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2:22 AM
(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
 
 
2 hours later…
3:55 AM
0
Q: Reasking a Question

Tanmay KulkarniWhat should I do if I find a question that I want to ask but does not have satisfactory answers, how can I make it active or re-ask the question without it being marked duplicate?

 
 
1 hour later…
5:00 AM
@WilliamBulmer Postulate?
 
 
3 hours later…
8:29 AM
Morning all!
 
9:14 AM
Hi guys, I have a little issue with matrix traces in quantum mechanics density matrices that I don't understand. Based on this discussion:
http://chat.stackexchange.com/transcript/message/31105713#31105713
If I do this "element-wise" time evolution, then the diagonal elements of the matrix are unchanged
but then the trace is constant!!! right?
The literature says that the trace of some time independent operator Tr(rho.A) is my expectation value
but this is then constant!!
The solution that we use in our calculations is that we sum ALL the matrix elements, and still call it trace. Why is that correct?
 
@MAFIA36790 You there?
 
user116211
@NoahP go on; bit busy now so maybe would be late to respond.
 
Ok - just a quick question. If I'm considering a universe which is radiation dominated, in order to find the time dependence of the density of matter, can I ignore cosmological constant? It makes it a hell of a lot easier @MAFIA36790
Just want to check im not commiting a cardinal sin or anything
 
user116211
9:30 AM
@NoahP Well, as I said, I'm not into cosmology yet; so can't comment. But if there is maths part, I may help.
 
Ahh okay, no worries. I think that im doing okay with the maths so far, thanks
 
@ACuriousMind Could you please look into my question, since you know the problem?
 
user116211
@yuggib: o/
 
\o
 
9:42 AM
@MAFIA36790 I've got to $p^{-1}dp=-4a^{-1}da$
It's not going to work if I integrate that is it?
 
user116211
@NoahP Why not?
 
Would I not get stuff to the power of 0?
 
user116211
@NoahP Elaborate your next step to solve the equation.
 
Wait, would I get $ln(p)=-4ln(a)$?
 
user116211
@NoahP looking good.
 
user116211
9:45 AM
Although lacking the constant.
 
Okay, thanks! And yeah, but I'm trying to find proportionality
 
user116211
@NoahP BTW, use \ln to get $\ln$
 
Ohh ok thanks
Do I then use () or {} for the terms?
 
Hello. Any one willing to explain me about the cosmological constant?
 
user116211
@NoahP if there is more than one term, yes, you have to use the latter one.
 
9:47 AM
Depends what you want to know @SwapnilDas ? I've learnt a bit over the past few days, I'm not sure who else here could help right now
Certainly there are people who can, I'm just not sure who is online
@MAFIA36790 Ok, thanks
 
Any thing you could tell, I must be indebted to you.
 
:31185348 I don't get the \o stuff
@SwapnilDas What do you know, whats your backround?
 
user116211
@NoahP oh, man; google emoticons.
 
@MAFIA36790 I've never seen that before though?
 
I am a tenth grader with interest in Cosmology
 
9:49 AM
Like :P ;) and everything yeah, O.o, but the \o is new to me
@SwapnilDas What do you know so faR?
 
Cosmological constant was Einstein's biggest blunders. It predicted a 'static universe model'
blunder*
 
I need someone's help on probability...
If I roll three dice, what is the chance of getting a result sum up to six only?
I can't think of a way other than list all the events....
 
@Shing Thats probably one for the maths chat, but a quick google would help you too
@SwapnilDas The cosmological constant refers to dark energy, most would agree
Though take everything I say with a pinch of salt - I'm not sure what grade I would be in (UK, year 13), but I'm 17 myself
And learnt all this in the past week
 
@NoahP um....okay, though I am just working on the problem set of statistical physics.
 
@Shing My solution would be to sum all the results in a table, but im sure they could suggest quicker, and its pretty quiet here atm
 
9:56 AM
@noa
Cool
 
I haven't studied probability for more than 15 years, but I think you can break your problem in two problems for simplicity. @Shing
 
I'm 15
 
Ok
 
If first dice come 1, then sum of the rest dices must be 5 and so on
2, 4
3, 3
4, 2
 
thanks, I list all the events and then obtain the correct answer, but then I realized what if the question is 100 dice? therefore, I am thinking hard of a way to calculate it, instead of listing all the events.
 
10:02 AM
I don't know about mathematics solution, but I think you can create an algorithm by the method that I said.
For computer solution.
(programming)
 
um... that's a way to do it, thanks!
 
Good luck!
 
@MAFIA36790 Shh, whisper it quietly, but I think I may have done Q5 without any (Apart from that quick integral) SE intervention...
 
11:01 AM
Hi all, just wondering about this matrix problem and require a bit of help. Here $J$ are symplectic matrices, $H$ is the Hessian of the Hamiltonian with respect to a point in phase space $x$ and $\eta$ and $\xi$ are phase space vectors. $(JJH\xi)\cdot \eta +(J\xi)\cdot JH\eta = -(H\xi)\cdot \eta +\xi \cdot (H\eta)$. I have no problem with the first term since $J^2=\mathbb I$, but how does the second term generate a negative sign! :)
*not generate a negative sign
 
11:14 AM
How do I filter out tags?
Ah found it
 
@lucas I think there is no direct method, at least binomial distribution doesn't work. since binomial only deals with "two outcomes". I couldn't really make the question into two outcomes.
 
@Shing I don't know of any way other than enumerating the possible outcomes either, but you can be a little clever and use a generating function: $(x + x^2 + x^3 + x^4 + x^5 + x^6)^3$. Expand that and add up the coefficients of the terms with powers of $x$ up to 6, then divide by the total number of outcomes.
 
11:36 AM
@DavidZ thanks for the help, I will try that out!
 
Make sure you don't leave the problem until you understand what that works.
 
okay :)
 
12:05 PM
Hey ... just a quick and stupid question (I haven't done QM in a while). For a superposition of eigenstates of a Hamiltonian in Schroedinger QM, the time evolution of the superposition is given by letting each eigenstate evolve with its own eigenfrequency, right?
 
operators are linear, so yes
 
yes, in general, for a time independent Hamiltonian
 
yep, time independent Hamiltonian, I forgot to add that; thank you both :)
 
@Shing I meant you can make your problem easier by breaking it. For example: 1/6 P(5)+1/6 P(4)+1/6 P(3) +1/6 P(2) and P(5) is the probability of sum of two dice to get 5 and so on. Then for P(5), ... you can do the same method.
And as I said I have completely forgotten that topic. It was a suggestion.
 
Jim
12:46 PM
A user accused me of behaving in a way that does not belong in the world of physics. I calmly explained why what he is arguing doesn't matter in the slightest, outlined a hypothetical (and likely true) situation that would show his motives as egotistical only, and explained that through my use of words I never had acted the way he claimed and that he should not repeat the claim without sufficient understanding in what I was saying. Then I dropped the mic and walked out of the conversation
#ThatllLearnHim
 
@TheQuantumPhysicist While the trace of $\rho$ is constant, that of $\rho A$ is not.
 
@ACuriousMind Hi :-)
 
@lucas Hello
@Jim Is this where I point out a flaw again? ;)
 
Jim
@ACuriousMind You must be a jedi, because every time I talk to you, I come out thinking I made a mistake somewhere
 
@Jim I'm...not sure how I feel about that
Never liked the Jedi much
(nor the Sith, for that matter)
 
Jim
12:53 PM
maybe they never liked you either
 
More of a Princess Leia?
 
Jim
for reference, here is the conversation:
@Jim 1) Analyze the very concept that all objects are moving across space-time at c, and that in turn the only change still left for each of these objects is a change in each of their specific directions of travel within space-time. 2) The inevitable outcome of this analysis is an independent discovery of SR, along with an independent deriving of all of the SR mathematical equations. Granted, some folk of today would say that this is merely a magical coincidental outcome. — Sean 21 hours ago
 
@Jim The feeling might well have been mutual
 
Jim
On a different note.....
here's my reaction when the computer all my unsaved work is on suddenly crashes:
user image
2
also when I come into work and suddenly realize it's Saturday
 
I don't want to see how you look on a Sunday, then :)
 
12:57 PM
May I ask what is your job?
 
Jim
physics lab coordinator
 
Ooh la la
 
@Jim That particular user has been at this for years. If you actually manage to change his mind, I'm beyond impressed.
 
I am unemployed:-(
 
Jim
@ACuriousMind change his mind? Doubt it. Win the conversation? Probably
 
12:59 PM
what is "winning a conversation"?
 
Jim
Stumping the other person momentarily and walking away before they have a chance to speak
without asking a question
 
I mean, if there is an audience, it works; but most audiences are dumb and can be won without being right, so I question the worth of these "victories"
that seems like a very mature way of handling conversation :o
 
Jim
makes you feel good
 
@Jim So, if I say "flargharble" so that the other person is momentarily confused and then walk away, I win?
 
1:01 PM
well, I'm happy it does that for you :)
 
Man, I've been playing this "conversation" game all wrong
 
Jim
@ACuriousMind I think you'd get a yellow card for that
 
愛天使伝説 <- I want a red card for that
 
1:16 PM
@ACuriousMind But we've agreed that element-wise multiplication is valid with $\exp(-i(\omega_i - \omega_j)t)$, right?
 
@Sanya ::flagged as inappropriate:: suspension pending upon approval by TPTB </joke>
 
@TheQuantumPhysicist Wait...what do you mean by "element-wise multiplication"?
 
I mean every element from one matrix with the corresponding one in the other
 
No
If I said that, it's wrong. Why would the time evolution change how the ordinary matrix multiplication behaves?
 
@ACuriousMind Because the components are independent
and the evolution doesn't mix them
because the density matrix is in the eigenbasis of the Hamiltonian
 
1:18 PM
You can evolve each element of $\rho(0)$ to $\rho(t)$ independently, but when you compute $\rho(t) A$, you have to properly multiply these two matrices
 
Interesting...
 
So the first entry of $\rho(t)A$ is $\sum_{i} \rho_{1i}(t) A_{i1}$, where $\rho_{1i}(t) = \exp(-\mathrm{i}(\omega_1-\omega_i)t)\rho_{1i}(0)$.
 
How has it been working with this guy all along... I wonder!
 
You're telling me that you have simulation code where $\rho A$ is computed as having $\rho_{11}A_{11}$ as the first entry? oO
 
Yes... and it works, and fits experimental data
 
1:20 PM
...I have no words
 
Neither do I xD
 
That sounds utterly impossible
 
Well maybe it's possible because the guy starts with a thermally populated state along the quantization axis
populating only the diagonal element of the density matrix
 
Ah
Yes
 
then he multiplies that with that operator, element-wise
 
1:21 PM
If $\rho$ is diagonal, that works.
 
and THEN, he evolves with element-wise evolution operator
hmmmmmmmmm
Seriously... worst code I ever dealt with that produced the best result
this is horrible!
So I guess to make this generally correct, I must do proper multiplication, and then evolve with element wise
 
But for diagonal $\rho$, I would say that $\mathrm{tr}(\rho A)$ is constant. Because diagonal $\rho$ in the Hamiltonian basis means that your statistical mixture is one purely made out of stationary states
 
or alternatively, multiply with the operator AFTER the evolution is done
 
And expectation values in stationary states should be constant.
So time-evolving such a diagonal $\rho$ seems pretty pointless to me
 
Actually it's diagonal along the quantization axis
 
1:24 PM
Because it doesn't evolve at all
 
but then is converted to the eigenbasis of the hamiltonian
that still makes it invalid, right?
 
Everything is in the same basis
maybe this is what makes it valid
 
Well, if he does the element-wise multiplication in the basis where $\rho$ is diagonal and then switches to the Hamiltonian basis and then evolves it element-wise there, I guess that works
 
@ACuriousMind morning
@ACuriousMind Did you know that $N(X\times\{a\};X\times X)$ is trivial?
 
1:27 PM
Not really... he:
1. Populates rho thermally
2. Converts the rho basis to the Hamiltonian basis
3. Converts the operator basis to the Hamiltonian basis
4. Multiplies rho with the operator, element-wise
5. Evolves using the element-wise evolution
6. Calculates the sum of all elements as the expectation value
And that works some how...
@ACuriousMind
 
@0celo7 I have no idea what that notation means
@TheQuantumPhysicist oO
 
@ACuriousMind Dude... I'm as surprised as you...
 
@ACuriousMind Relative normal bundle
 
If the sequence was 1-4-2-3, I'd believe that it works, but that way it just seems wrong
 
Well, it agrees with experimental data for some reason
it's probably a special case...
 
1:29 PM
@0celo7 Yeah, I have no idea what that is, so no, I did not know it is trivial :P
@TheQuantumPhysicist Have you checked that it actually does, i.e. the code you have produces the correct predictions?
 
Yes. In fact we wrote an extremely fast and efficient fitting code
and we use experimental data with it
and we fit data with mHz accuracy!
some data reaches sub-mHz!
which is a revolution in NMR
 
@ACuriousMind Let $Y\subset Z$ be a submanifold, then $N(Y;Z)=\{(x,v)\in Y\times \Bbb R^n\mid v\bot T_xZ\}$
 
I meant do you know that the source code you're looking at actually belongs to the program you are using? I'd suspect being given a wrong version or something seeing this strange code.
 
Well, I took his Mathematica, inefficient code
and I translated it to extremely efficient and optimized C++
and I fitted experimental data
and it works...
 
I see
 
1:31 PM
and I'm unable to understand it
 
this is very strange indeed
 
Agreed...
What would you do in this case?
 
Ask the author
 
He left... I'll have to try to reach him...
@ACuriousMind Thanks for the help though. I appreciate it!
 
What does "Mainz" mean in your profile? @TheQuantumPhysicist
 
1:33 PM
Actually converting my code to be 100% legal, by multiplying with the operating after the evolution isn't that expensive
@skillpatrol It's a city in Germany?
 
Proof?
 
@0celo7 proof of what?
 
Where even is Germany
 
On Neptune
 
huh
 
1:34 PM
@TheQuantumPhysicist Well, it might be worth a try just to see if you get the same results
 
I'm using quantum entangled signals to communicate with earth faster than light @0celo7
@ACuriousMind Yep! That's my next step!
 
I figured it was in Uranus
 
No, no. You're mistaken
It's Neptune
 
@0celo7 Thanks for answering a question I didn't ask :P
 
Also you can't use quantum entanglement to communicate faster than light
 
1:35 PM
^
 
@ACuriousMind Did you see my message about Milnor
 
@Slereah Well my people did it! We have a super-modern civilization
 
My prof didn't understand the proof either
 
You "earthians" are still immature :P
 
@0celo7 I wasn't really here the last few days, so I'm not sure which of these you are referring to
@TheQuantumPhysicist I think we're supposed to call them "humans"
 
1:36 PM
@ACuriousMind Well we call them earthians on our planet
 
Thanks for stopping by mr alien @TheQuantumPhysicist :-)
 
Calling them humans is racist
 
@ACuriousMind You were right, he hasn't opened the book in 10 years
 
@0celo7 Ah, that one. Yes, I read it, I'm not surprised :D
 
No problem! Always glad to take a peek at earhians @skillpatrol
 
1:38 PM
@TheQuantumPhysicist I'm in Germany, so it should be my planet too ;)
 
Unless you somehow only transported Mainz to Neptune...
 
@ACuriousMind Oh, nice to meet you here mate! We're communicating over a few light-hours
@ACuriousMind No, no. Whole Germany is involved.
 
Yeah, I guess it is a rather inefficient way of communication
 
@ACuriousMind xD
 
1:41 PM
How's the fly trap working? @0celo7
 
they seem to like it
hope it kills them
and they're not bothering me any longer
link?
 
huh?
why did you argue with shog
 
because I wanted to
 
1:56 PM
@skillpatrol holy crap there's two dozen dead flies in the trap
 
It does work.
Hi @JohnRennie
 
Afternoon
 
When is afternoon tea usually served?
 
@skillpatrol In the afternoon.
@NoahP ?
 
2:06 PM
It seemed more humorous when I thought of it than now
And I cant seem to delete it
Oh wel@JohnRennie Managed q5 and q7 with no intervention!
 
Cool :-)
Though perhaps I should reserve judgement until after they've been marked :-)
 
Well, one was a proof
And the others agreed with results I found online
In a book actually I think
An introduction to modern cosmologu, by Andrew Liddle
 
@Secret Thanks a lot.
 
@JohnRennie If you have a spare $8,888 going spare, you could always head on over to the Ritz Carlton in Hong Kong, as they serve the world’s most expensive afternoon tea.
@ACuriousMind :P
 
but tea is gross
 
2:12 PM
@skillpatrol The things I enjoy spending my money on include phones, tablets and laptops, but exclude trips to Hong Kong for tea.
 
@ACuriousMind how does "plead the fifth" get dubbed in German
 
@0celo7 Probably like vom Recht zu schweigen Gebrauch machen
 
They don't have "the fifth" ;-)
 
@JohnRennie Do you think you could point me in the right direction for q6? Last one!
 
The right to remain silent is a legal right recognized, explicitly or by convention, in many of the world's legal systems. The right covers a number of issues centered on the right of the accused or the defendant to refuse to comment or provide an answer when questioned, either prior to or during legal proceedings in a court of law. This can be the right to avoid self-incrimination or the right to remain silent when questioned. The right usually includes the provision that adverse comments or inferences cannot be made by the judge or jury regarding the refusal by a defendant to answer questions...
@NoahP Oh go on then :-) Are you going to upload a picture?
 
2:15 PM
Sure
 
Well 6b is a simple consequence of 6a, so getting 6a solves both. Have you worked out the relationship between $a$ and $z$ yet?
 
I havent derived it but I know what it is
$z=a^{-1}-1$?
 
OK, I don't know the answer to this offhand, and when faced with a new problem I find the best way is to poke at for a bit and see what falls out.
 
The question is already very sore if we're using that analogy
 
Firstly if we look at (6) it's a lot simpler than it seems. Do you know what the physical meaning of $\Omega_i(t_0)\rho_c(t_0)$ is?
 
2:28 PM
(Incase 5 or 7 help)
And I recognise the critical density and density parameter there?
 
Yes, $\rho_c(t_0)$ is the current value of the critical density, so what is $\Omega_i(t_0)\rho_c(t_0)$?
 
The current density?
Assuming $\Omega=\frac{\rho}{\rho_c}$
 
The current density of ... ? Remember that $\Omega_i(t_0)$ is the current fraction of component $i$.
 
I'm not sure what $i$ is
 
Ah, OK, fair enough, the question oesn't make that clear.
We have three components, light, matter and dark energy.
And the subscript $i$ is a just a convenient way to label them.
 
2:33 PM
Oh, so its all of them?
When you say light, you mean radiation?
Just checking
 
For example we can (arbitrarily) choose $i=0$ for radiation, $i=1$ for matter and $i=2$ for dark energy.
 
Ohhhhhhhhhhhhhhhh
I get it
 
it is best to use letters
 
So its basically the same equation for all three, but using $i$ instead of writing out the same equation with different letters
Normally it would be $r$, $m$, and $\Lambda$
But $i$ can represent any of them?
 
Yes. Actually I agree with Sam - I would have used letters not the subscript $i$.
But then I didn't set the paper :-)
 
2:36 PM
I'll pass on the complaint ;)
 
OK, so what is the physical significance of the term with $i = 1$ then: $\Omega_m(t_0)\rho_c(t_0)$
 
so that would be the density of matter
 
Yes, lets call that $\rho_0m$ (rather than the unnecessarily fussy $\rho_m(t_0)$.
 
@slereah what's the name of the theory that poses there are infinitely many world-lines?
 
Ok, sure
 
2:38 PM
instead of the closed loop world line of CTC
 
And you've already told me that $1 + z = 1/a$. So take the $i=1$ case, substitute for $1+z$ and what does the equation become?
Remember that ordinary matter is pressureless, which sets the value of $w$ to be ... ?
 
Branching spacetime
it's not much of a theory
 
@ACuriousMind btw, Hirsch has a really cool definition of Morse functions
A function is Morse iff it's exterior derivative is transversal to the zero section of the cotangent bundle.
exterior derivative/tangent map/pushforward
 
That would make $\omega=0$?
 
@slereah thanks
 
2:45 PM
Yes, so what does equation (6) simplify to?
 
Giving $\rho_i(z) = \frac{\Omega_{i0}\rho_{c0}}{a^3}$?
Where $i=m$
 
And we've already decided that $\Omega_{m0}\rho_{c0} = \rho_cm0$ i.e. the density of the matter.
So the equation is just telling you that $\rho = \frac{\rho_0}{a^3}$.
 
I derived that in q5
 
Which should be obvious.
 
So surely I can go from the equations I got in q5 then?
$\rho_m \propto a^{-3}$ and $\rho_r \propto a^{-4}$
 
2:52 PM
are we still doing this
 
@0celo7 'this' could mean many things
 
Basically yes. For radiation $\omega=4/3$ and if you put that into equation (6) you're going to get $\rho_r = \frac{\rho_{0r}}{a^4}$, which reproduces what you already know.
 
And for $\Lambda$, $\omega=-1$?
 
physics.stackexchange.com/questions/269218/… can someone highrep please close that threat for being inappropriate, badly formulated and stupid?
 
For dark energy $\omega=-1$ and that's going to give you $\rho_\Lambda = \rho_{0\Lambda)$ i.e. the dark energy density is constant (which it is).
 
2:58 PM
Yeah, okay
So, how would one go about proving the equation 6 then?
 
@Sanya 3 close votes and counting, it'll be closed soon.
 

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