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12:34 AM
@DavidZ & other mods, if this one gets deleted, please make sure it's converted to a comment instead
A: Why not use the magnetic scalar potential?

JERROLD FRANKLINI didn't see your question until just now. The magnetic scalar potential can be very useful, and is too often dismissed witout much consideration. I treat and use it in some detail in my textbook https://www.amazon.com/Classical-Electromagnetism-Jerrold-Franklin/dp/0805387331?ie=UTF8&kyeywords=...

@EmilioPisanty See also physics.stackexchange.com/a/268747
spking of QM+GR, ambitious/ wild conjectural framework
> If we perturb the state a little bit, how does the emergent geometry change? (Answer: space curves in response to emergent mass/energy, in a way reminiscent of Einstein’s equation in general relativity.)
12:54 AM
hello everyone
1:40 AM
1 hour later…
2:54 AM
yet ;)
@skillpatrol I killed one, I know they're real
3:35 AM
Get some spray @0celo7
3:49 AM
@skillpatrol no money
2 hours later…
5:26 AM
Nice day!
6:10 AM
What does mean the "OP"?
Original Poster
(of the question)
@skillpatrol Thanks a lot!
7:07 AM
@vzn : "...we’re building on ideas that have come before. The idea that spacetime geometry is related to entanglement..." Tsk. Einstein explained why light beams curved. General relativity is the best-tested theory we've got. Gravity is nothing to do with entanglement.
2 hours later…
8:40 AM
for those of you wondering if a time operator exists in non-relativistic quantum mechanics ;-)
\o @yuggib
@skillpatrol o/
@EmilioPisanty OK; that would have been better indicated by a flag (for future reference)
9:16 AM
Anyone around?
Hello! You wouldn't happen to know much/anything about cosmology would you? The deceleration parameter in particular.
@JohnRennie (In case you're around...)
@NoahP Sorry, I know nothing about that.
Bummer! No worries
@yuggib There are lots of questions on this very topic at PSE, on a footnote.
9:32 AM
This is talking about ultra-weak time operators.
9:44 AM
@NoahP Hi Noah
Just the man I was looking for! @JohnRennie
A universe with just dark energy has the de Sitter geometry
You've lost me there
The scale factor varies with time as $a(t) = e^{Ht}$ where $H$ is the Hubble constant.
So the universe expands exponentially with time.
A de Sitter universe is a cosmological solution to the Einstein field equations of general relativity, named after Willem de Sitter. It models the universe as spatially flat and neglects ordinary matter, so the dynamics of the universe are dominated by the cosmological constant, thought to correspond to dark energy in our universe or the inflaton field in the early universe. According to the models of inflation and current observations of the accelerating universe, the concordance models of physical cosmology are converging on a consistent model where our universe was best described as a de Sitter...
Okay, which would fit with what I said in the question?
9:48 AM
You would get $q_0 = -\Omega_\Lambda$ yes.
But i think your focus on things like the deceleration parameter is not helping you.
I'm going on a week long physics course next week around this, and have been given a set of questions to research
There are lots of ways to try and get a handle on an FLRW geometry, and the deceleration parameter is just one such handle, and I'm not convinced it's a partyicularly useful one.
I don't have to submit them, but I want to try and understand & be able to answer them all
@MAFIA36790 the questions on PSE are about mistaking time derivatives with an operator
OK, I guess if you've been given the problems to do you don't have a lot of choice in the matter.
9:50 AM
not much on the topic of time operators are they are explained there ;-P
@yuggib yup.
But really I would focus on understanding the form of a(t) and how it depends on various things.
@JohnRennie The questions seem to move on to that from now, for example:
Just asking, although beyond my scope, what did they mean by ultra-weak @yuggib?
Suppose the universe is spatially flat, and at some point was dominated by a fluid with a stiff equation of state $p=\rho c^2$
Then, solve the continuity equation for $\rho$ and obtain the scaling of $\rho$ with respect to the scale factor, a
@JohnRennie I'll probably be posting a couple more questions today
9:53 AM
That should keep you busy for a while :-)
Have you seen:
and the answers from Pulsar that I link to?
Q: How does the Hubble parameter change with the age of the universe?

John RennieHow does the Hubble parameter change with the age of the universe? This question was posted recently, and I had almost finished writing an answer when the question was deleted. Since it's a shame to waste the effort here's the answer anyway. Maybe this can be one of the canonical answers sugges...

There is lots of useful stuff there, and the Living Reviews article you linked is also quite helpful.
I'll have a look now, thanks for all your help
I'm in and out today, so I may not be able to respond immediately but if you ping me here I'll see it eventually.
@MAFIA36790 it is a ultra weak notion of time operator, based on how the canonical commutation relation with the hamiltonian is intended
Are you going to delete that last question now you have the answer?
(a time operator is one that noncommutes canonically with the hamiltonian, in a way that energy time uncertainty is satisfied)
9:55 AM
Should I answer it myself, delete it, or wait?
Well I can post an answer if you want, but it's kind of a one liner.
@yuggib Thanks for that.
@MAFIA36790 anyways the "hierarchy" of time operators, from ultra-strong to ultra-weak is explained inside the paper ;-)
@JohnRennie I'm just reluctant to delete it as it would be helpful to show my research next week
@NoahP There's certainly no obligation to delete it. It might be interewsting to see if anyone else answers.
9:57 AM
@yuggib Can't resist myself to read the paper; but yeh it's beyond my scope for now T__T
Whether asking other people counts as research is somewhat debatable :-)
@JohnRennie Okay, I'll leave it be for the time being then. I'm sure it'll be joined by more questions later. And I'd agree with you there, but plenty of research has been conducted beforehand! These concepts are a little difficult to grasp for a 17 year old who hasn't yet been taught beyond A-Level
(If you're in the US, Projectile motion is one topic I did this year, should give context), and has only been taught differentiation using the chain, product, and quotient rules - no implicit integration or anythinng
It didn't like that as one long message @JohnRennie
I have to say that the FLRW metric is a bit advanced to be teaching 17 year olds. You wouldn't normally do it until the later years of a Physics degree. Anyhow, good luck with it :-)
Yeah, this is all stuff I'm doing outside school, no help from my teachers within - the course is being taught by a University professor, and is intended to push I think. Thought it would be useful considering I'm applying to Oxford. Thanks!
@NoahP Re Oxford, for the admission, you just need to pass with high marks their admission exam which would be based on Physics and Maths if you are considering for majoring in Physics. The questions are not hard, in any way, but the problems are very lengthy though. Maths is comparatively easier.
10:06 AM
@MAFIA36790 Yeah, I'm going on a two week holiday near the beach soon, so plan on doing lots of past PAT papers, as well as BPhO papers. And it is Physics I'm applying for.
Thanks for the advice though! :)
@NoahP Olympiad questions are of numerical taste; but the admission exam would consist both of numerical and conceptual question although of varying numbers every year.
@MAFIA36790 Apparently theyve taken out the longer questions as of last year, so it's much less wordy
I doubt the entrance exam would be asking you about general relativity. I went to Cambridge not Oxford, but as I recall the exam was basically A /S level physics. It's just that the questions were more taxing than the A level.
10:19 AM
@JohnRennie That's my understanding of it too, but everything else, including my EPQ (If you remember all the black hole stuff a while back) will hopefully help in the interview stage
@JohnRennie No general relativity; OMG ;P
I also didn't encounter any SR question also.
@NoahP Maybe but not required, AFAIK.
@NoahP Just a basic entrance exam; not easy but really not hard to crack; 've to score really high marks as there is really a strong competition from all over the world.
10:28 AM
Ok, thanks for all the advice :)
Wow, this is ridiculous.
91 upvotes, lol
SE is offline.
@MAFIA36790 ?
@MAFIA36790 Ah ok. Havent had any issues myself
10:41 AM
@MAFIA36790 July 6
Would explain the lack of issues....
@Loong Latest.
@Loong Status blog is older June.
So, does that mean only I got that message?
Think so
10:44 AM
Hmm. Weird.
@JohnRennie You still around?
@NoahP Yes, still here :-)
I'm honestly at a complete loss as how to interpret this question
Yet if I post that, it'll be flagged as homework I expect - correctly so, although this isn't homework
What is the question? Is there a link to it?
10:49 AM
I can screenshot it
Go on then. You can upload the screenshot here.
It refers to the equations by odd names (different to those I find online), so I included the rest to give context
How should one upload a picture here?
There's an upload button next to the typing box
Oh! I am blind!
10:53 AM
Like so!
I got before! Thank you!
I never use those buttons. I always use Enter button on keyboard.
Well you're given the continuity equation in q3. You can just substitute for $p$ in that to get $\dot{\rho} = 6H\rho$.
What about a file? For example pdf
@DavidZ sure. Custom mod flag? I'd already flagged as NAA, I think.
@JohnRennie I presume you mean $p=\frac{\dot{p}}{-12H}$?
10:59 AM
Q4 says $p = \rho c^2$
I did that, but don't really see where it gets me. Also get nothing meaningful (IMO) when substituting into other equations such as acceleration, freidemann, or deceleration parameter
@JohnRennie Sorry, substitute p for rho in my message,
If you make the substitution eqn 4 becomes $\dot{\rho} = 3H(\rho + \rho c^2/c^2) = 6H\rho$
@JohnRennie This was a page back in my workings....
Well, that still doesn't change an awful much
Hmm, that's not what q3 says, and when I Google for the Friedmann continuity equation I get hits that give the same equation as q3 not your version.
Yeah I know, its an error of mine due to lack of coffee
Also, there is a typo on the sheet I sent ;the continuity equation should have a - before the 3H
11:10 AM
And $H = \dot{a}{a}$
So I get $H^2 = \frac{8\pi G \dot{p} a}{-18\dot{a}} + \frac{\Lambdac^2}{3}$
So that gives you $\frac{\dot{\rho}}{\rho}} = 6 \frac{\dot{a}}{a}$
Completely ran past that idea, that makes sense
Much more sense actually
If it's just matter then $\rho = \rho_0/a^3$, where $\rho_0$ is the density at the current time.
And $d\rho/dt = d\rho/da da/dt$
But is it just matter?
11:16 AM
Well the question says it's dominated by the fluid ...
Okay then
So $\dot{\rho} = d\rho/da \dot{a}$?
And that would be my final answer?
To be honest I'm not sure what the question means by solve the continuity equation and obtain the scaling ...
Yeah, neither am I
11:20 AM
But the next step is going to be to eliminate $\rho$ using $\rho = \rho_0/a^3$ then solve the resulting differential equation for $a(t)$.
How have you jumped to $\dot{\rho} = d\rho/da \dot{a}$?
Chain rule for differentiation
Differentiation of $p=\frac{\dot{p}a}{6\dot{a}}$?
No, the chain rule says that $dy/dx = dy/dz dz/dx$
For any $z$
Yeah, I get that, but what did you differentiate?
11:25 AM
Start with $\rho$, then from the chain rule $d\rho/dt = d\rho/da da/dt$
Assuming that $\rho=$..?
If we assume $\rho=\rho_0/a^3$ then $d\rho/da = -3\rho_0/a^4$
For some constant $\rho_0$ that gives the value of $\rho$ when $a = 1$
So $\dot{\frac{P_0}{a^3}}=\frac{-3P_0}{a^4} \dot{a}$?
I'm just thinking aloud, but if we've got as far as $\frac{\dot{\rho}}{\rho}} = 6 \frac{\dot{a}}{a}$ then if we can express $\rho$ and $\dot{\rho}$ as functions of $a$ and $\dot{a}$ then you're going to end up with a differential equation for $a(t)$.
@EmilioPisanty ah, gotcha. Yeah, a custom mod flag would have been ideal because then we'd see your flag message right alongside the NAA flags from others.
11:34 AM
That would seem right, but the keyword is if
I would have to get out some paper and start scribbling to get any farther, and I'm currently tied up with something else.
@JohnRennie Okay, no worries! I'm going to go for lunch now, and have a look at it after - I doubt I'll make much progress, in which case I may make a question of it, but will check here when I'm back. Thanks
11:49 AM
@DavidZ OK, will do next time.
Not sure if I can re-flag a post that already has an active flag.
But I'll flag a comment or whatever.
I'm also a bit weirded out by the review page on that one
Should've been deleted by Community if I understand the Recommend Deletion votes correctly.
12:07 PM
Q: Why was this answer not deleted at the sixth Recommend Deletion vote?

E.P.The last I heard, if an answer in the Low Quality Posts review queue gets six Recommend Deletion posts, it gets deleted immediately, and the post is marked simply as 'deleted on ...', such as this one (physics 10k+ or SE team only). However, I'm a bit confused by this review (again physics 10k+ ...

12:22 PM
Q: How to check on what's happened after I flag my question for migration?

uhohI'm pretty sure I flagged my question Why do they think WT1190F is low density? a few days ago, suggesting/requesting that it be migrated to Space Exploration Stackexchange. When I originally asked the question 8 months ago, I wasn't very familiar with either site, but now I can see that there is...

12:57 PM
This man says if you 'pack' your cigarettes (pound the pack hard against your palm), the ash stands firmer on the burning end and won't fall off easily when you move your hand. Does that make sense to you? Do you have any idea as to why? One doubtful possibility is that the minced tobacco inside the cigarette gets more intertwined and tangled in itself, hence firmer ashes.
1:22 PM
Q: Javascript applet for drawing

Hennadii MadanIt's best to see once than to hear (or read) a convoluted explanation a thousand times. I just wanted to answer one question and it was begging for a drawing. So I thought, OK, I draw it on paper, then photo, then send to myself , then edit, then upload. And at that point I decided it's not wor...

Okay. I guess that's because when you pack it, some of the tobacco comes out and the remaining tobacco becomes lighter, so the ash will be less dense too, lasting longer before collapsing under its own weight.
1:35 PM
@JohnRennie Not sure, but may have found something that helps
1:51 PM
@EmilioPisanty I don't think you can re-flag a post, I just meant that if you happen to think of it before casting a flag, you could use a custom flag instead of NAA to express your preference that the post become a comment rather than getting deleted outright. But if you think of it after having cast a flag, like this time, it's no big deal. A chat ping is better than nothing, I suppose.
@EmilioPisanty about tag edits, I know, but as I see there are more competent people for edits, more important with better feel what is appropriate on se.physics.
@MolbOrg Yeah. And we'll get to it when we get to it. In the meantime, the floor is open to anyone who wants to take a shot.
Tag wiki edits get peer reviewed anyway so you won't cause damage if the edit suggestion is bad.
And with tag wikis, the state of things is generally that any edit is better than nothing
@EmilioPisanty exactly, yes I know about per reviewed.
hm, for some reason it's hard to explain my point correctly
@DavidZ I see.
2:42 PM
@JohnRennie Would integration eliminate the time derivatives from $\rho=\frac{\dot{rho}a}{-6\dot{a}}$? I'm hesitant to even begin to contemplate that, let alone attempt it.
@JohnRennie When the question says scaling of rho with respect to a, it means: "What I mean by 'obtain the scaling of rho with respect to the scale factor' is that rho is dependant on a(t), and it's the proportionality of rho with respect to a. For example, what is the value of x if rho = c*a^x, where c is a constant."
3:02 PM
@JohnRennie Hopefully you can read my writing, but this is my ill fated attempt...
@NoahP Why not you use MathJax?
Did you just prove $1=2$ ;P
@MAFIA36790 Never used it before? I enjoy writing it out rather than typing, much quicker
And yes, I did
Any clues how?!
@JohnRennie Sorry for the spam, but the person who set the question also just told me " If you can see the 'trick' to the question, it's actually very straight forward to do, and can be done in a matter of just a few lines or so. Keep that in mind and you should be fine." Infuriating.
3:18 PM
@NoahP You should consider what the people you're asking to help you would prefer.
Besides, practice makes perfect: TeX doesn't have to be slower than normal writing for "normal stuff" (no extensive indices etc)
@JohnDuffield both agree/ disagree. its widely acknowledged that QM/GR are "not exactly compatible" in some limit cases but unf those limit cases are (so far/ mostly) out of reach of experiment. the basic examples are black holes and/or big bang. some unknown theory seems to remain to mesh them together coherently. the ref you cite does nicely summarize experiments that are "within reach". but alas one might say they are quite limited wrt the overall wildness of the universe.
@Danu Should it display as math in my browser? I just looked up a bit about MathJax, and thats what I understand, but its just comes up as raw Tex for me
Yes, it should. Use the link in the top-left corner of your screen.
You need to use a little bookmark button.
I can't see anything?
Under the "The h bar" header.
top-right, I'm sorry!
3:23 PM
Oh my god thats so much better
I was so confused why people wrote it in Tex on here
Now you should type out your questions in $\LaTeX$
Yeah, I had done so before, the picture was a one off
No problem ;)
3:37 PM
@Danu You still there?
Interesting, their idea is quite similar to a flashing thought I came across when reading the wikipedia article on quantum decoherence and von neumman measurement scheme (and also Susskind's book)

While currently I cannot really write out the full mathematical treatment yet to illustrate the idea (I am currently looking for a missing puzzle piece in this old article http://arxiv.org/abs/quant-ph/0702225 in order to understand how can entangled states disentangle)

But basically, if decoherence and einselection induced by the environment explain the apparent classical results, then using su
If $\dot{p}=-\frac{6\dot{a}p}{a}$, then $\frac{a}{-6p}=\frac{\dot{a}}{\dot{p}}$
Where $\dot{a}=da/dt$ and $\dot{p}=dp/dt$
So surely $\frac{a}{-6p}=\frac{da/dt}{dp/dt}=\frac{da}{dp}$?
Which would, when integrated (Using wolfram alpha as I have no idea how...) give $a=\frac{-a ln(p)}{6}$
The issue is that $a$ then cancels, which it shouldn't. Anyone see where I've gone wrong? @MAFIA36790 @JohnRennie ?
@vzn Regarding their speculation in that paper (have not read the actual paper yet), if space (or maybe spacetime?) is a result of a graph formed by a certain entangled state, then classical concepts like positions, momentum etc. will also be emergent qualities, e.g. as perturbations on the manifold formed by this graph of entangled state?
thus everything we know in physics might be just mutual information and correlations between states?

Need to read in more detail to see how they build the argument
@Secret actually havent read it either but skimmed/ read blog on it. not coincidentally carroll is a specialist/ expert on universe expansion models. think its all useful for expanding/ focusing the dialog in the area (QM+GR is a very old/ active area) but imagine it will all be received skeptically so far. nobody will get much excited without some predictions that match (currently unexplained/ ie anomalous) cosmological observations.
@Secret sounds like/ reminds me of wheelers "it from bit" philosophy, heard of it? yes many physicists are leaning more toward an information-centric view of the universe, and this is probably taken to the greatest extremes by digital physics enthusiasts eg maybe most prominently 't hooft etc
3:56 PM
@NoahP $$\frac{\mathrm d p}{p} = -6\frac{\mathrm da}{a}$$ The variables are well separated; just integrate them.
:31137390 $$\frac{\dot p~\mathrm dt}{p}= -6\frac{\dot a~\mathrm dt}{a} $$
Now, use substitution.
@NoahP oxford huh? ambitious.... think you have the "right stuff"? :)
@vzn I seem to remember reading some quote about aiming for the stars and landing on the moon?
And @MAFIA36790 I understand where you got the first line from, and the second, but no idea how to integrate either
I can do differentiation, but the integration module is next year...
@NoahP $$\int \frac{\mathrm dx}{x} = \ln |x| + \mathrm C\;.$$
4:02 PM
So $ln(p)+C=-6ln(a)+C$?
@NoahP You can use single constant for that.
Would I get $ln(p)=-6ln(a)$, or does the $-6$ effect the integration?
@NoahP A constant would affect integrating a function? No.
Ok, just wanted to be sure. So what I get is right?
@NoahP possibly.
4:08 PM
@MAFIA36790 You're sounding like my teachers with this reluctance to confirm answers :P
@NoahP You should learn how to integrate. Your result is currently not correct for the integrated equation.
Oh, I see you're getting help already.
@Danu I am indeed, but anymore would be appreciated. I'm just substituing into given equations
@NoahP $\rm C$ on both sides need not be same though.
4:27 PM
In figure a how to find pressure at pallet of mercury
And the air is present in column 20 cm
@JohnRennie I think, THINK, may have an answer
$\dot{p}=\frac{-6p\dot{a}}{a}$, letting \rho = p because its easier to type quickly
So $\frac{dp}{p}=\frac{-6da}{a}$
So $ln(p)+c=-6ln(a)+C$
Sorry the picture which is uploaded is inverted
By making $C-c=x$, $ln(p)=-6ln(a)+x$
So, taking $e$, $p=e^x \times e^{ln(a^{-6})}$
So, $p=e^x \times a^{-6}$, no? @JohnRennie
@NoahP Why did you $x$ for that?
Makes it a little tidier
4:34 PM
@NoahP You can use other letters too ;\
What did $x$ do to you?
@NoahP Nothing; but $x$ is generally used for denoting variable; but it's fine if you want to use it for constant though.
@NoahP looking good now.
@MAFIA36790 Maybe I do. Advice noted! Would the use of $y$, $e$ maybe be preferable?
And thanks!
@NoahP Just use $\rm C'\;.$
4:37 PM
No, $y$ and $e$ are also reserved for other stuff.
$C$'s and $K$'s are pretty normal.
Does no one have a sense of humour here..? @Danu
(and indeed, variations with $\bar C,\tilde K,\hat C,C'$ etc)
Can anyone help me
@NoahP Nope :)
@Danu Haha okay, I'm off now anyway. Hopefully I come back to a yes from John Rennie, rather than a NO!
5:33 PM
Lend me your ears
I don't get mollifiers
Apparently a big property of mollifiers is that
$$\int \rho(x) dx = 1, \forall \alpha, \int x^\alpha \rho(x) dx = 0$$
Which leads to the property
$$\forall q, \int f(x) \frac{1}{\varepsilon}\rho(\frac{x - a}{\varepsilon}) dx = f(a) + \mathcal O(\varepsilon^{q+1}) $$
How is that different from the delta distribution
Isn't a function with 0 moments always the same as the delta distribution
And isn't $\mathcal O(\varepsilon^{q+1})$ for all $q$ the same as 0
I know that the Hamburger moment problem isn't totally unique but I dunno what a distribution with all 0 moments looks like outside of the delta
5:51 PM
A suitable mollifier, not any boring old mollifier
As Colombeau calls them
@Slereah Distributions are not actually functions though, are they?
This is a mollifier
Mollifiers are 100% functions
$\rho \in \mathcal S$
Yes, but I am saying, distributions are not. That should be the difference.
Well yes, but then, can a function actually have all moments at 0?
And also my second question
I am not sure what "all moments at $0$" mean.
6:00 PM
$$\forall \alpha \in \Bbb N, \alpha \geq 1, \int_{\Bbb R^n} \prod_i (x_i^{\alpha_i}) \rho(x) d^nx = 0$$
With $\alpha = \alpha_1 + \alpha_2 + ...$
Its statistical moments are all 0
@Slereah I see.
Well, think of a small symmetric bump (e.g., $e^{-1/(1-x^2)}$) around $0$ which has support inside $(-1, 1)$.
yeah that's the standard example of a mollifier
I have never seen someone act so petulantly on this site
Wouldn't it have some variance?
And some kurtosis?
btw a "suitable" mollifier has to be 1 in the neighbourhood of 0
6:06 PM
Hmm, right, it has some even dimensional moments.
No, all the odd dimensional moments are zero.
@Slereah Sure, you can fix this.
Kurtosis is even
Degree 3 is assymmetry
(heheh ass)
Oh, yeah, sorry. I was thinking about skewness.
Not a statistics person.
@Slereah Yeah, that's kind of weird. Are you sure about "$\forall \alpha$" there?
Well that is what I am wondering
Clearly it fails for like the simplest example of bump function I know of.
Also the same with $\forall q$
If it was just that for all $q$ up to a certain order, it would be fine
But I find this odd
6:09 PM
@Obliv Participating on this site has taught me that about 10% of people are just completely crazy. Hiding under a thin veneer of sanity.
Right, "upto a certain order" sounds better.
The specific part :
@knzhou I guess it follows from the fact that a significant portion of the population is, indeed, crazy.
I think this should be a fair analysis question for the mathematics.SE.
6:11 PM
The OP was hiding behind large words but his argument was awful. I was just stating the obvious and he said I had poor reading comprehension :O! very rude
I dunno
I never get any answers at Math SE
Admittedly I am not an analyst so can't help you more on this.
@Slereah Make it to the point, put the question ahead of whatever you want to write and make your question as small as possible. People there are impatient and don't really want to read long questions.
K, I gotta go.
7:04 PM
@Slereah are you around
Hey everyone. Does anyone know a good QM book covers precisely Rigged Hilbert space?
I am
The first time I read about rigged Hilbert spaces was in Rovelli
I wouldn't recommend it
Although I will say that Rovelli is a magic book
All physics is in Rovelli somewhere
If you can solve this I will admit you are a gentleman and scholar
7:10 PM
Who's Rovelli?
Rovelli is the big LQG guy
He's the big dog
I see! one of the founders of the loop quantum gravity
@Slereah Thank you. what's the name of that book?
Quantum gravity
@Slereah Big dog, big nuts. Amirite?
It is certainly not a book for casuals
7:16 PM
@Slereah okay! what's the book?
What @0celo7 said
Quantum gravity
got it! thanks
7:43 PM
@0celo7 is it hard?
@KNP I've been at it for many hours
I can't seem to do it
7:59 PM
define j(t)=(cos t,sin t) that's continous. consider f o j (t) : [0,2\pi)->S^1 and j^-1: S^1->[0,2\pi). Let g: [0,2\pi) ->[0,2\pi) that is g(t)=j^{-1} o f o j. Then by definition of g, f((cos t,sin t))=[cos(g(t)),sin(g(t))]. Define g(2\pi)=g(0) and use hint to solve it. Is it true?
Is anything true, in the end
@Slereah I dunno what you think?
All we are is dust in the wind, dude
I remember you wanted to check GR theorems with metamath. you did it?
I did not
Been busy lately
8:09 PM
Is there anyone?
Nobody is here
@Slereah: are you a moderator?
Sure, why not
I've a following message when I'm trying ask: "Wait! Some of your past questions have not been well-received, and you're in danger of being blocked from asking any more."
Just kidding, I'm not a moderator
8:26 PM
Today I've no time, so see you tommorrow.
Q: Which duplicated question should be closed?

DiracologyThere were two questions marked as duplicated of each other in the closed votes list today: Why doesn't an electron's charge rip the electron apart? If like charges repel, why doesn't a charge break itself apart? The first one is older but apparently got few attention. The second was posted to...

@Jim While I applaud your efforts on
Q: What is the velocity of the photon through the fourth dimension x4?

Astrophysics MathPhotons are real, physical objects. The fourth dimension is a real, physical entity. Therefore, photons must have a relationship with the fourth dimension. They must have some velocity relative to it. What is the velocity of the photon through the fourth dimension x4? Do you agree with the f...

and wouldn't seek to dissuade you from writing similar answers in the future that user isn't—in my opinion—asking in hopes of getting an explanation: he trying to push he own, totally borked interpretation of what relativity means.
That said, a good answer to the question (no matter how bad the question is) is valuable in and of itself.
8:43 PM
I applaud it too +1
@dmckee Thanks, that's what I was going for
Kill'em with kindness.
kindness correctness
an eye for an eye?
(making the whole world blind)
9:02 PM
@hubot The physics mods are dmckee, David Z, Qmechanic, and Manishearth.
@KNP No.
