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8:00 PM
@Hippalectryon OK, I just got home. Did you solve it already?
 
@alarge Nope, I was doing something else
 
@0celo7 Hm, if you just had asked me "what does averaging over the direction of $p$ mean", I'd have said "integrate over the sphere with radius $p$, then divide by volume of that sphere". That LHS looks as if it is meant to be that, I'd assume a typo.
 
@ACuriousMind Ok, that's as I thought. He argues rotational invariance restricts the integral to have the form $\delta^{ij}$, and then the proportionality constant may be determined.
Why does rotational invariance restrict it like that?
 
Actually, it needn't be a typo - if you measure $\theta$ the other way, $\sin(90° - \theta) = \cos(\theta)$ (modulo sign), isn't it?
@0celo7 Hint: How many rotationally invariant 2-tensors are there?
 
True. I'd have to examine other surface integrals in the book.
@ACuriousMind Ok, I get that, but if I do this explicitly, I don't see how it works.
 
8:05 PM
@0celo7 Ah. Well, no idea how to do it explicitly, I think that's why one always goes with the tensor argument :D
 
Suppose I take $\vec p=(1,1,1)$.
 
I think I'll move on to the next chapter....
 
The integral is surely not zero if I multiply the various components and integrate.
Is $\vec p$ supposed to be a vector field?
 
@0celo7 Uh...you cannot take $\vec p$ as a constant, you are integrating over its direction!
 
Hmm, then I don't understand the problem exactly.
 
8:07 PM
You have to write $\vec p = \lvert p \rvert \times (\text{spherical unit vector})$ and integrate $p^i p^j$ of that, which will have the $\theta,\phi$-dependence of the spherical unit vector.
Not sure that was clearer than the exercise :P
 
@ACuriousMind One should then be able to verify the integral of the spherical unit vector components is zero, right?
 
@0celo7 Not exactly: The integral of the product of two different components of the unit vector should be zero.
 
@ACuriousMind That's what I meant ;)
 
Alrighty then :)
 
Seems rather tedious. @Icosahedron, you have a mission now. Verify that.
 
8:19 PM
@0celo7 The integral of the product of two different components of the unit vector should be zero?
I already knew that.
It's in griffiths E&m.
 
I knew it had an E&M connection.
It know it should, but you should do it explicitly.
 
That means I have to prove it?
That requires work.
 
@Hippalectryon Where did you get the equation from and what do you think the solution should look like?
 
Who starred my Zee comment? Do I recall correctly that @alarge is a fan of his?
 
8:31 PM
@Icosahedron Do you know where?
 
@0celo7 Near the beginning, though it was also in the lectures on mathematical physics by Carl Bender at PSI.
 
Near? I've skimmed the first 105.
It's possible I missed it.
 
let me see.
i know why.
 
Yes?
 
it might have been in purcell or shwartz
i'm looking now.
griffiths page 132 (3rd edition)
though that probably is not what you're looking for.
 
8:45 PM
Huh? Orthogonal functions?
 
yeah.
 
I thought we were talking about unit vectors.
 
Because the dot product of two orthogonal unit vectors is 0, then the integral of their dot product is 0 too, satisfying the orthogonality condition.
Is that bs?
 
Probably.
 
I read that from somewhere, or heard it in a lecture.
 
8:49 PM
We're not talking about products of unit vectors here.
We're talking about the product of components of the unit normal vector.
 
@alarge Do you need more information ?
 
@0celo7 Misread.
 
Maybe @Qmechanic starred my Zee post, I know he's read his books.
 
@Hippalectryon No. But as I said before, you should be able to just use the shooting method to solve the equation.
 
@StanShunpike Let's just continue talking here
 
8:50 PM
@0celo7 Why would he read his books? Isn't he far past that level? (assuming he read it recently)
 
The FLRW metric is an exact solution. Our universe is not actually the FLRW metric (since, for one, the matter content cannot be approximated by a homogeneous/isotropic fluid)
 
@Icosahedron So you only read books if you think they're harder than what you've previously read?
 
@Icosahedron One can enjoy and appreciate books even if one learns not much from them.
 
Our universe is not a known exact solution
(other than in the trivial sense that we live in it so it should be some exact solution :P although even that's debatable as GR may break down)
 
I'm sorry!
 
8:54 PM
@0celo7 I came to that conclusion a while ago ;)
 
@ACuriousMind I could have sworn one of the chat regulars was a fan of his.
Then I recalled one of Qmechanic's posts where he lauded QFT in a Nutshell.
 
@0celo7 : Caught me. Yeah, I starred your first sentence about Zee. However, if you are serious about your second sentence, I foresee a lot of disappointment coming your way...
 
@Danu Did I say the universe was an exact solution? Just trying to figure out if I said something that implied that.
 
@Qmechanic Of course not, it's an expression (I think).
 
@StanShunpike The question was quite unclear to me
 
8:57 PM
@0celo7 Have you considered being a detective?
 
@Icosahedron Many times.
Although, I'm terrible at finding various food articles in the fridge, freezer and pantry. Besides that, I'm pretty good at sleuthing.
 
@0celo7 You've mentioned the word Zee more than anyone else in this chatroom.
By far too.
 
@Icosahedron Zee.
If I had Tourette's, I'd say "Zee" before every sentence.
Zee
 
...93, 94, 95.
 
Holy crap, that's a lot of Zee!
Zee
 
9:02 PM
@0celo7: You seem a bit...zealous about Zee
 
2/10 for the pun haha
 
@ACuriousMind Have you come into contact with a certain Czech string theorist lately? Zee
 
@0celo7 lol, he doesn't have a monopole on that word
 
lol, Zee
 
Although he probably also has something to say about monopole zealots
@Danu Yeah, not my best, I'll admit
 
9:04 PM
@0celo7 No more creative pseudonyms?
 
@Icosahedron no, Zee
Zee is the best, isn't he? Zee
Zee is such a fun name, Zee
 
@Danu I think it was a bad question. It happens. Case closed.
 
@StanShunpike No worries :) I have stupid questions all the time ;)
 
9:27 PM
@Danu haha, well, if you stumble around learning as much as I do, its bound to happen :)
 
@Hippalectryon pastebin.com/rX790JSt there's the solution if you care. In Python this time. I didn't bother with many of the details, but anyway, there you go. Why don't you just solve the full eqns?
 
@alarge You mean solve them by hand ?
 
I mean that the paper said that those were an approximation of the full eqns on an earlier page.
 
Oh indeed, but since I didn't know how to solve the simplified ones, I figured out I'd first learn how to do that before doing the complex ones
 
Also, I think there might've been a typo in the equations (the integral end point), so I changed it. You can change it back, obviously, if you so feel.
 
9:30 PM
^ that's the original equations
 
Ok. Well the solution principle is the same, so you can see if you can get those
 
Yeah, I'll adapt your code, it's basically the same equations with more terms
Thanks :-)
 
No problem. There's a typo i n the cons function, the return is missing square braces
 
@alarge return array(h[-1], sum(h[:Ri]*rs[:Ri])*diff(rs)[0] - (1. + .5*R**2*h[Ri])) seems correct to me, where is a brace missing ?
 
look at the other arrays
 
vzn
10:01 PM
new bohmian-like fluid physics covered in NATURE!
> A wave of experiments is probing the root of quantum weirdness.
found via reddit
 
@vzn To be fair, it's Nature News and none of the cited articles were published in Nature.
 
vzn
@alarge sigh go ahead & continue to paper-cut it to death with fineprint etc... hey have long heard far worse slings & arrows in here & elsewhere... :|
...but maybe some of the researchers cited have been published in nature & various other highly esteemed/ prestigious journals :)
 
@vzn What's with the reaction? You wrote Nature in all caps, italics, boldface and an exclamation point at the end. I just pointed out that this is not a scientific contribution to Nature. I did not say that this would make the contents any less newsworthy or valid.
 
vzn
@alarge lol ofc its a scientific contribution but not a peer reviewed paper. its a peer reviewed survey/ essay :)
nature search results for author zeeya merali, 26 items
 
@vzn Well, most likely not peer-reviewed though. The most typical case is that the editors ask someone (whom they often know) to write an opinion piece about a certain topic.
Not that it really matters if it were refereed or not.
 
10:15 PM
@0celo7 tensorphobia is a thing.
 
vzn
@alarge its likely to be "peer reviewed" by editors who are generally expert/ highly qualified. but ofc agreed this is not (quite) same as scientific peer review.
 
I've never heard of an invited review paper being rejected (surely there must be cases, though), and those go through proper peer review. I think it is even less likely that an invited piece of news would.
 
vzn
@alarge "peer review" has both formal & informal meanings. the editors will sometimes suggest/ ask for revisions or quality concerns quite like referees. its not clear if the paper was "invited".
 
Yes, and Nature for example employs editorial rejections where papers that are not deemed worthy by the editor are not even sent forward. It's still not the same thing as peer review.
Given that almost all pieces of news are invited (as pointed out on Nature's website), it most probably was.
 
vzn
@alarge you are using the words "peer review" in a scientific sense. am using them in a less formal sense.
 
10:20 PM
Well, then it's a review process, but not really by your peers.
Regular newspapers, too, use internal reviews, of course.
 
vzn
its not explained how "invitations" work. typical magazines also have stories initiated/ "pitched" by authors. as for "peers", editors are roughly "peers" of writers. but not exactly scientific peers in the other technical sense of "peer reviews". ofc both the editors & writers have strong scientific credentials to write & edit in their chosen areas.
 
Well I suspect invitations in Nature work like in any other journal: The editors ask someone they know, someone who has contributed to their journal, or someone who has written a lot on the subject.
 
vzn
lol anyway this is all so narrowly/ minutely )( meta...
re search results, the author Merali is a longtime/ regular contributor. going back to 2002.
also previously writing on QM & particle physics, Higgs, etc.
 
@alarge I get 'find not defined', in what package is it supposed to be ?
 
@Hippalectryon I don't know. I program very little with Python: I only ever use ipython which loads all the packages.
 
10:30 PM
@0celo7 I don't understand I.4 tensor field line 5.
 
Looking.
 
I guess, though, "from numpy import *", "from scipy import *", "from pylab import *"
 
What exactly don't you understand?
 
find seems to be in matplotlib, i.e. pylab (so is plot)
 
@alarge Oh right I forgot pylab
 
10:32 PM
@0celo7 It doesn't seem obvious to me.
Can you explain why it is.
 
@Icosahedron What doesn't seem obvious?
The "implies" or the "thus"?
 
Thus.
 
Write the "implies" equation in components and take the derivative wrt. $x'$.
 
vzn
@ThomasKlimpel ^ =D
 
Hi - is it OK if I interrupt to ask an "etiquette" question?
 
10:35 PM
Sure.
 
@alarge I didn't edit any bracket and it seems fine..
 
@Icosahedron Does that help or do you want me to do it?
 
Thx... I have an extremely long threaded question on GR and a recent answer to Timeaus requires another long reply... should I add to what already exists or start a "new question" (with all the attendant hassle of cross referencing and quoting)?
 
@0celo7 I want you to do it please, I'm not used to indexes yet.
It's confusing to work with.
 
@Hippalectryon It worked for me as well, that's why I didn't notice it, but the return type is still off, so I'm not sure if it is actually doing the right thing or if you're just a bit lucky.
 
10:39 PM
@Icosahedron The implies equation is $x^k=\sum_h(R^T)^{kh}x'^h$, right?
That's just turning the matrix notation into indices.
 
Forgot the prime.
 
What do you mean by writing it out in components?
There are a lot of ways of doing that.
 
I mean do exactly what I just did.
 
I didn't follow.
Why is that true?
 
10:43 PM
Let $\{\vec e_k\}$ be a basis of $\mathbb{R}^n$. You know what that means, right?
 
Yes.
 
@Hippalectryon I checked now and fsolve only seems to care about the first element of the array. So the algorithm is a bit wrong (I guess you should be using an optimizer instead of trying to find the root)
 
Then any vector has the unique expansion $\vec V=\sum_k V^k\vec e_k$, right?
 
Yes.
 
(I'll omit the sums from now on.)
 
10:44 PM
@alarge An optimizer ?
 
So expand $\vec x$ as $\sum_kx^k\vec e_k$.
 
I'm not used to summation convention yet, can you not omit the sums?
ok.
 
Then the equation $\vec x=R^T \vec x'$ is $\sum_k x^k\vec e_k=\sum_{k,h}(R^T)^{kh}x'^h\vec e_k$.
 
@Hippalectryon Do you understand how the shooting method works?
 
@alarge kind of.
 
10:47 PM
Hmm, that might not be the best way of explaining that.
Let's try something different.
Write $\vec x$ as a column vector. Then $x^k$ is just the $k$-th component.
 
Why did you involve an h?
 
@Icosahedron Forget that. It's correct but confusing.
 
It was the k that confused me at first in the book.
 
@alarge What's bothering me most in the algorithm is that it requires you to give it $R$, which is a priori unknown
 
@Hippalectryon Well, because we have more constraints than free variables, you'll need to do something "creative".
How do you mean it bothers you?
 
10:48 PM
Do you understand that if $\vec x$ is just a column vector, then $x^k$ is the k-th entry?
 
Yeah.
 
@Hippalectryon Well, what's known?
 
@alarge R is supposed to be determined by the other variables
 
@0celo7 Though I was used to i as the indicy. (just nevermind what I said, continue)
 
We know alpha and beta
 
10:49 PM
So when we act with the transpose rotation operator on the column vector $\vec x'$, we get the new column vector $R^T\vec x'$.
 
According to the paper that is enough to find $R$
 
(Which happens to equal $\vec x$.)
 
By the standard rules of matrix multiplication, the column vector $R^T\vec x'$ has components $\sum_h (R^T)^{kh}x'^h$.
 
@Hippalectryon Ah, so R is free? Well, just make it free, then.
That makes sense, as then you have two constraints and two variables
 
10:52 PM
@0celo7 This might be a dumb question though why aren't you using the basis anymore?
 
@alarge How do I make it free in the algo ? (in the current code it has a fixed value and is never changed/determined)
 
@Hippalectryon The same way h0 is free. The way the shooting method works.
 
@Icosahedron It's confusing. Too many sums.
 
I understand up to here, now how do you get the implies from that?
 
@alarge But what becomes of Ri = find(rs >= R)[0] ?
 
10:54 PM
Ok, so you agree that we have $x^k=\sum_h (R^T)^{kh}x'^h$, right?
 
@Hippalectryon That's done where it's needed instead
 
Yes.
 
Apply $\partial/\partial x'^l$ to both sides of that.
 
Why is it $\partial/\partial x'^l$ and not $\partial/\partial x'^h$
 
Never ever have the same index more than twice.
Just do it, we'll talk about it more when you have the result.
 
10:59 PM
@alarge Hmm it's becoming really late here so I have to go, I'll probably try to modify it tomorrow. Since I'm kind of new to the method I'd be glad if you can modify the code so that it determines R (the simplified version, not the full one with additional terms), but if you don't have time or don't want to it's fine, thanks for your help so far anyway :-)
 
@Hippalectryon Fine. There: pastebin.com/GywE7Fwh
 
@Icosahedron Need help?
 
No.
It's $(R^T)^{kh}$?
 
Yes.
Did you look in the book or did you get that?
 
I got that.
 
11:03 PM
Actually, it's $(R^T)^{kl}$.
 
I wasn't sure if it was correct because.
yeah
 
Now here comes the magic of indices.
 
Why is it not h?
 
Simply replace the $l$ on both sides with $h$. All you're doing is changing a label, so this is permissable.
The index $h$ is tied up in the summation.
You can't introduce a third $h$.
 
Why couldn't you just use the same h?
 
11:05 PM
A lot of people are sloppy and will do it, but always keep in mind that we take the derivative wtr. $x'^l$ and then replace the $l$ with $h$.
I think Zee might explain that, one moment.
 
That's the part I don't understand.
Why does that work
 
It's simply a label that can be changed at will, provided you change the corresponding index on both sides of the equation to the same new index.
 
Technically it's not incorrect to use h in the process, though it's not entirely correct either?
Same result still.
 
Very poor form.
 
Why does that work with indicies though?
why isn't it just 0?
since it's wrt to l.
 
11:08 PM
Golden rule: $$\frac{\partial x^i}{\partial x^j}=\delta^i{}_j$$
 
Is that the dirac delta?
 
This works with all coordinate systems.
Yes.
 
@ACuriousMind kind of a weird color today
 
Thanks a lot.
 
So, we have $$\frac{\partial x'^h}{\partial x'^l}=\delta^h{}_l$$
 
11:09 PM
Is there a summation there?
 
Then the sum over $h$ knocks out the $h$ (turns it into $l$) and the result is $(R^T)^{kl}$.
 
Then you can switch back?
 
Yes.
 
...eh this needs some getting used to.
 
I found indices very natural to work with.
It's index free notation that takes getting used to.
 
11:12 PM
The only parts that seem unnatural to me are the switching and multiple indices.
Also how can you tell if there is a summation convention in an equation with indices?
Is it always?
 
Zee stresses the following point many times, but in case you skim: You can change summed indices at will. For instance, $V^i g_i=V^j g_j=V^k g_k$ as long as $i,j,k$ are all taken to run over the same set of values.
 
That suddenly made a lot of sense.
enlightenment
 
@Icosahedron The author will tell you at the beginning of the book if the summation convention is used. If it is used, he will tell you when it is not used before or next to the relevant equation.
 
@0celo7 Understood.
By the way, was this misunderstanding today really dumb?
If so, then I should really improve on this immediately.
 
It shows me that you're not ready for Hawking & Ellis, and I hope you see that too. But it's not dumb, I think many people are uncomfortable with indices.
HE doesn't introduce the summation convention, they just use it.
 
11:19 PM
I thought Hawking and Ellis only stressed basic calculus and algebra, it should be fine to read. $3x + 5 = 7$
 
I have no clue why they said that.
 
Another reason why not to read the preface of any book.
 
That's not the right lesson to take from this!
 
Perhaps the meaning of the word basic has changed over the years.
Imagine reading another book of theirs that requires advanced calculus and algebra.
 
The preface of BBS says you need some basic QFT, but they assume you are fully familiar with CFT, which is not in any intro QFT book that I know of.
 
11:23 PM
The preface of Penrose RTR says you need no background in mathematics or physics.
 
@Icosahedron Sachs & Wu GR for Mathematicians requires a "comfortability with geometry and topology fit for a second year mathematics graduate student" or something of the sort.
 
@0celo7 An honest preface.
I still can't find that book.
 
In the library?
 
uh....
of course in the library, it's missing
 
Lol.
I think I'll read it one day for the novelty, but I don't think it will deliver any real insight.
 
11:28 PM
Thx for considering the question... it's 01:30 in Spain so I'll look for an answer in the morning. G'night all.
 
@0celo7 And HE is found in course reserves in physics, and one is missing.
"I think HE will be fine for you, give it a try."
 
Aerospace?
 
Who knows what those Aerospace engineers are up to these days.
 
@Icosahedron Must have forgotten the equation.
 
@0celo7 Though it is completely reasonable for someone who has read Lee [ISM] right?
 
11:34 PM
Oh yes.
IRM wouldn't hurt, either.
Also, I don't think it's advisable to read it as an intro to GR.
It's not meant to serve as an intro to GR.
 
The preface disagrees.
 
There's a difference between not needing to know GR to read it and being able to learn GR from it.
 
Later, I'm going to try and finish this chapter then read shankar.
Thanks for the help.
 
Bye. You're welcome.
 
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