" physics is often equated to fallacies and fringe theories, including on this forum. Standard physics is rigorously categorized by specific historical scientific context, subject to unscientific beliefs of the time. As physics advances as science, it corrects itself. What was "standard

 
 
5 hours later…
5:45 AM
Hi @JohnRennie
First 0-60 km/h in 282 secs
then train stops
0-60km/h in 1sec
Then train stops
0-60km/h in 14.5 secs
0-60km/h in 28 secs
Seems wrong data?
This is speedometer app in iPhone. If anyone knows some better one please tell.
 
 
2 hours later…
7:28 AM
@cOnnectOrTR12 1. Why would it be wrong data? Trains, like cars and everything else, don't have to have constant acceleration, they can choose how fast they accelerate. 2. The precision of speed-measuring apps on a phone is limited by the available precision of the GPS position measurement.
 
7:45 AM
Train with constant acceleration would just end up at relativistic speed
 
8:01 AM
@cOnnectOrTR12 Sorry to be so long getting back to you. I had loads of questions this morning that I had to deal with first. I'm free now if you still want to discuss this.
 
8:33 AM
Based on its running 0-60km in 1 sec is quite fast.
The data is not consistent
From 282 to 1 to 14
 
1 second is indeed suspiciously short, but otherwise I don't understand what the problem is supposed to be
The train can regulate its acceleration just like a car with its gas pedal, why would you expect it to have "consistent" acceleration?
 
I'm sure some very serious people designed the train so that its acceleration would balance energy consumption, travel length, passenger safety and the wear of the engine
Possibly those people :
Also I forgot to look up Haag at home
Dang
being back to the office full time sucks
 
8:56 AM
@ACuriousMind these train can’t possibly accelerate from 0to 60 in 1sec. Too fast for a train in India. Plus I don’t understand the app how it works.
 
@Slereah no home office option?
 
A little, but still most of the week at the office
 
Does anyone know of a good app
 
@cOnnectOrTR12 ...that's why I said it's "suspiciously short". The only way an app that measures speed can work is by measuring your position at different times and computing a velocity from that, that's why I said it will depend on the accuracy of your GPS
if the GPS positioning wasn't accurate and your phone suddenly recalculates your position to be hundreds of meters away from where you "were" just a second ago, you'll get a ridiculously high velocity without actually moving
@Slereah ah, that sucks
 
maybe the train's trajectory is nowhere differentiable
 
9:00 AM
we have unlimited home office at least until the middle of '22, and it doesn't seem they even want everyone back in the office after that
 
@ACuriousMind does bad network can cause wrong results. I guess gps works via net
 
@cOnnectOrTR12 well, GPS is "a network" in the general sense - the GPS satellites continuously send out data that the GPS receiver in your phone then uses to calculate its position - but it is not dependent on your internet connection, no
there's a dedicated chip & antenna in your phone that do nothing but listen to GPS signals
 
@Slereah Masklesss non-socially distanced people all touching the same table/chart from the before times :z
 
but of course you can have bad signal quality in some places with that, too - it just doesn't have anything to do with a bad internet or phone network connection
 
@ACuriousMind so can I trust my app or not
 
9:07 AM
I don't know what sort of answer you want to that
I explained how that app likely computes your speed and for what reasons it can be inaccurate
the "from 1 to 60 km/h in 1 sec" is obviously untrustworthy, the rest of your measurements are at least plausible (doesn't mean they're correct!)
 
you could ask train people
There's no lack of train nerds
 
unless you have some other way you trust more to measure speed and compare it to your app, there's no way to tell how reliable your measurements are
measurements aren't as simple as "can I trust this device or not?"
 
How do we measure acceleration of a vehicle Normally. Is there is a device?
 
@cOnnectOrTR12 ...have you ever seen that thing called "speedometer" in a car?
 
9:10 AM
> Haag’s theorem (Haag 55, Hall 57) says that for a non-finite number of generators the canonical commutation relations do not have a unique (up to isomorphism) irreducible unitary representation.

> This is in contrast to the Stone-von Neumann theorem which says that for a finite number of generators the Schrödinger representation is, up to isomorphism, the unique irrducible unitary representation of the canonical commutation relations.
Presumably they mean creation and annihilation operators, or 'position-momenta' at each point vs. one everywhere?
 
@ACuriousMind what is that gun called which measure speed
 
Radar gun
 
but note that if you want to measure acceleration, then measuring speed and time as you did is not the only way - most phones these days have "accelerometers" in them with which you can directly measure acceleration (or force, rather)
 
Don't use a phone accelerometer you lunatic
You'd have more luck with a ping pong ball in some water
 
are they so bad? I've never really used any
 
9:15 AM
I know their compass sucks certainly
Can't imagine the accelerometer being much better
Also for weird liability reasons their GPS will not work if it's too high up and too fast, so that rogue dictators don't use them to guide rockets
terrible all over
 
@bolbteppa You can express this in many forms, but the simplest is probably in terms of c/a operators: Ordinary QM has finitely many $a_i = x_i + \mathrm{i}p_i$ (or something like that, I don't care about factors here), QFT has uncountably many $a_p$ as the Fourier modes of the field, and the CCR in both cases are $[a_i,a_j] \propto \delta_{ij}$ (with the appropriate $\delta$)
the finite case has a unique irrep, the infinite doesn't
 
Yeah it looks like that's what being said, another thing that just looks absurd because you can easily make normal qm look like that :(
 
what do you mean?
 
That's pretty good
In QM you set up 2nd quantization by considering a system of identical particles, if the spectrum of a single particle is discrete $E_1, E_2,...$ where $N_1,N_2,...$ are the occupation numbers and $\psi_{N_1,N_2,...} \approx (a_1^{\dagger})^{N_1} (a_2^{\dagger})^{N_2}...|0,0,..>$ are the stationary states of the system. If the spectrum is continuous it's the same except obviously $N_i$ becomes continuous.
It just doesn't matter whether $\sum_i N_i = N$ is fixed (conserved) or variable as far a the setup is concerned, it doesn't affect the abstract commutation relations, and choosing to work with a fixed $N$ is just an example/choice there's no reason why we couldn't choose to work with an $N = \infty$ example in non-rel QM (like we do even in classical mechanics)
 
9:39 AM
@bolbteppa I mean...Haag's theorem doesn't say you can't work with that, it says the representation of the CCR is no longer uniquely fixed just by the CCR themselves
if you've built the space up as a Fock space of excitations, the representation of your observables is very likely already fixed by the representations on the n-particle spaces. The point is that in interacting QFT your space of states is no longer a Fock space
That's another way of phrasing Haag's theorem in more physical language: Theories with infinitely-many degrees of freedom are not necessarily representable by a theory on a Fock space
 
Yeah maybe that's the way to think about it
 
I mean it can be a Fock space
Fock space is just a representation
Just that the operators won't act nicely on it!
You can use any garbage Hilbert space for it but don't expect the operators to act as if it's an ensemble of little particles
The field and momentum operators probably are not gonna act like little one particle field operators on the Fock space
 
10:07 AM
It's just very confusing
 
that it is
 
7
Q: The equivalence between Heisenberg and Schroedinger pictures

user929304In quantum mechanics, the two pictures of Schroedinger and Heisenberg are taken as equivalent, where in the former wavefunctions are time variants and operators are not, and in the latter it is the other way around. I think it is important to understand equivalences in physics in general, but thi...

and how does it affect this
 
QFT math is like being given a big list of particle positions and trying to guess if it's a duck or a human
It all kind of looks the same
 
App says 1g acceleration. How is it 1g when my phone is in rest. It increases when I throw my phone side ways or upward downward
Radar gun is costly
 
@cOnnectOrTR12 1g means you're on Earth :P
 
10:22 AM
I mean he could be in a rollercoaster
are you in a rollercoaster right now
 
well, only if the rollercoaster isn't accelerating
 
could be an accelerating rollercoaster on the moon
 
@Slereah why would be I in a roller coaster
 
some people think rollercoasters are fun (they're wrong)
 
@ACuriousMind so I have a acceleration of 9.8 when I am at rest
 
10:35 AM
the accelerometer isn't measuring acceleration so much as force
and if you stand on Earth, there's always gonna be the force of gravity acting on you, equivalent to an acceleration of 1g
I don't understand the question - what do you think it could mean if the force acting on something increases?
 
@ACuriousMind meaning earth is attracting me with a force that could accelerate me with 1g if I am not supported by the ground
How is force equivalent to acceleration if I am not 1kg
 
"equivalent" is not the same as "equal"
 
So what is it
 
11:01 AM
@cOnnectOrTR12 The accelerometer knows its own mass, so it can compute acceleration via $a=F/m$. Your mass is entirely irrelevant.
 
unless you're so heavy that you have a gravitational field
 
11:34 AM
Why do people post questions asking for the One Idea of a theory
Are they hoping that there is a secret trick to understanding it
Spacetime is really a tablecloth and all follows from there
 
@Slereah yes
 
I think what they really want to hear is that physics is secretly classical and there is just a thing to add
 
11:53 AM
You see the idea that there's a single well-defined secret that suddenly unlocks deep understanding once you know it everywhere - the "one TRICK to make QUICK MONEY" schemes, the "one BOOK to become successful" ads, etc.
 
One thing that betrays that cranks have never read physics is this idea that physicists are the ones pushing for weird unintuitive ideas
Like physicists don't spend 30 years trying to show everything is normal once something weird is discovered
Nobody like quantum theory, we just have to live with it
 
If there was even a single thing actually wrong, people would have jumped all over it decades ago, and calculus is probably the biggest barrier blocking the majority of them from being able to read the things that could have helped them see the flaws by themselves, though that doesn't account for backwards arxiv
 
 
1 hour later…
1:28 PM
@ACuriousMind If we measure acceleration 1g on app then it is done by measuring the force of gravity via some sensor then a=f/m where mass of the sensor is set 1
 
Why would it be necessary to set the mass of the sensor to 1?
the device measures some force $F$, and it knows it has some mass $m$ and the result of $F/m$ is simply 1g
 
Which will be 1g when mass will be 1 kg because force of gravity on earth is 9.8 N
but the device is not that heavy! Idk
You said yourself it doesn’t measure acceleration as much as force
 
Hello! in special relativity the relationship between infinitsmal distances between two intertial system is given as
$ds^2 = a ds'^2$
is is then abruptly shown that a = 1 with no rigiorous mathematical proof.
Do you know any article or paper, pdf , book , that refrences a proof to this fact? Thanks in advance
 
4
Q: Landau Classical Fields theory argument for invariance of $ds^2$

RosarioIn Landau's classical field theory, chapter 1, I'm getting mad with an argument used to derive the $ds^2$invariance from Einstein's postulate of the invariance of the speed of light. Landau says that if $ds^2= c^2dt^2-dx^2-dy^2-dz^2=0$ then we have to have $ds'^2=c^2dt'^2-dx'^2-dy'^2-dz'^2=0$ Tha...

 
Funny enough, i am reading landau right now as we speak and the same problem and question i have is posted in that link, thanks i will take a gander in there :)
 
1:37 PM
See my answer there
 
a very lenghty answer. i will read it now :)
 
I've been meaning to make the answer a bit clearer, e.g. do the 3D Eulidean analog more carefully, but it's still there and factors in the GR discussion they give in a later chapter for comparison
The video in my answer gives a closely related proof as well, but that really threw people off getting confused about quadratic forms and arc length parameters
I think this is one of the coolest SR things so if there's some real issue let me know
 
@cOnnectOrTR12 No, the acceleration due to gravity on Earth is 9.8 N/kg (or 9.8 m/s^2). If the sensor weighs 2 kg then the force that the sensor experiences (and measures) at rest will be 9.8 * 2 = 19.6 N.
 
2:38 PM
@NiharKarve so it measures force not acceleration but displays acceleration
Acceleration is fixed at earth surface and it does not depend on force of gravity
 
I thought that particles outside the Hubble sphere can move at superluminal speeds, but this statement "Comoving bodies at the particle horizon recede with velocity c." is confusing me because the particle horizon is expected to be enclosing the Hubble sphere.
The statement is from the cosmology book by Roos.
 
3:11 PM
No no wait! I got it
 
@ManasDogra it depends on what, exactly, the word "recede" means there!
the particle horizon recedes "at c" in conformal time, which is probably not the sort of time/velocity notion you're thinking of
 
so c is sort of a conformal velocity here?
Conformal time is defined as $\int dt'/a(t')$?
 
yes - the particle horizon is located at the distance corresponding to conformal time times c, and that's very likely what your source is referring to when it talks about "comoving bodies [...] receding at c"
caveat: cosmology is not my strong suit :P
 
But then Roos equates this c to $H\times d_{p,ph}$ where $d_{p,ph}$ is the proper distance to the particle horizon,i.e., scale factor a(t)*c*conformal time(the conformal time is integrated from the time of big bang to t)
@ACuriousMind strong is relative :p
 
The acceleration entirely depends upon what mass we use for sensor. According to set mass it experiences a force and the value of acceleration always comes as a 1g
So force increases with acceleration. When we accelerate like suppose in a train then the sensor measures more force , so increased acceleration? Right?
 
3:23 PM
I guess he equates that by using the Hubble's law but the velocity in Hubble's law is as far as I know is the proper or physical velocity, so how can that be c?
 
More importantly can we measure acceleration with this type of app? Like 1g at rest. So actual acceleration =measured-1g ?
 
Infact, we put c in Hubble's law for getting Hubble sphere's radius!
 
@cOnnectOrTR12 unless your "actual" acceleration is in the same direction as gravitational acceleration, it's not so simple (and if the meter only outputs a number you're out of luck, you need vectors)
 
 
4 hours later…
7:09 PM
0
A: The fine structure constant

TEJINDER PAL SINGHWhy do physicists look for a meaning to the fine structure constant? Because even if it differs from the measured value by a few percent, the universe would be very different. Stars would either not form, or very quickly collapse into black holes, thus making life impossible. So are we living in ...

I'm not sure what to make of this one. I looked at the arxiv post and it shows all the hallmarks of a crank paper, but it did get published and in a journal that has been around for 30 years.
 
@JohnRennie ah, so he finally got it published somewhere :P
 
Of course it's absurd but that hasn't been shown so in a sense it could be 'right' and it's now legitimized by a journal...
 
7:31 PM
the paper is classic in its incomprehensibility: On a total of nine pages, it pretends to "derive a quantum theory of quantum gravity and unification" and "derive" the fine structure constant, half of its nine quotations are just papers by the same author, but it's completely unclear what it's doing - formula drop out of nowhere, trivialities are explained in detail while the actual logic of what's happening remains mysterious, etc.
 
how this got past any reviewer who knows anything about QFT or GR is a mystery to me
@RyanUnger too slow, I still saw that :P
 
that was the intention
(removed)
 
ah, not a mistake, just a bit of cowardice, then :P
 
it's the unfortunate side effect of having a chat that is publicly accessible
 
7:34 PM
you're not even that wrong - this is indeed probably how many theoretical papers look to people not familiar with the jargon, too
 
It's basically a serious attempt (i.e. not backwards arxiv level) but still on the level of everyone's favorite attempt
 
fqq
@ACuriousMind I wouldn't be sure it did. I don't know about that journal in particular, but I once reviewed a paper for another journal in the series, the first round of review was a joke I recommended rejection and never heard from them again. The paper was published.
I think their standards are much lower than they used to be
 
7:50 PM
huh
that's a shame
 
 
2 hours later…
9:29 PM
@ACuriousMind there's few things which make me nope away from a paper faster than deriving the fine structure constant
like the whole Atiyah affair
claiming to have proven the Riemann hypothesis? seemed unlikely, but hey, it's Atiyah
claims to be able to also derive the fine structure constant? oh. oh dear
 
To be fair I don't think he left the realm of complex numbers unlike this one :p
 
So, here’s a calculation I’m checking for a course I’m grading
 
 
2 hours later…
11:33 PM
@Semiclassical ?
 

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