"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
@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.
@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.
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
@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
@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
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
> 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?
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)
@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
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)
@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
In 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...
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.
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
@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
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
In 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 :)
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.
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.
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"
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)
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?
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?
@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)
Why 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.
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.
@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
