One time I read about how in Einstein-Cartan, the rotational part of the gauge (Einstein tensor) is related to the translational part of the matter (Stress-energy tensor) while the translational part of the gauge (torsion tensor) is related to the rotational part (spin tensor)
I guess that an isomorphism with the tangent bundle naturally induces a natural bundle structure since if you compose that with the functor from the manifold to TM then you get your appropriate one for the isomorphic bundle
although while it makes a natural bundle, idk if it makes something a tangent bundle
What if I get one of those weird weighed tangent bundle instead!!!
There must be some extra condition on it
If the transformation law is something like $$\| \frac{\partial x}{\partial x'} \|^k \frac{\partial x}{\partial x'} $$
ChatGPT can now talk to Wolfram | Alpha. It still needs more training, but it already appears to be quite usable. writings.stephenwolfram.com/2023/03/…
I mean that Fock space forms a basis of the operators in Euclidean field theory. But these states wud no longer b eigenstates of the euclidean field hamiltonian, i think
So they wudnt represent observables, cuz they wont b energy
@ACuriousMind Fair point. ChatGPT makes WA a little more human-friendly, in both its input and output, and easier to use for people who haven't learned how to use WA or studied Wolfram Language.
My initial impression is that ChatGPT gets more benefit from this "symbiosis": it can now do complex calculations and access real-world data, so it's less likely to Just Make Stuff Up. However, the process is still fairly fragile, and requires a human that understands Wolfram Language to help ChatGPT & WA talk to each other, and to verify that ChatGPT's Wolfram queries are appropriate for the task at hand.
can any of you help me in understanding the basic difference between EM waves generated through electrons acc or any charged particle acc and light waves emerged from
Hi. As we understand, electrons can't be unmoving therefore electrons always moving or flowing. If the voltage is zero, where do the electrons move? Do they move everywhere? The air molecules are still moving around everywhere, so I guess same goes to electrons? So, zero volt reading on those instruments simply mean they can't detect electron motions that are too subtle?
@JackRod They are the same thing: EM waves are generated through acceleration of charged particles. When an electron in an atom changes to a different orbital it changes momentum.
Ok this is very interesting. To achieve this constraint naturally instead of artificially imposing it, Egdan uses an Euclidean theory on a Torus topology
He says these closed topologies like sphere/Torus r the only way to make sense of Riemannian QFTs. This means the momenta form a discrete spectrum
Then he looks for the momenta in the discrete spectrum that satisfy the constraint, for a fixed mass. He says for most values of $m$ and the dimension of the manifold, no solution may exist.
I think the reason he is working with a Torus is that the imaginary omega solutions are not even allowed on the Torus. So we dont need to artifically get rid of then.
Becuz a torus wud only admit cyclic solutions. Exponential growth/decay is ruled out
Very clever way to get rid of the bad harmonic oscillators
Ok but y this same idea is not applicable for the Higgs mechanism? Before the field re-definition, the quadratic term of the Higgs lagrangian has a coefficient of the wrong sign. This wud imply constraints on 3 momentum according to $p_0^2 = p_1^2+p^2+p^3^2 - m^2$ @ACuriousMind
So we cud possibly choose a suitable topology to impose this constraint like Greg does
honestly I don't find this very interesting - our universe is not Euclidean, there is no reason to expect that theories like relativistic QFT designed to work in the Lorentzian signature yield any meaningful results at all when applied to Euclidean signatures
the "Euclidean" theory that matters is the non-relativistic limit, not the theory you get from a simple switch to Euclidean signature
Again, I view the fact that our current theories don't work well in other signatures not as a sign that Lorentzian signature is somehow special, it's just a consequence of our theories being designed to work in Lorentzian signature but not designed to work in the others
if we lived in a Euclidean universe we would have designed theories of physics that could deal with whatever happens in that universe that presumably would fail utterly when we try to apply them to Lorentzian signatures
this sort of theory crafting is fine when you're Greg Egan and you want to write your very own idiosyncratic branch of hard SciFi, but I don't think it tells us anything about the universe we live in
1. When we initially discovered quantum theory, it was non-relativistic. So this argument reveals some compatibility between quantum principles and Minkowski signature
Now the thing that I REALLY find beautiful is : when canonically quantising classical field, we hav to choose the time direction. Now, this argument shows that ONLY ONE choice is good
I mean we gotta choose the dimension with the opposite sign frm the other three
that our quantization procedures treat time special is just another manifestation of that: we live in a universe where time is special and so all our physical theories treat time as special
that they stop working when you try to imagine a universe where time isn't special - where, indeed, it is even more unclear what "time" even means than in our universe - is utterly unsurprising to me
sure but Egan wants to write a story set in a universe so he needs to set up the universe such that there's a notion of time that allows him to tell a story
Can a telescope one day be able to localize the position of a galaxy in the celestial sky so accurately that the uncertainty principle needs to be taken into account?
@ACuriousMind 1. Because we also measure their momentum 2. The telescope is the detector in this case, which we use both to determine the position and the momentum / angular momentum
@ACuriousMind Okay - I understand that more or less. But beyond the technique used for measurement, doesn't QM also say that a body can't have a combination of momentum and position precisely defined below a certain threshold? So that would seem to mean that it's true regardless of the measurement apparatus?
I understand that we may be measuring the average rotation velocity, as you say with the stopwatch example. But the uncertainty principle also affect averages, because if it affects a $(x,p)$ measurement at any time $t$ it will also affect the averaging of those values along $t$ right?
I'm not saying that the uncertainty principle is the cause for "dark matter" or something, I don't think that's relevant for the sizes we are resolving. I'm just curious if a telescope may be able to one day resolve such tiny arc-lengths (solid angles) that this does come into play
@Amit My point is that in the stop watch example I'm not making a measurement of $p$ or $\langle p\rangle$, I'm computing the difference of two position measurements and dividing by $t$ and multiply by $m$ to get a "momentum"
but this is not the same as what QM means by a measurement of $p$ or $\langle p\rangle$
Something is bugging me here I can't put my finger on it :)
I guess my question can be reformulated as follows: suppose we forget about the fact that galaxies are these massive things, we only know about QM, and, all we know is that we are measuring bodies that live on this weird celestial sphere, and we are building instruments that can resolve them to a higher and higher degree of accuracy, hence defining their position better and better. Do we ever need to start worrying about the uncertainty principle :)
I don't understand how you can make a momentum or velocity measurement that doesn't intrinsically span a finite duration of time. As the time span approaches zero, the measured velocity approaches the instantaneous velocity, but as the span gets smaller, the errors in your measurements make a larger relative contribution.
Or say you're measuring the frequency of some wave. You need a lot of cycles to get an accurate value for the period (or wavelength) of the wave.
Heisenberg Uncertainty is important for particles. For big macroscopic objects the error due to HUP is totally tiny compared to other sources of error. For celestial bodies, forget it!
@PM2Ring That's why I said, suppose we don't know that they're macroscopic, we are making measurements of light sources that travel on a spherical shell.
I mean does it make absolutely no difference for example that I can localize a certain galaxy to occupy a very specific 0.0001 arcseconds of length on this shell?
Energy & frequency measurements of particles are equivalent, via $E=hf$. To have zero uncertainty in the frequency of a periodic wave you need to observe an infinite number of cycles.
@Amit But we do know they're macroscopic. And they aren't perfect point sources, they're always fuzzy.
The statement that they're macroscopic is a statement relating our local scale to their local scale, it's not clear how that affects the process of measurement, and ok, even though they're fuzzy we make them less fuzzy with better instruments...
@ACuriousMind Well that is kind of the question though, does it play a role. I can only point out that the one data point I have that suggests it might play a role is the increasing accuracy of determining the position in the celestial sky
But thank you I'll try to give it some more thought
and there's a general principle that the equation of motion of the field is also the equation of motion of a one-particle wavefunction (see physics.stackexchange.com/q/193237/50583) if you ignore all caveats about position operators and localizability and whatnot
rel. QM is just what you get when you restrict to the 1- (or 2- or whatever) particle space and forget all the rest
QFT should be just QM with more d.o.f., and we can often get away with pretending it is, but it is not clear to me that we should think of it like that
why do we use the word approaching in third law of thermodynamics ? why cant we use the system has constant entropy rather than approaching constant entropy , "the entropy of a system approaches a constant value as the temperature approaches absolute zero."
@ACuriousMind I think of it as : we have hand-vawy ideas but it'd make decades to them them rigorous. But that doesnt mean we shudnt believe in the hand vawy idea
I think a possible interpretation of all the technical issues with constructing QFT via quantization of classical field theory is that this idea is "sort-of right" but not actually the correct way to think about it
English isn't the language I text most people with so I have to make an actual mental effort to translate the textspeak to "real" English since I'm not used to it
@RyderRude I would say the opposite: The fact that all formalizations/axiomatizations of QFT that I know very much try to avoid ever referencing what I think you mean by "finite time QFT" is indicative that this is just a useful heuristic and not how QFT "really" works
like, the Wightman axioms talk about n-point functions and fields, not time evolution operators and states, there's the "local algebras of observables" people that also construct this very differently from QM, etc.
an independently of whether or not to take the Wightman axioms as the starting point there's theorems like QFTs that have the same n-point functions at all n are, in fact, the same QFT/action
meaning the entire theory is indeed in the n-point functions and the scattering "approximation" does not, in fact, forget information
The only problem I find with this approach is that relativistic QM would then need to be its own thing. This approach doesnt allow for unification.
I mean unification of both these applications of relativistic quantum theory
I agree that everything about the scattering theory is in the n-point functions. That aspect of the theory is self-contained. We can take that as the definition of that theory
the "relativistic wavefunction" for rel. QM. in momentum is something like $\langle p\vert \phi(x)\lvert 0\rangle$, write this as $\langle 0\vert a_p \phi(x)\vert 0\rangle$, this is just a slightly weird vacuum expectation value
I don't think it's as hard as you think to get to rel. QM
you just don't necessarily can motivate that limit by talking about some sort of underlying idea about how QM works
but again: intermediate steps in computations aren't required to make sense :P
I agree, mathematically, rel QM is still contained in the correlation functions. But the interpretation is now of finite time evolution, rather than transition amplitudes over infinite times.
I don't really think the "infinity" part matters so much, the physical idea is that "infinity" is just code for "stuff is far enough apart to not interact" anyway
so I opened "Pinciples of optics" by Born and wolf, 7th edition. the chapter of optics in metals
to my astonishment, the chapter starts with a sentence claiming that good absorbers of EM waves are also good reflectors, and that it's the case of metals. I was shocked, is this true?
i know the emissivity of polished metals is around 0.08, so this makes them absorb less than 10% of incident light, for example
@ACuriousMind I have one last thing. Maxwell's classical EM theory can b derived from finite time QFT if one takes the Ehrenfest theorem. This is one thing that I'm thinking isn't contained in the scattering QFT.
I agree Rel. QM is sort-of contained in the scattering theory
@ACuriousMind Tho I also think we need a better framework than finite time QFT to unify all of these aspects of relativistic quantum physics
Perhaps Algebraic qft. We are slowly getting there.
