The h Bar

uhoh

01:55

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Q: Two helium balloons meet a cocktail party. Do you think they might find each other attractive?

Here we are talking about the mutual force between the two balloons. This is in the context of Newton's law of gravity (so electrostatics etc. are ignored). You can assume that the cocktail party is held in a room that contains air. It is not necessary to assume that the Earth's gravitational fie...

0

Q: A system with effective negative gravitational mass but positive inertial mass

This post is related to an earlier question "https://physics.stackexchange.com/questions/589812/is-it-possible-that-antimatter-has-positive-inertial-mass-but-negative-gravitati?noredirect=1#comment1328882_589812" for which I received a lot of valuable and informative feedback - thanks! I'm hopin...

general-relativity gravity newtonian-gravity mass antimatter

The quick closing and suboptimal commenting on these illustrate the (perhaps inevitable) shortcomings of a large site like Physics SE.

Azmuth

04:11

Morning

Azmuth

05:39

@SirCumference morning! :-)

123

06:12

Morning All...

Ankit

@JohnRennie do you remember our chat for avogadro's law ?

You said that if we double the number of molecules , the pressure will be different and thus it is inappropriate to use avogadro's law since it is defined for constant temperature and constant pressure ..

@JohnRennie But according to kinetic gas equation temperature also depends on the number of molecules with this formula : (1/3)mNc² = RT , where N is the number of molecules. So double the molecule will double the temperature keeping other quantities constant. Right ?

John Rennie

@Ankit hi :-)

I'm answering a question in another room but I shouldn't be too long.

Ankit

@JohnRennie okay take your time :)

Ankit

06:57

@JohnRennie are you free now ?

John Rennie

@Ankit hi

Ankit

@JohnRennie can we start now ?

John Rennie

Shouldn't that be 3/2mNc² = RT ?

i.e. it's the equipartition of energy theorem that says the KE of each molecule is equal to kT.

Ankit

@JohnRennie no , KE = (3/2)RT

John Rennie

No wait, I got that wrong. The KE = 3/2 kT

So ½ mc² = 3/2 kT

Ankit

07:04

Yes

John Rennie

Rearranging we get 1/3 mc² = kT, and multiplying both sides by Na we get 1/3 Na mc² = RT

Which is your equation.

Ankit

@JohnRennie yes now can we say that doubling the total mass doubles the temperature for same c ?

John Rennie

In that equation Na is Avagadro's number so it is a constant. If you want to double the total mass the only way to do this is to double the mass of each molecule m.

And that would indeed double the temperature if we keep c constant.

Ankit

@JohnRennie but you defined it for one mole only . Actually it is n × Na . Right

?

John Rennie

If we use N to mean the number of molecules and n the number of moles then N = n x Na

Ankit

07:12

So it is m×Na or m×n×Na indicates the total mass of the gas . So we can increase the total mass by only changing the number of molecules . Right ?

John Rennie

You can take 1/3 Na mc² = RT and substitute Na = N/n. In that case you get 1/3 N mc² = nRT

But if you change N on the left hand side it also changes n on the right hand side.

Ankit

Yes thete is no problem to have unequal no. Of moles.

There *

@JohnRennie Wait for a sec . I am typing my question. It may take long :)

John Rennie

Right, but if you double N on the left side it also doubles n on the right side and T stays unchanged.

Ankit

@JohnRennie so increasing the number of molecules doesn't affect the temperature ?

John Rennie

No it doesn't

The temperature is proportional to the KE per molecule

Ankit

07:20

@JohnRennie so it can only change if we change the speed of the molecules or their mass ?

John Rennie

Yes

Ankit

@JohnRennie but why is temperature defined as KE per molecule and not total KE ?

John Rennie

That's the equipartition of energy theorem. It says every degree of freedom gets an energy of ½kT, and since there are three degrees of translational freedom it says the translational energy is 3/2 kT i.e. the KE per molecule is 3/2 kT.

Ankit

@JohnRennie sorry I didn't get that .

John Rennie

Equipartition theorem

In classical statistical mechanics, the equipartition theorem relates the temperature of a system to its average energies. The equipartition theorem is also known as the law of equipartition, equipartition of energy, or simply equipartition. The original idea of equipartition was that, in thermal equilibrium, energy is shared equally among all of its various forms; for example, the average kinetic energy per degree of freedom in translational motion of a molecule should equal that in rotational motion. The equipartition theorem makes quantitative predictions. Like the virial theorem, it gives the...

Nihar Karve

08:08

@JohnRennie any tricks for guessing which scalars don't vanish at a singularity?

John Rennie

Are any scalars defined at a singularity? Isn't the whole point that the geometry is undefined there?

Nihar Karve

08:49

@JohnRennie sorry, to check whether a point is a singularity or an artifact of the coordinate system

John Rennie

Ah, OK. scalars are coordinate invariant so all scalars have a well defined value at a coordinate singularity. It doesn't matter whether that value is zero or non-zero, only that it exists.

Nihar Karve

Wait a sec - I think I meant to ask something else (but along those lines)

I can't for the life of me remember - isn't there something where you have to keep checking higher and higher-order scalars to see if one of them blows up as we approach the singularity?

e.g. approaching r=0 for the Schwarzschild metric

@JohnRennie ping

Charlie

11:36

Just wondering, the procedure that moves us between state vectors and wavefunctions $\langle x|\psi\rangle$ involves the use of the elements of the extended, rigged Hilbert space, does this means that the original Hilbert space $\mathcal H$ is still isomorphic to $L_2(\Bbb R)$ but also that the larger rigged Hilbert space is isomorphic to $L_2(\Bbb R)$?

ACuriousMind

the larger space is not a Hilbert space

Charlie

oh

ACuriousMind

for one, $\langle x \vert x'\rangle$ is not a real number, so it has no well-defined inner product

Slereah

The larger (rigged) Hilbert space is the dual of the domain of $\hat{x}$

Which is uuuuh

The Sobolev space $H^{2,1}$?

That is the so-called Gelfand triple

You have the Hilbert space, a subset of that Hilbert space, and the dual of that subset

$$H^{2,1} \subset L^2 \subset H'^{2,1}$$

Or something to that effect

Charlie

ok ty, it's time to stop being lazy and just go try learn this :P

Slereah

11:43

The benefit of a Hilbert space is that you have Riesz' theorem

For a Hilbert space, the dual of a Hilbert space is isomorphic to the Hilbert space

so if $|\psi\rangle \in L^2$, you can say that $\langle \psi | \in L^2$

But this isn't true of the more general wavefunctions

That's why you have to be careful

Slereah

13:03

@ACuriousMind Why do people use the Lorenz gauge with the Proca equation

I thought it wasn't gauge invariant

ACuriousMind

@Slereah they're not gauging, the "Lorenz gauge" is an equation of motion for Proca theory

Slereah

Is it?

ACuriousMind

see Wiki

it's not directly the EL equation but you can split the E-L equation into the "gauge condition" and the more "ordinary" e.o.m. like Wiki does there

Slereah

ah yes

I'll look up a derivation

Nihar Karve

ok I refined my question a little - is there any better method than trial and error for checking if any curvature scalar blows up

you know, while trying to see whether a "pole" is a true singularity or just a coordinate-system gig

Slereah

13:10

I'm not sure, tbh

But at the very least, there's a limited amount of scalars to consider

Nihar Karve

Are you sure about that?

Slereah

Infinite, but you can check them, and as far as I know, it's rarely the case that you have to go very far to find a divergence

Nihar Karve

Why can't I contract arbitrary products of the Riemann tensor?

oh ok

Slereah

Usually you only have to check $R$ or $R_{ab} R^{ab}$ or $R_{abcd} R^{abcd}$

I can't think of any example where all of these are finite but a higher term isn't

There may be but it seems unlikely

Nihar Karve

probably pathological

but I'd be pretty disappointed if I had to check something like $R_{\mu\nu\rho\sigma}R^{\rho\sigma\lambda\tau}R_{\lambda\tau}^{\mu\nu}$

Although even that's one of the "small" scalars

Slereah

13:16

Also proving that every component of the Riemann tensor is finite would be enough to prove that no scalar diverges, but of course that would require finding an appropriate coordinate patch

Nihar Karve

oh yeah, I didn't think of that

Slereah

There's a method that only requires the Riemann tensor, but it's kind of worse?

You only have the Riemann tensor but then you have to do it on all the curves

Which is bad because some singularities are directional

ie they're not divergent on one curve but are on another

Nihar Karve

whoa, I'll check those out

Slereah

if you want an example it's uuuuh

The Israel singularities I think?

Nihar Karve

Werner Israel, right?

Slereah

13:19

Yes, not the country

Nihar Karve

because right now I'm getting search results for ethics in conflict :p

Slereah

The Israel-Khan spacetime

Nihar Karve

"two black holes held in equilibrium by a strut"?

Slereah

yes

Nihar Karve

nice

Slereah

13:22

But unless you're doing specifically studies on singularities, don't worry too much about it

Azmuth

@Slereah Good Evening! :-)

Slereah

No standard example of a spacetime has really weird singularities that can't be hacked out with just quadratic Riemann polynomials

Hello

Nihar Karve

in any case I'll be using FORM for practically anything related to big tensor contractions

Azmuth

sup?

Slereah

there may be theorems proving that such and such spacetime is non-singular given some conditions, but I can't think of anything general

Azmuth

13:26

Programmers often confuse between Halloween and Christmas because Oct 31 = Dec 25.

5

Nihar Karve

@Azmuth that used to be my favourite conversation starter :D

satan 29

@Azmuth lmao

Azmuth

😎

Azmuth

14:19

Does anyone has some solutions to Pimple/Acne problems...

?

My skin is a bit oily and I've to clean them every 5-6 hours (I live in a very high humid region), these pimples come and sometimes some pain too...

Nihar Karve

Could somebody motivate the Chern-Simons action?

This one: $\int_M A \land dA$

Slereah

14:36

something something topological invariant?

ACuriousMind

15:10

@NiharKarve that's not the Chern-Simons action, you're missing the $A\wedge A \wedge A$ term

The Chern-Simons form $\omega$ in $2n-1$ dimensions fulfills $\mathrm{d}\omega = F^n$ (insert $\mathrm{Tr}$s if appropriate), so the action $\int_M \omega$ is the boundary term of a Yang-Mills theory in $2n$ dimensions living on a bulk with boundary $M$.

Nihar Karve

15:29

@ACuriousMind yes, sorry, I meant the Abelian CS action

what does insert traces if appropriate mean?

ACuriousMind

in the non-Abelian case you have $\omega = \mathrm{Tr}(...)$ and $\mathrm{d}\omega = \mathrm{Tr}(F^n)$

Nihar Karve

why are so many of the top reps European?

seriously, in the top 11, excluding Kevin Zhou and Floris

ACuriousMind

15:53

that seems an arbitrary point for cutting off the "top"

JMac

@NiharKarve If you look at the top 12 at least 4 aren't European, possibly 5. All 5 might be from the US even. If you go to top 16, I think somewhere between 7-9 of the users aren't from Europe, and possibly from the US.

Physics Meta

16:31

0

Q: Why is it that people send questions from here over to worldbuilding even if they dont fit there either?

recently a user made a post on worldbuilding about making a flying saucer in real life stating that he was told to go to worldbuilding by user on physics stack exchange. Unfortunately i don't have enough rep to view the original deleted question or the comments on it, but it was decided that the ...

discussion

Nihar Karve

@JMac which four arent European?

ACuriousMind

@NiharKarve Floris (probably), Emilio Pisanty, knzhou, Selene Routley and no one knows where Qmechanic is from ;)

JMac

@NiharKarve Well Floris and DavidZ have their locations in the US clearly. knzhou is in the US, and I believe dmckee is from the US as well, or at least taught there.

ACuriousMind

ah yes, add dmckee and DZ

JMac

And that's mostly based off locations obviously, I don't know who is actually from Europe and who might just live there right now, same for US.

Slereah

16:41

The devil's heating mantle

ACuriousMind

I'd have thought there's no need to heating mantles in hell

Slereah

Turns out there is

And Jeulin provides them

Nihar Karve

"A Beautiful Mind" vs "The Theory of Everything" - go

ACuriousMind

I've seen neither :P

JMac

Out of those two, definitely Shrek 2.

Nihar Karve

16:46

I approve

FakeMod

@ACuriousMind Neither have I. I have window watched them a lot of time, but never have decided on whether to watch them completely or not.

ACuriousMind

what does "window watched" mean?

FakeMod

@ACuriousMind Just extend the definition of "window shopping" to watching movies. ("window watching" is not a real word, it's fake ;)

ACuriousMind

all words are made up

Nihar Karve

@ACuriousMind well, any other good biographies based on a scientific personality?

ACuriousMind

16:50

not a big biopic watcher I'm afraid

nor a reader of biographies

Nihar Karve

Feynman's?

ACuriousMind

nope

Nihar Karve

Surely you're joking! :D

bolbteppa

Zee takes $A \wedge d A$ as his Chern-Simons Lagrangian as part of talking about anyons in 2D and says it's a way to demonstrate fractional statistics, seems interesting idk what the higher D version means

Nihar Karve

I've actually never read much of Zee, how is he?

JMac

16:53

@ACuriousMind Do you read much fantasy? I know you've read Mistborn but that's about it lol

bolbteppa

Zee is good, worth checking

ACuriousMind

@JMac Yeah, mostly fantasy and sci-fi, though I'm reading far fewer books these days than I used to

Nihar Karve

@bolbteppa is he as chatty as Griffiths QM? (although I'll admit the main reason I turned to other books was because I wanted to see canonical quantisation first)

bolbteppa

There's 3 good Zee books, GR, groups, QFT, loads of good stuff in there - it's never complete but what source is, Griffiths aims to give a rounded course in what it does compared to Zee but at a more basic level

Nihar Karve

sorry, what was that last bit?

ah, I was talking specifically about the tone (kinda?)

but ok

JMac

17:01

@ACuriousMind If you're into fast paced books that are pretty straightforward but entertaining, I would recommend checking out Cradle by Will Wight. I wasn't sure if I would like them going in but I got really addicted by like the middle of book 1. It's not like a literary masterpiece or anything, but a lot of people find it super addictive.

bolbteppa

Skimming Zee and Tong they at least vaguely sync up on this

ACuriousMind

@JMac I don't really know what I'm into these days, I just know that many of the common plots and twists have started to bore me because I've seen them so often, so I'm mostly looking for stuff that's...unusual? I'm currently reading the Murderbot diaries by Martha Wells and liking it pretty well

John Rennie

I've read all the Murderbot novellas and loved them all, though some of the novelty wears off in the later ones.

I just loved the character of Murderbot.

And being novellas they only require an hour or two of your time.

JMac

@ACuriousMind I've read a lot of twisty series lately, and I think I was kinda burnt out on it too. I enjoy it, but like there's only so many in a row worth reading until it gets a bit much. Cradle isn't really one of those plot twist stories either. It's basically a Western take on Chinese "cultivation" novels, so it's not really a surprise that the MC constantly getting magnitudes more powerful basically and has crazy luck and stuff.

Also I've heard Murderbot Diaries is good, I might have to check that out.

Nihar Karve

Are they the ones on her website, marthawells.com?

John Rennie

17:10

Yes

Nihar Karve

what is this sorcery [EDIT: oops, I switched the () and [])

ACuriousMind

@JohnRennie most relatable inhuman killer machine I've ever encountered and its disinterest in the world at large keeps reminding me of that Pratchett quote that the one true hallmark of human intelligence is that we've managed to invent boredom in a world full of wonders.

John Rennie

With so many books having been written about AIs turning murderous I thought it was a marvellous plot device to have an AI that thought killing humans generally wasn't worth the effort.

Slereah

17:30

Do you know how complicated it would be to train an AI to kill humans

The robots would just end up shooting trees or stop signs

JMac

That's why the robots who run the internet give us captchas. They are slowly compiling information on how to tell the difference between stop signs, trees, and other non-target objects.

Nihar Karve

At least I get the first three rounds of captcha wrong (no need to thank me)

JMac

Oh snap it's starting to snow kinda hard. Time for the annual round of "whoops no one in the city has winter tires yet".

John Rennie