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1:20 AM
man we got two bangers today
but i'll just put the second one here
does light actually slow down in water??
 
im currently watching NDT talk in Newark nj about various things. It's pretty cool but learning about the artemis missions makes me skeptical about the apollo ones
I dont know the whole story but why are we taking 3 years to land with crew on the moon? How many attempts did it take in the 60's lol
I feel like our tech has advanced so much im surprised we aren't doing it faster
Why didn't we go back sooner anyway
hes in his socks lol
@SirCumference i bet u know!
The telescope that's going to take videos of the universe with AI seems really cool
Lol he's presenting and roasting inaccurate articles, with the writer name and everything
 
1:52 AM
Besides CMBR, what are the oldest and farthest things we can ever encounter in our solar system?
Sometime we have foreign asteroids visit us, but if I had to guess they werent formed very far away since the universe isnt old enough to have something travel from one end of our galaxy to the other (besides light and stuff)
 
2:05 AM
@RyderRude Thanks. It's a good video. IIRC, Martin Gardner explained the coin paradox in one of his columns. I remember spending a few hours one day when I was a little kid rolling coins around each other. :) FWIW, I answered a question about solar vs sidereal time a few hours ago. physics.stackexchange.com/a/790695/123208
Here's another circle(ish) paradox: on average, the closest planet is Mercury. astronomy.stackexchange.com/a/47345/16685
@Obliv Actually, the stuff in our galaxy is fairly well mixed. See astronomy.stackexchange.com/q/16311/16685
 
2:19 AM
When the andromeda galaxy collides with ours we'll be even more fairly mixed :P
As with light eventually being unable to reach us from certain locations due to expansion, i wonder what the radius of debris of varying size and speed is
 
@Obliv What has tech improvements to do with how long it takes to get to the Moon from Earth? The fact of the matter is that the energy and momentum needed to get from Earth to Moon is in roughly the same ballpark regardless of tech improvements, and the way chemical propellants work has not fundamentally changed.
 
Even without inflation i wonder how long it'd be before something billions of light years away visited us.
(not travelling at c obviously)
@naturallyInconsistent Im moreso speculating on why would we send a crewless ship to orbit the moon first, then a crew to orbit the moon, then the crew finally landing in 3 different annual missions
 
Consider a two subsystem system $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$
 
@SillyGoose oh no
 
When we say rotate the system $\mathcal{H}$, do we mean rotate each subsytem by the same number of radians?
LOL
 
2:22 AM
It'd be cheaper to just send a crew? Im not saying id be willing to take the risk but it shows a lack of confidence imo and considering we "landed" 60 years ago it seems sketchy.
 
@Obliv oh, such decisions are done by committee. Maybe you can even find the progress reports detailing why they chose to do it this way.
 
I am thinking that a physical transformation of space should not change how subsystems are entangled with each other, so by rotating the composite system, we should mean to rotate each individual subsystem accordingly
but im not sure
 
This generation is soft. just shoot them via trebuchet like in the old days is what I say
 
@SillyGoose You keep thinking you can compose different subsystems. You kinda have to be careful about that. For example, pretending that an electron on atom A is different than the electron on atom B is technically nonsense because in QFT it is just one electron field. (Nonetheless, the approximation is physically sensible so it mostly works great).
But more importantly, you need to state what your subsystems are. For example, if you say spin-half of electron 1 v.s. spin-half of electron 2, then yes, when you rotate, you rotate both.
 
Hm okay let me state more precisely
 
2:27 AM
But if you say the subsystem is "spin-half of electron, iso-spin of nucleus", then the iso-spin space is independent of the spatial rotation space, so it will not be rotated
 
@Obliv The budget constraints were different 60 years ago, in the middle of the Cold War. Also, relative costs have changed. Computing power is a lot cheaper, but the cost of wages for skilled trades people & scientists & engineers is much higher.
 
Let $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$ be the state space for an arbitrary two subsystem system. Is a physical rotation implemented on this composite system by a unitary operator of the local form $U_1 \otimes U_2$
where $U_1: \mathcal{H}_1 \rightarrow \mathcal{H}_1$ and $U_2: \mathcal{H}_2 \rightarrow \mathcal{H}_2$
 
Roy Kerr is 89, but he's just posted a pre-print on ResearchGate:
4
Q: Why does Roy Kerr claim that the Kerr black hole does not contain a singularity?

noir1993In a preprint posted on a popular forum, Roy Kerr claims that there is a widespread misunderstanding related to the singularity inside the black hole that bears his name. Can anyone explain his argument in "simple terms"? Are there any pitfalls in his reasoning? Here's a link to the preprint:...

 
In particular is not implemented by a general operator of the form $U: \mathcal{H}_1 \otimes \mathcal{H}_2 \rightarrow \mathcal{H}_1 \otimes \mathcal{H}_2$
 
Now I think you are correct, as long as you remember that, say, the spatial rotation operator on iso-spin, for example, is the identity operator.
 
2:30 AM
Right right
okay
excellent :D ty
 
I think you can derive this in QFT language; it would just be a load of commutators between U and the field operators.
Like, this is another part of why the QFT mathematical language is vastly superior to QM 1st quantised language.
 
@PM2Ring that's surprising, considering science precedes defense in everyway.. a moon exploration could have progressed our science incredibly far relative to the cost it would have took. NDT mentioned the percentage of the budget is like .04 for nasa and space exploration
for india its like .03 and they have a much smaller budget obviously and yet they were able to visit the moon
What even was the point of the space race?
Did we believe there were secret moon guns over there?
if we were interested in general science then, it makes no sense to me why it'd take so long to go back...
we generally make things more efficient and better overtime..
 
2:48 AM
@Obliv No. However, if you have the technology to put something in orbit, you have the capability of delivering a warhead to anywhere on Earth. The USA were uncomfortable that the USSR beat them to the 1st object in orbit (Sputnik), and the 1st man in space. So they had to do something to clearly demonstrate superiority.
 
@naturallyInconsistent Hm, what do you mean by this in this particular instance?
Separately, does anyone know of a nice visual of quantum entanglement? Like a picture you would use to explain the concept to someone without misleading them.
I find the concept a little difficult to explain without invoking the bra-ket formalism of QM
 
@Obliv The surface of the Moon is mostly pretty boring. And it's expensive to go there, especially if you want to do a round trip. A one-way trip to Mars is cheaper, and the surface is more interesting to explore.
Besides, we've been doing useful science with the Moon ever since Apollo 11, with lunar laser ranging. en.wikipedia.org/wiki/Laser_Ranging_Retroreflector That data not only helps us model the Moon's interior, it also improves the accuracy of our model of the Earth & its rotation.
 
3:43 AM
@SillyGoose The way we deal with spacetime symmetries in QFT forces all the fields to handle the symmetries in the exact same manner (via their respective representations, which can be different), such that there is no possibility for extra entanglement that you were worried about. And it all comes out as commutators, no need for extra shenanigans. I think it is vastly superior to the QM notation, where it could possibly entertain some extra entanglements.
 
3:59 AM
@LeakyNun Yeah, had a really difficult time with this too, way back when I was learning this. Actually, she should have coded the transient solution in the Fourier decomposition form too, so that then she could Fourier recompose the pulse case. She is also wrong on the conception of what a plane wave is, but that kind of pedantry is not helpful for how beautiful her videos and 3B1B's videos are on the topic
@Relativisticcucumber Do note that there are so many different ways to get it. ACM is doing one integral and one IbP to extract the behaviour. I'm actually advocating for a double integral. In particular, I am saying you could have chosen to do $$\int\mathrm dxe^{-ipx}(\partial_t^2-\partial_x^2)\int\mathrm dp^\prime\phi(p^\prime,t)e^{ip^\prime x}$$
In the above double integral, first being the differential operator into the integral on the right, exchanging the x derivatives for p primes, and then perform the x integral to obtain a Dirac delta distribution, which then annihilates the momentum integral, leaving you with p and no more p primes.
 
4:32 AM
@PM2Ring I appreciate your replies. This is very insightful. I just assumed that because the moon was so close (relative to mars) and was a part of earth at the time of creation that we'd want to explore its structure a bit more
the fact that it's so boring is kind of interesting, since there is no erosion to worry about all the history is basically preserved right there.
And impact craters could have foreign objects and stuff mixed in
 
@naturallyInconsistent Ah i see. Yes, what originally made me realize that I never formally answered the question I asked to you in my head is one of my thesis readers essentially asked the question "Is this structure that is fundamental to your thesis invariant under physical transformations?", which is a question at least extremely emphasized to me by another prof in the context of learning about QFT (from the Weinberg end of the spectrum) but not QM per se
i cannot wait until next semester...all courses in quantum theory or topology :D and continuing research in my thesis topic; or I might drop one of the quantum courses to also do a tiny bit of math research :D
 
is qmechanic still active on the site?
i remember he seemed like a robot almost, like he'd edit my questions seconds after I posted it. and he'd do that for so many questions every day
 
4:56 AM
@Obliv Yes, we do want to explore its structure more. But we brought back a lot of Moon rocks. To get more in-depth info would require visiting a lot more sites, drilling deep into the surface, and doing seismological stuff (like setting off explosions and looking at the resulting waves). That's not cheap. And not easy to do with robots on one-way missions.
Once you're in space, with enough velocity to leave near Earth orbit, it doesn't take a huge amount of fuel to get to Mars, compared to the Moon. And with Mars, you can do aero-braking.
Also, because of the lack of erosion, Moon dust is horribly abrasive. The particles are very jagged. And the fine dust gets into everything.
 
How soon do you think we'd have people on mars? And would it be a one way trip?
 
Mind you, Mars dust isn't great either: it's chemically toxic. But at least the particles have been eroded by wind action and ancient water action. And the annual freeze-thaw cycle. IIRC, around 20-25% of the atmosphere freezes in the Southern winter.
I don't think you'd get many sane people volunteering for a one-way trip. ;)
 
@PM2Ring It's so cool that we have pages on lunar soil with speculation, analysis etc from the lunar missions
beyond cool
Oh sorry I meant to say it's a round trip!
LOL
It's mind-boggling to think mars had an atmosphere and deep oceans a few billions years ago or earlier
like what happened? I don't think it had plate tectonics and perhaps its core/mag field wasn't powerful enough and it just died out
Mars has a winter? Do all planets have "seasons" based on tilt like earth?
 
@Obliv I have some info on the water of Mars: astronomy.stackexchange.com/a/32485/16685
@Obliv Well, the tilt of the axis of Mars is currently fairly similar to Earth's. So that gives it seasons. But its orbit is much more eccentric than ours. And that makes the southern hemisphere winter much more severe than the northern winter.
17
Q: On Mars, why are the seasons "strongly amplified" in the southern hemisphere and masked in the northern hemisphere?

d_eIn the Darian calendar entry on Wikipedia we read (emphasis mine): The Martian year is treated as beginning near the equinox marking spring in the northern hemisphere of the planet. Mars currently has an axial inclination similar to that of the Earth, so the Martian seasons are perceptible, thou...

 
5:13 AM
This might sound ridiculous but would its orbit have changed much over the billions of years?
I'm curious if its eccentricity remained pretty much constant perhaps contributed to the lack of development of life
although there are probably a bajillion reasons why life never developed on mars
it seems like earth just has the perfect conditions..
Elon Musk suggests we'd want to melt that dry ice and artificially recreate its atmosphere.. global warming would be a good thing in the case of mars since it's further away from the sun
the eccentricity thing would need to change too.. How about we move it closer to earth
perhaps via a very large magnet
 
@Obliv It's got a much smaller mass, so it didn't develop as much heat during formation, and it cooled down faster. So its interior is more solid than Earth's, and there's less heat to drive tectonic processes. As far as we know, Mars has virtually no tectonic activity.
 
At least that means more records. IIRC the ocean floor can only be 180MA old before the lithosphere subducts
continental lithosphere preserves very little since there is a lot of activity on earth
zircons are basically the best way of determining age for rocks I think
 
Zircons are good. But there are also a few types of dating using isotopes with long half-life.
It takes insane amounts of energy to change the orbit of a planet. And anything we try to do with Mars, Jupiter will just undo. ;)
 
I believe that's what zircons are used for.. I think they have a high blocking temperature so whatever isotopes are in there we can tell how much has decayed via the preserved daughters
Jupiter is kool. during the NDT talk he showed a video of the "giant fireball" on jupiter that some amateur astronomer captured. It was due to an asteroid that was roughly 500m in diameter
it was actually quite massive.. no wonder the dinosaurs went extinct.
And the layer of heavy metals preserved in the line of clay 60 Ma suggests such an impact killed them
or at least contributed greatly to global warming & reducing food sources
 
@Obliv True. But there are other things, like potassium-40.
 
5:32 AM
oh i meant to say the minerals in that 60 million annums line have certain structure that we could reproduce through atomic bomb testing or something but it is pretty much uniform throughout the planet
basically suggesting some violent explosion
even if we don't have a crater we can basically find this stuff everywhere
 
Another quasi-moon was identified earlier this year: en.wikipedia.org/wiki/2023_FW13 It's only 10-20 m in diameter, though.
 
In archean rocks, we have conditions in that period that were very hot and so it's difficult to find preserved stuff so zirchons are useful in that way
 
@Obliv Yeah. That's how it was first found: it's an anomalous layer of iridium, a platinum group metal
 
@SillyGoose that is a confusing statement to read. I think this is just a bad disconnect between physics education; One thing I really like about Ballentine and Sakurai is that both authors decided to treat spacetime transformations in their introductions to QM, which then brings QFT a little closer. It is not that you cannot do it in QM, it is that it is not so urgent in QM, whereas it becomes really important in QFT.
 
@PM2Ring That's amazing
My sister gushes anytime she sees hawaii in anything so I'll be sure to send that to her.
she's worked and visited haleakala. Also was mentioned in the NDT talk. It has great conditions for astronomical observations
Since you're from Au @Pm2Ring did you know Australia has some of the oldest dated rocks on the planet? I have no idea why I think the beaches just have super old quartz
I couldn't find a good map but this basically conveys the continental cores
So in order to date said rocks, we'd need minerals with very high blocking temps (like zircons) so that the insides aren't compromised
K-40 is found as well as thorium and uranium etc in them
sorry just had to info dump stuff i learned in my intro geology class lol
 
5:47 AM
@Obliv Of course! And almost all pink diamonds come from Oz, albeit from the other side of the continent. ;)
 
I didn't even know pink diamonds were a thing.
Gonna have to talk to the rock guy about them
 
They almost all come from the Argyle mine. You can tell they aren't cheap, just from the vibe of the website. argylepinkdiamonds.com.au
 
 
2 hours later…
8:22 AM
@SillyGoose Impossible to answer in general, since not even all physical systems carry a representation of the rotation group (consider a particle confined in 1 dimension). This is not an academic idea, it really depends on what the decomposition of your Hilbert space represents: Formally it's perfectly valid to say that a particle in 3 dimensions is composed of the subsystems "particle in the x dimension", "particle in the y dimension" and "particle in the z dimension":
$L^2(\mathbb{R}^3) = L^2(\mathbb{R})\otimes L^2(\mathbb{R})\otimes L^2(\mathbb{R})$
since rotation mixes the x,y and z coordinates, the representation on $L^2(\mathbb{R}^3)$ does not come from individual representations on the $L^2(\mathbb{R})$
 
9:12 AM
@PM2Ring very counter intuitive
 
9:33 AM
also, in cases where both H1 and H2 carry a non trivial rep of the rotation group, the rep on H1xH2 need not be the sum of the individual generators becuz there is no SvN theorem for the rotation group, which means mathematically there cud b other reps that r unitarily inequivalent to L_1 ^ i + L_2 ^i
but physically, the generators on H1xH2 tends to work out to be L_1^ i + L_2^i
physically, we r looking for the Poincaire algebra which requires rotation to commute with the Hamiltonian. The Hamiltonian on the joint system is a sum of the Hamiltonians plus an interaction term. so rotation generators also tend to be sums of the individual generators
however, it doesnt work out this way for the boost generator on the joint system. it only works out for rotation and translation
becuz boost depends on the full Hamiltonian, including the interaction term, in relativistic theories
 
9:55 AM
or rather, the SvN theorem is irrelevant here. either of $L_1 ^i$, $L_2 ^i$ or $L_1 ^i+L_2 ^i$ cud work as the rep of rotation on H1xH2. it's physical arguments that pick the last one
@Monty hi. the information gain upon observing an outcome is solely dependent on the probability of that outcome. the information is defined to be $-log (p)$. en.m.wikipedia.org/wiki/Entropy_(information_theory)
it is independent of the number of possibilities
in case of coutably infinite discrete, the information gains r finite but their expectation happens to diverge sometimes
in case of continuum, both the integrand and the integral diverge
becuz $log (pdx)= -\infty$
so it has to do with infinitesimal probabilities in continuum
 
10:11 AM
@ACuriousMind Hm oh mayn. Well I would like to restrict to $n$-qubit systems, that is Hilbert space that is isomorphic to some tensor product of 2D complex vector spaces.
 
@SillyGoose Well, in that case, if you have a group that acts on the space of a single qubit, the only consistent thing is indeed that the action on the n-qubit space is the usual tensor product representation
 
Right but I guess the alternative is that perhaps there is some notion of "rotating the composite system" that is different from rotating each individual subsystem
 
that would be inconsistent
if you have a notion of transformations of subsystems, the transformation of the full system has to always be the transformation of all the subsystems individually
 
(that's not some rule of quantum mechanics, to me that's just part of the consistent definition of what a transformation in physics is)
 
10:16 AM
okay I see
 
this consistency condition rears its head in various places: If you have more than one way to think about a transformation, the result of all the ways needs to be the same
 
@ACuriousMind hm wait so would you call this consistency condition physically motivated
 
I would call it logically motivated :P
if you have a thing of n parts and you have a notion of how to apply a transformation to each part individually, in what sense would the thing consist of the n parts if it weren't the case that applying the transformation to the thing means applying it to all the parts?
 
note that at the very least u cud take two non interacting electrons, call it a system and then rotate it. for the predictions to b correct, the result wud hav to agree with the individual rotations @SillyGoose
if it didnt agree, u cudnt consider them as isolated electrons. u wud hav to consider the whole universe while doing any experiment
 
@RyderRude oh, cud it?
 
10:21 AM
whether the systems are interacting or not is completely irrelevant; this is not about dynamics
 
@ACuriousMind i do not trust in my ability to use logic to answer this with certainty :P
 
this is just to motivate it @ACuriousMind
yes. i think we r not even considering the time evolution
u can talk about predictions simply after doing a rotation
 
@ACuriousMind so you see, e.g., the consistency condition of transformation of a field (via unitary conjugation or finite dimensional non-unitary poincare) as a rearing of this head?
 
u r right
 
@SillyGoose yes, that's another instance of it!
 
10:23 AM
hm okay. well this is nice to be stated explicitly. i have not encountered such an explicit statement elsewhere.
 
or e.g. the derivation of the representation of momentum/angular momentum as in this answer of mine - you know how the vector $\vec r$ rotates, so the result of the rotation of $\lvert \vec r\rangle$ must be $\lvert R\vec r\rangle$ no matter how you try to represent that rotation
 
we should note that boosts do not follow this rule. the interacting Hamiltonian matters for the boost generator on the combined system
this is because boosts r dependent on dynamics
when u boost, u r implicitly also mapping future evolution and past evolution to the present, because of relativity of simultaneity
 
anything involving time translation simply fails the hypothesis: "If you have a notion of how to apply a transformation to each part individually" is false for subsystems that interact
 
yes. for the trivial example, this fails for the Hamiltonian itself
 
@ACuriousMind it might have been right around this time last year that I was reluctant to accept this sort of relationship :P
or maybe that was during the recent summer...
 
11:28 AM
@RyderRude gotcha
 
12:02 PM
do u think physics is supposed to solve consciousness? why or why not?
 
 
1 hour later…
1:12 PM
Didn't you ask this yesterday? :P
18 hours ago, by Ryder Rude
do u think physics is supposed to describe consciousness someday? why or why not?
 
QED Mr. Feynman :P
∞ many hours later...
 
1:32 PM
yeah...
 
what is the role of the $\chi$ here ?
 
1:45 PM
@Monty it is to make the argument of log dimensionless, as $p(x)$ is a probability density which is dimensionful
 
2:06 PM
@Monty In your text as written, it's something that controls the spacing of the points $x_m$, i.e. it's the "density of events" on which the probability density $p(x)$ lives. Think of it as a parameter of this continuum limit: There's a difference if you start out with evenly spaced points $x_m$ or if they're all bunched up around one end of the intervals so one of the bins is much larger than the rest.
Often such details about continuum limits drop out when you perform the limit, but not in this case.
 
yes that is what the link Ryder sent is talking about
( them dimensionless bit I didnt follow though )
the*
 
i mean it makes the entropy co ordinate invariant
tho the point density thing is what it literally is. it just also happens to make entropy co ordinate invariant (which was a problem with Shannon's continuum entropy)
 
@Monty The point is that functions like the logarithm or the exponential need dimensionless arguments. Otherwise the result has no definite dimension, since they're in terms of power series and adding $x$ to $x^2$ only makes sense if $x$ is dimensionless. But a probability density isn't dimensionless, it has units of "per length" (or "per volume" or whatever it's a density over).
so purely from that standpoint you need something with units of length - your $\chi(x_m)$ so that the product $p(x)\chi(x)$ is dimensionless
 
note that the distribution need not be over physical space. so the dimension argument is weird. the co ordinate invariance argument is the correct one
cuz one can just say that the distribution is over a mathematical space rather than involving units
the dimension consistency requirement in physics is basically also co ordinate invariance under scaling
 
in the end this kind of problem (and its solutions which have to argue strangely about dimensions or coordinates) is the price we pay for not having set up a proper "theory of densities" in the mathematical sense; most people encounter this for the first time with Wien's law where if you try to treat the spectral density as an ordinary function and try to determine the density in terms of wavelengths from the density in terms of frequencies, you get nonsense
but as I also said, in this case the function $\chi$ is not some kind of stop-gap solution to "fix the entropy", it's a genuine physical parameter of the continuum limit
(you may decide in many cases to not worry about it too much, but it has a direct physical meaning)
 
2:34 PM
no I do want to worry about it
this isnt a question of quantization is it
 
2:45 PM
i think we can use differential geometry here. p(x) is a scalar density. so p(x)dx is co ordinate invariant. -log(p(x)) isnt
so somethig is required there to multiply with p(x). and this thing has the interpretation of being the point density function in the discretization scheme
 
@Monty No, this is a classical continuum limit, nothing to do with quantum mechanics at this stage
 
@ACuriousMind I meant quantisation in the sense of approximating a continuous probability measure by a discrete one
 
@Monty wikipedia has clarified this. basically, if u had a discrete distribution with a bunch of points in a region with point density approximately $m(x)$, then the continuum entropy formula approx gives u the Shannon entropy of that discrete distribution
 
@Monty I wouldn't call that quantization, that's what's called a discretization (or "putting the theory on a lattice")
 
people call it quantization apparently.
 
2:55 PM
wikipedia calls it quantization, but it's wrong terminology
 
ok noted
 
yes, this terminology is popular
 
and yes, this is very much related to the limit from the discrete to the continuum: The discrete probabilities are just probabilities $p(x)$, but in the continuous case, the probability is $p(x)\mathrm{d}x$ with $p(x)$ a density, so the kind of function with which the two theories deal is different
 
but what exactly is $S$ in the line above 3.13
in that picture I sent
its the entropy of ... what distribution?
I think* I get where the $\chi$ is appearing, from the link Ryder sent. My thought is : if we want to approximate a density with a discrete distribution we need to pick a set of $N$ points $\{x_i\}_{i=1}^N$ such that as $N\to \infty$ these points densely fill the space
and this is a choice, so could choose them such that in the limit these points have a density $\chi$
 
@Monty it's the entropy of the continuum limit of the specific distribution defined by the $\chi$ and the $p(x_m)$
as I said, often in physics we're lucky and such choices drop out of all physical quantities in the end, but you often have such parameters for such limits in intermediate stages
in this case we're unlucky and have to admit that the entropy remains dependent on the $\chi$ even after taking the limit
I would expect your text to discuss this in more detail after this derivation
(or say that for what they're going to do it doesn't really matter :P)
 
 
1 hour later…
4:16 PM
what is the general classification of bundles on a physics manifold
cuz we can have non tensorial quantities like densities and Christoffel symbols
is there a more general thing than tensors under which these can be absorbed
 
4:31 PM
@ACuriousMind This answer of yours is pure gold
 
4:42 PM
@Mr.Feynman you linked a question :P
 
why do you have to nitpick everything D:
 
because it's funny :D
 
I agree, I do that too. I guess it's the kind of thing that's funny as long as you're the one doing it :P
 
And I wrote that answer briefly after discovering Trautman's papers and having one of these "why didn't anybody tell me about this?" moments. It reflects my best understanding of the problem complex around GR, gauge theories and "diffeomorphisms", but I'm not quite happy with the balance of conceptual explanation to technical details in it
@Mr.Feynman I'm not pretending that I was saying that for anyone's amusement but my own
 
@ACuriousMind So it was recent, like a year ago
 
4:53 PM
yup
May 1, 2022 at 17:00, by ACuriousMind
yeah, I think Trautman already figured out exactly the view on GR I was always sure there had to be but never really saw anywhere...how did I never find this before?
 
As per the conceptual part: there is a couple of things still unclear to me but I'm not sure there is any way to make something so abstract less technical
 
GR as a gauge theory is a little tricky
And it is done in quite a lot of different formalisms which doesn't help
 
Meanwhile... I'm dealing with quantum phase transitions and it turns out that the partitition function can be expressed as
Where $\tau$ is imaginary time and $\beta$ is inverse temperature as usual
Why do they claim that in the $\beta\to+\infty$ limit (zero temperature) the integral becomes $\int_{-\infty}^{\infty} d\tau$?
That is an important point because that's where the mapping to a $d+1$ dimension classical system occurs (since imaginary time becomes another spatial dimension)
> At zero temperature ($\beta=\infty$), the $\tau$ dimension becomes of infinite extent in the classical picture, and we are integrating over the whole $\tau$-space $\int_0^\beta d\tau\to\int_{-\infty}^{+\infty}d\tau$
 
5:09 PM
it's a bit of a parlor trick but I think the idea is that $\beta$ here controls the size of the compact "Euclidean time" dimension $S^1$
and we customarily write the integral over an $S^1$ in terms of integrating from the 0 angle to the $2\pi$ radians
when you take $\beta \to \infty$, the topology changes from $S^1$ to $\mathbb{R}$ and we customarily write an integral over $\mathbb{R}$ as $\int_{-\infty}^\infty$
 
5:21 PM
@ACuriousMind It makes sense. Why did you think of $S^1$ though? Because of the boundary conditions of the path integral (and thus the periodicity of the $x_i(\beta)$ functions?
 
@Mr.Feynman yes
A function that's periodic with period $\beta$ is the same as a function on a circle with circumference $\beta$
 
Ok, so the reason why all this is happening is that we're basically moving to a new boundary condition of the form $x_i(-\infty)=x_i(\infty)$
I mean, my last statement is kinda vacuous but it makes this change of topology more intuitive to me
After all this is not so different from when one takes a formal infintie period limit from a Fourier series to a Fourier transform and the integration domain shifts from $[0, T]$ to $\mathbb{R}$ for the coefficients
So I could just make sense of it writing my path integral with the boundary condition $x_i(-\beta/2)=x_i(\beta/2)$
 
yep, exactly
 
Thanks
 
 
1 hour later…
6:32 PM
I need to ask because I am now getting confused on different wordings I'm seeing. By "a Lie algebra valued one-form" , call it $\omega$, in the context of a connection on a principal fibre bundle $P$, do we mean a map: $$\omega:T^*P\rightarrow \mathfrak g$$ or do we mean a differential form who's components "are Lie algebra elements", as in $$[\omega]=(g_1,...,g_{\text{dim}P})$$ with $g_i\in\mathfrak{g}$ I always thought it was the former, now I'm thinking it might be the latter
 
It's a one form so it takes values in a vector space [the tangent bundle] and outputs elements of a Lie algebra
so $\omega : TP \to \mathfrak{g}$
 
Right, ok
Ah right yeah sorry that should be $TP$ I wrote there not $T^*P$
 
The lie algebra valued part is really because you're considering the vertical bundle, which is a Lie algebra
 
Is that a sub-bundle of $TP$ in some sense?
 
And a connection is just a projection $\omega : TE \to VE$
 
6:35 PM
Is $E$ here the total space?
 
Yes, in this case it's $P$
The vertical bundle is the part of $TP$ that's "vertical" to the fiber, so it's roughly $TG$ for $G$ the Lie group at that point
since a bundle is roughly a product you have that $TP$ is kind of $TG \oplus TM$
 
I thought the definition of the vertical subspace involved it being the kernal of the projection or something
 
@Charlie I'm not sure what your notation $[\omega]$ is supposed to denote
 
That is also true yes
 
Or the kernal of the pushforward of the projection
Just the "matrix elements" of a linear vector/operator @ACuriousMind
 
6:38 PM
but at the most basic level a "$V$-valued form" is just a form $\omega = \omega_{i_1\dots i_p} \mathrm{d}x^{i_1}\wedge\dots\wedge\mathrm{d}x^{i_p}$ where $\omega_{i_1\dots i_p}\in V$ instead of in $\mathbb{R}$
this is the same as saying it's a map $\omega : TM^{\otimes p} \to V$
 
Well that's what I was thinking, because I was reading today and it reminded me of your description of the "operator valued vectors" in QFT, namely thinking of the Dirac operator as the tensor product of the endomorphisms over $\mathcal H$ and the spinor space.
I was specifically reading this line
 
and it's also the same as saying that instead of a section of $\Lambda^p T^\ast M$ it's a section of $\Lambda^p T^\ast M \otimes B_V$ where $B_V$ is some bundle with fiber $V$
(strictly speaking this last one is a bit more general and the other two formulations kind-of assume $B_V$ is the trivial bundle)
 
@ACuriousMind At this point the expansion in the $dx^i$'s is very formal now right
 
not any more formal than in any other context
 
I'm not sure I follow the equivalence after thinking about it for a minute, if we just deal with 1-forms then in what sense are $\omega_i\mathrm dx^i$ and $\omega(\vec v)$ related? if $\vec v$ is just some vector in $T_pM$
I'm not distinguishing clearly between working with a field and just working at a point
Though I don't think it matters
In terms of just figuring out the concept
Uh actually maybe that does make sense
 
6:48 PM
for any choice of coordinates $x^i$, the $\mathrm{d}x^i$ are basis of 1-forms, so for any 1-form $\omega$ you can write $\omega = \omega_i(x) \mathrm{d}x^i$ where the $\omega_i(x)$ are just $\mathbb{R}$-valued functions
saying the form is "$V$-valued" just means instead of $\mathbb{R}$-valued functions you have $V$-valued functions there
 
Ok I follow that
When I said it's very formal I just mean that $f(x)\mathrm dx$ makes sense but only if $f(x)$ is something that can be "multiplied" by a form
Like a real number
 
That is what components are
 
@Charlie we're just defining everything to make sense here
 
@Slereah Well but because the $\mathrm dx^i$ live in a vector space they come with a notion of scalar multiplication, but if you "let the components be $V$-valued" you'd need some definition of $V\cdot \mathrm dx^i$ is all I mean
 
It's the form that's vector-valued
 
6:52 PM
there's a very general concept that making something "$V$-valued" just means tensoring with $V$
 
so it is the $dx$ that is vector valued here
 
@Slereah no, in my formulation it's really the components
 
Yeah we had the discussion on the tensor product stuff when someone asked about Dirac spinors a while back
 
Ah yes
 
and that's what's happening here: You're taking a function $\omega : TM \to \mathbb{R}$ and making it into a function valued in $V$, $V\otimes \mathbb{R} = V$, so you just get a function $\omega : TM \to V$
conversely, in the section view, you have not $\omega : M\to T^\ast M$, but $\omega : M \to B_V \otimes T^\ast M$, where the tensor product is fiberwise - the fiber of $T^\ast M$ is $\mathbb{R}^n$ (the $n$ components of $\omega$), which becomes $V\otimes \mathbb{R}^n = V^n$ (the $n$ $V$-valued components of $\omega$)
we don't need to worry about having to define the product of elements of $V$ with $\mathrm{d}x^i$s or whatever, the tensor product just does it for us
 
7:06 PM
It's almost there for me, all of the words make sense but I just can't put it all together just yet
I need a return tram ticket to Heidelberg :P
I will continue to think about it periodically, I'm only looking at principal fibre bundles tangentially
 
@Charlie I owe ACM a couple of beers
Or a couple of bears
@ACuriousMind which would you prefer?
 
uhhh...beers?
definitely beers
 
7:22 PM
If you tame your bears you can get free beers
 
@Mr.Feynman why not both?
 
@Charlie If you have troubles with that to start with maybe you should look into vector space stuff before going onto bundles
Get used to using the various spaces and maps
 
The thing is I'm not really super interested in how exactly your make it super formal to the point that I would go out of my way to read at length about the topic :P
 
Well why are you even reading about connections then :p
Connections aren't the ideal topic for that
 
Because I'm watching my favourite series on geometric physics by Frederic Schuller and I was reading the wiki page about PFB's at work while some stuff was loading and I got to the bit about connections and it reminded me about this lie algebra valued form stuff lol
 
7:28 PM
@Loong where did you get that?
 
I remember the last time I touched it maybe a year ago it made virtually no sense but now I'm a bit older and wiser it makes moderate sense
I do appreciate your answers and help though
 
@user85795 from any grocery store in Finland? Just buy it before 9pm.
 
Thanks pal.
🍺🍺🍺🍺
Help yourselves everybody :-D
 
@skullpatrol Don't drink and drive!
 
7:43 PM
 
8:00 PM
@Loong good point
@Loong I need more context to this :P
 
@Mr.Feynman It's a Finnish beer and in Finland alcohol is only allowed to be sold by stores before 9 pm
 
If you try after 21:00, they will take it away from you at the checkout. Don't ask how I know.
 
@ACuriousMind Conjecture: you are a store owner :P
@ACuriousMind why do you know Finnish laws
 
8:15 PM
@Mr.Feynman There's plenty of laws like that in several European countries
 
something about Mogwais and Gremlins
 
It just needs to be said, physics is one of the most challenging things there is, my god...
 
There is a beer brand named Moretti here in Italy. It reminds me of Prof. Moretti every time I see it :P
 
8:32 PM
At this point, usually someone is about to mention the pale lager from Fugging, Austria.
 
I don't condone it (any more), but if you want to put some Austrian hair on your chest
Stroh Austria GmbH (Austrian German: [ʃtroː]) is an Austrian manufacturer of rum, especially spiced rums and high-proof rums used in warm drinks and cooking. The Stroh Rum brand is one of the best-known spirits from Austria. The name is widely used as a generic synonym for spirits with a similarly high alcohol content in Germanic speaking regions. The company is privately held by Austrian owners. == History == With roots starting in 1832, Stroh began large-scale liqueur and brandy manufacturing in Klagenfurt in 1857. The company was named after its founder, Boštijan Stroh, who started the...
 
8:49 PM
how long have you abstained for
 
Since Last Christmas, it's rare
 
cool
 

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