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3:00 AM
Modular functions stuff yeah
 
right
Theta functions fit into there as well, but uh
that stuff is beyond me
 
Same
 
3:16 AM
Hi!
Got a phymath qn
When $\frac{v}{c} << 1$ then K.E of amovinf particle is $<< m_{0}c^{2}$
I thought of it like this -
K.E = $\frac{mv^2}{2} = \frac{m_{0}c^2 v^2}{c^2.\sqrt{1 - \frac{v^2}{c^2}}}$
 
No. That is not how kinetic energy works in special relativity.
 
Long time though I had a touch with relativity @Semiclassical , can you help me proceed?
 
It's so atonal that it feels like it'd be downright hard to sing and play the guitar at once
 
Well, the rest energy of a particle in SR is $m_0 c^2$
Do you remember what the expression for the total relativistic energy of a particle is?
 
3:24 AM
$E^{2} = p^2c^2 + m^2c^4$ ?
 
sure, that should do
so $E=\sqrt{(p c)^2+(m_0c^2)^2}$
If $p=0$, that gives $E=m_0 c^2$ as expected
 
$p=0$ implies $m = m_{0}$
 
But if $E$ is the total energy and $m_0c^2$ is the rest energy, what should $E-m_0c^2$ represent?
@BAYMAX that only works if you talk about relativistic mass, and one really really doesn't do that.
It's $m_0$ in the above expressions.
 
So,$E^{2} = p^2c^2 + m_{0}^2c^4$ ?
 
right.
 
3:28 AM
@Semiclassical that should represent Kinetic energy?
 
@bwDraco I think you're a little confused. I was the one who was flagged.
 
Right, $K=E-m_0c^2$.
 
@bwDraco I am friends with jasper.
 
@Dodsy welcome back
 
@bwDraco I was flagged for standing up for Jasper due to twink calling him "crazy" for having mental illness.
@Semiclassical thank you :) gfs family visiting today.
 
3:30 AM
One moment please...
 
If we rearrange that a bit, we get $$E^2=p^2 c^2+m_0^2 c^4=(K+m_0 c^2)^2=K^2+2K m_0c^2+m_0^2c^4$$
 
You literally told me.
you linked my post
and said that was what was flagged.
I thanked you and then left to avoid a ban...
 
@Semiclassical yes
 
@bwDraco I've been banned ~6 times on this chatroom due to duck.
 
Sorry, my bad. But clearly, someone is responsible for this flag abuse...
Is @TheGreatDuck present?
 
3:32 AM
No.
 
le sigh
 
He goes by the name Typon and left.
 
He was here just a bit earlier
 
but really I avoid this drama.
that's why i left.
I was banned just last week for saying "omfg"
 
As in, within the last page up on the transcript
 
3:32 AM
Ah.
 
the amount of drama/paranoia/jumping to conclusions I've seen here is getting really tiresome
 
Okay... @Typhon, was there a good reason for this flag?
 
Dodsy, I realize the suspicion that it's Duck but until there's hard evidence from Shog or something it's best to not make outright accusations
 
It probably was not him.
 
@Daminark Quite.
 
3:33 AM
@Daminark agreed.
 
If there's actual evidence that said flag was associated to him, that's one thing.
 
Sorry that I blamed you by mistake, @Dodsy. But someone needs to find out just who is responsible.
 
I don't see any $v$ term ?@Semiclassical
 
Maybe so. But you're not in a position to investigate it.
 
No problem, I was just a little bit angered to come here and see that I was being accused of flagging myself.
 
3:35 AM
I'm getting really tired of people drawing conclusions based off of inference and speculation.
@BAYMAX indeed not. we'll get there, though.
 
@Typhon when did I accuse you of having duplicate accounts :o
 
ok
 
I really hope we can get to the bottom of this. If @Typhon is responsible (he was not actually chatting at the time), we'll get him suspended.
 
if we look at the second and fourth expressions in those equalities, we can see a common $m_0^2c^4$ term
so let's cancel that off to get $p^2 c^2=K^2+2Km_0 c^2$.
 
Jun 5 at 5:09, by Shog9
@Avery @bwDraco well, I looked. Turns out @TheGreatDuck and @Dodsy have been trading flags here for a couple of days. Latest volley was courtesy of Dodsy, but TheGreatDuck still leads by a fair margin for raw numbers pointless flags.
 
3:37 AM
$p^2c^2 = K^2 + 2Km_{0}c^2$
 
@bwDraco
 
Yes
 
also that is being taken out of context.
 
Who else can we call to definitively find out who's responsible for the stuff?
 
Yep. At this point, we use the fact that the kinetic energy should be small compared to the rest energy.
 
3:37 AM
We had not been "trading flags there for a couple days"
I was banned for saying "titties"
 
@BAYMAX So which is more relevant, $K^2$ or $2Km_0c^2$ ?
 
and everytime somebody got banned, guess who was present?
and I had so many issues with the guy
 
IIRC Typhon was in fact in the room, but not actually saying anything.
 
I flagged 3 of his (non offensive) posts and when a mod looked into it, he said that.
 
@bwDraco That was from months back
 
3:38 AM
It was
that's old news.
 
Oh! actually we have $\frac{v}{c } << 1$ and have to prove $K << m_{0}c^2$ @Semiclassical
 
@Dodsy The right thing to do is to use a mod attention flag.
 
I really don't want to start this up again.
 
This room needs more moderator/room owner presence.
 
3:39 AM
Then I'm going in the wrong direction, alas.
 
Guys there's nothing actionable yet, I realize this is a load of shit going on but nothing we say here now matters
 
I don't get why twink said "it was probably dodsy"
he seems to not like me.
 
Just gotta alert the mods or appoint new ones
 
He could have been the one who flagged me
 
That's what threw me off.
 
3:39 AM
There's quite a lot of attitude going around.
 
in the context, we were in an argument.
 
But, again, I don't think paranoia helps much.
 
Find out who's doing this, shut everyone down, and learn to chill
 
No it doesn't.
I need to stay out of this, I hate this drama bs.
 
Dodsy I thought you had been suspended
 
3:40 AM
which is why I promptly left after being flagged.
Sigh...
For what?
 
If someone reasonable becomes a room owner the first job should be to ban the trolls
 
@Twink and @Dodsy look starting this discussion again is just gonna take things downhill, can we just reconcile, say it was a misunderstanding, and cut it out?
 
@Daminark yes but I don't understand the hate.
and why would I flag my own post?
 
@BAYMAX Given that they want you to start from here, you probably don't want to start from $E^2=(pc)^2+(m_0c^2)^2$.
 
@Daminark who are you?
 
3:41 AM
I'm not blaming you here, in this case I think you were wronged
But these discussions are just getting out of hand
@Twink I'm Daminark
 
yes
I agree
 
Someone turn this into a mathematical discussion, now.
 
You know what? Somebody should make me a room owner, and I'll probably use the right tools to keep this issue in check.
 
@LeakyNun What do you think I'm doing ?!?
 
Yes@Semiclassical
 
3:42 AM
:/
 
Probably not a good idea, bw..
 
@Semiclassical sorry :P
@bwDraco and you'll accuse someone of flagging himself?
 
If I see spurious flags, I'll probably place the room in timeout.
 
There's another expression for $E$ in special relativity, though.
 
@Semiclassical ... I thought that one is the canonical one lol
 
3:43 AM
I won't kick-mute anyone without clear evidence, but we seriously need more moderation here.
 
Yeah let's not make a non-regular a room owner.
 
@Leaky this isn't gonna help
 
$E^2=p^2c^2+m^2c^4$
 
@LeakyNun Teaching Chaos. Day 1 of class: asked students for a function to use as example rather than using one whose behavior I knew ahead of time. Talk about working without a wire.
 
@BalarkaSen Ah well.
 
3:43 AM
Okay look I hereby declare that this room will be dedicated to this problem
 
lol
 
Let $G = (V,E)$ be a loop-free digraph
 
@nitsua60 you teach Non linear dynamics?
 
@nitsua60 what were you trying to teach?
 
@LeakyNun well, the trouble is that that's in terms of $p$ and not $v$
 
3:43 AM
@LeakyNun Like I said, this was a mistake.
 
@Semiclassical btw why $m_0$?
 
To emphasize that it's the rest mass.
It's not really necessary, but eh
habit
 
@Semiclassical instead of?
 
@LeakyNun The notions of iterations, fixed points, orbits, periodicity....
(dynamical systems)
 
so what did they give you lol
 
3:44 AM
well, an old way of thinking was to define the relativistic mass as $m=m_0\sqrt{1-v^2/c^2}$
 
Show that there exists some $F\subset E$ with $|F| \le \frac{1}{2}|E|$ such that $(V,E\setminus F)$ is a directed acyclic graph
 
@nitsua60, any access to information on who raised the flags?
 
@Semiclassical :)
 
now, this is something no one does nowadays because it's not a helpful way to look at it
 
3:45 AM
@bwDraco I, ten minutes before going to bed, have no interest in digging into you-all's drama. Sorry.
5
 
@nitsua60
 
Also, can you clear the stars on or delete this?
 
@nitsua60 Good idea.
 
but $m_0$ does get used instead of $m$ because of that af times.
 
@bwDraco ?
 
3:45 AM
It's a bit silly, but eh
 
Somebody flagged that?
 
@Dodsy It was a mistake and should be removed.
 
Oh i see.
 
@Semiclassical Yes
 
@Dodsy s/flags/stars/
 
3:46 AM
@bwDraco deleted at owner's request
 
Not sure how I mixed those up.
 
@Semiclassical even now?
 
Thanks.
 
well, take a look at the Wikipedia page on invariant mass: en.wikipedia.org/wiki/Invariant_mass
 
I appreciate those words, Draco.
 
3:47 AM
Oh, stars. Hmm... I might have been able to do that. Ship's sailed on that one, though =)
 
@Semiclassical hmm
 
in the bit on particle physics, it's $m_0$.
I think it's more a holdover of previous notations, tbh
 
And I appreciate you having it removed.
 
but old habits die hard, especially when people keep using them
2
 
@Dodsy Look, I seriously hope that whoever is flagging your messages is ultimately caught and suspended.
 
3:48 AM
@Dodsy but why were you flagging yourself?
 
I'm leaving. I love the math here, but you revel too much in your own drama. Have a nice night.
 
"messages" is an interesting term. Suggests that this isn't the first time.
 
...huh, I hadn't even realized how pertinent that comment was
 
@nitsua60 ah, sorry for putting you off.
 
3:49 AM
I think inverse trig functions are interesting.
but what are they for?
 
@nitsua60 Night.
 
@Semiclassical This is beautiful
 
it really is
 
@Dodsy for inverting trig functions
 
:o
but what is their application.
 
3:50 AM
solving trig function equations
 
there's a few answers possible
 
yesterday, by Dodsy
@Abcd I just assumed you were typhon because you asked me a lot of personal questions yesterday, sorry about that.
yesterday, by Jasper
Is abcd = typhon?
 
@bwDraco paranoia.
 
the most obvious is: suppose someone tells you the proportions of a right triangle, and you want to find out the angle of it. you'd use inverse trig functions to find that
 
yesterday, by Leaky Nun
@Abcd how do I know if you're Typhon
 
3:51 AM
Oh right.
 
@bwDraco it means I refuse to comment.
 
@nitsua60 perhaps in case you missed it! as i saw you told you are teaching Chaos theory so thought of telling you of this chat room,it would be great if you step by some time! Dynamical systems and chaos theory
 
@Semiclassical I should have known that but I had a different sort of idea in my mind when asking the question. :P
 
along the same lines, if someone tells me the x-y components of velocity then I can inverse trig functions to determine what my heading is e.g. how many degrees north of east
 
Now do we have good reason to believe that someone is socking?
 
3:52 AM
I just learned today how you limit the domain of the function to find its inverse.
 
waitttttt
 
?
 
1 hour ago, by Typhon
@Dodsy @Abcd @LeakyNun I choose to leave the site and you people send me a bunch of email notifications excusing me of having duplicate accounts? If you have an issue contact a moderator. I'm not your babysitter and it's not me job to mediate when you people decide to get in an argument. I've asked you to leave me alone, and that wasn't a joke. Stop trying to contact me. Thank you.
we never tagged Typhon. We just mentioned him by name.
 
>_>
 
drama
 
3:53 AM
How is he getting those email notifications?
yesterday, by Abcd
@Typhon These people think I am you :(
nvm, solved
 
@LeakyNun team@stackexchange.com?
 
@bwDraco ?
 
What is socking btw?
also please can we leave this all alone, Draco?
 
No love for my problem :'(
 
@Dodsy Operating multiple accounts for some illegitimate purpose.
 
3:54 AM
@Dodsy for a fun way of looking at inverse trig functions, suppose you take the graph of $y=\sin x$ and flip it along the x=y line i.e. draw the graph of $x=\sin y$
 
We really don't need to be sloshing through old postings.
right. @Semiclassical
 
@BAYMAX Thanks--I might just bug people there for ideas =)
 
It's actually a lot of fun
 
@Semiclassical now we are talking about before clipping it?
 
3:55 AM
yeah
 
Alright then. I think I'll step back for a bit and watch what happens. Hopefully, we won't see any more flag abuse.
 
that graph will badly violate the vertical line test, since any vertical line that passes through it will do so an infinite number of times
 
A one liner if you know a fact about directed acyclic graphs
 
so then it's range would be $(-\infty,\infty)$
 
But even then it took me hours to come up with the solution
 
3:56 AM
or simply $\mathbb{R}$
 
right
 
@nitsua60 actually I don't understand what you said?
 
so if you want to get a function, you'll have to consider only a finite portion of that graph
 
13 mins ago, by Daminark
Let $G = (V,E)$ be a loop-free digraph
12 mins ago, by Daminark
Show that there exists some $F\subset E$ with $|F| \le \frac{1}{2}|E|$ such that $(V,E\setminus F)$ is a directed acyclic graph
 
One cool thing is that the domain of an inverse function is the range of the original function.
:o
 
3:57 AM
I'm going to step out. See you all later.
 
@bwDraco okay thank you for your help.
 
The other cool direction for inverse trig functions is to say how one generalizes the logic to get inverse elliptic functions i.e. elliptic integrals
I don't know that story so well, though
 
@Semiclassical the portion of the sine graph is $[-\frac{\pi}{2},\frac{\pi}{2}]$, right?
 
right
 
@BAYMAX Knowing there's a room dedicated to the topic I'm teaching is super-helpful, thanks. One use I can imagine putting it to is to say "hey, folks--I'm looking for a two-dimensional system that's a nice real-world example of saddle-behavior. Any ideas?"
 
3:58 AM
:o elliptic integrals ooooo
sounds very cool.
 
$p^2+c(1-x^2)^2$ :)
 
So today I handed in my hw and noticed a mistake.
 
Where are my dragons?!
 
I said that for the function $f(x)=3x^2+2$ that at $x=1$ , $y=14$ :(
 

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