« first day (3817 days earlier)      last day (30 days later) » 

12:01 AM
That balloon question is now on the HNQ...
12:19 AM
1 hour later…
1:45 AM
@AndrewMicallef The Hot Network Questions list that's shown on every page (unless you're using the mobile view).
3 hours later…
4:46 AM
Q: For those of us who are in academia, is there a category of questions that you use PSE for?

Dvij D.C.I am not sure if this exactly fits into the intended scope of the meta but I will try it out. This is clearly an explicitly subjective question with no universally correct answer, but I suspect that the rules on the meta are fairly less stringent than on the main to possibly accommodate this kind...

5:22 AM
Is the formula for the potential of an ideal dipole not at the origin the same as that the ideal dipole at the origin ?
3 hours later…
8:52 AM
4 hours later…
12:24 PM
Hello. There are the quantum mechanical hyper-virial relations, which are deriving from commutators. For example [H,x] = ip, leads to <a|p|b> = i (Ea-Eb) <a|x|b> for momentum p and position x operators. I wonder if a relation to the differentiation rules for the Fourier-transformation exists? Theses are
Hello World..
Pls see the above link and solve the puzzle.
Hello @ACuriousMind pls see the link
@123 Please don't ping people unless you have a specific reason to believe they might be interested. I'm not in the mood for homework, sorry.
@RaphaelJ.F.Berger sure, the Fourier transformation of a wave function in position space is the corresponding wavefunction in momentum space. It's a popular exercise to show that $[f(x),p] = \partial_x f$ and $[f(p), x] = \partial_p f$ holds (maybe with some factors of $\mathrm{i}$ I'm too lazy to look up).
@ACuriousMind ... Ops you are in harsh mood today.
12:51 PM
@bolbteppa Arnold in the Proceedings of the Gibbs Symposium
2 hours later…
3:13 PM
What has the equation $f = x^3 + y^2 + \lambda_1 x + \lambda_2 = 0$ got to do with $\mathrm{su}(3)$ (or $f = x^2 + y^2 + \lambda_1 = 0$ with $\mathrm{su}(2)$) ($x,y \in \mathbb{C}$)
@bolbteppa in what context? i.e. why do you think there's a connection at all?
Sorry it's discussed in the reference right above haha
Q: Are there any books/articles that apply abstract coordinate free differential geometry to basic thermodynamics?

TysonThe mathematical structure of thermodynamics by Peter Salamon (pdf) would be an example, but i would like a more abstract natural formulation of application of differential geometry or even geometric algebra to for example Maxwell relations in thermodynamics that does not use coordinates.

@bolbteppa uh, I don't see any of the equations you wrote down in that picture?
The equation is in the reference above, which that picture came from that was posted above yesterday
You can see the link between those diagrams and thermo vaguely in that $dE = TdS - PdV$ involves tangent (hyper)-planes so in the manifold of tangent planes you can imagine them intersecting in ways that produce configurations that look like the root diagrams of Lie algebras and so somehow express this in terms of the reflection groups
@Slereah this seems to say this was the first paper to write some of that stuff down, such as these polynomials
3:32 PM
Hey guys! at what level does the fluid stress energy tensor work as an approximation in general relativity?
Like at what level of zooming in or zooming out does the fluid approximation work at?
I'm tempted to ask this question on the main site but I'm hoping I can formulate it correctly first
Don't know what you're talking about
@MoreAnonymous It works whenever the results are within tolerance for whatever you want to compute, like every approximation :P
approximations need not be about "zooming", they can be related to a host of other factors, it always depends on your situation
@ACuriousMind Yes I think I should be more specific and ask what error does this introduce to the geodesic equation and how are these error bars calculated (or how are these errors bounded)?
"There exist interesting connections between the simplest degenerate critical points of functions and the simple Lie algebras $A_k$, $D_k$, $E_k$ (or at least their Weyl groups). In the present paper it is shown that critical points which are simple (without moduli) are classified by the series $A_k$, $D_k$, $E_k$."
@MoreAnonymous why would using the fluid stress tensor introduce an error "to the geodesic equation"? if the concrete situation you want to model is that of a universe filled with fluid, there's no error
the error comes from the ways in which whatever situation you want to model is not entirely like the fluid-filled universe
3:45 PM
@ACuriousMind Except when I zoom in I don't see a fluid I see planets stars dust and lots of vaccum
@MoreAnonymous ah, you didn't tell us you wanted to model the real universe!
@ACuriousMind What's the point of physics otherwise! :P
Like sometimes I get very confused
@MoreAnonymous well, but we don't always need to be interested in the whole universe at once, and there are certainly subsets of the universe filled with fluids
@ACuriousMind Agreed I should have been clearer
so while the approximation might work "very zoomed out", there's nothing that would prohibit it from also working "very zoomed in" if the part you're zooming in on has a lot of actual fluid
3:48 PM
@ACuriousMind So this is what puzzles me ...
In GR I could always zoom out more
and claim what too is a fluid
@MoreAnonymous that's not specific to GR, you can always zoom out. The relevant question is whether you thereby lose track of whatever you actually are interested in or not
but both those situations are not the same since if I took the solutions of one id have to integrate with volume which is dynamic
Physical models don't just exist because we like pretty models, they exist to answer specific questions we have about the world. Certain approximations lends themselves well to answering certain questions, others don't
How do I calculate the order of that number?
@MoreAnonymous rotation curve of what? a nebula? a galaxy? a supercluster? but in any case I don't know anything about cosmology so I can't tell you how to actually do that in practice
3:53 PM
@ACuriousMind I was referring to rotation curve of a galaxy
But okay :/
Gonna ask on the main site then
"How does cosmology predict the rotation curves of galaxies?" is certainly a well-defined question for the main site and much better than a vague inquiry about fluid stress tensors
and I'm sure we have enough experts to answer that rather quickly
@ACuriousMind I was going to ask how does one decide at what scale the fluid approximation works in the case of rotation curves of galaxies?
Are you trying to ask when is it okay to e.g. set the energy-momentum tensor in GR equal to $T^{\mu \nu} = (\rho + P)u^{\mu} u^{\nu} + P g^{\mu \nu}$ as one does assuming a perfect fluid
Since I was typing up my question:

So general relativity strictly speaking is a theory where only fields are allowed. To make calculations in it we often use the fluid approximation and think the matter surrounding us instead of stars, planets, dust, vacuum, etc as a continuum. Since I have used an approximation on the side of (the stress energy tensor) Einstien's field Equations (which introduces an error) what error is introduced in the einstien tensor. Specifically what does error does this introduce to the geodesic equation
What is the fluid approximation, is it the setting of $T^{\mu \nu}$ equal to that or something else or are you talking about some different thing
4:07 PM
@bolbteppa I think that $T^{\mu \nu}$ is an approximation
Is it the fluid approximation you're talking about, e.g. it's literally the model of a perfect fluid
@bolbteppa Yes
Its what enables us to do the calculation in the first place since "only fields are allowed"
Right - why wouldn't you include the explicit formula you're talking about to help the people trying to understand what you're asking about, if you say something vague you could be talking about anything
@bolbteppa I like to think with words more but will heed for next time
Here's a thought, in electromagnetism when you're discussing a particle moving in an electromagnetic field, you write $S = - \int mds - e \int A_{\mu} dx^{\mu} - \frac{1}{4} \int d^4 x F^2$ where the $F^2$ term is irrelevant in this case. If you want to talk about the fields in response to the source charges, you re-write the middle term as $- \int A_{\mu} J^{\mu} d^4 x$ and then the first term in the action becomes irrelevant.
The fluid model kind of constructs the EM tensor axiomatically, what it should look like, that traces back to the sources of particles and charges, and so for these you can construct the `geodesic equation' for those particles which is just their EOM in a gravitational field, so I'm not sure how it affects the geodesic equation if this is what the latter part of what you're asking means
In other words, the model of a perfect fluid assumes the underlying charges and source particles move in a way such that the energy-momentum tensor constructed from them takes the simple form above, which obviously physically just means they move `ideally'. If you start complicating how those particles and charges behave e.g. including interactions the EM tensor will get crazy fast, but if you're looking on the scale of solar systems who cares about a penny dropping on the surface of one planet
4:23 PM
So general relativity strictly speaking is a theory where only fields are allowed. To make calculations in it we often use the fluid approximation and think the matter surrounding us instead of stars, planets, dust, vacuum, etc as a continuum. I'm thinking in terms of the FLRW metric:

$$ds^2 = dt^2 - a^2(t) (\frac{d \bar{r}^2}{1-K\bar{r}^2} + \bar{r}^2 d\Omega^2)$$

The stress energy tensor is given by:

$ T^{\mu \nu } = (\rho c^2+P)u^{\mu}u^{\nu}+Pg^{\mu\nu} + \text{Error}$


$ \kappa G^{\mu \nu} = T^{\mu \nu} = (\rho c^2+P)u^{\mu}u^{\nu}+Pg^{\mu\nu} + \text{Error}$
I agree the order of the error will be small ... But my question is how small? Like how sensitive a machine will I have to build to detect this?
I'd say something like that is in some cosmology book in a perturbation section
TIL the pronunciation of "chiral" is "kairal"
I've been saying "sheeral" my whole life
am I French
@NiharKarve bruh
"sheeral" is close to how you pronounce it in German, but the 'sh' is a 'ch' sound further in the back than 'sh',
interestingly, there are some German dialects that substitute 'k' for 'ch', so perhaps that's an evolution similar to how English arrived at their pronounciation there
the ancient Greek pronounciation was likely an aspirated k, though, so maybe the 'k' is actually closer to the "original"
4:43 PM
Feynman famously mispronounced "Bernoulli" as a freshman at MIT
I think we all know what I'm getting at here
@NiharKarve Feynman didn't care about linguistics much as well as far as I can tell from the anecdotes
@NiharKarve sure, but can you play the bongos?
@ACuriousMind I play the piano and compose my own music
sets up queue to start sharing music links
Any other composer here?
@ACuriousMind I famously cannot play the bongos
I guess I now have to concede the fact that Feynman is just a little bit better than me
not a feynman, then
4:52 PM
You can get the stress energy tensor of a fluid from the geodesic equation
you can get the geodesic equation from a fluid
But those are kind of different processes
@ACuriousMind I more of a couch potato does that make me a near hawking :P
@Slereah Care to elaborate a bit more? In case its too long ive asked the question on PSE
(I have several other replies to that but none is appropriate for this room :P)
you can describe a fluid as the statistical mechanical limit of free particles in GR
You can also look at how a self-gravitating fluid (like a star) behaves
in some limit, it's like the geodesic equation
that's why we can pretend the sun is a point mass most of the time
4:55 PM
@ACuriousMind Ah the good ol days of banter are no where to be found now excuse me while I ponder about the universe with Slereah on my chair (which I will add wheels to)
@Slereah Reference?
Errrr the only one I can think of is Synge, I guess?
There are probably more modern ones
but I mean it's basic stat mech stuff
except the point masses move along geodesics
So my question is what is the error term in this approximation?
Like if tomorrow someone says you know it can interfere with a particle physics experiment what would you say?
the error on what
also computing the error of an approximation is very hard and rarely done, alas
Except as a vague O notation
"you can describe a fluid as the statistical mechanical limit of free particles in GR"
which is a copout
5:05 PM
@Slereah I accept vague O notation
good enough for me
Depends on the fluid you describe, then!
Probably gonna be okay up to the scale of whatever the point particles are interacting like
Let's assume dust for simplicity (a universe filled with dust) in the FLRW metric
as I said, I think Synge does that derivation
Lemme check
oh god
He calls it "The material continuum"
it is a bit dated
reading it
chapter 6 if you need it
5:09 PM
you mean 4?
The GR fluid
5:25 PM
So this is basically my answer:
I'm not gonna lie, I won't read all the old timey GR to work it out
Synge has a lot of nice stuff but it is very of its era
It's the Duffield era, back when they only wrote down the tensors they saw in the sky
he was a very physics oriented guy
no differentials, you just travelled along a path a little way
1 hour later…
6:53 PM
Do all internal lines in a Feynman diagram represent virtual particles?
What if I have a Feynman diagram that starts with an electron and a positron which undergo pair annihilation followed by pair production followed by pair annihilation followed by pair production. The intermediate e+ and e- produced can be observed, right? If yes, then the internal lines representing the intermediate e+ and e- are not virtual?
So if I am trying to observe intermediate particles, then I am supposed to split the big Feynman diagram into two where the intermediate particles are external lines?
Is it possible to have two photons emerging from the same vertex in EM interactions?
What does "mass shell" mean? I often see it in phrases like "particle off mass-shell" or "particle does not lie on the mass shell". I understand what is meant by it but I don't get how this terminology came to be. What is a shell here? What is a mass shell?
1 hour later…
8:17 PM
@Yashas The definition of "virtual particle" is "line in a Feynman diagram". It doesn't mean anything else. I rant about that at length e.g. here
@Yashas no, there is nothing about the single internal line that would make it a "real" particle - as an internal line it can still have off-shell momentum
@Yashas the "mass shell" is the hyperboloid(s) in momentum space formed by the equation $p^2 = m^2$
"shell" is just another word for "surface" in this context
@Yashas no, because photons are not charged, they cannot interact with one another directly - the lowest order interaction between two or more photons is the box diagram
9:13 PM
Noob here. My question: physics.stackexchange.com/questions/629862/… has four 'close' votes against it. What should I do?
I thought close votes can only be cast when a question has been downvoted to minus two or lower. Am I mistaken?
@MatthewChristopherBartsh People are voting to close as opinion-based, so you should edit it to ask an objective question about physics. "Why has no one ever done X?" can conceivably have completely non-physical answers like "no one has ever funded that", "no one finds that interesting", etc.
@MatthewChristopherBartsh any question can be close-voted by users with more than 3k reputation
the score of the question does not matter for that
the only votes that are restricted by the score of the question are delete votes, and manually cast delete votes are pretty rare anyway
Is there another forum that would deal with a question about how how astronauts think, and how their supervisors think or what laws and/or rules govern the behavior of astronauts?
I'm new to SE, so I don't know much about the parts I haven't visited.
When a question, is closed, how long after that will it get deleted?
Only one of the four answers the question has received is opinion based, and it has been heavily voted down. Why not vote to delete the opinion based answer instead?
So how would I edit it to ask an objective question about physics?
@MatthewChristopherBartsh Questions with at least one positively-scored answer will never be deleted
@MatthewChristopherBartsh I don't really think you can do that, tbh - not all questions are on-topic here. That doesn't mean they're bad questions necessarily, just that this isn't the place for them
@ACuriousMind Oh. That IS good news.
About the nondeletion rule
@MatthewChristopherBartsh Space Exploration is closer to that, but I don't know enough about that site to tell you what's on-topic there
9:27 PM
Does a closed question get seen less?
I hadn't heard of Space Exploration. Another 'SE'.
Can a closed question be voted up or down? How about the answers to it?
voting on closed questions works normally, they can also be reopened (either if the question is edited or if simply the community disagrees over whether it is on-topic or not)
9:47 PM
Is there a physics education part of StackExchange (SE)? What about if asked an answered my own question, "What is an entertaining way to illustrate for physics students air resistance and the absence of it in a vacuum" or "What is an entertaining way for space explorers to educate the public?" or something like that?
10:35 PM
is there a way to rigorously derive that the squared mass of a field in a Lagrangian the second derivative of the potential with respect to the field at the vacuum, $m^2=\frac{\partial^2 V}{\partial \phi^2}_{\phi=\phi_0}$? Or can it be only reasoned qualitatively?
10:47 PM
@Bohemianrelativist hos do you define mass here?
11:16 PM
@fqq I don't know. The textbook of particle physics just tells us to look at the quadratic term, the term containing $\phi^*\phi$ and compare (equal) its coefficient, say $c$ with the term $\frac{1}{2}m^2\phi^*\phi$, then solve $c\phi^*\phi=\frac{1}{2}m^2\phi^*\phi$ for $m$ to get $m=\sqrt{\frac{c}{2}}$, which is called mass. I found this very ad hoc.
sorry, it should be $m=\sqrt{2c}$.
I read the above way of finding the mass long time ago. Then recently the lecturer in Electroweak Interaction said:"you probably have seen in a lot of places $m^2=\frac{\partial^2V}{\partial \phi^2}_{\phi=\phi_0}$". But I have never seen it anywhere, though I can see it qualitatively, the mass, which is the charge for gravity, is related to the curvature.
11:40 PM
the "curvature" here is the curvature of a potential around a minimum, it has nothing to do with gravity
anyway yes, that's sort of the definition of mass in this context
in a free theory $\mathcal{L}=1/2\partial_\mu\phi\partial^\mu\phi+m^2/2\phi^2$, one-particle states with momentum $p$ have energy $E^2=p^2+m^2$. So it makes sense that $m$ is the mass of the corresponding particle
If you have a generic potential $V(\phi)$ you can in principle expand around a minimum, and to perturbation theory around the free theory with $m^2=\frac{\partial^2V}{\partial\phi^2}$

« first day (3817 days earlier)      last day (30 days later) »