The h Bar

Akash. B

04:39

@JohnRennie no scope for physics?

John Rennie

@Akash.B anyone who's good will have no problem getting a job. It's rare that employers will only consider someone with exactly the right experience. If you're smart than most employers will be happy to give you time to get up to speed.

dmckee

The hard part is getting your foot in the door for the interview.

Talks I had with HR people suggest that many shops are relying more on expert systems and machine learning to do a lot of front-end filtering these days.

I've also talked to people at specifically technological places where the forbid HR to put any but the most trivial filters (Do they claim to have the right level degree? Can they work in the country? Do they have a body temperature above 30 C?) between the slush pile and the hiring manager, so that is not universal.

Anyway. Most physics majors really do have the right stuff, so once they get in front of human beings they do OK.

Captain Bohemian

07:02

but real physics jobs are vacant before PhD.

what's the difference between the highest degree and the most advanced degree?

Novalium Company

I just heard about a thing called Artificial Womb/Uterus.

Just imagine growing human and other species like we grow food in the future. lol

I'm entering some dark place, lemme stop. But this is how the matrix began so...

Also I made this reddit.com/r/marvelstudios/comments/bdd40h/…

Tell me what ya think

user199721

07:50

heard something interesting a while ago: say we have two optical media 1 and 2(1 is optically denser) and light travelling in medium 1 is incident at their boundary at an angle greater than the critical angle. Then does the light enter the second medium at all? i used to think the answer was 'no' but a physicist i know has told me otherwise. any comments?

John Rennie

@user199721 the light penetrates into medium 2 as an evanescent wave.

user199721

yeah, that is fine

but no transverse waves in medium 2, right?

John Rennie

Correct

user199721

hmmm... i will see if i can post a link to the diagram i was shown...

user199721

08:11

https://imgur.com/DJKLlJO
@JohnRennie
This is not an accurate picture. But pretty much exactly what he drew when he was talking about it

did not make any sense to me... was he probably talking about evanescent waves?

Novalium Company

08:35

A device that simulates tastes on the tongue is a thing, wtf?!

I want one plez

John Rennie

08:55

@user199721 yes, I would guess that's an evanescent wave.

user199721

ah so nothing to be scared of i guess. Probably why i could not find any sources where they explained this kind of total internal reflection.

user8718165

09:11

Hii @JohnRennie

Can I ask here

kyle campbell

Am trying to prove that for a paricle in a potential V(r) the rate of change of
the expectation value of the orbital angular momentum L is equal to the expectation value of the torque: $\frac{d}{dt} L = N$ where N is r cross the negative curl of V. Any suggestions?

Lozansky

@kylecampbell Maybe try Ehrenfest? Just a guess

John Rennie

@user8718165 yes, though I'm currently busy in the other room. Someone else may answer though.

user8718165

@JohnRennie I'm sorry but I don't know why am I not able to get it... Need a lot of help :-(

@JohnRennie could you please draw a diagram telling why all the forces should be equal and cancel out rather than just the horizontal and the vertical components being equal and cancelling out. I just want to know this thing.

kyle campbell

09:34

@user8718165 If the sum of the forces in the vertical and horizontal direction is 0, then the sum of the forces is just 0. Your forces are equal in magnitude and opposite in direction, and in the picture, they are not acting at any angle. Therefore, if the picture is appropriately drawn, the net force acting on your particle is 0.

@Lozansky Thanks

John Rennie

@kylecampbell Ehrenfest's theorem was the one I was thinking of. If you google something like ehrenfest theorem angular momentum torque you'll find lots of articles on it.

user8718165

@kylecampbell if they are like $F_{left}=-F_{right}$ and $F_{up}=-F_{down}$ then that point is at rest...

kyle campbell

@user8718165 Let's say that your forces are equal in magnitude. Then $\sum F_x = F_{left} - F_{right} = 0$ and $\sum F_y = F_{up} - F_{down} = 0$ which $\implies$ equilibrium.

user8718165

@kylecampbell yes...I agree

@kylecampbell but that doesn't mean that $F_x=F_{up}$ or $F_x=-F_{down}$...correct?

kyle campbell

No, it doesn't imply that. As a matter of fact, if you only had one force acting, you could not have a net force $=0.$

I think maybe you meant $F_y$ for both of those.

user8718165

09:50

@kylecampbell well...that should't make much of a difference

I think

kyle campbell

What shouldn't make a difference?

user8718165

@kylecampbell oh sorry...it does....if it weren't $F_y$ then the forces wouldn't add upto zero...

kyle campbell

If your particle's net force is non-zero, its acceleration is non-zero, and your particle will move (it won't be in equilibrium anymore).

user8718165

@kylecampbell yeah

kyle campbell

Yes, but I'd be careful what you mean by $F_y.$ There is more than one force acting in the y-direction.

That's usually why we say $\sum F_y = ma_y$ and $\sum F_x = ma_x.$

user8718165

09:53

but why do they say that all forces acting at a point on a liquid are the same? like here physics.stackexchange.com/questions/31822/…

that's where I get stuck...

doesn't that imply $F_{up}=F_x$ and $F_{down}=F_y$?

kyle campbell

That's just a result of Pascal's law, and it turns out that a pressure change is transmitted undiminished throughout a fluid. It's the principle behind hydraulic lifts.

user8718165

@kylecampbell so does Pascal's law suggest that $F_{up}=F_x$ and $F_{down}=F_y$

kyle campbell

I can't seem to find where they say that on the page you linked. Do you mind quoting directly?

user8718165

@kylecampbell oh the statement of the main question that itself confuses me

kyle campbell

Right, so the force due to $P_s$ in the picture is the only force with two components. In order for the cross-section of fluid to be in equilibrium, we require that $\sum F_y = P_y - P_{s,y}$ and $\sum F_x = P_x - P_{s,x} = 0.$

This is a static fluid, by the way.

user8718165

10:03

@kylecampbell basically, I wanted to know what does the user mean when they say
**Why is pressure in a liquid the same in all directions?** in the question I linked

kyle campbell

The surrounding fluid exerts a force on that cross section, so that cross section exerts a force on the surrounding fluid (in each corresponding direction) by Newton's 3rd law.

user8718165

@kylecampbell could you please elaborate a bit

kyle campbell

Do you mean for a static or a dynamic fluid?

user8718165

I understand 3rd law

bolbteppa

@Slereah Thoughts? ("I think that all young people thinking as theoretical physicists who are interested in black holes should simply buy this new 2019 book...", this is similar apparently)

user8718165

10:06

@kylecampbell I hope the user is talking about static fluids

@kylecampbell anyways yes static..

kyle campbell

Well, if the fluid is static, then you can separate the fluid into small cross-sections like the link you gave earlier. Since the forces must balance on the cross-section for the fluid to be static, the sum of the forces on any one particle (or cross-section) of fluid is 0.

i.e. $\sum F = 0$ in every direction

for any particle you choose

user8718165

@kylecampbell yes...got it

@kylecampbell one last thing... what does $P_x = P_y$ mean in the statement $$P_x =P_y=P_s$$ in the article I linked... I understand that $P_x = P_{s,x}$ and $P_y= P_{s,y}$

kyle campbell

"Since dx, dy, and dz are very small quantities, dxdydz is negligible in comparison with other two vertical force terms, and the equation reduces to,"

That's where it comes from

Their argument is a bit trickier since they're using an argument with differentials, but you could make the same argument just considering a very small cross-section, summing the forces on it, and then dividing by the area.

user8718165

@kylecampbell so if the point we consider is infinitesimal dimensions and ignore gravity...then all the the horizontal and the vertical forces are equal (ignoring the sign)...did I get it correct?

and if we consider the signs...the net force is zero

kyle campbell

10:21

Well, since we're thinking in 3-D, the forces are all equal in magnitude wrt their components in the x-y-z plane. But yes, that's the idea.

user8718165

ok thanks a lot

Slereah

@bolbteppa monday after a harsh week end

Not sure I have thoughts yet

bolbteppa

Secret

10:37

In the plane wave packet example, where will the bohm particle go when it follows the critical trajectory between the family of reflected trajectories and family of transmitted trajectories?

user8718165

10:55

@JohnRennie I'm stuck with the proof of Archimedes principle(just a bit). Could you please help me out?

John Rennie

@user8718165 what's the problem?

user8718165

@JohnRennie problem solving room please...its simpler

Abhas Kumar Sinha

Hey, Landau and Lifshitz's mechanics have a serious problem, in the first section, it says that $L$ is a function of $q, \dot q, t$. After that, it says in the next section that Intertial frame of reference is a function of only velocity square and nothing else $L(\dot q^2)$, after it says in the later part that it is the difference of KE and PE or $L(q, \dot q)$. Can somebody explain that to me?

user8718165

@JohnRennie imgur.com/3oUxrEx this image

John Rennie

@user8718165 OK, that looks clear enough. What's the problem with it?

user8718165

11:08

@JohnRennie can we move to the problem room please

John Rennie

@user8718165 OK ...

bolbteppa

@AbhasKumarSinha need to read it more carefully

Abhas Kumar Sinha

@bolbteppa hmmm.... Okay, one last try.

bolbteppa

@AbhasKumarSinha Maybe 50 tries over a year or three would be better ;)

Abhas Kumar Sinha

@bolbteppa That book seems of that level :P

bolbteppa

11:31

@AbhasKumarSinha Section 4 is on free particles, section 5 is on particles which are not free, and the form one ends up with is based on Galilean relativity thinking as in section 3, some of this will change when you go to special relativity, the whole thing changes when you go to QM

Abhas Kumar Sinha

@bolbteppa any improvement needed? Now is that correct? physics.stackexchange.com/a/472044/208739

bolbteppa

Just need to fix a bracket in one of the lines but otherwise fine

Abhas Kumar Sinha

@bolbteppa That's very good to hear...! :D :) XD :P ^_^

@bolbteppa It doesn't says anything why $L = KE - PE$?

bolbteppa

It does

Abhas Kumar Sinha

@bolbteppa "The second term on the right of this equation is a total time derivative only if it's a linear function of the velocity $v$" ref: $$L(v'^2 = L(v^2) + \frac{\partial L}{\partial v^2} 2 \vec{v}. \vec{\epsilon})$$ as $$L' = L(v'^2) = L(v^2 + 2\vec{v} . \vec{\epsilon} + \vec{\epsilon}^2)$$

What that means?

bolbteppa

11:43

@AbhasKumarSinha if these basic things are too much for now, maybe try easier material?

Abhas Kumar Sinha

@bolbteppa easier?

@bolbteppa I mean why there is no $\vec {\epsilon}^2 $ in that Taylor expansion?

bolbteppa

@AbhasKumarSinha It literally says "Expanding this expression in powers of $\varepsilon$ and neglecting terms above the first order" in the sentence above that expression, not even reading it at this stage :\

Emilio Pisanty

12:32

oh wow

> how I think it would work, I believe we live on the exhaust side of the blackhole. I think that it's just a big warp drive engine compressing space on one side as positive energy as matter and expelling it out as heat and radiation as negative energy and expanding space. Just a thought.

that's a remarkably low bar

good call by @ZeroTheHero

Semiclassical

12:49

@Secret within the approximations that Norsen is making, I think that such a trajectory would leave the particle at the barrier for all time

Secret

ah so the particle will stick to the barrier if they follow the critical trajectory

but if that's the case, we should see a pile up of particles on the barrier

Semiclassical

Right. Sort of like an unstable equilibrium

Well, keep in mind that while those trajectories would exist mathematically

They’d only occur with probability zero

Secret

ok

Semiclassical

If the particle starts infinitesimally forward of the critical position, it’ll end up moving forward ie transmitting

And if it starts infinitesimally behind, it’ll ultimately move backwards ie reflect

That said, I do think this kind of question is significant in the context of arrival times

As in, if the particle’s trajectory is near the critical trajectory, then you’d expect it to take longer to arrive at the relevant detector than one which starts farther away

Secret

yeah, it will be swept by the wavefunction a bit less the closer they are to the critical trajectory

Semiclassical

12:58

That said, I know there are subtleties as far as time-of-flight measurements go

So I don’t know how that all works out in practice

Secret

well, regarding surreal trajectories, how fast the measurement pointer respond is important in determining when the particle arrive

Semiclassical

Right

You start to run into surreal trajectories when the response time is slow (relative to the motion of the particle I guess)

That’s my vague recollection of what the Gisin paper said anyways

one aspect to emphasize I guess: pilot-wave theories really do think of particle positions as being physically real. But that doesn’t mean that position measurements are unequivocally taken to simply reveal that reality

Whether they do depends crucially on the experimental setup you’re using

Secret

I tend to use the trajectories as a guide on which portion of the probability density are travelling to where

Semiclassical

Yeah

Secret

I am a lot more comfortable dealing with field-like entities than point particles

In fact, before I actually knew bohm, I do have a suspicion that some of my quantum questions back in 2013 may be inspired from the notion of a pilot wave

Semiclassical

13:09

Oops

Secret

That is, I consider wavefunctions are some real entity which provide an envelope on how likely events with certain values of properties can be found, and the result of thinking about that, is I start to ask weird questions like how fast the wavefunction itself travels, which makes no sense in orthodox quantum mechanics

The wavefunction in my understanding, thus acts like some kind of field that constrain where particles and events can be found in where

This is actually not very far from the pilot wave picture of bohm. All that is needed is to add the requirement that the particles follow some deterministic dynamics and you are done

0

Q: How fast a quantum state updates in a quantum experiment when the experimental setup itself changes. Are there known experiments on this?

Consider the following Sagnac interferometer setup: where $B$ is a beam splitter which can be raised and lowered with an adjustable frequency $f$, $D$ is a detector which does not click when $B$ is up and clicks when $B$ is down. $S$ is an electron beam source which fires single electrons with...

The thinking in this past question of mine for example, talks about things in terms of a wavefunction as if it is a potential energy surface for probability and other observables, leading to weird equations that makes no sense as Acuriousmind pointed out

In the orthodox picture, each particle carries the state with them and depending on the interpretation, the state really do have a physical meaning, or just an accounting tool

Semiclassical

here's at least one attempt to deal with time-statistics according to pilot wave: arxiv.org/pdf/1802.01863.pdf

i have a lot of respect for that kind of effort, though I can make no judgment as to its veracity

oh, nice. the initial-state preparation method they discuss in the conclusion is the same one I had thought of

Secret

Likewise, for this other past question of mine, there is an implicit assumption that the wavefunction behaves like a potential surface. The answer then revealed my assumption is wrong, thus sending me back to the drawing board: physics.stackexchange.com/questions/288971/…

Semiclassical

The tricky thing, of course

is that while you can certainly choose to give physical status to the integral curves you get

that's not something which is forced upon you

you could just as well regard it as a useful fiction, suitable for providing intuition, while nevertheless refusing to think of these trajectories as being for actual particles

For my part, I find the question "are these trajectories real" to be less interesting than "are these trajectories useful"

Secret

13:33

Well, they are indeed useful, but I do wish they (whether trajectories or wavefunctions) are real because then we can start talking about directly manipulating them

Right now, the tools we have is we can only indirectly manipulate these entities by e.g. changing the voltage, changing the initial conditions etc.

Semiclassical

Here's a question in that spirit. Suppose you want to build a working quantum computer. In what sense, if any, is it useful to think about Bohmian trajectories?

my sense is that, when it comes to making quantum algorithms, the answer is basically "there isn't one"

for the simple reason that the name of the game there is finite-dimensional hilbert spaces, e.g. a two-level qubit

Secret

I think bohm trajectories are useful to work out which portions of the probability density routes to which part of the computer, because the critical trajectories divide the outcomes into regions

Semiclassical

On the other hand, when it comes to designing actual hardware

that's where I wonder if the Bohmian POV could be useful

As in: If you're figuring out how to design an algorithm, you usually don't have any reason to care about how transistors work

Secret

If one can control where the critical trajectories are with the hardware, then there is great control on where the bohm particles (or in the orthodox view, what states the particles are likely to be in) will end up

Semiclassical

but if you're designing the actual circuits being used? you might care substantially more

but that's really just speculation on my part

Secret

13:41

One of my wild wishes that I hope quantum mechanics can make it real, is something like: I poke the position p in the hilbert space, and then I made a node there in the wavefunction. I want to be able to modulate the wavefunction directly without using indirect methods like electric fields, magnetic fields and initial conditions, the same way we can control electric fields to point anywhere we want with machines

Semiclassical

that's my hunch, tho: BM isn't terribly important for quantum software, but it may be useful for quantum hardware

Secret

hmm...

What actually controls where the critical trajectories are. The potential step and tunneling example suggests it is controlled by the experimental setup itself

Curio

(CH3)3CBr + OH- → (CH3)3C+ + Br-
(CH3)3C+ + OH-→ (CH3)3COH
v = k [(CH3)3CBr]

Semiclassical

@Secret that sounds right to me

Curio

What is the fastest step of the reaction?

Semiclassical

13:49

it fits with the spirit of "you need to consider the specific experimental setup if you want to interpret the trajectories"

Secret

which means we don't have as much direct control on where they are, unlike how we handle electric fields and stuff

Semiclassical

yeah

user199721

in the first step, we are getting a carbocation(tert-butyl).carbocation formation has a high energy barrier and is hence slower
so the second step is the faster one @Curio

Curio

How could I discover that? @user199721

Secret

Jul 24 '18 at 16:10, by Semiclassical

hmm, now it is easier to understand this diagram

user199721

13:55

@Curio it is a typical SN1 reaction: en.wikipedia.org/wiki/SN1_reaction

Secret

That the critical trajectory tends towards well 2 is still a bit strange though given it is a symmetric well

Curio

@user199721 is the first one slower because we have to break a bond instead of the second one?

Semiclassical

No disagreement there

user199721

yes @Curio

Curio

@user199721 many thanks!

user199721

13:58

no problem! :)

Secret

Apr 5 '18 at 13:31, by Secret

This bulge still looks very mysterious

Semiclassical

yeah

i suspect the honest answer there is that, if you're looking for something which a given two-slit can actually measure

you look to the far-field trajectories

i.e. not near the bulge

if you try to get up close, i wouldn't be surprised if some of the assumptions made start becoming nonsensical

Secret

yeah, because the orthodox predictions is there will be no bulge both near and far field

Semiclassical

right

so the bulge would reflect certain idealized assumptions

Secret

I guess bohm may have issues with near field phenomenon, which might explain why there is such repulsion regions in the trajectories

Semiclassical

14:04

that'd fit with the whole surreal trajectories business I guess

near-field also means short times

and therefore any lag in the position pointers is going to create issues

Secret

yeah

Btw in my uni, one of my physics postdoc collegue is secretly a bohmian believer, thus I recommended him the papers you recommend me and boteppa to read

Semiclassical

nice

Secret

Currently going to continue to read the surreal trajectory paper after tidying up some PhD calculation files

Semiclassical

i'm a bit too muddled to regard myself as a bohm "believer". but I do think that a lot of the reasons for dismissing it are based on ignorance

Secret

I am not quite a bohm believer, but I am enchanted by its nonlocal treatment of things

Semiclassical

14:08

that doesn't mean there aren't good reasons to be skeptical of it, of courrse

Secret

One reason I like nonlocality may have to do with my personal insecurity: Imagine if you screw something up, and you can hide like schroedinger's cat, then anyone who tried to find you will get a blur instead

Semiclassical

within the domain it was created for---namely, several non-relativistic particles---I think it's entirely coherent

but life is not just nonrelativistic QM :)

Secret

Bohmian mechanic's main challenge is a way to get to a Relativistic Bohmian Field Theory

and that remains my only skepticism

Semiclassical

I'd pose it a little differently: the challenge is to make sense of what on earth relativity can mean in such a theory

Secret

I used to not like Bohmian because it is deterministic, but at that time I do not know about how it recovers probabilistic dynamics from ignorance of initial conditions thus I was pretty ignorant back then a long time ago in my 2nd year undergrad

Semiclassical

14:12

my sense is that, if you're only concerned with getting agreement with what relativity predicts

then the Bohmian difficulties aren't so hard to overcome

But Bohmian mechanics certainly seems to reject the spirit of relativity

Secret

but isn't it has a way to prevent nonlocal signalling like how the orthodox does?

Semiclassical

well, that's what I'm getting at

at a technical level, it's not so hard to forbid nonlocal signaling

Abhas Kumar Sinha

@bolbteppa it's an hour and I've read the whole chapter again, but still, I didn't get what that means? "Expanding this expression in powers of $\vec \epsilon$ and neglecting terms above the first order, we obtain"

Semiclassical

but the nonlocality still has to be present, albeit in a hidden way

and it's hard to square that with the spirit of relativity theory

c.f. arxiv.org/abs/1307.1714

Abhas Kumar Sinha

@bolbteppa Oh okay, I got that... But that took a lot of time for me to understand that...

Secret

14:18

well, entanglement are also nonlocal correlations, and that works fine with relativity. I think nothing can really break relativity if the nonlocality is hidden thus avoiding the signalling. It also does not seemed far fetched to propose the wavefunction controls the foilation. In fact, I suspect it might be necessary to think about such dependence, as that 2017 paper that showed a contradiction when trying to do an extended wigner friend experiment shows

Semiclassical

also, note that Norsen is also an author on the paper I just linked

Secret

there really still not enough people to work on bohm stuff

Abhas Kumar Sinha

There is a very very big problem with quantum mechanics, that's why i hate it...

Secret

@AbhasKumarSinha what problem?

Semiclassical

What I like in that paper is the suggestion that the kind of foliation you need for BM to make sense is a foliation that'll exist in any field theory (bohmian or otherwise)

Abhas Kumar Sinha

14:21

@Secret the only problem which other theories don't have. It's an contradiction to common sense.

Semiclassical

In which case the issue is not so much "why do you need the foliation" so much as what status do you attribute to it

Secret

@AbhasKumarSinha I love things that contradict common sense, because it shows how big the world is

Semiclassical

but I genuinely don't know

Secret

I will need a lot more GR background to even start reading deeply into reference frame stuff in other fields, though

Abhas Kumar Sinha

@Secret Contradiction to common sense isn't something very big, but misleading too. Remember, "how big the workd" isn't for quantum mechanics ;)

Semiclassical

14:24

same, hence why I stay away from it @secret

Secret

and arguably, my GR intuition is even worse than quantum mechanics, which Johnrennie found it strange because he found GR more intuitive for him

Abhas Kumar Sinha

Atleast GR makes some sense....

Secret

One of the things GR throw away is absolute distance, that is quite a dealbreaker in my current mode of thinking because when I construct all these conceptual mental spaces that help me to reason with abstract concepts, I rely alot on the notion of distance to not change as I move to another position in the space

But in GR, space itself is changing

thus I don't have easy reference points to check my orientation

Abhas Kumar Sinha

@Secret Is mathematics complete nonsense? NO, Abstract nonsense is complete mathematics ;) en.wikipedia.org/wiki/Abstract_nonsense

Captain Bohemian

if our spacetime is globally a parabolic manifold, rather than a hyperbolic manifold, what will it look like?

Secret

14:27

though, with the help of Akiva posts on hyperbolic space animations, I am starting to regain a new kind of intuition on dealing with GR spacetimes. Thus I might soon figure out how to navigate them without getting lost

Quantum field theory arguably is a lot harder for me to not have misconceptions though, because no longer you can thing about "two objects bump into each other and interact", but can only talk about "in this region, something and something else happens, and then this is the distribution of results"

I also argue that some metaphysics actually help me to expand my intuition to very counterintuitive concepts:

in The Symposium, Apr 1 '18 at 13:25, by Secret

So "spacelessness" will sound like something that lack dimensions of any kind, and yet is not a point, hmm... it sounds really structureless. Even sets have a notion of cardinality...

Captain Bohemian

I think if our spacetime is a parabolic manifold, that means space and time are of the same signature, then can we still distinguish space and time?

Secret

by having some idea on what spacelessness and timelessness is like, suddenly GR spaces looks a lot more straightforward to understand

@CaptainBohemian well, the time axis may still show an arrow, and causality will flow along that direction, making it distinguishable

Abhas Kumar Sinha

What's the difference between the Special relativity and general relativity?

Secret

SR deals with inertial frames, no curved spacetimes hence it is the easier of the two

Abhas Kumar Sinha

oh okay

Captain Bohemian

14:32

@AbhasKumarSinha special relativity is the limit of general relativity when gravity vanishes.

Abhas Kumar Sinha

GR must be harder than SR?

Captain Bohemian

@AbhasKumarSinha of course, far harder.

Abhas Kumar Sinha

But, doesn't anyone agrees that the name SR should be assigned to GR and GR name to SR because of the fact that special case of (gravity) is GR?

@CaptainBohemian your name reminds of a story of a soldier of Bohemia, is is inspired so?

who used rocks to cook food in village

Captain Bohemian

@AbhasKumarSinha actually I misconsidered you to be that person who often posts about coding here when I first saw your message because your avatar looks similar to his/hers.

Abhas Kumar Sinha

oh okay

@CaptainBohemian brainly.in/question/5117978

Captain Bohemian

14:40

@Secret causality doesn't require spacetime to be hyperbolic?

Secret

I don't know the full answer, but it seems this PSE gives some info on how the metric signature and topology is insufficient to ensure there is causal separation:

7

Q: When does causal separation imply no spacelike separation?

(See here for notation.) In Minkowski space, if $p\prec q$, then there is no spacelike curve $c:[0,1]\to \mathbb{R}^{n-1,1}$ with $c(0)=p$ and $c(1)=q$. This is obvious from a spacetime diagram. Here a "spacelike curve" is a $C^1$ mapping from a nondegenerate inverval into a Lorentzian manifold w...

general-relativity mathematical-physics differential-geometry causality

also, back in newtonian spacetimes, everything are (+,+,+,+) and still there is clearly a time direction singled out

(and in a way, newtonian spacetime is what happens when the light cone squishes flat because $c \to \infty$)

Kyle Kanos

14:59

It's disappointing that when you come back from being suspended, they don't give you your rep back in the new rep window :( it'd have been nice to see 20+ rep in that box (at least until I accidentally clicked it)

Pernk Dernets

15:09

Do you guys now why it is so effective to cool off coffee by blowing? It's much mor eeffective than stirring the coffee with a spoon?

I think it's because by blowing you remove the air above the cup which have a very high humidity. Then a new balance will have to settle, causing much more evaporative cooling. Is my thinking correct?

user199721

i think it might work that way @PernkDernets
https://physics.stackexchange.com/questions/127309/why-does-blowing-on-hot-coffee-cool-it-down

Pernk Dernets

uh nice! Thank you

Captain Bohemian

16:10

@AbhasKumarSinha I mean @enumaris

Novalium Company

16:55

Pff I hope Thanos lives.

Maybe a redemption.

This my 2nd almost 3rd year in this chatroom let's celebrate. Yay me.

M.N.Raia

17:11

Good morning/afternoon/night for everyone!

The East Wind

I finally found a mistake on a wikipedia page

M.N.Raia

Consider then the 3-velocity components and two reference frames $S$ and $S'$ in standard configuration. I know that concerning a change of coordinates, the primed coordinates are functions of the non-primed ones, i.e.: $x'(t,x,y,z)$ $y'(t,x,y,z)$ $z'(t,x,y,z)$ $t'(t,x,y,z)$. But in order to construct the velocity in a particular frame (say, $S'$) the derivative of the position are taken on a function $x'(t')$.

The East Wind

Clausius–Clapeyron relation

The Clausius–Clapeyron relation, named after Rudolf Clausius and Benoît Paul Émile Clapeyron, is a way of characterizing a discontinuous phase transition between two phases of matter of a single constituent. On a pressure–temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of the tangents to this curve. Mathematically, d P d ...

Under the ideal gas approximation section, the third equation should read L upon t delta v

not delta s upon t delta v

M.N.Raia

But we found very often the relation with the chain rule: dx'/dt' = (dx'/dt)(dt/dt')

The thing is that, in order to apply the chain rule, we must consider this function: x'(t(t'))

And I'm losing my mind trying to understand why a primed quantity (x') should depend on a non-primed one (t)

ACuriousMind

17:31

@M.N.Raia But...you already wrote that you know how the primed coordinates are functions of the non-primed ones?

In unrelated news: Meetings where everyone just waffles around until I give up and say I'll just do it suck. Arrrrgh.

Loong

Meetings ~~where everyone just waffles around until I give up and say I'll just do it~~ suck

ACuriousMind

@Loong I think that's usually a sign of a dysfunctional meeting culture - many places tend to have lots of meetings for issues that could've been dealt with more efficiently in writing, but it's possible to have productive meetings.

M.N.Raia

17:47

@ACuriousMind I didn't understand your point. The thing is, x(t) something which you derive with respect to time to obtain the velocity. In primed coordinates it's the same thing instead (of course) you're using primed coodinates x'(t'). But in order to give some significance for texts books when they said "here we use the chain rule", the only way I found was to write x'(t(t')), but this seems to be something wrong.

ACuriousMind

@M.N.Raia You already know how to write $x'(t)$. The whole thing also applies in reverse, so you can also write $t(t',x',y',z')$. Which means that you can apply the chain rule to $\mathrm{d}x'/\mathrm{d}t'$ to write it as $\mathrm{d}x'/\mathrm{d}t \cdot \mathrm{d}t/\mathrm{d}t'$. It's not very obvious why this is useful, but I don't get what confuses you about the principle of it.

M.N.Raia

@ACuriousMind I don't understand the meaning of x'(t) or x(t') or even t(t'). On the other hand x'(t,x,y,z) is quite clear.

ACuriousMind

@M.N.Raia Ah! A velocity is along a path, that is, when you're computing a velocity you have functions $x(t), y(t), z(t)$ and $x'(t'), y'(t'), z'(t')$, and you're computing the derivative of $x'(t')$. Since you also have $t(t',x',y',z')$, along a path that is $t(t', x'(t'), y'(t'), z'(t')) \equiv t(t')$. Then you also have $x'(t) \equiv x'(t, x(t), y(t), z(t))$.

M.N.Raia

18:10

@ACuriousMind x'(t') \equiv x'(t'(t))?

ACuriousMind

I used $\equiv$ there to denote that I'm "forgetting" that the original function had four slots and just consider it as a function of the parameter all four slots depend on

M.N.Raia

if I compute the chain rule with, x'(t'(t)) I get dx'(t'(t))/dt which is wrong because I we have a primed quantity with respect to a non -primed quantity. I want to reach this formula here dx'/dt' = dx'/dt\cdot dt/dt'

ACuriousMind

Why do you leave out the dollar signs? Don't you have ChatJax activated to render MathJax? (see the room description in the top right corner for a link for how to do it)

M.N.Raia

nop kkk I'm using a tablet

ACuriousMind

Oh, I don't think it works on mobile

M.N.Raia

18:17

oh ok I'll see here.

ACuriousMind

@M.N.Raia I wrote above how to get functions $x'(t)$ and $t(t')$. If you plug the second into the first, you get $x'(t(t'))$. Now applying the chain rule gets you where you want.

M.N.Raia

@ACuriousMind yes! This is the right function $x'(t(t'))$ which gives you the chain rule. But how can a prime quantity depends on a non-primed quantity? It will be the end of my doubts if you say that $x'(t(t')) \equiv x'(t')$

I mean why not $x'(t'(t))$ ?

ACuriousMind

@M.N.Raia We have three functions: $x'(t')$, the naive path in the primed system, and two functions along the path $x'(t)$ and $t(t')$ derived from the coordinate transformations.If you plug $t(t')$ into $x'(t)$, you get a function of $t'$ as $x'(t(t'))$ along the path. But in order for the coordinate transformations to be consistent, this - as a function of $t'$ - has to be the same function as the naive $x'(t')$.

I'm afraid I don't know how to explain it better

M.N.Raia

I did here the calculations

Thank you

I understood properly.

The whole thing deals with a parametrized path

and the a priori knowledge of coordinate transformations

As you said since the begginjng

Novalium Company

19:04

Where can I read free articles on topics like physics, neuroscience, neurotech and so on...

Kyle Kanos

sciencedaily.com is one place, I guess

RIP Notre Dame Cathedral

enumaris

20:09

@CaptainBohemian what?

Loong

0

Q: 2019 Moderator Election Q&A - Question Collection

Physics is scheduled for an election next week, April 22nd 2019. In connection with that, we will be holding a Q&A with the candidates. This will be an opportunity for members of the community to pose questions to the candidates on the topic of moderation. Participation is completely voluntary. ...

discussion featured election

dmckee

@ZeroTheHero Ask and you shall receive, it seems.

ACuriousMind

Heh, I can only post one answer per 60 seconds!

Jake Rose

Anyone done astrophysics at cam?

Physics Meta

20:27

2019 Moderator Election Q&A - Question Collection

2

ACuriousMind

@PhysicsMeta Tsk, you're late

enumaris

feels like all I'm doing lately is reading research papers...

every time I think I came up with a cool idea, I find that someone has already done it in the literature...

ACuriousMind

@enumaris Would you rather duck tape new functionality all over haphazardly designed heaps of old code? I'll trade.

enumaris

I'm staying the hell away from ABAP

ACuriousMind

Aw, the Allgemeiner Berichtsaufbereitungsprozessor doesn't bite :(

(it has no teeth)

enumaris

20:34

wat

is that what abap stands for lol

ACuriousMind

@enumaris Nowadays, it's "Advanced Business Application Programming", but that up there was the original German phrase it stood for

enumaris

the second "a" comes from Aufbere...?

ACuriousMind

@enumaris It's a compound noun: Allgemeiner Berichts-Aufbereitungs-Prozessor

enumaris

I see...o.o

ACuriousMind

It means (roughly) "Generic processor for the preparation of reports"

enumaris

20:39

sounds like the backronym is not a good translation lol

ACuriousMind

Well, it...grew. At the beginnings it was little more than a little script language for the R/2 system. Then it became a fully fledged procedural programming language (+ built-in SQL dialect and UI framework). Then an object-oriented dialect grew on top of it, functional-style expression, etc...

At some point along the way, the original name simply wasn't an accurate description for what it did anymore

enumaris

I'm still staying away

ACuriousMind

:'(

enumaris

I'm pretty much as far away from core SAP technology as it's possible to get within SAP lol

ACuriousMind

I literally belong to the sub-unit called "The Core" :D

enumaris

20:44

XD

We are very far away from each other then

ACuriousMind

It's not the tiny garage shop it once was ;)

enumaris

quite a bit bigger now...

Kyle Kanos

21:11

@ACuriousMind you may need to revive the answer you deleted... Jnat's post says the ones under the bar will only be shown if there's <8 questions

If we don't get that, it'd be fine to keep it deleted; if more one may want to consider it

ACuriousMind

@KyleKanos I'll think about whether I like it enough that I still want it if we get 8 other questions.

Kyle Kanos

Roger Roger

ACuriousMind

@KyleKanos If you like it enough, you can of course just post it yourself

Kyle Kanos

I could. 10k peeps might find it odd given my comment on your.

I also imagine a bunch of people would just DV mine because it's me vs others posting the same thing

So I'll just stick with voting & commentating

ACuriousMind

@KyleKanos I undeleted it with an explicit acknowledgement that it's also in JNat's questions

Kyle Kanos

21:23

o7

Kyle Kanos

21:35

Okay, I lied. Found a question I liked, so I copied it

Sir Cumference

Crap the election is in a week

Gotta get rep

@obe give it back please

enumaris

22:01

uh...

ACuriousMind

@KyleKanos Bit of a tricky one, what if they have no poorly received meta posts? (For instance, when I nominated, none of my meta posts had a negative score)

ZeroTheHero

22:31

@dmckee I did ask for a very large sum of $$ but still waiting... the whole vote is quite secondary to my first ask.

ACuriousMind

@ZeroTheHero You only get the swag if you're elected moderator ;)

enumaris

welp, finished reading like 5 papers today...dunno if they contribute much to my work...but I dunno what I should be working on atm

I have no access to data...I'm a data scientist with 0 points of data... cries in data science

ACuriousMind

more like no-data scientist

Kyle Kanos

@ACuriousMind a good point. I suppose i'd counter with picking a low-score one w/ (much?) higher scored opposing answer

enumaris

22:46

@ACuriousMind cries harder

danielunderwood

23:15

@enumaris you're a theoretical data scientist!

enumaris

23:33

loool

danielunderwood

gotta hire an experimentalist to work with data!

kyle campbell

23:53

Is there a way to determine how long a damped pendulum would swing for before it comes to a halt?

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