If we have an uncountable power set $2^S$, and we consider a subset $H\subseteq 2^S$ for which everything in $H$ is disjoint, is it possible for $H$ to be equinumerous with $2^S$?

Might want to ask it in the maths chat @SirCumference

Is there any research paper which tells behaviour of impulsive forces during collision

Hello people

I am basically heading towards studying GR (finally!).

Could you tell me anything about the pre-requisites and material I need to start from?

I am in 2nd year, so you can safely assume the pre-reqs like Classical Mech, QM-1, EM-1, Maths-III covered.

From which text should I start, and how. Please enlighten me, thanks :)

@SwapnilDas special relativity and a bit of lagrangian mechanics might be helpful

can't really recommend a good book because mine is godawful

@SirCumference Ohk lol.

spec rel itself requires some linear algebra knowledge

yeah true.

@SirCumference Hey, you're interested in astrophysics right?

yeah

Wow, even I want to go into the field. Is this the right way?

by "this" do you mean just learning GR?

No I mean the all the stuff I just wrote and GR

Is this the right order, and what else should I read?

what kind of astro are you interested in?

Hmm, GR type.

so like galactic scales rather than stars?

Yeah.

I don't like nuclear physics.

hmm, galactic astro is probably the subfield what i know the least about, but fourier analysis is used a lot

@SirCumference I see.

@SirCumference What kind do you like?

i'm mostly interested in either stellar astrophysics or cosmology

still kinda on the fence but currently doing research on cosmology

Yeah cosmology is something I'm inclined to.

@SirCumference Woah cool.

What could you recommend me for cosmology?

kinda depends how deep you wanna go, if you do computational work then GR and knowledge of waves would be very helpful

if you do theoretical work then diff geo and dynamical systems would probably show up

@SirCumference dynamical systems? Never thought that.

From where did you study differential geometry?

there are some uses for it iirc, mostly in determining the end of the universe, but i mainly focus on the early universe

@SwapnilDas i had an undergrad course with another godawful textbook, can't really recommend one

@SirCumference Ohk lol.

my GR book and diff geo books were probably the worst i've ever used

well on second thought Townsend QM takes that honor

You must have heard of Schutz, how's that?

@SirCumference :P

@SwapnilDas i havent lol

i "learned" diff geo from spivak

and by that i mean it was a confusing quagmire

@SirCumference oh I see.

but if you wanna do analytical work in most fields of physics, probably the most important thing to know is wave stuff

@SirCumference Yeah, I know. Always pops up from somewhere lol.

So given your undergrad struggles, how did you perfect your knowledge?

I guess the closest thing I've done to "perfecting" it is getting a strong intuition on most things i've learned

Hmm I see.

even very strange things like the uncertainty principle in QM can be made intuitive

Ohh..

@SirCumference My Prof teaches QM from Sakurai, do you think it's fine?

unfortunately i don't know that one, i used Townsend (terrible) and Shankar (very good)

@SirCumference Yeah Shankar is a famous one.

shankar does a great job bridging classical mechanics to QM

makes it seem less like QM falls out of the sky

I guess I should try studying from the same.

Is it just intro or all of it (QM)?

depends how you define "all of it", it's a good book for a two-semester undergrad QM course

Hmm, then it's fine.

@SwapnilDas helps to know basic hamiltonian mechanics before shankar, but it's not necessary

@SirCumference Oh ok. Thanks for telling.

np

@sheltonBenjamin That's a very very important question. It has already apeared 3 times in JEE Exam

How to prove it?

Azmuth do you know a book for forces in rotating frame?

For practicing

@sheltonBenjamin Try doing yourself, it's an easy question, try to draw coodinates straight from radius and perpendicular to it, then, find the tension components on both axis, rest you can do from here

Insane Algorithmist! Incredible answer!

@SirCumference For every $X\in H$, pick some $x\in X$. Since all the $X$ are disjoint, the map $H \to S, X\mapsto x$ is injective, so the cardinality of $H$ is smaller or equal than that of $S$, hence smaller than that of $2^S$.

@ACuriousMind Yeah, thought about it and realized something similar. Thanks tho

@AccidentalBismuthTransform hi, are you around?

If you are we can discuss your comment

> Fred brings home 100 kg of potatoes, which (being purely mathematical potatoes) consist of 99% water. He then leaves them outside overnight so that they consist of 98% water. What is their new weight?

> Answer in edit history

Banach–Tarski potato paradox?

Yes, sir.

I added a possible tag to Emilo's question sir.

Asked 8 years ago?

:-/

Hello @JohnRennie yes I am around but I removed the comment instantly, I have reread your sentence and it makes sense to me

heterogenous nucleation on the wall of the container costs more in energy than onto small bubbles in the middle of the liquid (which is also heterogeneous nucleation)

that is (mostly?) correct ?

or not...

damn, I get confused. looks like not to me because the surface to volume ratio is greater for spherical bubbles than for half spheres on walls

@JohnRennie i would like to know that when we say about the kinetic energy of a body isn't it the kinetic energy of all the atoms constituting that body ?

I need an answer in yes or no !.

@ACuriousMind do I smell the axiom of choice

@RyanUnger Yes! :D (I wouldn't be surprised if this fails in a world without choice)

@Ankit Unless you're looking at a body at 0K degrees, the atoms have lots of kinetic energy/thermal motion even when the body at large is perfectly still.

@Ankit yes

How can i solve $d^2theta/dt^2=mgsinthetafracR$

I wanted to find time it takes for a pendulum to cover angle theta

In vertical circular motion

@ACuriousMind @JohnRennie thanks for the response ☺️.

@JohnRennie so when we say kinetic energy of a body it is the combined kinetic energy of all the atoms in it. Am I right ?

Or is it incorrect?

@Ankit suppose you have a body moving at a speed v, then the atoms inside that body have an average speed equal to v. Yes?

@sheltonBenjamin that differential equation doesn't have a simple solution.

@sheltonBenjamin for small angles we can use the approximation $\sin\theta \approx \theta$, and we get the pendulum equation. But for large angles like moving in a vertical circle this approximation isn't valid and there isn't a simple solution.

I've found this wonderful QFT introduction: itp.uni-frankfurt.de/~philipsen/smws10/ral.pdf

Contrary the Peskin & Schröder, I understand it. :-)

@peterh-ReinstateMonica great :)

enjoy:)

@peterh-ReinstateMonica These notes are excellent!

When I was a beginner, I had to study them through more advanced books, wasn't that lucky and aware of these notes :-)

@Azmuth If a scalar quantum field is static in time, does not make it far more simple to calculate the VeV of its (space-only) gradient?

@peterh-ReinstateMonica VeV?

well.. expectation value

the answer is still no :P

$\langle grad \phi \rangle$

@ACuriousMind Thanks :) :) :)

@ACuriousMind What that means? I still didn't get it...

@Azmuth Please stop trolling other users in this chat room by pretending to know more about what they're talking about than you actually do. The concept of VeV (= vacuum expectation value) is a standard concept of QFT and if you don't know what it is you definitely haven't studied it through any "advanced books".

@ACuriousMind I don't know about it... certainly, haven't studied very depth of QFT..

@peterh-ReinstateMonica note that the gradient is a vector and in a Lorentz-invariant theory you therefore will always get 0 for its expectation value. Only Lorentz scalars can have non-zero VEVs.

but this has nothing to do with the field's expectation value being static or not

Has anyone tried Ice-Cream with Noodles or Pizza with Milkshake ?

LMAO!

hehehe XD :)

Coffee and soup

sandwich with cupcake

ROFL!

@ACuriousMind Thanks! But, the nabla-term in the Higgs hamiltonian is a scalar, because it is an energy density; more clearly it is the sum of the squares of 4 vectors. So it can be already have non-zero VeV while it is lorenz-invariant?

@peterh-ReinstateMonica I don't know what you mean, it occurs in the Hamiltonian only as a square, i.e. as the norm of the vector $\nabla \phi$.

note that the expectation value of a vector can be zero while the expectation value of its norm is not, a classic example is momentum in atomic orbitals

@ACuriousMind So $\langle | \nabla \phi | ^2 \rangle$ can be non-zero?

@ACuriousMind Btw, the field I described is not a vacuum state. The vacuum state has the lowest possible energy density, this field has not.

@peterh-ReinstateMonica Well, you said VeV. If you're not looking for the expectation value in a vacuum state, then it's just an eV :P

@ACuriousMind Right, thanks :-)

And $\lvert \nabla \phi\rvert ^2$ isn't a Lorentz scalar either, since the energy density is the zero-th component of the 4-momentum and not a scalar

but anyway, in non-gravitational QFTs the energy density in vacuum is divergent anyway and needs to be renormalized, so I think whatever you're trying to do is probably not what you think you're doing.

@ACuriousMind How can then the Higgs energy density have a not lorentz-scalar term then? It would mean that switching reference frames, we would have a different energy density?

@peterh-ReinstateMonica Yes. Energy is not a frame-invariant, it's the zero-th component of 4-momentum.

(you're not the first to fall into the trap of thinking energy should be invariant, see physics.stackexchange.com/q/154842/50583)

@ACuriousMind I understood, if we move in this space-varying higgs field, then we see a different field, with a different energy density. Ok, but I just want to calculate the energy in a specific frame of reference. For example, if the field is in a device in a lab and everybody moves with v << c.

Hi everyone !

hi..

@ACuriousMind ok

→ 6 messages moved to Trash

Imagine you are sitting in a big hallway temple.

During morning

and full on meditation there.

@peterh-ReinstateMonica I don't know what "the field is in a device in a lab" means. Also, you seem to be thinking about this very classically - a quantum field does not have a definite value, all it has is an expectation value for a specific quantum state.

Someone comes and starts singing this - twitter.com/vonbrauckmann/status/1283273767498571776?s=20

If you're talking about the idea of "false vacua" for the Higgs field with your "space-varying field", then what varies there is the local vacuum state and hence the VEV of the field itself, there is no need to talk about its energy density.

@ACuriousMind Ok, I remark it.

@ronakjain typo?

> But I have come to know that if the length of the slit is finite than there is no variation in intensity along the length

Should that be infinite not finite?

@ACuriousMind I think, because the energy density is not lorenz-invariant, a result can be still calculated, but it will depend on the frame of reference?

@JohnRennie should we continue this discussion in our room

Tomorrow ...

Ok . Bye :-) good night

@peterh-ReinstateMonica As I already said, in a non-gravitational theory the value of the energy density in vacuum is both divergent and meaningless

it's a renormalization parameter you can set to whatever you like, not a meaningful observable

@bolbteppa Good Evening Sir!

The only place where the value of vacuum energy becomes relevant is in the "cosmological constant" part of the EFE.

And the prediction from QFT gets within $10^{120}$ of the experimental value :-)

@ACuriousMind Here I got an energy density, additionally without hardcore QFT, how is it possible?

@JohnRennie It's already very tiny... String Theory sounds lololol.. They cannot be verified...

^Till date ofc.

@peterh-ReinstateMonica Well, G. Smith just plugged the VEV of the Higgs field into the classical expression for the potential. That's not a QFT computation, and it is unclear what these values mean physically (but it is likewise unclear to me what value you are actually interested in).

note also that if you want to do this properly in "full QFT" it's subtler still because the VEV of the field is not actually exactly the minimum of the classical potential, see physics.stackexchange.com/q/75845/50583

@ACuriousMind I was interested in the 2.45×10^45 J/m3. If something could create such a zero higgs field, it would explode with this energy.

Probably as a huge bunch of photons.

(Assuming that the answer is correct)

it's certainly a good first approximation to the energy that would be set free/required for the phase transition

@peterh-ReinstateMonica The answer is "correct" in that the energies involved in a Higgs phase transltion will be huge. It's not "correct" in any specific detail :P

I am hunting for a similar calculation, but this time with a field where the squared-sum of the components is 246GeV and is static in time, but the components themselves vary. Thus, only the nabla-term has energy density.

@peterh-ReinstateMonica Again, a quantum field does not have a value, it only has a VEV.

The field configuration you quote there is not a vacuum configuration because a field with constant value and non-varying components will have smaller energy than the spatially varying field because its gradient is zero,

(this is the general argument for why the VEVs of fields with standard Lagrangian that consist of a kinetic field + a potential that doesn't involve derivatives are always constant)

but of course you can plug any field configuration you want into your formula for the energy density (it just won't have anything to do with QFT or VEVs), what exactly is your problem with that?

That I never did this in my life, and until now I had no idea, how to even begin it :-) But now I already have some...

If I substitute it into the formula, how realistic will be it?

@ACuriousMind What do you think, will be it in an order of magnitude?

"Realistic" in what sense? What real-world situation are you trying to describe?

@JohnRennie I use vim since 1995 and eclipse since 2008. I tried a lot other editors, none of them could do better than them (although I think, eclipse requires a HUGE refactor what will likely never happen).

@ACuriousMind Yes. The result of the calculation, vs. the measured energy of a real-world situation.

@peterh-ReinstateMonica But what is the real-world situation here?

As I said this field configuration is not the vacuum expectation value of a Higgs-type theory, so it doesn't model false vacua, it's just a random field configuration

@ACuriousMind I have an impression, if the experimental setting is enough big, possibly the energy density won't be so astronomical. We have the square of a gradient...

@ACuriousMind And an experimental device is many, many times bigger than the planck-length.

@ACuriousMind I have no idea, how big should it be to be achievable in a lab. Possibly the result will be light years of more. Possibly it will be a femtometer. I have also no idea, how could such a marco-sized higgs field be produced. It is just a... layman daydream ;-)

But I am already on the track.

I don't feel that you've answered my question.

The real-world situation: imagine a cube $(0,0,0)-(\Delta x,\Delta x, \Delta x)$. $\phi(\underline{x},t)=\rm{246GeV} \cdot e^\frac{i x}{\alpha}$.

If there would be some trick (technology) to affect the higgs field, maybe this cube could be produced also in a lab. At least, we would have enough energy for that.

(Now I am not sure if we understand the same on "real-world situation")

But what is $\phi$ there? I'll repeat myself one last time, it's not a vacuum expectation value, and if this is the expectation value in a non-vacuum state, then all you're doing is computing (part of) the energy needed to create this state from the vacuum. What's interesting about that?

@JohnRennie I am also considering to learn emacs... since some decades, possibly the eclipse refactor will first happen :-)

@ACuriousMind It would be a new, macro-sized field. The first one which can be played in labs and not the EM field.

@ACuriousMind You are the first physicist with whom I could talk about it and understood what I am saying :-) Thank you very much!

To be honest I still don't really understand what you think you're doing, but if this conversation helped you I'm fine with that :P

Verschlimmbessern -> Software Development.

I'll die of laugh

That's too hard joke for me...

I'm serious, who writes such questions...

Oh my god, my cheeks hurt this time..

Thank you for this nice demonstration of the target audience for trolling.

@ACuriousMind ..?

Lockdown has made people more creative...

Hi!

hi

no idea

- taken from Introduction to Classical Mechanics by David Morin

cool book :)

@FakeMod Morin's good at explaining stuff

Sorry for throwing random unspecified math symbols, let me just assign the physical meaning of all of them. The above analysis is of the solution of the following ODE: $$\ddot x +2\gamma \dot x + \omega^2 x = F\cos (\omega_d t)$$

@SirCumference Sure, he definitely is, but in this case, I think (forgive my audacity) he is wrong.

::shocked audience noises::

::More Noises::

@ACuriousMind Do you know Laplace transform and those cool things?

Now my concern is that when $\gamma=0$ (the special case being talked about in the above excerpt), the solution of the differential equation becomes $$\frac{F}{\sqrt{\left(\omega^2-\omega_d^2\right)^2+\left(2\gamma \omega_d\right)^2}}\cos(\omega_d t - \phi) + A\cos (\omega t)+B\sin(\omega t)$$

Sorry, I'm wrong, solution is right.

where $\phi=0$ when $\omega>\omega_d$ and $\phi=\pi$ when $\omega<\omega_d$.

Hmm....

So, where's the problem?

So, Morin goes on to say the rest of that excerpt, i.e. "...The motion is therefore either exactly in phase or out of phase with the driving force, depending on which of $\omega$ or $\omega_d$ is larger."

@FakeMod Page number please?

@Azmuth Page 115

1 sec

They already quoted the passage above. Can't you just let FakeMod finish formulating their question?

But, when Morin says the above excerpt, he ignores the rest of two terms, i.e. $A\cos (\omega t)$ and $B\sin(\omega t)$. Those two would definitely change the particle's motion such that the motion is no longer either completely in or out of phase with the driving force.

@FakeMod Section 4.5? It's different in my version of book..

...which does make the last sentence of that excerpt wrong. It would be great if anybody here could confirm my suspicions. Thank you!

@Azmuth Just above that.

@FakeMod Example question?

@Azmuth Not really. It is under that example. I mean, literally just above section 4.5.

(BTW, if anyone is willing to answer, my question/doubt is over, you can proceed and answer it, thank you)

@FakeMod He's probably only talking about the first term because the second and third terms are transient when there is damping. Consider first the approximation that we're at very late times (so that the two terms are irrelevant) and second the approximation that $\gamma$ is very small/zero

@ACuriousMind But when we say that $\gamma=0$, don't the last two terms become (for the lack of a better word) omnipresent rather than just being transient? They do, right. If yes, then I don't see any reason to ignore them at later times.

Yes, you're right that what's written there is wrong if you just do $\gamma = 0$.

But it isn't wrong if we're doing an approximation where $t>>0$ and $\gamma << 0$, and I think he's just sloppy about that when he says $\gamma = 0$.

@ACuriousMind Alright, that clears all of it. Thank you! As for the condition $\gamma \to 0$ (but $\gamma\neq 0$), what should be the correct analysis?

@Azmuth So that's about 860 TBq of H-3? Yes, releasing it to the sea is the best you can do.

@FadedGiant Wow! You are really an expert!

It would have been released to the sea anyway by now without the accident.

@FadedGiant Aquatic animals?

like sharks? won't radiation harm their genes?

What if they become like real anaconda as in the movie?

@Azmuth not significantly compared to the natural background

@FadedGiant how?

(Ah, I see you answered it already, so thanks) But, when $\gamma\to 0$ and $t\to \infty$, doesn't the exponential term $\mathrm e^{- \gamma t}$ become indeterminate?

@Azmuth Ok, I will not discuss such nonsense.

@FakeMod Yeah, you have to stipulate that $\gamma t$ is large, too.

I.e. we're at "small" $\gamma$ (whatever that means in the concrete situation) and we have waited until the transient terms have died down.

@ACuriousMind Gotcha' thank you!

Just watched the first episode of the anime "Psycho Pass". I must say it's amazing. If someone is interested I can share the link.

@FakeMod i take it back lol

@SirCumference It's just a minor fluke that slipped through, not major enough to discredit Morin ;-)

@ACuriousMind Asked a set theorist, apparently $H$ still won't have the cardinality of $2^S$ even without choice

interesting

@KenzoTenma I've heard good things about it (specifically the extended edition of S1) but am hesitant to watch it, since people don't seem to like the many other seasons

@SirCumference The concept that they introduce in season one is really making me quench for more. BTW, you can consider the following: The first season has a total of 22 episodes and the subsequent seasons (2 and 3) contain a total of 19 episodes (neglecting the many movies of it). So you can consider those two as a single season, which can be easily watched within 19 days (one ep per day).

I heard the same thing for the series "Death Note" and hence never watched further than 1st ep. But this time I am willing to invest my time into it and see for myself if it turns out to be a good one or not.

Well, the first half of Death Note is pretty good :P

second half of death note never bothered me much, aside from being a lot more complex than the first half

like there were fake death notes that were disguised as real death notes that we later found out were actually real death notes

I'd rant about what's wrong with the second half but that'd spoil it completely :P

@KenzoTenma you'd probably really like Steins;Gate, it's got a sciency atmosphere and great plot twists

just be prepared for the term "world line" to come up a lot and have a totally different meaning from spec rel

BTW, why hasn't anyone here watched "Monster" anime? I mean it is the best of all the mystery thriller anime. It is considered to be a master piece.

too many episodes for me, nowadays i can barely squeeze in time for a 12 episode show :/

@KenzoTenma code geass is basically a better version of death note fwiw

@SirCumference it's worth spending time on it. You won't regret even a second of it.

@SirCumference that is on my "yet to watch list".