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3:01 PM
did @Lagranian get the right answer in the end?
 
Was the small animal snuff film Exploding Varmint
 
im confused
Does the lack of clothing enhance his hunting skills?
 
Hey, everyone. Can you guys help me with this problem of Non-linear dynamics?
It says to find a mechanical system that obeys dx/dt=sinx. And Then explain intuitively why x=0 and pi are unstable and stable fixed points.
But, I am having trouble finding any mechanical example of this kind!\
Can you guys give me some hinta?
 
3:25 PM
Consider for example an harmonic oscillator, what is the equation of motion?
a pendulum*
to be correct
 
How about measuring circular motion against only one axis
which is analogous to a harmonic oscillator
 
@Phase This still doesn't give you the first derivative
 
What do you mean?
 
He needs dx/dt in the equation, which doesn't appear in the case of oscillatory dynamics.
 
wait
Nevermind
 
3:36 PM
Sigh
-7
Q: Gravitational field

Good ole boyAccording to science without gravity we would be thrown off of the earth. So why are we still here scientifically we should not be standing on this planet but yet we are. Something doesn't add up, can someone help ease my mind with there input.

Yet he asks the exact same question a few days later, and it gets hot network
1
Q: Earths rotation

user19600The Earth is spinning at about a thousand miles per hour. Shouldn't we be thrown from its surface? Yes I know someone would comment "gravity" but physically we should be thrown from Earths surface. Scientists say that gravity is holding us here but that seems to contradict physics. Maybe someon...

 
i mean
To be fair
The second is formulated slightly better
 
> but that seems to contradict physics
?????
 
i know i know but at least he said "seems" rather than "scientifically we should not be standing on this planet, something doesn't add up"
 
I...guess
Still surprised by the HNQ
 
yeah but i mean
It's not as if I don't regularly see things that shouldnt be there, there
 
Anonymous
3:49 PM
@EmilioPisanty The half-circle contour gives a value which is half of the actual value
 
Anonymous
I think it is necessary to use the keyhole contour here
 
Anonymous
I don't know why
 
Anonymous
Perhaps it is because $(z)^{1/2}=e^{\frac1 2{\ln(|z|)+\arg(z)}}$ repeats itself after $4\pi$ (for same magnitude of $z$) while $\ln(z)$ does it for $2\pi$.
 
in The Periodic Table, 26 mins ago, by Abcd
Can someone explain this statement to me: "Can you please explain this statement: "The oxosalts of group 2 metals are less stable to heat than the corresponding group 1 salts because the metals and their hydroxides are less basic than those of group 1?"
Unabke to find satisfactory answer since i) only group trends I can find and ii) Group 2 metal cations should be smaller than there anion counterparts, not larger, thus polarisability arguments don't work
 
Jeez in 3 days the temperature dropped from 63° to 25°F (~17° to -4° C)
What is with this weather
 
3:57 PM
You're all intelligent people
And at the risk of airing dirty laundry where it shouldn't be:
What's the best way to deal with an Ex messaging you most days of the week
Without being cruel or hurtful
 
@Phase Depends, is it obvious she still wants to be together or is she legitimately trying to be nice?
 
She's asking me "if I'll ever want to be with someone else"
and shit like that
 
the overbearing stuff that made me end it in the first place
 
I mean I don't have all the details, but if it sounds like maybe you should talk to her again. Make it clear she needs to move on
 
4:03 PM
I have
 
And if you bring up the stuff she says in a non-harsh way, she should back off
 
the conversation evolved a bit and I wrote
"It's fine to be sad but not necessarily regret things, it's best if we go our separate ways"
Things are so annoyingly complicated sometimes
 
@Phase Sounds like a problem I wish I were having around now...
 
How come?
 
Have only had one relationship in my life, for a month or so, in college
 
4:06 PM
I'm currently in a complicated relationship with my hangover.
 
Still haven't really figured out how to approach other girls
Sounds childish, I know...
 
nothing wrong with that. But if you want a relationship, only way is to just do stuff. Play social video games or whatnot
Or if your friend suggests something that involves meeting new people
Go for it
And most importantly
 
^simple, yet sage, advice
 
Rejection is meaningless, the sadness passes extremely quickly compared to the potential regret of never asking
Make your feelings clear as soon as is appropiate
 
@Phase I don't really know how to ask
 
4:08 PM
What about something as casual and cliche as "wanna go see a film?"
phrased to be more in tune with how you talk obviously
 
That might not hint at anything
 
Does it really have to? If you have a great time then it's a great time
Chances are if she enjoys it as much as you then you can go for another
And make your feelings clear when you feel its a good moment
Or you could just go straight out the gate with "Hey, You seem nice / i like you, how about a date to ________"?
Idk dude what's important is that you try innit
 
@Phase jeez, isn't that out of nowhere
 
I guess so but i cant give a phrase that works in all situations
How you approach it is less important than the fact that you do approach it imo
 
11
A: Unique limits of sequences plus what implies Hausdorff?

Michael GreineckerFirst countable is enough. Let $x\neq y$ be two points in your space that cannot be separated by neighborhoods. Let $O_1,O_2,\ldots$ form a neighborhood base of $x$ and let $U_1,U_2,\ldots$ form a neighborhood base for $y$. Choose a sequence $(z_n)$ such that $z_n\in O_n\cap U_n$ for all $n$. Now...

 
4:10 PM
Don't let feelings go unsaid
 
Why is $(z_n)$ a converging sequence here
 
@Phase I think this is really the most important thing, but then I've got a feeling everyone's going to disagree with me here...
 
@Phase This is more complicated than physics somehow. There needs to be a formula that always works.
 
@0celo7 @ACuriousMind @BalarkaSen
 
Relationship is a beyond PSPACE problem
 
4:11 PM
@SirCumference It's more complicated because people let/make it more complicated :P
 
don't even think there is any algorithm to solve it
 
At the end of the day @SirCumference if your date doesn't realise it's a date at that point, if you just roll with it and enjoy the day as it happens then not only will you know better if you're compatible, but you can then bridge the gap into "hey today / tonight was great, how about we do it again" and throw in bits alluding to it being a date
 
@Mithrandir24601 FWIW, I completely agree.
@Slereah Because the definition of the limit of a sequence is "point that contains an element of the sequence in every neighbourhood".
And since one $z_n$ lies in every neighbourhood basis set, one lies in every neighbourhood.
 
@Phase Again, this is me being me, but I still don't really get the difference between two people going and doing something together and two people going on a 'date'
 
Ah yes
Thx
Although...
 
4:13 PM
@Mithrandir24601 Welp me watching a movie with a family member is not equal to me dating a family member
 
That's in every neighbourhood of the basis
Is that equivalent
 
@Mithrandir24601 well, not any in the actual day spent but there's a bit in terms of "clear feelings". They could be interested but not in any way more than a friend
 
@Slereah Well, every neighbourhood is a union/intersection of the basis, so yes.
 
seriously though rejection is meaningless. You'll feel better in a day or two.
 
Aight
thx
 
4:15 PM
@Mithrandir24601 The difference is that it's labeled a "date" and at least one side has romantic intentions :P
2
 
@Phase Sigh, working up that confidence and getting rejected will probably hurt more...
 
Nah
Well
Ok
 
@ACuriousMind This is by far the clearest description I've seen yet
 
It WILL but all subsequent rejections and courage-mustering will be a lot easier
No-one likes rejection but my point is that after the first time, you'll realise how much better it is to be rejected than just letting an opportunity pass and regretting it
 
What if you don't know if you've got 'romantic intentions'? (OK, yes, I'm just being pedantic now :P)
 
4:16 PM
Would we really notice the difference if there was a preferred reference frame, in the sense of experiencing observers, experience everything simultaneously on that frame's simultaneity axis?
 
When I was but a wee lad I had a crush on a girl for years, and felt like I could never tell her about it. It was a shitty few years. Opened up for the first time, got rejected, had a pretty hellish time but after everything was way better
 
A machine would not notice a difference. As clocks would slow down relative to that preferred reference frame, so would everything within the machine. Only a conscious observer MIGHT experience the difference, should such a preferred reference frame exist and only if the experiencing entity is outside the physical realm
 
@Mithrandir24601 Then you're human and have to figure it out along the way ;)
 
something one could actually test for scientifically
One would have to accelerate some conscious observer to relativistic speeds to find out
 
Wait hold on, is rob a professor?
 
4:20 PM
The only time the "date" descriptor really matters is if one person thinks it's a date and the other thinks it isn't (and both think it's a meaningful distinction)
 
^^^
but non-mutual dates are harmless most of the time [if harm is deemed as counter to your intentions]
 
@ACuriousMind Excellent test for you an AI
 
It's really hard to define my previous sentence the way I want it to mean.
Basically as long as you're getting to know each other, or explore possibilities, it's not the worst thing in the world to have a 'date' not be seen the same way by both parties. Just know when to move on romantically / emotionally if they don't become interested / respond positively to yours
AAAAAAAAAAAAAAAAAAA
edit hell
 
having a preferred reference frame allows us to actually tell if someone far away is dead or alive
 
@ACuriousMind what's your profile pic again?
 
4:25 PM
and it does not change any local predictions in the context of relativity
 
@Phase Kaelyn the Dove from NWN2: Mask of the Betrayer
 
Oh
Also an obsidian game
damn you must be a pretty solid fan
 
@Phase Well, it can get kinda awkward :D Had a lengthy conversation about defining friendship once that started with "You know this isn't date, right?"
@Phase I am!
 
wait
I can't even define friendships that I have
let alone as a general term
 
I'm also a fan of what they did before they called themselves Obsidian
 
4:28 PM
What did they do?
oh baldurs gate nice
 
Baldur's Gate, Planescape: Torment, Fallout 2
 
0
Q: Does this question belong in stack exchange physics or philosophy?

Anant SaxenaI recently had a question which I wasn't sure if this would be the correct site for as it seemed too speculative. But by the time I completed it, it seemed, more concrete than I'd imagined. I would like a second opinion of which site it belongs to before I post it (and delete the other as cross p...

 
Ah, fond memories of the six CDs Baldur's Gate 1 came on
 
How alpha decay works?
Is it important?
 
Constant switching. So annoying, but worth it
@Lagranian Important for what?
I'm confident many people go their whole lives without knowing how $\alpha$-decay works.
 
4:32 PM
@ACuriousMind So like
Hmm
Is it a important knowledge
 
@Phase That's not too bad - I'd say two people are friends if they: enjoy spending time with each other, are able to openly express any feelings/emotions with each other and are willing to go out of their way to help each other is (off the top of my head) a pretty solid basis for a good friendship
 
it seems like a bit chemical
 
@Lagranian All knowledge is important for some purposes, and useless for others.
 
@Lagranian did you get the right value of $a$ in the end?
 
I have a very sophisticated tree to define my various levels of friendships with other people
 
4:33 PM
@ACuriousMind That was great ;)
 
@Mithrandir24601 I think I'm a bit touched by the aut[redacted]m fairy, and that might be why
 
@Phase you mean equator problem?
 
@Lagranian I guess so
did you get like 0.3
 
@Phase No, it starts with 0.9
 
its wrong then afaik
 
4:35 PM
from $\omega^2r$
are you sure this equation is true?
 
@Phase Ah, that can cause issues :/
 
I wasn't being entirely serious Dx
 
lol
I'm interested in alpha decay equations
 
@Lagranian yes that's true.
You should get like 0.3 m/s^2.
Give me the value for the radius again?
 
@Phase Oh, yes.
 
4:36 PM
that you used
 
6371
 
km
 
yh
 
@Mithrandir24601 Ah, there's much nuance to that. Particularly the expressing emotions bit is tricky.
 
yeah. 0.33 @Lagranian
 
4:37 PM
@ACuriousMind Is it allowed to ask our questions here?
 
If you have any other value you calculated it wrong.
 
@Phase We bet it then
 
We bet what?
 
@Lagranian You can ask anything you want here in chat, but no one is obligated to answer.
 
@Phase I've found 0.33
 
4:37 PM
I feel like maybe you forgot to put pi in your calculation
I thought you said you got 0.9?
 
@Phase yes however
Now I got 0.33
 
ok
 
Same with you
 
However, if you have many questions about homework-type problems, the Problem Solving room might be better for you.
 
@ACuriousMind Is that active?
 
4:39 PM
Looks pretty active from the transcript to me.
 
@Lagranian Alpha decay is important if, for instance, you want to look at radioactivity
 
now I know what the alpha decay is. However, don't have any idea about how to write the equation for it.
 
well
$A^n_m \to A^{n-4}_{m-2} + \alpha$
or you could write it as a Helium ion
 
Let's discuss about it.
why is it $n-4$ and $m-2$
 
because alpha emission is basically shooting out an electron-less helium atom
which is 2 neutrons and 2 protons
so the mass number has 4 taken away, and the atomic number has 2 taken away
 
4:44 PM
I got it. wow
what hack you using
<3
So the formula is
$n-4$ and $m-2$
I'm gonna try solve a question now
 
I guess but you need the rest of it
 
$\phantom{Gd}^{149}_{64} \mathrm{Gd}$
 
right?
how does it decay
just single alpha emission?
 
$\phantom{Gd}^{145}_{62} \mathrm{Gd}$ + $\alpha$
 
then yes it's just $+ \alpha$
 
4:47 PM
Great!
Also the $\alpha$ means Helium Ion
 
@ACuriousMind I just realised how much I hate that notation. why is an electron $-1$ proton number? If it's just a measure of charge instead, why is there no distinction between ions and atoms?
yes
 
What if it become beta decay?
What's different?
$\beta$
 
Well, you lose a proton and get an electron instead. It depends on the sign of the beta though
 
oh so beta is harder right?
 
Not really, but $\beta$ and $\beta ^-$ are different processes
basically in particle interactions all that matters are conserved numbers
 
4:50 PM
@Lagranian That is impossible.
 
RIP, is it something about Gd? @Loong
 
Gd has 64 protons.
After alpha decay, it would be Sm.
 
Oh right lmao
How didn't I realise that
Brains slow today.
@Lagranian when I wrote the general form with $A$ i wasn't trying to mean that it's the same atom on both sides
I should've wrote $A \to B$ instead
 
Hm, by contradiction, if a sequence has points in every set of the basis but doesn't converge, there's a neighbourhood $U$ of $p$ such that $\forall n, p_n \nin U$, but $U$ is either the union of two basis element, in which case there's a $p_n$ in it, or the intersection of two, but I'm not sure how to prove it for intersections
An intersection could contain no such point
 
my god
my brain is firing atm
where did i do mistake
 
4:54 PM
literally just the symbols
 
oh right
I know the alpha decays now.
 
An element is determined by it's Proton number
so if you have some atom like Gd, after the process it'll be a different atom
 
wait, what?
what do you mean by it will be a different atom?
and why is it becoming $Sm$
What does determine it?
 
i told you
the proton number
If you change the number of electrons, it becomes an ion
if you change the number of neutrons, it becomes an isotope
If you change the number of protons, it becomes an entirely different element
 
Yes but, how do we know what element it will be?
 

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