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1:09 PM
@MAFIA36790 $\omega=1$
can anybody help me
why nobody is replying
to me
that is clearly a homework question
can i tell my try
they you will answer
@Slereah I have the proof:
Actually, not quite
When I assume $\lambda$ is real, I need to show it's positive
Not sure if that's possible...
@ACuriousMind An orientation-preserving iso can still have negative eigenvalues, right?
1:25 PM
If it has an even number of them, yes
Do you know how to save the argument?
@Jim Well, the score/answer ratios also can mean that these users only bother to answer popular questions. Every metric you can conceive of is flawed somewhere :P
@MAFIA36790 @Jim
1:46 PM
One more exam to go :D :D :D
@Danu Can you help? :)
I'm not in the mood for helping with your work, sorry.
@Danu You are the only person I know to use ":D" to describe the fact that exams are not yet over :P
@Danu no worries!
@ACuriousMind 4 out of 5!
@3750 Penrose has fallen into the rabbit hole that is understanding consciousness... And he came out changed :P
@ACuriousMind Plus I've got like 8 days now to prepare for a single exam
The luxury! :D
2:03 PM
@ACuriousMind You always have to point out the obvious logical flaws in my arguments, don't you
@NoahP where is the cosmological constant term?
@Jim Exactly! :)
No term, no worries?
I suppose it would be a valid approximation for early times, but you would still need to actually perform the integration
Can I show you the question I'm actually trying to answer? @Jim
sure, but I'm at work in the labs now, so I don't make promises on helping out with it in the near future
Ok, fair enough
Its 4
2:33 PM
I see. So what was the solution to the continuity equation?
If you see the maths chat, I just did it, but TLDR; $\rho=Ca^{-6}$. Though the mathsy people are arguing that $H$ could be negative
Also, equation 4 on the sheet is missing a - in front of the $3H$
2:51 PM
H won't be negative unless the universe is contracting
What I said
However, despite that, no closer to a solution
$$-\nabla p -\rho\nabla \phi = 0$$ where $p$ is pressure and $\rho$ is density; $\phi$ is a scalar potential;
Feynman says, "it has no solution".
Why is it so?
BTW, that is equation of hydrostatics.
2:57 PM
because we don't have enough information to constrain a solution?
@Jim ah.
He says there can be solution only when $\rho$ is constant.
Otherwise, convection current would start up. I didn't get that.
never did fluids myself
Damn Feynman; the all-rounder T__T
3:01 PM
"Suppose that the Universe as spatially flat, and at some stage in its evolution, is dominated by a fluid with a so-called sti ff equation of state: $p=\rho c^2$"
Does that mean I can say $\Lambda=0$?
@ACuriousMind I fixed the proof, no thanks to you
You know I suck a linear algebra
@NoahP no. $\Lambda$ dominates at late stages. The question just says the fluid dominates at some stage. The universe being flat means only that $k=0$
@Jim And with that message, so dissapears my last hope
@NoahP: hi, I'm here though a bit tired after driving all day.
Hi John! You have come in my hour of need!
@JohnRennie Though if you're too tired then I understand!
3:34 PM
Now you you have $\rho = \frac{\rho_0}{a^6}$ it should be easy from here, or am I missing something?
I've tried all sorts on the Maths SE, and on here. I can't seem to get anywhere further than some horrible equations
You're told the universe is flat so $k = 0$, and there is no dark energy so $\Lambda = 0$.
woah woah woah
Are you sure no dark energy?
The question says that "at some point it is dominated by a fluid"
Im very happy for $\Lambda=0$ to be true, but not sure
3:37 PM
Well start with no dark energy and see what happens
In that case the first Friedmann equation becomes:
$\frac{\dot{a}}{a} = \sqrt{\frac{8\pi G \rho}{3}}$
OK so far?
Yep, got that
then substitute in $\rho$?
Substitute for $\rho$ and you get:
$\frac{\dot{a}}{a} = \sqrt{\frac{8\pi G \rho_0}{3a^6}}$
if John's got this, then I can relax. I've been coming back to check the computer intermittently every 10 min or so. Not enough time to guide you through math solutions though
3:39 PM
Which simplifies to:
@Jim Its been very much appreciated
$a^2\dot{a} = \sqrt{\frac{8\pi G \rho_0}{3}}$
This is where we horrify the mathematicians by treating $\frac{da}{dt}$ as if it was a fraction, and we multiply both sides by $dt$ to get:
$a^2 da = \sqrt{\frac{8\pi G \rho_0}{3}} dt$
And now we just integrate both sides to get:
3:42 PM
$\frac{a^3}{3} = \sqrt{\frac{8\pi G \rho_0}{3}} t $
So $a(t) \propto t^{1/3}$
should get $a\propto t^{1/3}$... ^ yeah that
@Jim ha, beat you! :-)
I could marry you
Maths SE had been floundering for around an hour with that
3:43 PM
serious? standard result
No joke
If you add back in the dark energy then I don't think the integral has a neat closed form solution
That may be why you stunned the mathematicians
You sure about $\Lambda=0$?
But note that dark energy only dominates at late times, so if your matter is dominant at some time then dark energy can't have been significant before that time. It can only start having an effect at later times.
I'm very hesitant to now go anywhere near 4c
3:46 PM
@NoahP well I didn't set the question, but it has a simple answer if $\Lambda = 0$ and no simple answer if $\Lambda \ne 0$.
Unless the chap running the course is a sadist I'm sure he meant no dark energy.
@JohnRennie It would make sense
And they are a dark energy cosmologist
So fingers crossed
Im going to go with it anyway
The Hubble parameter is just $\frac{\dot{a}}{a}$, and now you know $a(t)$ it's straightforward to calculate H.
Likewise $\rho(t)$ since it's just $\rho_0/a^6$
well, you know what $a$ approaches at late times
3:49 PM
@Jim Yes!
:31164197 oops, typo.
Aha ok
I must admit I had got stuck at $\frac{\dot{\rho}}{\rho} = -6\frac{\dot{a}}{a}$, so your maths friends helped out there.
Yeah, collective effort
So $\dot{a} \propto \frac{t^{\frac{-2}{3}}}{3}$?
I dont know how to do proportionate sign
3:52 PM
See, general relativity is easy peasy. I don't know what all the fuss is about.
$H \propto \frac{\frac{t^{\frac{-2}{3}}}{3}}{t^{\frac{1}{3}}}$?
@MAFIA36790 I was panicking looking for excess { or } there, thanks
@NoahP: Why not you work with time-dependent $\omega\;?$ It would be a good exercise.
$H \propto t^{-1}$ isn't it?
3:56 PM
@MAFIA36790 Believe it or not, that is the next question
@NoahP I can help then.
@JohnRennie How are you getting rid of that 3?
@NoahP O.O
Indices addition.
@MAFIA36790 ?
3:57 PM
The factor of 3 is just a constant. If we're only looking at proportionality we can ignore constant factors.
Ahhh ok
@MAFIA36790 I didnt mean the indices
@NoahP ooh.
wikipedia for Lamdba-CDM model gives the correct answer for $a(t)$
complicated. Uses $\sinh$
And finally, $\rho \propto t^{\frac{-1}{3}}$?
ignores radiation (which is bad for rad-dominated era, but that was a lame era anyway)
4:00 PM
@NoahP are you sure about that?
I forgot the power to the 6 thing
So no
Its been a long day at the library for me!
Actually I think the FLRW metric is the ugliest of the simple analytic solutions. The black hole solutions are far more elegant. Can't you get your chap to set you problems on black holes?
The most elegant is Minkowski @JohnRennie
And if you can prove it's a stable solution, you will be pretty good at GR
Well, OK, I guess that's true :-)
I wish, enjoy learning about black holes, what I did my EPQ on. And $\rho \propto t^{-2}$?
4:02 PM
@NoahP that's what I get.
Great! I'm going to leave 4d for now then
Thanks for all the help everyone
The scary thing is, the other 3 doing the project were stuffed at q2 - if you think I'm struggling, then you can only imagine the rest
@MAFIA36790 That's very different from "no solution"
Well q2 is a bit rubbish. It's a plausibility argument for the form of the Friedmann equations but it's a fake.
It's what the book we've been given uses
It relies on doing the integrals in just the right way to get the answer you want.
4:06 PM
Yeah, but I'm going nowhere near all of that tensor stuff with einstein
That isn't the way Friedmann did it :-)
@ChrisWhite As I said, I've no idea what he wrote. He then went to other topic without discussing it further. Maybe a single line was sufficient but at least not for me ;/
I can imagine!
Tensors are warm cuddly things that only want to be your friend if you treat them nicely.
You'll meet a lot of tensors if you do a physics degree :-)
There is already a fine physics y introduction of tensor and tensor ellipsoid in Feynman Lectures; it is intuitive and quite lucid and you can try that.
4:10 PM
I hope to do a physics degree, but no tensors yet please!
What is q2
What is wrong with tensors
@NoahP yeh, yeh I know that paper; you bored us with that many times ;P
@0celo7 asked!
just kidding
4:12 PM
And its double sided ;)
I can't remember how the barotropic parameter is defined, though I'm sure it's only a Google away, and I doubt 4d offers any great challenge.
@JohnRennie I'm thinking its just $c^2$?
From the definition given of $p=\rho c^2$
No, it's a number in the range -1 to 0.5 or something like that.
@MAFIA36790 I'm not sure that statement can even be interpreted to be correct. Sure, there exists a phenomenon of convection, but that doesn't mean you can't momentarily satisfy that equation with no forces (convection is an instability, not an externally driven phenomenon).
It can be any number
4:15 PM
Moreover, there exist convection-free fluids! The stratosphere is by definition stable against convection.
Oh, maybe 1
I'm not sure
Thats for tomorrow
The general condition as to whether convection occurs in that equation is whether the entropy gradient has the same sign as the potential gradient. (You'll more often see things like the Schwarzschild and Ledoux criteria, but these are restatements of that same principle.)
@NoahP Aha, yes, $w=1$. See this section.
@ChrisWhite: He says if $\rho$ is constant, then only the potential term becomes pure gradient and the solution would be $p +\rho\phi = \text{const}.$ What does he mean by pure gradient?
woop woop! @JohnRennie
4:20 PM
@MAFIA36790 If $\rho$ is constant you can move it inside the gradient, and then you have $\nabla(\rho\phi + p) = 0$.
@ChrisWhite ohh. Thanks.
@ChrisWhite: So, only if $\rho$ is constant, then can only the equation has solution?
@MAFIA36790 That seems to be what you're saying Feynman is saying. I disagree. Density is not constant in the stratosphere, nor in the radiative zone of the Sun, nor in large parts of the ocean that are not convecting.
They are all in hydrostatic equilibrium though, and in the Sun (and to a lesser extent the stratosphere) this is despite the convective destabilizing effect of a heat flux.
Right, Im off now then. Thanks all!
@JohnRennie @Jim @MAFIA36790 Special thanks for putting up with me
4:27 PM
@NoahP you're welcome. It's always humbling to discover how much I've forgotten about the Friedmann equations.
@ChrisWhite Well he at least says so here; you can check that if I missed something:
> the equation of hydrostatics. In general, it has no solution. If the density varies in space in an arbitrary way, there is no way for the forces to be in balance, and the fluid cannot be in static equilibrium. Convection currents will start up. We can see this from the equation since the pressure term is a pure gradient, whereas for variable ρ the other term is not. Only when ρ is a constant is the potential term a pure gradient. Then the equation has a solution.
@NoahP Don't marry all of us ;)
I really have no idea what he's up to.
4:44 PM
what's this people are talking about physics in the h bar?
@3750 Here only one man talked physics ;) We are just chatting.
I will feel honor if ACuriousMind votes one of my posts up, but I think I never will see that because he/she is too much fastidious! :-)
Went from 4 votes to close and downvotes, to no votes to close, 6 upvotes, comments and answers (on the adult site), ;) physics.stackexchange.com/questions/267574/…
5:06 PM
@CountIblis I have a small quantum field theory conceptual question, maybe you can help me?
why microcausality will require spacelike seperated field operators to either commute or anticommute?
It is naive question.Why is the mathematical expression of a the phase difference of superposition of 2 sinusoidal waves is different from that of two progressive wave?
Do you mean $sin(\theta_1)+sin(\theta_2)$ vs $sin (k_1 x-\omega_1 t)-sin (k_2 x-\omega_2 t)$?
@Bernard o/
@MAFIA36790 Howdy
5:12 PM
Any interesting stuff you doing @Bernard?
@MAFIA36790 Slowly and painfully dying (aka. living)
Jokes apart I'm working on a pretty cool project
I think it's cool at least
@BernardMeurer Well, acquainted with such situation; just drink lots of coffee/beer.
@BernardMeurer coolio ;P
@MAFIA36790 No money for beer, the abstinence is already kicking in
5:15 PM
My home stash is running low
Well, I'm off for now; good nite @Bernard and others.
@MAFIA36790 See ya mate! Goodnight!
5:47 PM
Assuming the two waves are emitted from the same source. Then the path difference is given by the difference between the distance of the two waves

For ordinary sinusodials of the form $sin(\theta_1)$, where $\theta_1$ is $\omega_1 t$, the period is $T_1=\frac{2\pi}{\omega_1}$ radians. Therefore the wavelength is given by $v_1=\frac{\lambda_1}{T_1}\Rightarrow \lambda_1=T_1v_1$ where $v_1$ is the speed of the wave (for light, $v=c$, for other waves, it depends on the medium that the wave propagate in)
Corrections: $\Delta (x)=\lambda_2-\lambda_1=2\pi \left(\frac{v_1}{\omega_1}-\frac{v_2}{\omega_2}\right)$ and $\Delta(x)=(\lambda_1+\lambda^*_1-\left(\lambda_2+\lambda^*_2\right))=2\pi \left( \frac{v_1}{\omega_1}+\frac{1}{k_1}-\frac{v_2}{\omega_2}-\frac{1}{k_2}\right)$
6:25 PM
@Secret Can I convert path difference into phase difference by multiplying with k?
1 hour later…
7:28 PM
@Secret Thank you very much for your help.
2 hours later…
9:56 PM
Hi all
hey guys, whos an experimentalist here
This is driving me crazy. I cannot remember a specific word.
What is the adjective for a theory or first principle that actually can be derived from more fundamental principles?
An example would be Kepler's laws or the Rydberg formula
Or the periodic table before the selection rules of QM were discovered
I am only a lowly 3-4th year undergrad student
@WilliamBulmer What does that have to do with anything
what does what?
You being an undergrad
10:04 PM
Oh, I was responding more to user507974
@ACuriousMind Prof's words: "wow I don't remember Milnor using that much algebraic topology"
Is anyone here a neophyte like me?
@WilliamBulmer axiom?
actually no, axiom would be that one step more fundamental
ab initio
is waht you are looking for
@user507974 Thanks, but I am looking more for an adjective for something was taken AS an axiom, or maybe was simply just based off of observation Kepler's laws, but was later explained by more fundamental principles
Rydberg's formula is a good example, but perhaps a better example would be the E and B fields of EM being replaced with the covariant formulation of EM via SR
I think the word starts with a P
10:54 PM
@WilliamBulmer Sure
11:21 PM
In QM the state of a system can be specified by the eigenvalues of a complete set of

commuting observables. Suppose that you have such a complete set of commuting observables

for a local system and you then consider another system that is spacelike separated from the

first system. If the observables for that system do not commute, then that would mean that

some observables for the second system can be expressed in terms of observables of the first

system. So, you could measure some property of the second system using only a measurement
@skillpatrol got a fly trap
Q: Probably HQ user but with LQ start, I suggest tolerance

peterh(related PSE post) Here is a link to a VLQ answer from a new user, who may be capable later to produce high level content. Here is the site which he refers in his post. This time it seems to me not as the "I found the Holy Grail, gimme Nobel Prize now" line. It seems to be a prof emeritus, but ...

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