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5:01 PM
@Kaumudi.H sausage rolls
 
@BernardoMeurer Reminded me of this: ece.ubc.ca/~sasha/papers/eurosys16-final29.pdf
 
@JohnRennie Nice :-)
 
@JohnRennie they look awesome
 
They were very tasty. I ate them with a chili barbecue sauce/
 
man, now I'm hungry
 
5:08 PM
@JohnRennie Cool :-) What are those pins in the background?
 
@Kaumudi.H Pins? The yellow and orange thing?
 
Yeah, those.
 
It's a coffee mat made by my niece when she was about six.
 
huh, those look familiar
are those the beads you put on a template, and then you iron them and their plastic, so they "melt" together?
 
@heather Yes, that's exactly it.
 
5:10 PM
@JohnRennie Ah, OK :-) It looks very nice.
 
@JohnRennie ah, that makes sense then. I did those when I was 6-7.
 
@Kaumudi.H It would be more than my life's worth to not be using it when my niece comes to visit, but in fact it is the perfect size for my coffee mug.
 
:-) I see.
 
Yes, it does look very pretty :-) I couldn't imagine making that now, let alone when I was six.
 
5:13 PM
@heather I suspect it's a common pastime for six year olds both sides of the Atlantic :-)
 
@JohnRennie I have no recollection of playing with beads like those in childhood. Then again, I suck at crafts of all sorts :-|
 
Me too! But then ... well ... PHYSICS!! :-)
 
:-) Yes. PHYSICS!!
I should go to bed. Goodnight :-)
 
Bye
 
o/
 
5:34 PM
Me reading some physics
 
heheheh
 
me reading some physics
@DanielSank, hello
 
Still ten episodes left before I have watched all of star trek
end my life
 
@Slereah Which series remains?
@heather Yo.
 
Ten episodes of season 3 of TOS
 
5:40 PM
@Slereah Ah. That one is at least not as bad as Enterprise.
Enterprise got super weird.
 
Yeah, Enterprise was kind of a chore to get through
Star Trek the animated series on the other hand
Short
 
Uhhh yeah... time travelling space Nazis...
 
I think I watched it in one day
 
@Slereah I forgot that existed!
 
How could you
It's the one where Spock summons Satan
 
5:42 PM
o_O
 
woah
I have some watching to do.
 
It's not very great but it's fun in a weird 60's way
 
To complete all of Star Trek, I need to finish roughly the second half of TOS and apparently watch a cartoon!
 
Also there's a new Star Trek series coming out soon-ish
 
5:44 PM
!!?!?
(Oh, and I'm not counting the movies. Should I count the movies?)
(The latest one was horrible)
 
Star Trek 2 through 6 are great
I even like 5
 
I don't think I've seen 1, 5, or 6.
 
6 is a great way to end TOS
 
hmmm ok
 
Unlike Star Trek 10
pretty bad way to end TNG
TOS has way too many episodes with people in togas
2
 
5:49 PM
Yeah Nemesis was dumb and I never saw Insurrection.
 
This is the future, quit showing me ancient rome
 
@Slereah very true.
 
I'm guessing they got a lot of ancient rome sets and costumes for cheap
 
Oh yes.
What do you think of the Abrams reboots? I think the first one was ok and the others were awful.
awful
 
Eeeh
None of them were really that good
 
5:50 PM
Yeah the first one was passable.
 
The big problem of Abrams is that he doesn't understand what Star Trek is
Or doesn't care
though that was also true of most of the TNG movies, too
Star Trek isn't an action piece
 
Yeah that was my complaint too.
The new ones are action movies. And really, they're not even good action movies.
The Matrix is a good action movie.
It's the gold standard, IMHO, for sci-fi action.
 
Star Trek isn't even that great at being science fiction
 
@Slereah ?
 
Star Trek is mostly about morality stories
Well how many episodes of Star Trek actually deal with the consequences of science and technology on society
A few, but not that many
Star Trek barely has consistent science because every episode something breaks what they know of science
 
5:54 PM
@Slereah Ah, well, TNG is, I would agree there.
 
Most series, really
Hell especially TOS
 
@Slereah Again, that's only TNG.
DS9, Voyager, and Enterprise do considerably less of that.
 
I dunno, what would you say Voyager is about
I mean, on an episode to episode basis
Not the overarching plot
I mostly remember episodes about moral dilemmas and such
Like that time the doctor had some help from a death camp doctor
 
@Slereah Voyager was a mix. Some sciency "oh no the organic black hole has got us" episodes, some "the prime directive blah blah" episodes, some "does the doctor have a conscience" episodes...
 
Star Trek is never really about science, though
It's always technobabble
 
5:57 PM
IMHO science fiction does not mean the story is "about science".
 
Well prime directive episodes are such kind of morality stories
 
Like... Ghost in the Shell is sci-fi but it's not about science.
 
What would you say science fiction is
 
@Slereah Yeah, for sure.
 
Just a setting?
 
5:57 PM
@Slereah Yeah, more or less.
 
Would you say Star Wars is science fiction
 
@Slereah That's a great point. I would call it "sci-fi/fantasy".
Now I have to think about why.
I think I see where you're going.
 
My point is mostly that if you can change the setting without changing the story, I wouldn't consider that science fiction
 
@Slereah That's a good point.
Is The Matrix sci-fi?
 
Hm
Kinda?
 
5:59 PM
I'm starting to think, just sitting here talking with you, that sci-fi often does involve some kind of morality issues...
 
You could change the setting I guess
Plato's cave or something
It would be harder, though
Well people like moral dilemmas better than they like science
 
I have to run, but I'll think about this more.
 
Me I had a lot of fun just reading the description of the neutron star people in Flux
 
@Slereah Well yeah, without some kind of human-level conflict you don't have much of a story after all.
@Slereah Oh interesting.
 
Pop quiz: Does Jaba The Hutt have any siblings?
 
6:00 PM
Yes.
Give me a hard one.
I know all the Star Wars.
 
@skullpetrol no, because the expanded universe isn't canon anymore
 
Actually wait a sec maybe he didn't.
Crap.
::consults Wookiepedia::
 
wtf why?
 
As male individuals were not biologically meant to nurse their offsprings, they sported much smaller breasts and produced no milk. Mandalore the Lesser (then a gladiator),[9] Aron Peacebringer (a planetary leader)[10], and Anakin Skywalker (in certain circumstances, such as on Nelvaan) would freely exhibit them.
 
6:08 PM
What's that got to do with Jabba?
 
All things are interconnected
 
Interconnected are all things ;-)
 
@alarge That is awesome! Thank you for linking me to that!
 
6:27 PM
Did you decide which physics textbook you're gonna get?
 
@skullpetrol No, I should do that now actually, thanks for reminding me
@Skyler How old are you again?
 
Take your time; it's an important decision.
 
@BernardoMeurer just turned 23
 
@Skyler Good, so you're fresh
Also, Happy belated birthday :)
 
6:31 PM
Thanks
And fresh? Interesting word choice
 
@Skyler I'm picking a book for my first course in mechanics
 
@BernardoMeurer lower division right
 
Happy birthday.
 
Kleppner & Kolenkow has the advantage that @DanielSank knows it, and since he's my main support when I'm fucked and need answers that's good
@Skyler Wat?
"Fundamentals of Physics" is recommended by the prof
 
@skullpetrol thanks, its been 5 days so its already almost time to stop counting it as new
 
6:33 PM
@Skyler happy birthday =)
 
@BernardoMeurer like this is a first course in mechanics
@BernardoMeurer doh, you said it here
what math background?
 
@Skyler I went through Liner Algebra and Analysis I (All of Calc I + some or all of US Calc II I think)
 
@heather thanks heather
@BernardoMeurer and any physics textbooks you know you like already
 
@Skyler Would you count Kama Sutra as a physics textbook?
 
Which book did you use last year?
 
6:38 PM
@skullpetrol For what? It's my first ever physics course
I mean, apart from High School BS
 
@skullpetrol me?
 
Which physics textbook did you use in high school? @BernardoMeurer
 
physics.stackexchange.com/questions/313117/… - with the section on "For higher energy electrons the loss of energy as bremsstrahlung becomes increasingly important" would it make sense to think of it as objects with greater energies have a harder time braking, so they lose more energy?
 
@skullpetrol No clue, some Brazilian one
 
Is the Theoretical Minimum set up at an upper division level? It might still be good as a reference since it sets up some good frameworks early on
 
6:40 PM
I guess even with that understanding I still don't get the 10 MeV limit or how $Z^2$ works exactly
@Skyler by Leonard Susskind?
 
Was it calculus based? @BernardoMeurer
 
@heather yea
 
@Skyler I have a copy of that book...it aims to start from the basics, and it includes a review of calc.
it's not too bad, in my opinion.
 
@heather how advanced does it get
 
6:42 PM
Gets into lagrangian/hamiltonian if i remember right, plus poisson brackets and such
let me grab the copy
 
oh not bad
 
But, but... You're in grade 8? @heather ;-)
</end sarcastic joke>
sorry
 
not sure i see anything worth apologizing for...
also @BernardoMeurer. Ever heard of Libgen
 
To a gifted student.
 
my biggest regret is not actually caring to look for that site until my last quarter as an undergrad, use it
my BIGGEST, may be an overstatement
my biggest regret is probably not doing something like what heather is doing
 
6:49 PM
ah, okay
the lecture titles (and interlude titles):
1. the nature of classical physics
interlude 1. spaces, trig, and vectors
2. motion
interlude 2. integral calc
3. dynamics
interlude 3. partial differentiation
4. systems of more than 1 particle
5. energy
6. the principle of least action
7. symmetries and conservation laws
8. hamiltonian mechanics and time translation invariance
9. the phase space fluid and the gibbs-liouville theorem
10. poisson brackets, angular momentum, and symmetries
11. electric and magnetic forces
and finally an appendix on central forces and planetary orbits
 
Suskind does a "Theoretical Minimum" series of lectures on YouTube.
 
@heather I have made a quick example plot. This is how it looks like for water:
You see that at high electron energy, the contribution of bremstrahlung wins.
 
@Loong yeah, i know that's true. I was asking if my intuition for why that happens is correct.
 
7:05 PM
 
10 MeV also doesn't seem to have any particular significance in that graph...i wonder why that was mentioned in my book...
 
@heather I don't know. However, the most important high-energy beta nuclide is N-16, which as maximum beta energy of about 10 MeV.
 
Also does your intuition include a logarithmic scale?
 
@skullpetrol my intuition generally doesn't include graph paper, logarithmic or not =P so no.
@Loong what does "beta energy" mean?
 
@heather the kinetic energy of an electron that is the result of a radioactive beta decay
 
7:16 PM
@Loong like the energy after the decay happens?
 
Yes.
 
@heather The decay energy is distributed between the nucleus, the beta particle (i.e. electron), and the antineutrino. Therefore, beta particles show a continous spectrum from about zero to a characteristic maximum beta energy.
 
I'll put this here for future reference: the HNQ sidebar currently has
13
Q: Why is a neutron in free state unstable?

user145764A neutron is a neutral particle which is merely some times massive than an electron what makes it so unstable outside the nucleus that it has a half life only of about 12min?

and
32
Q: Can a neutron star become a black hole via cooling?

user289661How much does thermal expansion affect neutron stars? Would the loss of temperature cause a neutron star to be more densely packed and thus collapse into a black hole?

as its physics representatives
This is how it should be.
4
 
7:32 PM
@Loong what is the exact mechanism that determines how much each one gets?
 
@Loong i'm sorry, i don't understand your explanation. what do you mean by "beta particles show a continuous spectrum"? Is this the spectrum of all possible energies for the beta particle?
if so, then i understand what you said prior.
 
How was Chicago? @0celo7
 
@heather If the decay was to proton (or other heavy remnant) + electron there would only be one way to share the energy between the two particles that conserved momentum.
So a measurement of the electron energy spectrum would show a single isolated peak (modulo instrumental blurring and scattering tails).
 
@0celo7 it's mainly conservation of momentum, but you need a relativistic momentum for the beta particle
 
But with three particles there are many ways to share out the energy depending on the relative angles of the particle emissions. The result is a broad electron energy spectrum.
You should be able to convince yourself of this using a classical model of an object blown apart into either two or three discrete pieces of prearranged masses by an explosive that delivers fixed energy to the parts. No need to bother with relativistic energy and momentum to get the idea.
@BernardoMeurer I have a decided loathing for Halliday's books. Have since the 90's. Unfortunately for you there are some very good scientists who consider them very good. So you're facing mixed recommendations.
 
7:45 PM
@Loong conservation of momentum does not give a distribution...
 
@0celo7 Do you mean the shape of the distribution?
 
@Loong of course
 
@0celo7 It is the need to conserve both momentum and energy that forces your hand, because the energy available comes from the mass difference and so is fixed for any given decay. Work the problem in the frame of the parent particle for simplicity.
 
@dmckee so it is the spectrum of all possible energies for the electron/beta particle?
 
@dmckee Picking books is hard
 
7:49 PM
@heather Well, we only measure the electron energy (neutrino escapes undetected and the recoiling proton or nucleus is much lower energy and therefore hard to detect).
@BernardoMeurer I used Giancolli in college and didn't hate it. I have a desk copy of the newest edition and find it no worse than the competition.
There is something to be said for Shankar's books: amazon.com/… REeatively readable, so you might like them.
 
@0celo7 If you want the exact shape of the distribution, you could use the Fermi theory of beta decay. However, you should ask a physicist about it. ;-)
 
There are no end-of-chapter exercises, but they go with his open course through Yale.
 
"For energies up to 10 MeV, electrons lose their energy to the detection medium mainly by excitation and ionization of the electrons of the medium, as in the case of heavily charged particles." @Loong that was the quote involving the 10 MeV threshold
 
@heather The energy of this approximate threshold depends on the absorbing material (e.g. air, water, lead). Do they mention the material in your text?
 
let me check again
 
7:58 PM
@dmckee I think I'll use random.org
 
It does mention that the "loss of energy as bremsstrahlung becomes increasingly important and the intesity of this varies as $Z^2$ where $Z$ is the atomic number of the medium."
otherwise, that's literally the beginning of the section.
 
::chuckles:: We have the same problem in trying to decide what books to adopt for courses. Unfortunately, the school makes us stick with anything we choose for at least three years, so the process is fraught.
@heather Depending on your goals in this reading it might (or might not) be worthwhile to just take a few things on faith at this point and come back to the subject when you have accumulated a little more knowledge of how these ideas are used.
 
@heather @heather My above-mentioned plot is for water. In that plot, the threshold looks more like about 100 MeV. But for a high-Z element like lead, a value of 10 MeV could be correct.
 
Sometimes knowing the context in which ideas will be applied makes it easier to see what authors have in mind.
 
it does mention lead in a later example @Loong
wow, then I have two out of three of my confusions about the passage cleared up
thanks @Loong
and @dmckee
 
8:03 PM
@heather ok, I have made a quick plot for lead:
10 MeV looks good
 
how do you calculate these so quickly?
 
I anticipated that your example would be lead. ;-)
 
Just judging from the way the plot is present he may be using ROOT, which has some surprisingly good facilities for that kind of thing which makes it relatively easiy to generate plots.
 
okay, another question: when it says that "the intensity of this varies as $Z^2$ where $Z$ is the atomic number of the medium" what does it mean? is $Z^2$ literally the point at which the energy loss due to bremsstrahlung equals the energy loss due to ionization?
that makes the most sense intuitively, but it doesn't make much sense mathematically - lead's atomic number is 82, and $82^2$ is 6724 - not sure how that relates to 10 MeV.
 
@Slereah These "theorems" often restrict very strongly the class of theories they are talking about. I'd bet that e.g. gauge theories coupled to matter are not in the class of theories forbidden here, otherwise the Yang-Mills mass gap question wouldn't be so hard.
 
8:11 PM
mb
still
QFT is all bogus
Throw it in the dump
 
@heather "$x$ varies as $y$" normally means that $x=ky$, for $k$ (roughly) a constant
 
Let's go raise goats instead
 
so there's a constant that gets 6724 to 10 MeV @EmilioPisanty? (thanks for the terminology translation
 
so "varies as $Z^2$" means that if you took something with half the atomic number, then the intensity of (whatever) would be 1/4
 
@Slereah Have fun with that :)
 
8:13 PM
@EmilioPisanty sorry, i'm confused by your example - what do you mean by "half the atomic number" - half the atomic number of what?
 
You could make your own goat tallow :P
 
Here is me
 
@heather as in, if your target was made of niobium ($Z=41$) instead of lead
 
$41^2\neq\frac{1}{4}$
 
@heather $k \,41^2 = \frac14 k \,82^2$.
 
8:16 PM
 
It's hard to make sheep scary
 
@EmilioPisanty where'd the 1/4 come from?
 
@Loong that's the creepiest thing I've seen this month
2
 
They are fluffy little clouds
 
@Loong that is scary
 
8:17 PM
@heather $42=\frac12 82$
$41^2 = \frac14 82^2$
 
also they couldn't find their way out of a shoebox
So I'm not very terrified
 
okay...so 1/4 = k?
sorry, i'm just confused by this.
 
@heather No. The statement says that you have some quantity of interest $I(Z)$ that depends on $Z$ and they're telling you that it goes like $Z^2$, which means that there is some constant $k$ such that $I(Z) = k Z^2$.
This is only useful if you're willing to consider different media with different $Z$
 
right...
 
and, in that case, the dependence $I(Z)=kZ^2$ tells you how the intensity will change as a function of $Z$.
 
8:21 PM
yeah
so where does your thing come in?
 
So, if you have some medium with $Z=Z_1$, and you want to compare the quantity of interest $I(Z_1)$ in that medium with, say, some other medium with $Z=Z_2=\frac12 Z_1$ (such as lead as medium 1 with $Z_1=82$ and niobium as medium 2 with $Z_2=41$), then you would have $$I(Z_2)=kZ_2^2 = k(\tfrac12 Z_1)^2 = \frac14 kZ_1^2 = \frac14 I(Z_1).$$
i.e. "if you halve the $Z$ then you get a quarter of the response".
 
so using a different $Z$'s equation to solve for another $Z$?
sort of?
so $k$ changes from element to element?
it says 9 MeV electrons in lead lose as much energy due to bremsstrahlung as ionization...but wait, Loong said it was 10 MeV and that's what it implied earlier....
and how does the energy of the electron come into play in that equation? it has too..
this book =/
 
8:40 PM
You know so far in that GR book I'm doing
I have not written down the EFE
 
is there a way I can get the latex in this chat to format?
 
@StevenSagona RTFM
18
A: Any chance of MathJax in chat?

Ilmari KaronenAs a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

 
@StevenSagona Look in the upper right corner of the chat room.
It's called MathJax here.
 
thanks
 
@ACuriousMind this game has "terrible boss battles" syndrome
 
8:48 PM
I just returned from one of the most excellent RPG sessions I've ever had, I've no mind to discuss video games right now, sorry
 
hmm, i think i can move on from this section once i understand one thing: how does the electron energy play a role in the equation $I(Z)=kZ^2$
 
@heather maybe $k=k(E)$?
 
@0celo7 you mean $k$ equals the electron energy?
i'm not sure that makes sense
 
@heather my Knoll says $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{NEZ(Z+1)e^4}{137m_0^2c^4}\left(4\ln\frac{2E}{m_0c^2}-\frac43\right)$, but I haven't checked it.
 
@heather no, that's the whole point, $k$ is a constant and does not depend on $Z$.
it might depend on a bunch of other variables, though
the point of $I\propto Z^2$ is that if you keep everything equal and change $Z$, then you know how the response will change
 
8:57 PM
@EmilioPisanty sorry, what does that open infinity sign mean?
 
@heather "proportional to"
\propto
a.k.a. "varies as"
 

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