The h Bar

user228700

11:00 AM

Right, and this is because..? (Ik it seems obvious but sometimes, my brain assumes things, which leads to misconceptions)

John Rennie

Ok, this is going to be a rather handwaving argument so tread carefully ...

user228700

OK...

John Rennie

If you have a mixture of A and B and you evaporate off a molecule of B there are going to be more A-A interactions. This is simply because if you have an A-B pair and you evaporate off the B you leave behind an A in the liquid.

If A is less volatile than B (I know this is the opposite to your diagram but I've assumed this all through this morning's chat) then A-A bonds are probably stronger than B-B bonds.

user228700

Yes...

John Rennie

Evaporating away a molecule of B means there are more A-A bonds while evaporating off a molecule of A means there are more B-B bonds.

user228700

11:04 AM

Yeah...

John Rennie

But if A-A bonds are stronger than B-B bonds then we get more energy from forming extra A-A bonds than from forming extra B-B bonds.

So evaporating a molecule of B lowers the total energy more than evaporating off a molecule of A.

QED

user228700

QED?

John Rennie

Q.E.D.

Q.E.D. (also written QED) is an initialism of the Latin phrase quod erat demonstrandum, meaning "which is what had to be shown" or "thus it has been demonstrated." The phrase is traditionally placed in its abbreviated form at the end of a mathematical proof or philosophical argument when the original proposition has been exactly restated as the conclusion of the demonstration. The abbreviation thus signals the completion of the proof. == Etymology and early use == The phrase quod erat demonstrandum is a translation into Latin from the Greek ὅπερ ἔδει δεῖξαι (hoper edei deixai; abbreviated as ΟΕΔ...

user228700

And u thought I'd know that :-P Hang on, lemme digest what you've told me, lest my brain effs up again.

user228700

Okay. That really is pretty hand-wavy but I accept it because it makes some sense :-P

user228700

11:10 AM

Right. So wait, below the bottom curve, there is only liquid...what above the curve at the top?

John Rennie

16 mins ago, by John Rennie

Now suppose we start with some vapour at a very high temperature and a composition $x$.

user228700

Okay.

John Rennie

As we cool this vapour we move down a vertical line on the diagram, that is the vapour composition stays constant and the temperature falls.

When we hit the upper line is where our vapour starts condensing.

user228700

Riight...

user228700

And there's...an equilibrium in the middle...

John Rennie

11:16 AM

All the points in between the lower and upper curves are compositions that are unstable and immediately separate into some liquid and some vapour.

This is what I was getting at first thing this morning when I posted those diagrams.

user228700

Oh, so it's not an equilibrium :-|

John Rennie

No. The space in between the curves is effectively a forbidden zone where no stable compositions exist.

user228700

Okay, say the liquid's done boiling.

user228700

And now the temperature is increased, and we enter that forbidden zone.

user228700

Oh, hang on...

user228700

11:19 AM

How is it that the dew point is above the boiling point?

John Rennie

OK suppose we start at $x = 0.5$

Start at low temperature and steadily increase the temperature

user228700

Okay.

John Rennie

When we hit the lower curve our 50/50 liquid mixture starts to boil.

user228700

Yes...

John Rennie

But the vapour doesn't contain 50% B and 50% A because B is more volatile. The vapour contains more B than A.

user228700

11:22 AM

Yeah...

John Rennie

So that means as the vapour forms it takes more B than A out of the liquid. And as a result the composition of the liquid changes. The value of $x_a$ increases and $x_b$ decreases.

user228700

Oh, yeah...

user228700

And then that boils too...

John Rennie

So now we've moved to the right on the phase diagram. Now we hit the lower curve at a higher temperature i.e. the boiling point has gone up.

@Kaumudi yes, and as we continue boiling $x_a$ keeps increasing, $x_b$ keeps decreasing and the boiling point keeps going up.

user228700

Hang on, what does that mean when we hit the lower curve again, as we move towards the right?

John Rennie

11:26 AM

We started heating a 50/50 mixture and we hit the lower curve at some temperature $T_{50/50}$.

If we has started with 51% A and 49% B then we would have hit the lower curve at a slightly higher temperature $T_{51/49}$

user228700

Yeah, and that's the temperature at which the mixture boils...then the boiling point keeps increasing because the more volatile substance leaves the mixture...

John Rennie

Exactly

Mew

@JohnRennie

did you hear the law of momentum was violated?

John Rennie

If we keep boiling and boiling we'd end up with pure A left

user228700

So once the the mixture boils, it stops boiling?

John Rennie

11:29 AM

@Kaumudi yes, in the sense that to make it boil we need to up the temperature a bit.

Mew

@Kaumudi did you hear about the EM drive?

user228700

And we keep increasing the temperature...and we keep going right? o.O

user228700

@Mew Nopedy nope.

John Rennie

@Kaumudi Yes

Mew

@Kaumudi it changes everything

user228700

11:30 AM

That's...not a line, that's a curve!

user228700

I mean, okay, maybe not.

John Rennie

I think most of use the terms line and curve interchangeably ...

If we mean a straight line we'd say a straight line

user228700

:-P Alright. But okay, after boiling, if we started out with some composition, then we'd be traveling sideways...

John Rennie

@Mew no-one outside Harold White's group thinks the EM drive actually works.

Mew

@JohnRennie it's the year of disrupting the establishment

"Noone" thought Brexit, "noone" thought Trump, and then "noone" thought EM DRIVE!!!

John Rennie

11:33 AM

@Kaumudi yes. Boiling preferentially removes the more volatile component so the liquid left behind moves to the right on our graph.

user228700

Yeah, okay...and then where do we hit either of the curves again?

John Rennie

Start at the point where we first hit the curve - at $x = 0.5$

user228700

Okay...

John Rennie

by boiling away some B we move horizontally to the right, so we are now at a point slightly below the lower line.

user228700

Yeah...

John Rennie

11:35 AM

So to make the liquid boil again we bung in some more heat to raise the temperature i.e. move vertically back upwards until we hit the lower line again.

user228700

Yeah, OK...

John Rennie

And the same happens again. We boil off some B and move to the right, then we have to move vertically upwards to hit the lowe curve again.

user228700

Yes...

John Rennie

And this keeps happening until we've moved all the way to the right and got pure A.

At which point the boiling point is obviously just the boiling point of pure A.

user228700

Aaaaah!!!

user228700

11:38 AM

OK, sir, what dyou want in return for these past few hours in which you've so generously helped me? Seriously, I'm not kidding about how I'm infinitely grateful.

Mew

The EM drive generates more thrust than photon thrusters

John Rennie

@Mew really? I only skimmed the paper, but the impression i got was that th force measured was less than the radiation pressure if they'd just fired the micowaves into empty space.

user228700

If u give me an address (of someplace sufficiently far away from your home so that u don't get stalked or whatever and close enough that u can go pick it up), I'll send u a bunch of sweets for ur birthday next year. I mean, okay, I don't know how much that would cost but I could check to see... @JohnRennie

Sanya

@JohnRennie their discussion claims it's two orders of magnitude above light pressure

John Rennie

@Sanya OK, I'll read the paper more carefully. But we're still talking about tiny forces for large amounts of power.

Sanya

11:40 AM

I find it hard to gauge the paper by just reading it - as long as no one else can reproduce it, I don't want to think about it too much either

Mew

I think I"ll do a PhD in EM drive

Sanya

@JohnRennie I'm not saying it's true, just to be clear about it

John Rennie

@Kaumudi gosh no, don't do that. Mailing costs from India to the UK would be prohibitive.

Mew

@Kaumudi I'll PM my bday and addy tho

John Rennie

@Kaumudi I thoroughly enjoy talking about physics so you're encouragaing me to do something i enjoy. No other reward is necessary.

user228700

11:42 AM

:-( What can I do then?

Mew

I'll have John's candy

John Rennie

@Kaumudi Ace the exams! :-)

Sanya

but the efficiency that they claim was somewhere in mN/kW if I remember correctly - which apparently is better than using a laser to propel something away ... I'm waiting for other experiments about it for now, no use in cracking my head over it

Mew

@Sanya why wait for others to do the work for u

Sanya

@Mew cause there's more rewarding and interesting things to do

Mew

11:43 AM

oh

John Rennie

It's just conceivable they've genuinely discovered something interesting. After all the Casimir effect seemd pretty weird at first, but it proved to be real.

Mew

I agree John

user228700

@JohnRennie -__- I'm serious! Okay, maybe I'll go to London someday and then if u want, I'll mail you sweets from there! Or anything else. Seriously, you've no idea how grateful I am.

Sanya

anyway, anyone feels like more thermodynamics of mixtures? I've got a small question myself that seems pretty simple to me but I keep on not being sure about it

John Rennie

I'm not sure we understand QFT well enough to be absolutely confident there are no other weird effects lurking in the undergrowth waiting to jump out and surprise us.

Mew

11:44 AM

Go ahead Synay

user228700

@Mew I'm a teenager and I don't understand ur lingo, man :-| What are u saying?

John Rennie

He'll Private Message you his birthay and address ?

Mew

@Kaumudi I syd I'll PM (personal message) by bday (birthday) and addy (address) for my candy (and i'll take John's too)

John Rennie

NO MAN TAKES MY CANDY!!!

2

Mew

lol

user228700

11:46 AM

xD Ah, okay. Didn't understand what "addy" is supposed to be :-P

John Rennie

@Sanya what's the question?

user228700

@JohnRennie If I ever have the opportunity, I'll send u some sweets someday. If I make it to college next year, you'll be one of the handful of reasons why.

Sanya

if I have a binary mixture of components A and B and the Gibbs energy of mixing $G$ is negative above the whole range of possible mixtures $x_A$ and $x_B = 1 - x_A$, the two components could still not mix if there are some local maxima at some composition with value $x_A$, because by demixing into two phases, the system could reach a minimum for each of the phases

user228700

Maybe I'll write you a letter of gratitude after my exams, ooh, that'll be best-seeing as I really can't do anything else :'-(

Sanya

now if the second derivative of $G$ is strictly positive, there will not be any maxima, therefore, this possibility for unmixing does not exist

John Rennie

11:48 AM

> the Gibbs energy of mixing $G$ is negative above the whole range of possible mixtures $x_A$ and $x_B = 1 - x_A$

That just means the free energy of the mixture is less than the sum of the free energies of the pure compounds.

Sanya

yeah sure

John Rennie

It doesn't mean that the free energy of the mixture is less than some other combination of mixtures.

Sanya

yep, that is my trouble

usually, the conditions for possible mixing in any composition $x_A$, $x_B$ that I have found are the free Gibbs energy needs to be negative for all values and the second derivative needs to be positive for all values

John Rennie

Hmm, I'd have to sit down and think about what restrictions that sets on the values of $d^2G/dx_a^2$ ...

Sanya

while I understand that this will basically give me a Gibbs energy of mixing which has one global minimum and two maxima at $x_A=1$ and $x_A=0$, I'm not sure why a demixing into of $x_A, x_B$ into two mixed phases with lower mixing Gibbs energy is in general not conceivable

@JohnRennie the positivity is such that there are no local maxima in the $G$-landscape which would always lead to unmixing

John Rennie

11:55 AM

Well such a system is always locally stable.

That is the process $x \rightarrow (x-\delta) + (x + \delta)$ always raises the free energy.

Sanya

@JohnRennie I'm not sure I understand what you mean by that

John Rennie

Suppose you have a single phase with the composition $x$.

And you ask what happens if we phase separate this into two phases, one with slightly lower $x$ and one with slightly higher $x$

Sanya

yeah

one of them might have a smaller $G$ of mixing, one of them a higher one (if I am not in the local minimum initially)

John Rennie

If the $G(x)$ curve is everywhere concave then this process always raises the free energy

Sanya

ah ok - that was the part that I did not realise/understand

John Rennie

12:01 PM

Suppose we split the composition $x$ into two phases with compositions $x+dx$ and $x-dx$

Sanya

that is basically the definition of concave, is it?

$f(x(1-\alpha)+\alpha y) > f(x)(1-\alpha)+\alpha f(y)$

wait, isn't it convex?

$f(x(1-\alpha)+\alpha y) < f(x)(1-\alpha)+\alpha f(y)$

@JohnRennie sorry - I'm listening, just thinking out loud, but that was rude

we split it, ok

John Rennie

To be honest i'm thinking as I type as well. Let me draw a diagram ...

Kyle Kanos

Ice cream for breakfast makes you smarter, Japanese scientist claims

John Rennie

Is that what you had in mind?

Sanya

you beat me to it :D I was just plotting the same

John Rennie

12:08 PM

OK if you take any point $x$ on that curve then the gradient $dG/dx$ is higher on the side with increased G than on the side with decreased G.

Sanya

yep, I understand - and that is the reason no demixing will occur

John Rennie

So if you start at a composition $x$ and split it into two phases with compositions $x+dx$ and $x-dx$ the total value of $G$ always increases.

Sanya

thank you :)

John Rennie

My only concern is that I've given a local argument. Is it necessarily true globally?

I guess it is because $d^2G/dx^2$ is positive everywhere.

Sanya

yeah, that's the point I think

John Rennie

12:11 PM

God I always hated thermodynamics :-)

Sanya

I actually think it is one of the coolest theories physics has :p

in any case, thank you a lot for the help :)

koolman

@JohnRennie me too

Sanya

"A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended its area of applicability. Therefore the deep impression that classical thermodynamics made upon me. It is the only physical theory of universal content which I am convinced will never be overthrown, within the framework of applicability of its basic concepts."

:D

user116211

12:35 PM

So what is your question here? Perhaps this question should be closed because it is "unclear what you are asking"! — sammy gerbil 7 hours ago

user116211

o sammy; you didn't get it :(

Kyle Kanos

Well to be fair, there isn't actually a question in Sanya's post

Sanya

there isn't - and that should be obvious

pointing it out seems to be useless unless you claim that this post clutters up meta

which at the post rate in our meta is ... a bit weird, but well ...

user116211

@KyleKanos wait, this is not a question.

Kyle Kanos

IMO: you should pose the question "What types of questions are considered as homework here"

And then answer it with the links

user116211

12:38 PM

@KyleKanos isn't it evident?

Kyle Kanos

@MAFIA36790 No. It's literally a statement "I did this work"

Sanya

I wanted to keep the discussion and this apart, I was actually hoping that if anyone answers, they would answer with "I did the same"

maybe I should state that

Kyle Kanos

I disagree with that, but don't have time to get into it

GTG

Sanya

oki~ see you

user116211

@KyleKanos Happy Thanksgiving.

user116211

12:42 PM

What is the difference between \begin{eqnarray} \end{eqnarray} and \begin{align} \end{align}? It seems they render the same thing here.

Emilio Pisanty

@MAFIA36790 ... prooobably, I think?

koolman

Hii @Sanya are you indian

Emilio Pisanty

As I recall, current guidance is to use align

user116211

okay.

Emilio Pisanty

127

A: eqnarray vs align

Although eqnarray may seem to work "well enough", Avoid eqnarray! Avoid eqnarray! Avoid eqnarray! Use align and the rest of the ams environments. See texdoc amsldoc (PDF) or the short math guide for LaTeX for documentation on how to use them.

user116211

12:45 PM

WoW! thanks @Emilio; reading.

Sanya

@koolman no

koolman

Okay

Jim

@JohnRennie Obligatory xkcd reference. I AM GNOME ANN!

Bernard Meurer

1:05 PM

@JohnRennie My laptop wrecks everyone else's in circuit simulations for FPGA programming

I feel like a boss

:P

John Rennie

1:19 PM

@BernardMeurer go, go, go!

A friend of mine is speccing up an awesome workstation. I think you saw the Facebook post. But if you look closely at the single thread execution speeds even the latest fastest 22 core, yes 22 core!!!, CPUs aren't that much faster.

So the quad core gen 3 i7 in your laptop is not shamed even by the very latest CPUs. Well, not unless you have software that can exploit massive parallelisation.

heather

hello

user116211

> The organic chemistry course on the other hand introduce you to concepts which are very strange, like the COOH group and its charges, the PO4 group and its negativeness, the idea of electron transfer, the mechanism of breaking CC double bonds, and all that black magic that makes Chemistry so interesting.

user116211

source

Bernard Meurer

@JohnRennie I did see that! What a beast of a machine!

heather

Happy Thanksgiving everyone!

user116211

1:32 PM

Speaking of Thanksgiving, @JohnRennie, I saw Harvest Festival in UK predates everything.

Qmechanic

@DavidZ and other reviewers: Looking at the post again, I still think the present version of the post is not useful for the community, and should be closed, but I don't insist. OP should do a better job high-lighting his conceptional question, which is rooted in confusing work and impulse. Also his should include his (supposedly wrong) calculation of work because it plays a central role.

Jim

@MAFIA36790 That would make sense. Settlers in North America that created Thanksgiving were from (among other places) England. It'd make sense if they started Thanksgiving to mimic a festival they were already used to celebrating

@heather A bit late aren't you? Thanksgiving was over a month ago

heather

@Jim, what...?

Jim

@heather turkey, cranberries, pumpkin pie. That holiday. All over a month ago

heather

@Jim, please stop messing with my head =P

it's early in the morning

Jim

1:39 PM

october 10th, to be precise

heather

for a second you actually got me to believe you

Jim

I'm serious

heather

uh-huh

wikipedia link.

user116211

@heather October 10 is the Harvest Festival.

heather

@MAFIA36790, oh, okay, gotcha.

Jim

1:41 PM

@heather Thanksgiving

heather

@Jim, Thanksgiving may come from the harvest festival, but that doesn't mean it is the harvest festival.

Jim

no, I mean thanksgiving. I celebrated it october 10th

as did everyone else around me

Physics Meta

1:59 PM

1

Q: Misuse of custom close reasons

I was going through the close review queue, I saw the question Satellite is brought to the earth’s surface, it hits the earth with a velocity of 4 km/s - Determine the work done by friction [on hold] and voted to close as homework. However, before doing so, I saw the close reason "I'm voting to ...

user116211

2:10 PM

@PhysicsMeta I thought he was quoting a sentence from the post; but didn't find the exact line in the post.

user116211

Yo @bernardo.

Bernard Meurer

VHDL is the language of the devil

That's all I have to say

Slereah

Guys

What's an example of a non-compactly generated Cauchy horizon

All I can think of is like

Big ctc spacetimes + some molified step function

Not sure that would even Ork

Work

user116211

@Slereah 0celot might know.

Slereah

@0celo7 get your shiny red arse in here

user116211

2:18 PM

Googling the term, linked a paper of Krasnikov on time-machine.

heather

0celo is banned @Slereah...

user116211

noticed; thanks @heather.

user116211

@heather he can be pinged, can't he?

heather

@MAFIA36790, I think so, yeah.

The mods don't exactly appreciate it though.

Slereah

2:34 PM

Unban @0celo7

At least he knows what he's talking about

heather

@Slereah, I wish

ACuriousMind

@Slereah It will not ork but it might goblin.

4

Jim

2:51 PM

@ACuriousMind You seem to be making it troll

Slereah

I am starting to wonder if time machine production is at all possible without fucking things up

Already the Cauchy horizon has to be past incomplete geodesics generated

Pretty bad sign

ACuriousMind

@Jim This is ettin ridiculous.

Slereah

If it's compactly generated apparently there's a pretty general theorem regarding the propagator and the base points

I'm not quite sure if making it not compactly generated would help that much

Jim

@Slereah how exactly could one even have a practically useful time machine that doesn't violate local conservation of energy?

ACuriousMind

@Slereah Why would you want to make a time machine if not for fucking things up?

Jim

3:00 PM

Point: ACM

Slereah

Time machines don't violate conservation of energy

Jim

@Slereah so if I go back in time, it's okay that now the matter that makes up my body exists in two places at one time? Or how about the fact that before I arrive, there's nothing there and after, there I am. A big flouncy ball of energy that was previously non-existent

Slereah

No more than any other spacetime, anyway

It is okay

Jim

Even jumping forward, where's the energy gone between jumps?

Slereah

Because the second body was already there

There's no jumping involved in general relativity

It's all a continuous trip

The real big violation is unitarity

Jim

3:05 PM

Also, if I go back with a solar panel, absorb some photons and then use that energy to go back slightly further and absorb those same photons before I previously absorbed them (wash, rinse, repeat), I'm using infinite energy from the same photons

Slereah

I'm not quite sure you understand how things work in general relativity

Jim

Forward in time, no prob. Back in time, You can't do it in a useful way without making discontinuities in energy, momentum, or position

Slereah

This isn't Back to the Future

Jim

thank god

Slereah

Your time machine doesn't run on pepsi

Jim

3:07 PM

coke?

ACuriousMind

@Slereah Is it a snob and only takes afri cola?

Slereah

Pepsi probably paid them more to do it

It was the era of the cola war

Jim

Jumping isn't necessary. When you go back in time (forward in your own proper time, but backwards in coordinate time), what existed in your location the moment of coordinate time immediately before you arrived? Not you because you didn't travel back to that time

Slereah

It was actually you

Time travel in GR isn't that complicated if you stop overthinking it

Jim

So you travelled further back than you wanted to?

Slereah

3:11 PM

I mean it is terribly complicated

But not metaphysically

You have a geodesic

You follow it

That's about the size of it

Jim

but that geodesic doesn't intersect with that location and coordinate time.

Slereah

It's no more complicated than doing it in flat space

Jim

...

Slereah

The complicated part is

THE CAUCHY PROBLEM

Jim

You are missing the point

Slereah

3:12 PM

Dun dun duuun

I think you are under the impression that the past existed differently before the trip

Jim

no

Slereah

That is the error not to commit

Jim

Any practically useful time travel geodesic will allow you to visit a time in the coordinate past while not requiring you to travel through all of the universe's history to reach it

Slereah

Then I have no idea what you're talking about

What is "the coordinate past"

Jim

which means at some coordinate time, you will not exist (the matter/energy that currently makes you up will, but it hasn't travelled back yet, so it doesn't count). Then, bam! You exist

Slereah

3:16 PM

What

Jim

Proper time vs coordinate time. Practically useful time travel doesn't involve going backwards in proper time, because that's just stupid and impossible to verify. Therefore, it must be coordinate time you travel backwards through

Slereah

You can go back in "coordinate time" without time travelling

Time travel is defined coordinate independantly

ACuriousMind

@JohnRennie lol, do you want to start an interpretation war there? :P

John Rennie

@ACuriousMind it's true! (John says indignantly whilst trying not to grin :-)

Actually I should delete those comments - they aren't especially constructive.

Slereah

dropbox.com/s/fmkco5g31yqkej1/Strongly%20causal.jpg?dl=0

Example of going back in coordinate time with no CTC

ACuriousMind

3:24 PM

@Slereah ...the worldline of the time traveller is going to hit that removed part, isn't it?

Slereah

I wouldn't if I were him

ACuriousMind

Well, but he can't avoid it, can he?

That is, does this spacetime even make any physical sense?

Slereah

Do you know what happens to geodesics that bump on removes point

Define "make physical sense"

John Rennie

Should this be reopened?

-2

Q: Finding eigenvalues in quantum mechanics

How do you find the wave function, $\psi(x)$, when given the Hamiltonian, and the equation: $$ \hat{a}\psi(x) = \alpha\psi(x) $$ Where I know the operator $$\hat{a} = (1/sqrt(2))*((\hat{x}/sqrt(h-bar/mw))+i(\hat{p}/sqrt(mw*h-bar)))$$ And the Hamiltonian, $$\hat{H} = \hat{p}^2 /2m + (mw^2 \hat{x...

quantum-mechanics homework-and-exercises operators wavefunction eigenvalue

Slereah

It's probably not a good model of our universe

John Rennie

3:25 PM

It's been edited to make sense, but now it seems like a homework question.

Slereah

It's fine withî GR though

ACuriousMind

@Slereah In this case, "can we tell a coherent (if crazy) story about what the time traveller sees"?

Slereah

Sure

The time traveller won't see anything time travelly

Since no causal curve is closed

ACuriousMind

Well, what does happen when they hit the removed part, then? Do they just cease to exist?

Slereah

Yes

ACuriousMind

3:26 PM

Well, that's a shitty time machine. :D

Slereah

That's what happens when you hit a singularity

Well it's not a time machine

The spacetime is 100% causal

It's not strongly causal though

ACuriousMind

@JohnRennie Why should it?

We don't reopen things that still fall afoul of a policy, even if it's not the one named in the close banner.

Slereah

I mean having singularities isn't usually considered that crazy

Kerr has spacelike singularities

Those are pretty bad

ACuriousMind

@Slereah But they're usually not naked!

John Rennie

@ACuriousMind yes, leaving it closed makes sense. It seems silly to reopen it just so we close it as homework.

ACuriousMind

3:29 PM

I think most people would consider a naked singularity unphysical

Slereah

They are in the extreml regime!

Sir Cumference

I'm sorry, I'm confused. If light is redshifted by the expansion of the Universe, wouldn't that violate conservation of momentum?

Slereah

Well naked singularities basically have the same problem as CTCs

Loss of unitarity

@SirCumference no conservation of energy in GR

John Rennie

@SirCumference no, because the red shift is isotropic.

Light coming from all directions is redshifted equally.

Slereah

The FLRW metric isn't static

ACuriousMind

3:30 PM

@Slereah He asked about momentum!

Slereah

No reason for energy to be conserved

Oh

ACuriousMind

FLRW certainly is spatially homogeneous and so momentum is conserved, no?

Slereah

Then yes that

John Rennie

@Slereah these part time relativists ... :-)

Slereah

The average momentum should be 0 anyway

At least in he comoving frame

There's a relatively simple example of the problem of singularities in QFT

Eletron orbiting a black hole

Time evolution isn't unitary

It corresponds to the electron crashing in the singularity

Naked singularity is even worse

It can act as a source or sink for fields

ACuriousMind

3:36 PM

@Slereah Well, luckily, that take infinite coordinate time for us outside observers, right ;)

Not sure violation of unitarity in $+\infty$ years is really bothering me

Slereah

Think of the poor infalling observer

Hés gon'a die and witness a violation of unitarity

John Rennie

@ACuriousMind I'm not sure it does take infinite coordinate time for a naked singularity ...

It certainly does if there's a horizon, but I don't think it does when there is no horizon.

Slereah

He's gonna BE a violation of unitarity

ACuriousMind

@JohnRennie I didn't respond to the naked one, that's just distasteful.

Slereah

Terrible end

ACuriousMind

3:39 PM

@Slereah Sounds kinda awesome, actually

Slereah

Well you know where to die

Though that probably disappears with quantum gravity

ACuriousMind

"Violated unitarity" is a one-of-a-kind epitaph

Slereah

Also the epitaph wouldn't be that

John Rennie

Here lies the body of S.M. Lereah - 1985 to $\infty$. The probability he violated unitarity is 110%.

4

Slereah

"Will violate unitarity at timelike infinity"

If on earth

ACuriousMind

3:43 PM

@JohnRennie Wait, how did you recover the body?

Slereah

You can't even have a grave

Since you don't ever die

John Rennie

Details, details ...

ACuriousMind

Is this tombstone from the year $\infty + 1$? oO

Slereah

More of a lost at sea monument

John Rennie

From the year $\aleph_1$

ACuriousMind

3:44 PM

That's a cardinal, I think years should be denoted by ordinals ;)

Slereah

Year $\omega + 1$

You know I wonder what the green function of a space with a naked singularity looks like

heather

I'm so confused right now, but I'm still laughing =P

Slereah

ae100prg.mff.cuni.cz/pdf_proceedings/Chirenti.pdf

"Moreover, a naked singularity created from some exotic initial data conditions should become quickly neutralized (classically, or via quantum pair production). "

No fun at all

"Despite of this nice property of the scattering problem, the spacetime is non-globally hyperbolic and the time evolution of the fields is not unique "

The horror

"This means one has to specify additional boundary condition at the singularity to obtain a fully unique time evolution."

"Is there any intuitive physical condition that we can further impose on the fields, that will uniquely select the appropriate Green’s function? At least to get the geometrical optics continuous extension of the black hole case one can impose the condition that nothing falls in or out of the singularity. This means there is neither absorption nor superradiation in the scattering and the S-matrix of the K-G field is a unitary operator."

ie the most boring condition

Secret

What happens if I touch the material that made up the naked singularity, according to theory?

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