The h Bar

heather

12:20 AM

well, if you're okay with something a little more on-edge you could sci-hub it. (i don't have access, though.)

Sir Cumference

12:39 AM

Howdy

heather

hello

Bernardo Meurer

::coughs::

Sir Cumference

@BernardoMeurer Need a cough drop?

Bernardo Meurer

@SirCumference Does it have alcohol in it?

Sir Cumference

@BernardoMeurer Jeez

rob

1:07 AM

@JamalS Interlibrary loan?

JamalS

@rob I'm not sure how to arrange that - I'll have to ask someone at my library

rob

@JamalS ILL is great. Most academic libraries have an online request form, since the ILL backend is electronic as well.

JamalS

@rob My university definitely has that feature, I just have to find a library that has the book I guess

rob

@JamalS No you don't --- they do that for you.

JamalS

@rob Oh, awesome

rob

1:13 AM

@JamalS You just send your library a solid reference and you get an email after a while that your book is ready.

JamalS

@rob For free?

rob

(My grad school library would even deliver to the department mailboxes --- I got spoiled.)

@JamalS The cost is that your library also sends ILL books out. So, no cost to you.

JamalS

Oh, I see

rob

My public library also does ILL but with a limit of one book per patron per year or something. But the university libraries: you want a book that exists, you don't need some special irreplaceable edition, you'll eventually get a hold of it.

JamalS

@rob I have to get authorisation from my supervisor, and then I get it for free

Though tbh I'd offer to pay for it myself just to save time

rob

1:17 AM

@JamalS Depends on how long you expect to need the book, I guess.

Physics Meta

3:04 AM

1

Q: Can I refer to a previous question I asked within the content of a new question?

I am thinking about a new question that is based on a premise that I stated in (I did not claim that it - the premise - was a new idea), and comments and answers I received in response to a previous question that I posted. Can I (and, if so, should I- ethically/ or in good taste etc.) identif...

Yashas Samaga

3:41 AM

the Best of PSE thing is dead? lol

vzn

3:53 AM

@BernardoMeurer interesting, coincidentally was looking into options related to this recently. is this a nearly similar product category? ever heard of it? was close to what wanted... us.creative.com/p/sound-blaster/sound-blaster-e1

Kyle Oman

4:14 AM

I did a "radio" (podcast) interview about my research and grad school experience with my campus station, if anyone is curious! (There is one blunder I should correct: the DDO catalogue is from David Dunlap Observatory, Toronto, not Dominion Observatory, Victoria).

vzn

@KyleOman nice! do you have any online pgs? any interest in a speaker session here? meta.physics.stackexchange.com/questions/7783/…

Bernardo Meurer

@vzn I really dislike the soundblaster line of products

vzn

@BernardoMeurer really? whys that

Bernardo Meurer

Do you just want a USB DAC? What's your price range?

They're overpriced for what they are, they really just don't deliver for the most part. Even their PCI/PCI-E Sound Cards have been forgotten in favour of Asus' counterparts nowadays

vzn

@BernardoMeurer ended up buying something not very expensive but still 48k sampling & 24 bit digitizer :) ... use it with a ps4. its also 7.1, but the sony doesnt seem to support it (does support 2 tho). also doesnt support the sound adjustment! arg! oh well, it half works :|

Bernardo Meurer

4:22 AM

@vzn Ah, I was going to tell you to get this, the reviews have been amazing

vzn

@BernardoMeurer its amazing how many there are, what a range, seems a bit staggering, do like the high end ones... do you think you can tell the difference between eg 48k/ 96k or 24/16 bit?

Bernardo Meurer

@vzn 48 vs. 96? No; nada

24 vs 16? Yes, sometimes

and only when I use my very good headphones (Sennheiser HD650), with a USB DAC (Fiio X3II), on a quiet environment

So, is it significant for the listening experience? Not very much.

Is it significant for archival means? Yes.

I'll still prefer to pay a bit more for the 96/24 version of an album than the 44.1/16

heck I'll pay more to get the DSD mastering a lot of the time

vzn

@BernardoMeurer sounds like some nice equipment :) ... yeah have spent quite a bit on speakers/ hifi/ receiver/ etc around here too

Bernardo Meurer

@vzn It works well enough, I want a proper USB D/A now, have my eyes on that Fulla 2, it's all I need pretty much

DAC/Amp combo

now my money is going into my secondary setup, turntable and speakers

What have you got going over there?

vzn

4:47 AM

@BernardoMeurer family gave us some nice big fat speakers ~300w each or so, got a hdmi receiver, 400w subwoofer recently... 4k tv... like to play rock band/ guitar hero, have bought a lot of dlc... unf its hard to get the older songs in the new v4... arg... think these guys have cool stuff too :) ionaudio.com

Bernardo Meurer

Damn, 300W each?! What are these?

vzn

@BernardoMeurer good question gotta go look em up! theyre old but big/ loud! lemmee see...

oops a little )( off! jvc sp7700 200w ea... my brother-in-law used to be a huge hifi geek & these are his hand me downs :P =D

DanielSank

Could a mod plz unfreeze chat.stackexchange.com/transcript/45780

@ACuriousMind @dmckee @DavidZ @Qmechanic @rob

Bernardo Meurer

@DanielSank I'll ask it here

DanielSank

(and maybe up the timeout. Bernardo and I come back to that room from time to time)

Bernardo Meurer

4:59 AM

Let $B=(u,v,w)$ and $B_1=(u+v,-w,u-v)$ be bases of $\mathbb R^3$ and $x_B=(2,-1,1)$. What will be the coordinates of $x$ on base $B_1$, $x_{B_1}$?

DanielSank

@BernardoMeurer Two vectors can't be a basis of $\mathbb{R}^3$.

Yo, @rob, help a brother out?

rob

@DanielSank Okay, done

Bernardo Meurer

@DanielSank Huh? Yeah they can. Individually

DanielSank

@rob Thanks, bruh.

Bernardo Meurer

There are infinite possible basis for any given field

@rob God bless

DanielSank

5:01 AM

@BernardoMeurer wtf you can't span a three dimensional vector space with two vectors.

Bernardo Meurer

@DanielSank Move rooms

rob

Those two expressions of gratitude are ... different ... from each other.

DanielSank

@rob One South American, one Californian.

vzn

@BernardoMeurer havent heard of DSD much is music sold in that fmt?

Bernardo Meurer

@vzn yeah, it's growing slowly, maybe it will die, maybe not

The masterings are amazing

vzn

5:06 AM

@BernardoMeurer (reading about it on wikipedia...) do you have a player? or a computer?

Bernardo Meurer

I have a DAC that supports it

and use my PC

vzn

@BernardoMeurer do you buy the music digitally? when you say the DAC supports it, is it mainly the 96k sampling rate/ 24bits? which DAC?

Bernardo Meurer

DSD is not PCM

it's 1 bit but with 2.8MHz frequency

or more

I either buy it digitally, or rip it with a hacked playstation 3

hdtracks.com sells it

vzn

so the DAC has 2.8MHz/ 1 bit input signal?

Bernardo Meurer

uhum

Don't ask me how it works, I think it soudns the same as PCM

I just like the mastering

vzn

5:11 AM

which DAC are you using?

oh Fiio X3II ok

Bernardo Meurer

Yeah

it's a portable player that doubles as a DAC :)

hdtracks.com/dsd-albums

vzn

@BernardoMeurer in other words usb interface to computer? geez amazon doesnt even mention that, think its hard to buy/ compare this stuff sometimes when the key details/ specs are unlisted... amazon.com/X3-II-Resolution-Music-Player-Generation/dp/…

rob

5:36 AM

39

Q: Why don't we use weights to store energy?

One of the main reasons why we haven't switched to clean energy is the lack of efficient storage methods - But, why aren't we using dead weights to store energy and draw it back later when needed? As an example of what I mean:

energy power weight

Sigh ... this turned into a "list question" and is accumulating a bunch of answers that say "oh! another counterexample."

Secret

smbc-comics.com/comic/a-group-project

rob

@Secret That guy is brilliant.

"Son, it's time for ... the quantum computing talk." smbc-comics.com/comic/the-talk-3

Sir Cumference

So, what's up?

@ACuriousMind The entire subject is just so darn complicated

Also *cosmologist :P

rob

6:02 AM

Huh. This is well-done: smartereveryday.com/toiletswirl

John Rennie

@SirCumference remember that the universe has lots of components, matter, dark energy, photons, maybe phantom energy ...

The equation of state for the whole universe is a combination of the EOS of these components.

Sir Cumference

@JohnRennie IK, but I'm getting a lot of conflicting answers

John Rennie

So phantom energy can be present at a low level but the overall EOS still results in a Big Crunch

Sir Cumference

ACM said up there that phantom energy's pressure does not change with volume

John Rennie

@SirCumference no he didn't

Sir Cumference

6:06 AM

Starting from here

8 hours ago, by Sir Cumference

I'm a bit confused

John Rennie

I read that. ACM was pointing out that strictly speaking ther EOS relates energy density and pressure

But my comments about volume are still correct

Sir Cumference

OK, so let me try to get this straight

Say we have a closed matter-dominated universe filled with phantom energy, but the phantom energy has an extremely low energy density

Can the Big Crunch still happen?

John Rennie

Yes

Sir Cumference

@JohnRennie Also, doesn't ACM seem to contradict that?

8 hours ago, by ACuriousMind

But I find talking about "volume" here confusing because in FLRW universes the energy density does not scale with volume except for non-relativistic ordinary matter.

@JohnRennie OK, so once the scale factor reaches 0, will phantom energy's pressure continue to grow more and more negative?

John Rennie

With phantom energy the energy density $\propto a^n$ for some positive number $n$. And the energy density of niormal matter $\propto a^{-3}$ i.e. $\propto 1/V$

Sir Cumference

6:17 AM

@JohnRennie Wait, $n$ is positive?

John Rennie

So as we approach $a=0$ the density of regular matter becomes infinite and the density of phantom energy becomes zero

YES

Sir Cumference

No idea if this is talking about the same $n$, but here's what ACM sent me a while back

John Rennie

THAT'S THE WHOLE POINT

Sir Cumference

phy.mtu.edu/apod/work/friedmannadventures.pdf

John Rennie

Those $n$ are the negative of my $n$. After all, matter density obviously doesn't scale as $a^3$ does it?

Sir Cumference

6:19 AM

...yes, that just hit me...

John Rennie

Ordinary matter scales as $a^{-3}$ i.e. $1/V$

That table shows the values of $n$ in the equation: $$\rho(a) = \frac{1}{a^n} $$

Sir Cumference

@JohnRennie Huh...so any possible significance of phantom energy is gone in a Big Crunch...

John Rennie

Phantom energy increases in desnity as the universe expands. So as long as the universe doesn't expand too much it can stay small.

But there will be a critical value of $a$ where the phantom energy becomes dominant. If the universe ever expands that much the phantom energy will drive it to a rip.

But a universe with PE can expand, stop and recollapse provided it never exceeds the critical value for $a$

Sir Cumference

@JohnRennie I assume that depends on the mass density/curvature of the universe, etc.?

John Rennie

Yes

Sir Cumference

6:24 AM

Huh

Mathematically, I understand why $a$ would reach infinity in a finite time, but intuitively it just sounds confusing.

Matter receding from me at a finite speed eventually becomes infinitely distant from me

If it weren't for the math, I'd say that makes no sense at all

rob

@SirCumference So many things are like this.

John Rennie

GR becomes more and more intuitive the more you learn about it. Beginners to GR usually find it utterly bizarre.

Sir Cumference

6:51 AM

@JohnRennie Oh sorry to ask, but I have just one more question

My prof says that a Friedmann universe with $\rho > \rho_{\text{crit}}$ expands "slower than the speed of light", one with $\rho = \rho_{\text{crit}}$ expands "at the speed of light", and one with $\rho < \rho_{\text{crit}}$ expands "faster than the speed of light"

No idea what "expanding faster than the speed of light" means. Does he mean that in the other universes, a Hubble sphere does not exist?

John Rennie

The statements are meaningless because the expansion velocity is $\dot{a}$ and that isn't simply a velocity. It's a velocity per unit length. So it can't be directly compared to the speed of light.

You'll have to ask your prof exactly what he means by the statement.

Sir Cumference

@JohnRennie OK, that's what I was thinking

Secret

7:53 AM

Cholesterol as signal molecules through the cell membrane

sciencedaily.com/releases/2017/01/170117140005.htm

Interestingly, people tend to found me less wordy when I speak the 1000 word paragraph compared to typing it all out

user228700

Hi, everyone :-)

Secret

It draggss behind

user228700

I've a quick homework-tsy Math question.

Secret

Swearing implies honesty?

user228700

I'm trying to find the solution of this trigonometric equation: $\sin \alpha + \sin \beta = -2a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}$

user228700

8:01 AM

Rearranging (and applying an identity, etc.) this equation, I find that either $\sin (\alpha + \beta)/2 = 0$ or etc. (The other possibility is not relevant to my question)

user228700

By my textbook hasn't even considered this case, they have directly jumped onto the other possibility. Does anybody know why this is..?

user228700

(Please ping me if u know...)

anonymous

@Kaumudi.H Which variable are you solving for ?

user228700

How does that matter?

anonymous

@Kaumudi.H You said I'm trying to "find the solution of this..."

user228700

8:07 AM

Yes, but that is not what my question is. I'm not exactly looking for a solution, no--I only need a condition.

user228700

What does it matter? My question is different from "I need to solve this"...

anonymous

@Kaumudi.H Hmm, in that case $\sin(\alpha/2+\beta/2)=0$ is a correct condition. Nothing more can be said...

user228700

.__. I don't understand why my textbook has ignored this condition!

Secret

$\sin \alpha + \sin \beta = -2a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}$
Using sum to product formula:
$2 {[sin(\alpha+\beta)/2] \times [\cos(\alpha-\beta)/2]}= -2a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}$
Simplify
${[sin(\alpha+\beta)/2] \times [\cos(\alpha-\beta)/2]}= -a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}$
If $[sin(\alpha+\beta)/2]=0$ then both sides becomes $0=0$ which is always true and not useful. Hence only the $\cos$ case need to be considered, which corresponds to $[\sin(\alpha-\beta)/2]=0$

anonymous

@Secret What do you mean by "not useful" ?

It is a useful condition and absolutely true.

user228700

8:09 AM

Yes, why is it not useful? It is a valid solution of the above equation.

Secret

$0=0$ is a statement that is true for any solution. However the equation in question may have a solution that only holds for some values. Remember that the solution set of an equation is the intersection of all possible cases combined

user228700

No, no, $0=0$ only if $\sin (\alpha + \beta)/2 =0$, which is the condition!

Secret

Put it more simply, because $[sin(\alpha+\beta)/2]$ appeared on both sides, setting it to zero means it is true regardless of what $\alpha$ and $\beta$ are

and by solution, I guess the equation want you to solve for a condition for $\alpha$ and $\beta$. This condition is given by the $[sin(\alpha-\beta)/2]=0$ case

anonymous

Even for the equation $x^2=x$, $x=0$ is true. It makes no sense when you say $0=0$ is true for any solution. No matter what, we don't neglect the possibility that $x=0$ is a solution.

Secret

in these equations, we are solving for $\alpha$ and $\beta$, not $sin(\alpha+\beta)/2$

of course if the solution ask for the variable $sin(\alpha+\beta)/2$, then it =0 is indeed a solution

user228700

8:14 AM

Now I'm confused as to why we do this in the first place.

user228700

Oh, no, never mind.

anonymous

@Secret If you are ignoring $sin(\alpha+\beta)/2$ you may be ignoring some possible values of $\alpha$ and $\beta$. Moreover, how are you solving for both $\alpha$ and $\beta$ from a single equation?

user228700

@Sec: I'm unable to understand ur logic.

Secret

Ok nvm I stuffed up my logic, rewriting stuff now

user228700

@Sir: Hello. Unable to sleep?

user228700

8:25 AM

@Secret ?

Secret

Still typing: (the full solution is kinda long)

user228700

What? NO.

user228700

Why are u typing the full solution? Man, my question was different from "Help me solve this" :-(

John Rennie

@Kaumudi.H I think Secret's point is that with some manipulaion the equation becomes:

$${[sin(\alpha+\beta)/2] \times [\cos(\alpha-\beta)/2]}= -a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}$$

user228700

Yes, it does, and there are two possible solutions for this equation, one of which is $\sin(\alpha + \beta)/2 = 0$

user228700

8:30 AM

My textbook has implicitly assumed that this solution is not valid and Why is it not valid? is my question.

John Rennie

You can divide through by $sin(\alpha+\beta)/2$ because it appears on both sides

So the equation is independent of the value of $sin(\alpha+\beta)/2$

user228700

Only if it's not zero. So to put my question another way, why isn't it zero?

Secret

No you cannot do that if it is zero

The full solution set looks like this:
$2 {[sin(\alpha+\beta)/2] \times [\cos(\alpha-\beta)/2]}= -2a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}$
$2 {[sin(\alpha+\beta)/2] \times [\cos(\alpha-\beta)/2]} +2a{[sin(\alpha+\beta)/2] \times [\sin(\alpha-\beta)/2]}=0$
$2 [sin(\alpha+\beta)/2] ([\cos(\alpha-\beta)/2] +a[\sin(\alpha-\beta)/2])=0$
$ [sin(\alpha+\beta)/2] ([\cos(\alpha-\beta)/2] +a[\sin(\alpha-\beta)/2])=0$

This means

$[sin(\alpha+\beta)/2]=0$ or $([\cos(\alpha-\beta)/2] +a[\sin(\alpha-\beta)/2])=0$

But I think anoymous's question is a valid one, are we solving for $a$ or the $\alpha$, $\beta$?

user228700

Dude. I very much appreciate your effort but why did u solve it?!

user228700

8:33 AM

^

Secret

Because I have no idea whether we are solving for $a$ or $\alpha$, $\beta$ thus I end up running through all cases in order to figure out what exactly we are solving and why

user228700

They've taken $x=\sin\alpha$ but I took $x=\cos\alpha$. I don't think it makes any difference.

user228700

@Secret You could've asked me, you know...

user228700

Since they went another way, the method will obviously be different but as seen above, they haven't considered the case in which $\sin (\alpha + \beta)/2 =0$

Pissed off layman

What does "Aliter" mean at the bottom?

user228700

8:39 AM

"Different way to solve it".

Pissed off layman

thanks

Secret

no , let me think about this harder...

user228700

It certainly doesn't evaluate to zero.

user228700

$dy/dx = - \sqrt{1-y^2}/\sqrt{1-x^2}$

user228700

...which, indeed, is (still) not what we want but we can't ignore the solution simply because it's not what we want, can we?

Secret

8:43 AM

yeah, I think it is a valid solution. The question does not really rule out that the two surds cannot be negative, thus this case is still valid

user228700

It's not, however, what we want :-/

Secret

unless the convention that $\sqrt{}$ is taken to be positive

user228700

@Secret Hm?

user228700

@Pissedofflayman No problem :-)

Secret

well, I think $dy/dx=\pm \sqrt{1-y^2}/\sqrt{1-x^2}$ as the negative case will be covered if one of the surds is negative. It is equally valid for the original equaiton to be $\sqrt{1-x^2}-(-\sqrt{1-y^2})=a(x-y)$

user228700

8:49 AM

Hm, OK. Conclusion: My textbook has been sketchy, yet again.

user228700

@Sec, @Anonym, @JohnR: Thanks very much :-)

anonymous

@Kaumudi.H No, your textbook is correct. I told you that you should always upload the full question instead of a sub-part. Now please listen as to why they ignored the other case.

user228700

Um, I don't recall you having told me that, but OK..?

anonymous

The other case is a boderline one at which the $\frac{dy}{dx}$ becomes undefined.

alpha and beta lie between -pi/2 and pi/2.l

Now depending on whether a is positive or negative we will take the condition for cos(\alpha2+\beta/2)

=0

Either alpha and beta both have to be pi/2

or both have to be -pi/2

For $\cos(\alpha/2+\beta/2)=0$

Ok till here ?

user228700

I was trying to understand why $\sin (\alpha + \beta)/2 \ne 0$, not $\cos$.

anonymous

8:54 AM

@Kaumudi.H In the picture you uploaded

cos(\alpha/2+\beta/2)=0 is the other case

user228700

In the case of $\sin$, one could be $-\pi/2$ and the other $\pi/2$ which would give us the required condition.

user228700

20 mins ago, by Kaumudi. H

They've taken $x=\sin\alpha$ but I took $x=\cos\alpha$. I don't think it makes any difference.

anonymous

@Kaumudi.H Even in case of sine, that would be a border case

where derivative is undefined or infinite

that is why it is ignored

try using the other condition and check

user228700

Undefined? How?

anonymous

$\sin(\alpha/2+\beta/2)=0$ what are the possible values of $\alpha + \beta$ ?

user228700

8:56 AM

No, never mind. I understand.

user228700

Thanks.

anonymous

given that $\alpha$ and $\beta$ lie between -pi/2 and pi/2 ?

@Kaumudi.H Really ?

user228700

Yes.

anonymous

Okay.

user228700

.__. Why in the world would I lie?

anonymous

8:58 AM

@Kaumudi.H To utilize your time on something more important. Maybe ? :)

anyway

good that you got it

bye then

user228700

Nope, I really did understand, thanks.

user228700

OK, bye :-)

Secret

@anonymous In that above case since $\alpha + \beta = 2n\pi$,$\alpha, \beta$ will be of the form (\pi/2,\pi/2), (\pi/2,-\pi/2) and their permutations, which is really lying at the boundary point of the inverse trig functions (and undefined for arctan)

anonymous

@Secret Exactly!

Secret

O man I hate cases, I always miss at least one

this is why I like to solve questions under the philosophy of full solution sets

anonymous

9:05 AM

@Secret There were only two cases :)

Secret

But I forgot that boundary points only have left or right derivatives defined and that is the gist needed to rule that case out

I really need to pay more attention to bounday points in functions...

anonymous

Yeah, and that is why it is important to quote the whole question in maths. You never know what you are missing out unless you see the whole question!

Secret

exactly!

anonymous

Anyway, i enjoyed the problem. It was nice nevertheless.

Secret

I am glad my high school maths skills have not become too rusty

Secret

9:09 AM

Predicting quantum decoherence before it occurs

Bhargav

9:19 AM

would appreciate some help with this physics.stackexchange.com/questions/305853/…

anonymous

@Bhargav It will be a damped harmonic oscillation (if I got your question correctly). Use en.wikipedia.org/wiki/…

Secret

10:37 AM

Movement: Walking along the curve, shift: translating the curve

Pissed off layman

11:28 AM

in Homotopy Theory, yesterday, by Charles Rezk

https://www.nidirect.gov.uk/articles/how-apply-your-higher-category-theory-and-p‌ractical-driving-tests

Secret

11:41 AM

has nothing to do with high category theory

Meanwhile this is necessary but very boring

heather

12:26 PM

hello

anonymous

@heather holla!...submitted your GR project?

heather

@anonymous yup

(well, sort of - we turned out to have an extra day)

but it's done.

anonymous

Good :)

heather

i'll probably keep improving it on my own time though =)

anonymous

made the videos? :D

heather

12:31 PM

i talked to the teacher and she said i didn't have to do them, though i did end up making a quick introductory, sloppy, did it in a hour or so video in powtoons.

anonymous

never heard of powtoon. googling!

powtoon.com

wow..seems great.

heather

here's the video

powtoon.com/online-presentation/blBtPlYTIRq/?mode=movie#

it's bad, I know.

and my voice sounds terrible, voiceover does that.

anonymous

@heather man, you made a video on GR! That belittles all those little mistakes :) I will watch the video today and let you know. Thanks for the link btw :)

got some work now...cya !

heather

@anonymous see you

it's only like 3min, fyi, it doesn't say much

it took me an hour to make because i was procrastinating

Sir Cumference

Morning

heather

12:43 PM

good morning

Sir Cumference

@Kaumudi.H Yep...

Sir Cumference

12:57 PM

@heather You have a GR teacher?

Jim

Stupid winter with stupid snow and stupid ice. Honestly, what's snow and ice ever been good for except hockey and raising Earth's albedo?

heather

@SirCumference no, social studies

lol

Sir Cumference

@Jim Raising Earth's albedo counts as a good thing?

@heather Oh XD

I was about to say I have only ever met 3 or so people in person who know GR well

Jim

@SirCumference do you not think it does? It reflects more incident light from Earth, thus lessening the amount of heat that can be trapped by greenhouse gases

Sir Cumference

Meanwhile I know more than 20 who know QM well

I guess GR is just that much less known

Jim

1:02 PM

@SirCumference Yeah but GR is research level whereas QM has been replaced by QFT for the most part in frontier physics. So how many have you met that know QFT well?

Sir Cumference

@Jim Huh, never actually learned about that.

@Jim Zero.

@Jim But yeah, I wish I knew that instead of learning about the differences between geometric and Bond albedo

@Jim But darn dude, cosmology is so complicated. I have huge respect for guys like you who can intuitively grasp it.

Jim

@SirCumference I don't know if you have to learn it so much as put 2 and 2 together from the description of what albedo is and understanding the greenhouse effect

@SirCumference intuitively? Maybe after lots of practice

Sir Cumference

@Jim Yeah, more or less not thinking

Jim

It also helps to have a weak grasp of reality

Sir Cumference

@Jim Weak grasp of reality?

Jim

1:05 PM

@SirCumference you gotta live in your own little world

Sir Cumference

@Jim Hmm...see, Newtonian mechanics wasn't a piece of cake to grasp initially in high school, but it always seemed to make sense, at least qualitatively

Now I find myself saying

7 hours ago, by Sir Cumference

If it weren't for the math, I'd say that makes no sense at all

Jim

when it comes to cosmology, all you gotta do is know enough math to get a handle on tensor calculus using einstein notation, get used to the terminology, and embrace the idea that whatever you think at first is probably wrong.

Sir Cumference

> embrace the idea that whatever you think at first is probably wrong.

Huh...

Jim

Intuition and experience-based reasoning has been wrong so many times in cosmology that we come to expect the first theory about any new topic is probably way off

Sir Cumference

Hm, I'll keep that in mind. Thanks then

Jim

1:14 PM

I've never met a group of people who pay less attention to their gut feelings than cosmologists. Not saying they're unbiased. It's more like reality has come along and gut-punched their ideas so often that it's gone numb. No gut feeling. They'll sometimes pursue data or math with a confirmation bias, but there's so little self-confidence in instinct that they'd be skeptical thinking up is the opposite of down without the math to back it up

Sir Cumference

Well, it's at least relieving to know this stuff isn't expectable to anyone. Maybe that adds a somewhat valid reason for my struggling.

Jim

::Jim just realized he has been speaking about himself in the third person. Jim does not care. After all, there's no "I" in "Jim"::

4

anonymous

@heather It was nice. Saw it. But I think you could add some more stuff and animations. At present it looks like a slide show with voice rather than a video :)

@Jim Hi Jim! Nice to meet a cosmologist on SE. Where do you work/research presently ? (If you don't mind saying :))

Secret

1:39 PM

0

Q: Is The following Schem Real?

Nuclear fusion in a liquid Electrolytic Environment The main necessary to consider the formula of the nuclear reaction, the following This is the nucleus of the helium fusion reactions nuclei of deuterium and tritium reagents. The occ...

nuclear-physics fusion

Cold fusion?

Jim

@anonymous I work at the University of Toronto Mississauga. Unfortunately I'm not doing active research at the moment, but I'm keeping my ear to the ground for opportunities.

Secret

I don't like memorisation: It is a pain to work out which is the correct number without memorising

Kyle Oman

3:14 PM

I did a "radio" (podcast) interview about my research and grad school experience with my campus station, if anyone is curious! (There is one blunder I should correct: the DDO catalogue is from David Dunlap Observatory, Toronto, not Dominion Observatory, Victoria).

5

Pissed off layman

No need to memorize @Secret I'd say 3 meters front and back because parallel parking is the toughest thing to master.

Try doing it on a hill :-/

rob

4:01 PM

@Pissedofflayman Three meters front and back? That's practically enough space to park an additional car

But then, I've parallel parked in spots so tight that people assumed I had been blocked in by the people around me.

user228700

@heather: Nice video! :-) (Also, you sound exactly as I've imagined :-P)

user228700

Huh. Dyou wear braces? It sounds like you've a lisp (I used to wear braces so I'm adept at recognising these things :-P)

user228700

@JohnR: Evening! :-)

John Rennie

Hi :-)

user228700

@JohnRennie Wadup?

Bernardo Meurer

4:08 PM

@JohnRennie Howdy

John Rennie

@BernardoMeurer Hi

@Kaumudi.H Not much really. I've just been fixing a friend's (Android) tablet, and we may have had a drink or two afterwards :-)

user228700

Oh, OK :-)

user228700

I used to have an android tablet.

user228700

Elocity.

John Rennie

4:12 PM

"Used to" ?

user228700

One of my dad's friends used to work for the company. They eventually stopped manufacturing them and had a bunch of excess tablets and one of them I got.

user228700

@JohnRennie Yeah, it died. It was brutal.

John Rennie

@Kaumudi.H you make it sound like it exploded! Or maybe burst into flames!

user228700

:-) No no, its death was slow and painful.

user228700

It didn't even have Google Play so it was doomed from the beginning, really...

user228700

4:17 PM

I still have it but it won't switch on--no charge and the charger's broken.

Bernardo Meurer