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Charlie
Charlie
12:16 AM
oh ok, so we're not introducing a new field of operators
 
 
2 hours later…
Bohemian relativist
Bohemian relativist
2:17 AM
is quantum field theory the first place to encounter functional integrals in physics? I seem to have never encountered this before. I just wonder why the lecturer use square brackets on the integral variables which are functionals, e.g. [dA], the integral over the gauge potential - he said that's because it's a functional integral. What's special about a function integral so that [] is used? What does the brackets [] enclosing dA denote?
 
 
3 hours later…
Nihar Karve
Nihar Karve
4:51 AM
someone's gone and downvoted 10+ questions in a row on the main page
will those be reverted?
it's reasonably clear who did it
 
 
4 hours later…
undefined
undefined
9:14 AM
Hi, if I want say that 'This has a probability of x to happen'. Do I always have to give a time scale/frame too? like 'chance of x to happen in n time'?
Or other way asked: Can i express a probability without specifying a time in which the given probability can happen?
 
More Anonymous
More Anonymous
@undefined In quantum mechanics?
 
undefined
undefined
@MoreAnonymous yes and in macroscopic too
 
More Anonymous
More Anonymous
@undefined If I ask what is the probability of heads after flipping a coin
 
undefined
undefined
I tried to google it but google didn't understand what I asked and/or I asked the wrong way
 
More Anonymous
More Anonymous
its always 50% no time interval involved
As for quantum mechanics you might find this interesting
4
Q: Conditional probability between time parameter and operator in quantum mechanics?

More AnonymousQuestion and Background So I came across a question on conditional probability in quantum mechanics: How is conditional probability handled in quantum mechanics? There's an interesting comment which tells why this does not work for "the non-commutative case" I was wondering, however, since the...

quantum-mechanics quantum-information probability
 
undefined
undefined
9:18 AM
example question would be 'the probability that the pencil will fall down is 10^-2'
or do I have to say 'the probability that the pencil will fall down is 10^-2 in 365 days'? Or is it without the days meaningful too?
 
More Anonymous
More Anonymous
So in classical mechanics probabilities arise from lack of information of initial conditions
For your example question ... Lets say I don't know the inital state of the pencil
But I do know the coefficient of friction and I assume all angles between 0 and 90 degrees to the table are equally likely
 
undefined
undefined
yes
 
More Anonymous
More Anonymous
Then this is a classical mechanics problem
Which can be solved
 
undefined
undefined
yes but does solving it involve a time frame in which it will be likely that then pencil will fall?
or is the probabilty time independend?
 
ACuriousMind
ACuriousMind
@Bohemianrelativist It is just to make clear that this is no ordinary finite-dimensional Lebesgue or Riemann integral.
 
More Anonymous
More Anonymous
9:27 AM
@undefined The probability you will find the pen lying flat on the table after time time t
is a time dependent probabilty
 
ACuriousMind
ACuriousMind
@undefined If we're doing classical mechanics, there is only a probability to begin with if we are uncertain about the initial state of pencil, and then the probabilities essentially only depend on that initial distribution of possible values for the pencil state. In any case, asking "What is the probability that the pencil will fall during time t?" is just a different question than "What is the probability that the pencil will fall at any time in the future?"
these are just two different probabilities, the latter the limit of the former when $t\to\infty$
 
More Anonymous
More Anonymous
@ACuriousMind Ah sorry my bad
Didnt notice that
wording
 
undefined
undefined
@ACuriousMind ah that makes sense. So 'will fall at any time' is a different question and has a different propability than asking 'will it fall during time t'
thank you @MoreAnonymous and @ACuriousMind :)
 
ACuriousMind
ACuriousMind
you're welcome :)
 
More Anonymous
More Anonymous
Ditto
 
 
2 hours later…
PM 2Ring
PM 2Ring
11:12 AM
@NiharKarve It's highly unlikely that any automatic process checks for that kind of vandalism. You can mod flag one of the questions with a brief explanation. I just asked about this here: chat.meta.stackexchange.com/transcript/message/8793618#8793618
 
Nihar Karve
Nihar Karve
11:26 AM
@PM2Ring thanks, will do
 
 
1 hour later…
Charlie
Charlie
12:35 PM
@Bohemianrelativist It's just notation, and QFT is probably the most standard starting place to encounter functional integration, unless you're learning other things on the side.
 
 
1 hour later…
ZeroTheHero
ZeroTheHero
1:59 PM
@tpg2114 are you there?
 
tpg2114
tpg2114
Hey, good morning!
 
ZeroTheHero
ZeroTheHero
:D
I'm good.
thanks for stopping by.
I've been reading around those notes you posted mathematik.uni-dortmund.de/~kuzmin/cfdintro/lecture10.pdf
In the specific example given in there, the 3rd order derivative occurs as a result of a discretization scheme but I reasoned the general theory also applies to any 3rd order term.
 
tpg2114
tpg2114
Yeah, that's generally true -- in the modified equation analysis they are doing, they are asking the question "My discretization is only approximately equal to the original PDE, but what is another PDE that my discretization is exactly equal to?"
 
ZeroTheHero
ZeroTheHero
Also in the example of the square pulse it so turns out that its Fourier decomposition contains only odd harmonics, but I presume the wiggles would still appear for - say - a sawtooth signal.
(which contains both even and odd harmonics)
 
tpg2114
tpg2114
Yes, in fact you can get wiggles even in smooth signals, like a Gaussian pulse
 
ZeroTheHero
ZeroTheHero
2:07 PM
I gotta try that...
Ok next question (if you don't mind the rapid firing).
 
tpg2114
tpg2114
Let me see if I can dig out an example picture for you while you're firing away
 
ZeroTheHero
ZeroTheHero
Most of the "old textbooks" I have (Crawford and others) deal with dispersion from the perspective of a wave equation, which would be double-derivative in time. Does the interpretation of the third order derivative in space depend on the degree of the time derivative?
 
tpg2114
tpg2114
It might depend on the time derivative -- in fact, I was pondering this exact thing just before you got here because I thought it might come up. When I said odd terms are dispersive, that was coming from the position of a first-order time derivative.
 
ZeroTheHero
ZeroTheHero
hmmm... interesting.
 
tpg2114
tpg2114
Do you have library access to get papers?
 
ZeroTheHero
ZeroTheHero
2:11 PM
yes.
very much so.
 
tpg2114
tpg2114
 
ZeroTheHero
ZeroTheHero
also I was wondering if the interpretation remains if you have more than one spatial degree, i.e. $\partial^3\psi/\partial^2 x\partial y$...
 
tpg2114
tpg2114
This is some analysis we did for a particular purpose (we're in the process of writing the journal version now), but we dig into the dispersion and dissipation of various methods
 
ZeroTheHero
ZeroTheHero
hold on... I can't find through the doi directly.
right... we don't get that journal... I'll get it through interlibrary loan.
 
tpg2114
tpg2114
I can send you a copy another way if you'd like
If you are okay sending me an email, I can respond with the paper
Not at the address in the paper -- I'll give you one here... hah
 
ZeroTheHero
ZeroTheHero
2:16 PM
sure...
go ahead.
 
tpg2114
tpg2114
Got it? I'll delete it
 
ZeroTheHero
ZeroTheHero
YGM
 
tpg2114
tpg2114
Anyway, in that paper we were able to show that a purely dissipative operator applied to the solution gives you dissipation on the solution, but a dissipative operator applied to du/dt and then integrated forward in time gives you purely dispersive behavior in u
 
ZeroTheHero
ZeroTheHero
Ok I gotta read this.
 
tpg2114
tpg2114
The dispersion shows up because d^2/dx^2 applied to d/dx gives you d^3/dx^3 -- pure dispersion
 
ZeroTheHero
ZeroTheHero
2:20 PM
I'm struggling with such terms because I'm working with vector fields in a non-standard coordinate system - a choice had to be made of either simple functions and complicated coordinates or simple coordinates and complicated functions...
and we think these terms are an artefact of the coordinate system.
nevertheless they might have some interpretation in terms of the "easy" set of functions.
I got the paper thanks.
 
tpg2114
tpg2114
As for whether it happens to mixed derivatives, that's also an interesting question. I think it will because if we think of it like d/dy (d^2/dx^2) then it's going to look like a convection of dissipation, which behaves like dispersion I think
I'm not positive on mixed terms though, those don't show up in Navier-Stokes so I haven't given it a ton of consideration
 
ZeroTheHero
ZeroTheHero
Ok I have a good place to start and I have your email. Do you mind if I eventually send you additional emails?
 
tpg2114
tpg2114
Sure, that's no problem
 
ZeroTheHero
ZeroTheHero
next time I'm in your neck of the woods I'll reach out.
I don't go there much (flyover country for me) but I do get there once in a while. There's a good school there...
 
tpg2114
tpg2114
It is a good place to fly over... haha
 
ZeroTheHero
ZeroTheHero
2:24 PM
Thanks mon. I gotta go as I'm subbing for my chair in a meeting with the Dean in 2 minutes.
 
tpg2114
tpg2114
Although the museum is worth going to, if somebody is going to be here anyway
 
ZeroTheHero
ZeroTheHero
:D
 
tpg2114
tpg2114
No problem -- enjoy the meeting ;)
Feel free to ping me here any time, or shoot me an email if a longer form is better
 
ZeroTheHero
ZeroTheHero
impossible task in such meetings. Stay well and thanks again.
I will do that.
 
Mauro Giliberti
Mauro Giliberti
Hey everyone! I've got a question but I don't really think it fits in the site. If you had the metaphysical power to change one and only one (mainstream physics')formula (in any way you want, changing powers or order of derivation, adding extra terms) what would it be and why?
 
ACuriousMind
ACuriousMind
2:30 PM
unfortunately, most of the things wrong with the world cannot be fixed by changing physics :P
 
Mauro Giliberti
Mauro Giliberti
You're sadly right, but I meant in a "that could be cool" perspective
 
ACuriousMind
ACuriousMind
then let me adjust one of the gravitation/force laws so that humans could jump vast distances compared to their physical size (like some insects)
 
Mauro Giliberti
Mauro Giliberti
I think that while lowering c might be an obvious choice it would be really neat
 
ACuriousMind
ACuriousMind
@MauroGiliberti the problem with changes in dimensionful constants is that they're meaningless if you don't say what else you're changing/keeping constant
 
tpg2114
tpg2114
 
Mauro Giliberti
Mauro Giliberti
2:36 PM
Mmmmh that's true
 
tpg2114
tpg2114
@ZeroTheHero There's a picture showing oscillations in a Gaussian pulse. But, looking back through my notes, it's because it's a nonlinear set of equations and a pure gaussian pulse in the species (methane pulse in air background) is nonlinear
So the wave front steepens, the back side wiggles, and because it's species mass fractions it gets renormalized to get rid of negative mass fractions... which funny enough makes this case turn into a "flame" eventually even though it's non-reacting. Pesky numerical errors!
If I remember correctly, if I go through and actually define the Gaussian profile in the characteristic equation and convert that back into the conservative equations, then the numerical dispersion is much much smaller and it doesn't steepen.
 
Mauro Giliberti
Mauro Giliberti
@ACuriousMind what if one wished for all the formulas where c appears to be substituted by c*10^-7 m/s instead? And assuming everything else stays the same
Wait I think there was a MIT game about this
 
ACuriousMind
ACuriousMind
@MauroGiliberti in that case you get a bunch of formulae with the wrong units ;)
 
Mauro Giliberti
Mauro Giliberti
@ACuriousMind oh gosh you're right
I should get some sleep.
 
tpg2114
tpg2114
Although if one were to use a purely central scheme (no dissipation terms in the truncation error, only dispersive ones) and one did not include any artificial dissipation, it would show plenty of oscillations and blow up. It's those disperisve truncation terms without any dissipative truncation terms that make forward-time, central-space unconditionally unstable
 
ACuriousMind
ACuriousMind
2:42 PM
There's a good question here about why it's difficult to say what "changing c" even means
@MauroGiliberti yeah, that simulation is very neat, but I'm not sure what kind of different physical laws it really represents - they just changed the value of $c$ in time dilation and length contraction, but since they don't have to worry about e.g. chemistry they don't have to figure out what this means for stuff like the fine structure constant
maybe atoms aren't even stable in a world where time dilation sets in "so soon"!
 
Mauro Giliberti
Mauro Giliberti
15 mins ago, by Mauro Giliberti
You're sadly right, but I meant in a "that could be cool" perspective
 
ACuriousMind
ACuriousMind
Dec 1 '15 at 16:41, by ACuriousMind
Always remember, we hate fun.
2
but yeah, sorry, you probably need someone else for this kind of speculation - I like my sci-fi "soft", i.e. mostly unrelated to actual physics ;)
 
Mauro Giliberti
Mauro Giliberti
Touché
@ACuriousMind However, I guess that if one wants to keep track of all the interconnected formulas out there, there is probably little to no room for any change, would you agree?
(whatever your answer is, don't tell it to Tegmark)
 
ACuriousMind
ACuriousMind
3:05 PM
@MauroGiliberti I really don't know - I'm not sure what happens if you e.g. keep the fine structure constant constant, lower $c$ and compensate by adjusting e.g. $\hbar$ correspondingly
but that's probably also not very well-defined - you would probably need to make a list of all dimensionless constants and then figure out how to adjust them to get the closest to an effective "lower c"
 
Mauro Giliberti
Mauro Giliberti
I'm not sure that a comprehensive list can me made
 
ACuriousMind
ACuriousMind
well, the list depends on your fundamental theory of course - there's a different bunch of constants for some classical theory than e.g. for the Standard Model
 
ZeroTheHero
ZeroTheHero
3:21 PM
@tpg2114 Interesting. I will try to play with simple things to get a feel for what’s going on. I’ll try to reproduce the wiggles for the square wave and then a sawtooth wave. Thanx.
@ACuriousMind hey there. Is there a way for me to “save” the previous conversation with @tpg2114?
 
ACuriousMind
ACuriousMind
@ZeroTheHero Try "create new bookmark" in the "room" menu below the room description
 
ZeroTheHero
ZeroTheHero
thanks.
 
Physics Meta
Physics Meta
0
Q: Why is an answer sometimes given in the comments?

Mauro GilibertiAfter answering this question I noticed this comment that says pretty much the same thing I said. Should I delete my answer because the comment was posted first? Should the commentator have written the comment as an answer instead? Personally, I think that if a statement answer the question, even...

discussion
 
tpg2114
tpg2114
@ZeroTheHero No problem! In a funny coincidence, my co-author just sent me the revised draft of the journal version of that paper about 2 minutes ago... heh. I asked him the question about d^2u/dt^2 and we're working through it now. He's the more math/analysis person, I try to keep him grounded into applications
 
B. Brekke
B. Brekke
3:52 PM
I am considering the symmetry and irreps of a Hamiltonian $\mathcal{H}$. The energy eigenfunctions of $\mathcal{H}$ are always basis functions that generate irreps of $\mathcal{H}$. But my question is, are all basis functions also energy eigenstates?
 
ACuriousMind
ACuriousMind
@B.Brekke What do you mean by the "irrep of a Hamiltonian"?
 
B. Brekke
B. Brekke
@ACuriousMind My Hamiltonian has a number of symmetries, and these form a group which yields a number of irreps
 
ACuriousMind
ACuriousMind
4:14 PM
@B.Brekke ah, then yes - every irrep is an eigenspace for $H$, otherwise $H$ would not commute with all the group's generators.
 
fqq
fqq
4:52 PM
@B.Brekke as ACM said, if the Hamiltonian commutes with all the group, it must act as a multiple of the identity on each irrep (Schur's lemma)
 
Sir Cumference
Sir Cumference
5:28 PM
if anyone knows: is Scientific American a pop science magazine or a journal?
because the wikipedia page says magazine but it also gives it an impact factor for some reason
 
B. Brekke
B. Brekke
Okay, thanks!
 
tpg2114
tpg2114
@SirCumference Articles there are a mix -- it's not a research journal by any means, but they do have some good writeups of science
FWIW, Science is also a "magazine" but is obviously pretty high IF.
They have prose-type articles as well as research articles in Science. Scientific American is all prose, but kind of sits between pop-sci and what would show up in something like Science
More towards pop sci though
@ZeroTheHero So thinking about the second-order wave equation makes my head spin a bit, but here's what I think we've figured out... for the second-order wave-like equation u_tt = something, spatial derivatives of order 2^n for n > 1 are dissipative. All other even orders are dispersive. Odd-order derivatives result in fractional derivatives in u_t and I don't have any idea what that means
For example, d/dt of u_t = u_xx (purely dissipative) becomes u_tt = u_4x; u_t = -u_xxx (purely dispersive) becomes u_tt = -u_6x; u_t = -u_4x (purely dissipative) becomes u_tt = -u_8x, and so on.
On the other hand, u_tt + u_xx + u_xxx = 0 can be factored into (d/dt + sqrt(1+d/dx)d/dx)*(d/dt - sqrt(1+d/dx)d/dx) u = 0, which becomes a system of two equations: u_t + sqrt(1+d/dx)u_x = v and v_t - sqrt(1+d/dx)v_x = 0. And I don't really know how to wrap my head around what a wave with a fractional-derivative wavespeed behaves
Probably need to work through the Fourier transform to find the relationship between frequency and wavenumber to get a better idea of what's going on there
 
RewCie
RewCie
6:06 PM
Hey, how's everyone here?
Heard y'all missed me? :P
@JohnRennie Hello Sir, how's everything?
 
ZeroTheHero
ZeroTheHero
@tpg2114 yes I’ve also been thinking about this. if you have a wave equation of the type $\partial^2 \psi/\partial t^2=$ (stuff) then you
oups...
Then you’d think it must have even powers of the derivative else you’d get a dispersion relation of the type $\omega^2$= stuff with odd powers of $k$.
 
Sir Cumference
Sir Cumference
@RewCie Ey long time no see
 
RewCie
RewCie
@SirCumference college, startup stuf
you?
How you are doing?
 
ZeroTheHero
ZeroTheHero
so in principle unless you could have negative values of $\omega^2$ unless the coefficients of the odd powers are constrained. In fact (absence of proof is not proof of absence) I couldn’t find dispersion relation for wave where $\omega^2$ was a function of odd powers of $k$.
 
 
1 hour later…
Physics Meta
Physics Meta
7:15 PM
0
Q: Is it okay to get into a long to and fro discussion in the comments?

silverrahulIs it okay to get into a long to and fro discussion in the comments , with the guy asking the question ? If that is not okay, then where can we discuss ? i have seen some comment threads are imported into chatrooms by mods. Is that something only mods can do ? Or can we create own chatrooms to di...

discussion
 
 
4 hours later…
Bohemian relativist
Bohemian relativist
10:46 PM
how to see that a fermion in the fundamental representation of SU(N) in 4-dimensional spacetime has 4N components?
 
ACuriousMind
ACuriousMind
@Bohemianrelativist The fundamental of $\mathrm{SU}(N)$ is just the representation with dimension $N$ by definition where $\mathrm{SU}(N)$ is just the special unitary matrices
"fermion" in 4d is not specific enough to say anything, but if you really mean Dirac spinor, then a Dirac spinor has 4 components in 4d, and if it additionally transforms in the fundamental of SU(N), then it has $N$ components in that fundamental, meaning 4N components total
in general something transforming in rep $V$ of group $G$ and rep $W$ of group $H$ transforms in the rep $V\otimes W$ of group $G\times H$, and $\mathrm{dim}(V\otimes W) = \mathrm{dim}(V)\cdot \mathrm{dim}(W)$.
 
Bohemian relativist
Bohemian relativist
11:25 PM
@ACuriousMind are the 4 components here related to the Lorentz transformation? I doubt, but the lecturer said this.
 
ACuriousMind
ACuriousMind
@Bohemianrelativist I'm not sure what you mean by that - a Dirac spinor is a certain (projective) representation of the Lorentz group, and it has $2^{m/2}$ components where $m$ is half the number of dimensions of spacetime rounded down.
of course this is in some way "related to the Lorentz transformations", since they are elements of the Lorentz group
 
Charlie
Charlie
Do the isometry groups of higher dimensional spacetimes have analogous representations to that of 3+1D?
as in, there is a rep of the isometry group of N+1D spacetime which corresponds to a higher dimensional bispinor representation?
I'd never really thought about it
 
tpg2114
tpg2114
@ZeroTheHero If the powers of k are odd, then they are also imaginary no? So it would be something like \omega^2 = -i*k^3 for example. I don't think there's anything that prohibits it mathematically, but I don't know how to interpret it physically (or "physically" meaning the impact of numerical discretization on the solution). Is the vector space in which you are working giving you second-order-in-time equations where this becomes important to figure out?
Wolfram Alpha tells me that sqrt(i*k^3) is (-1)^(1/4) k^(3/2) and I don't even know how to think about (-1)^(1/4)
 
Bohemian relativist
Bohemian relativist
@ACuriousMind what does "rounded down" here mean?
 
tpg2114
tpg2114
Although I haven't actually derived the dispersion relation on paper, just in my head quickly, so I could be wrong about the imaginary k's floating around
 
ACuriousMind
ACuriousMind
11:39 PM
@Bohemianrelativist uh, literally just round down the number you get - if your dimension is e.g. 7, then half that is 3.5, and rounding down gives 3.
 
Bohemian relativist
Bohemian relativist
@ACuriousMind ok, just as I guessed.
 
ACuriousMind
ACuriousMind
@Charlie The "Dirac spinor" has generalizations to all dimensions in being the unique irreducible representation of the corresponding Clifford algebra.
well, it's not unique in odd dimensions, there are actually two there differing by something similar to parity
 
Charlie
Charlie
ah yeah I forgot about the clifford algebra stuff
 
ACuriousMind
ACuriousMind
in general the spinor representations in arbitary dimensions are pretty annoying because physicists as usual tend to kind of mess the math up, see this question of mine
they're also extremely annoying if you want to think about Wick rotation properly
I'm still not sure if anyone has figured out how that's supposed to work
 
Charlie
Charlie
11:56 PM
I wonder if Wick rotating your spatial coordinates is of any use, or if you just end up with $d-1$ more problems
I guess it would have the same effect though, you'd just get an overall sign depending on your metric
 
ACuriousMind
ACuriousMind
@Charlie $\mathfrak{so}(n,0)$ and $\mathfrak{so}(0,n)$ are largely equivalent, except for the notion of "pseudo-Majoranas"
 

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