The h Bar

Charlie

00:16

oh ok, so we're not introducing a new field of operators

Bohemian relativist

02:17

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?

Nihar Karve

04:51

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

undefined

09:14

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

@undefined In quantum mechanics?

undefined

@MoreAnonymous yes and in macroscopic too

More Anonymous

@undefined If I ask what is the probability of heads after flipping a coin

undefined

I tried to google it but google didn't understand what I asked and/or I asked the wrong way

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?

Question 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...

undefined

09:18

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

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

yes

More Anonymous

Then this is a classical mechanics problem

Which can be solved

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

@Bohemianrelativist It is just to make clear that this is no ordinary finite-dimensional Lebesgue or Riemann integral.

More Anonymous

09:27

@undefined The probability you will find the pen lying flat on the table after time time t

is a time dependent probabilty

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

@ACuriousMind Ah sorry my bad

Didnt notice that

wording

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

you're welcome :)

More Anonymous

Ditto

PM 2Ring

11:12

@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

11:26

@PM2Ring thanks, will do

Charlie

12:35

@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.

ZeroTheHero

13:59

@tpg2114 are you there?

tpg2114

Hey, good morning!

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

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

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

Yes, in fact you can get wiggles even in smooth signals, like a Gaussian pulse

ZeroTheHero

14:07

I gotta try that...

Ok next question (if you don't mind the rapid firing).

tpg2114

Let me see if I can dig out an example picture for you while you're firing away

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

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

hmmm... interesting.

tpg2114

Do you have library access to get papers?

ZeroTheHero

14:11

yes.

very much so.

tpg2114

arc.aiaa.org/doi/abs/10.2514/6.2019-1645

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

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

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

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

14:16

sure...

go ahead.

tpg2114

Got it? I'll delete it

ZeroTheHero

YGM

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

Ok I gotta read this.

tpg2114

The dispersion shows up because d^2/dx^2 applied to d/dx gives you d^3/dx^3 -- pure dispersion

ZeroTheHero

14:20

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

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

Ok I have a good place to start and I have your email. Do you mind if I eventually send you additional emails?

tpg2114

Sure, that's no problem

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

It is a good place to fly over... haha

ZeroTheHero

14:24

Thanks mon. I gotta go as I'm subbing for my chair in a meeting with the Dean in 2 minutes.

tpg2114

Although the museum is worth going to, if somebody is going to be here anyway

ZeroTheHero

:D

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

impossible task in such meetings. Stay well and thanks again.

I will do that.

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

14:30

unfortunately, most of the things wrong with the world cannot be fixed by changing physics :P

Mauro Giliberti

You're sadly right, but I meant in a "that could be cool" perspective

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

I think that while lowering c might be an obvious choice it would be really neat

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

Mauro Giliberti

14:36

Mmmmh that's true

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

@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

gamelab.mit.edu/games/a-slower-speed-of-light yes that's it!

ACuriousMind

@MauroGiliberti in that case you get a bunch of formulae with the wrong units ;)

Mauro Giliberti

@ACuriousMind oh gosh you're right

I should get some sleep.

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

14:42

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

15 mins ago, by Mauro Giliberti

You're sadly right, but I meant in a "that could be cool" perspective

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

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

15:05

@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

I'm not sure that a comprehensive list can me made

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

15:21

@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

@ZeroTheHero Try "create new bookmark" in the "room" menu below the room description

ZeroTheHero

thanks.

Physics Meta

0

Q: Why is an answer sometimes given in the comments?

After 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

@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

15:52

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

@B.Brekke What do you mean by the "irrep of a Hamiltonian"?

B. Brekke

@ACuriousMind My Hamiltonian has a number of symmetries, and these form a group which yields a number of irreps

ACuriousMind

16:14

@B.Brekke ah, then yes - every irrep is an eigenspace for $H$, otherwise $H$ would not commute with all the group's generators.

fqq

16:52

@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

17:28

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

Okay, thanks!

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

18:06

Hey, how's everyone here?

Heard y'all missed me? :P

@JohnRennie Hello Sir, how's everything?

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

@RewCie Ey long time no see

RewCie

@SirCumference college, startup stuf

you?

How you are doing?

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 $k$.

Physics Meta

19:15

0

Q: Is it okay to get into a long to and fro discussion in the comments?

Is 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

Bohemian relativist

22:46

how to see that a fermion in the fundamental representation of SU(N) in 4-dimensional spacetime has 4N components?

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

23:25

@ACuriousMind are the 4 components here related to the Lorentz transformation? I doubt, but the lecturer said this.

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

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

@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

@ACuriousMind what does "rounded down" here mean?

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

23:39

@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

@ACuriousMind ok, just as I guessed.

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

ah yeah I forgot about the clifford algebra stuff

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

23:56

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

@Charlie $\mathfrak{so}(n,0)$ and $\mathfrak{so}(0,n)$ are largely equivalent, except for the notion of "pseudo-Majoranas"

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