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12:00 AM
but differentiating this expression twice gives -w^2r instead of -wr
am i missing something?
 
@BalarkaSen Yo I need JEE help
 
?
do you have to integrate something horrible
 
is it okay if i post a link to a question here?
 
@BalarkaSen I have an alpha particle hitting 9Be, and the result is 12C + a neutron. Can the neutron have an angle greater than 90 degrees to the path of the incoming alpha?
 
idk bro
that sounds hard math
 
12:08 AM
By "greater than 90 degrees" do you mean directed forwards?
Isn't that what you'd normally expect?
 
no, directed backwards
I think it's possible it can go backwards
 
how are you drawing the 90 degrees
 
because the target is heavier than it
 
0
Q: Finding shape of orbit of an isotropic oscillator

Mohammad Areeb SiddiquiThis is a particular solution to a differential equation of an isotropic oscillator with the potential energy $V = \dfrac{1}{2} k(x^2 + y^2)$ $$\vec r(t) = \vec r_0\cdot cos(\omega t) + \dfrac{\vec v_0}{\omega}\cdot sin(\omega t)$$ According to this derivation of the equation of motion for thi...

 
All you have to do is conserve momentum $\vec{p}=m_\alpha\vec{v}$ and energy $E = \frac{1}{2}m_\alpha v^2$
 
12:10 AM
$\theta$
@enumaris the velocity is relativistic
 
oh, then use the relativistic formulae
 
well that doesn't tell me what $\theta $ is
 
well, if you can find values of final velocities in a configuration that conserves both quantities and makes an angle >90 degrees then you know that angles >90 degrees are possible
maybe assume theta = 180 degrees and see if you can find a set of v's that work
everything is co-linear so the equations should be pretty simple: $\gamma_\alpha m_\alpha v = -\gamma_n m_n v_n+\gamma_{C} m_{C} v_{C}$
sorry that should be C
Just work in the rest frame of the beryllium
wait
are there other particles involved in this reaction
does it give off light
or a neutrino
if not, then the second equation is then $\gamma_\alpha m_\alpha = \gamma_n m_n +\gamma m_C$
give some values for $v$ and you're all set :D
 
If I have an uncertainty in data and I want to plot the log of that data is the plus/minus just log(plus/minus value)?
 
it's 180
the book says so
good enough for me
 
12:24 AM
I thought u were a math person, why are you dealing with particle collisions all of a sudden
0
Q: Calculate initial kinetic energy given spring constant, mass, and distance

EFeds7A 2kg block slides with no friction and hits a spring with spring constant k=2000 N/m. The block compresses the spring in a straight line for a distance 0.15m. The block’s kinetic energy, in J, at that point is 0 J. What was its initial kinetic energy, in Joules? So far I used potential energy =...

how can I possibly answer this without giving away the answer
Conservation of energy says the final potential energy must equal the initial kinetic energy...so...you already have the answer...
 
cant we just $T = kxd$ in this question?
the kinetic energy converts into the workdone?
 
huh..
You don't even need to do that. The energy is conserved (assuming no friction) so the initial kinetic energy, the thing he is looking for, is simply equal to the final potential energy. The kinetic energy got converted to potential energy.
 
@enumaris engineering major
 
In other words, he already solved the problem when he solved for the potential energy
oh
 
12:35 AM
I'll let someone else figure out how to answer that without giving away the answer
 
how do i work out the equation of motions given V = 1/2(x^2+y^2) and the particle is in 2 dimensions?
 
Lagrangian methods?
 
the lagrangian isn't introduced yet so i as thinking of using f = - dv/dx
 
give up
 
hence mx'' = -kx
 
12:40 AM
or, you can do something more clever like....realizing the problem is separable in dimensions so the answer should be essentially a superposition of the x-direction and the y-direction answers
yeah that works
 
the general solution of this equation is confusing tho
 
where'd the k come from though
 
f = - dv/dx = -kx
 
your V doesn't have k in it
 
oh i forgot to put it in the equation xD it is
typo
 
12:41 AM
I wish latex worked for me in firefox
then I could latex out a nice answer
but you're pretty much there
 
it isn't working for me in chrome
 
mx'' = -kx and my'' = -ky
solve separately
You need some initial conditions to get the full answer though of course
 
x = Acos(wt) y = Bcos(wt) right?
 
you're missing some initial conditions
x=Acos(wt+delta), y=Bcos(wt+epsilon)
 
the conditions aren't given ill assume them to be at t=0, x(0) = x_0
 
12:44 AM
use initial conditions x_0, v_x0 and y_0, v_y0 to get the
 
why the delta?
 
answers
initial conditions
if delta is not there then t=0 means X=A
and y=B
 
why couldn't that be?
 
it could be
but you didn't specify those were the initial conditions
I'm giving you the most general answer
 
ohhh yes
so with delta
x = Acos(delta)
 
12:46 AM
no
x = Acos(wt+delta)
 
at t=0 i meant
 
right
so now X can take any value -A<x<A
at t=0
 
A is the amplitude of oscillations, the amplitude will depend on the initial velocity
so you need the initial velocity and initial position to specify A and delta
is this homework btw...am I solving a homework problem...I'm not supposed to be solving homework problems...
 
no its not xD
im reading theoretical minimum
and was confused by one of the derivations for an elliptical orbit of an isotropic oscialltor
still am actually
 
12:50 AM
you can get a variety of solutions
including oscillations in x only
oscillations in y only
circular orbits
ellipitical orbits
diagonal lines
 
yes this is actually one of the exercise to prove that there are circular and non circular orbits
 
It depends on how the phases delta and epsilon align and the amplitudes A and B
delta, epsilon, A and B are then determined by the 4 initial conditions
 
the book stated the general solution of -w\vec r = \ddot r is c1cos(wt) + c2sin(wt)
 
uhm
yes, that's another way to put it
 
isn't this the general solution for -w^2\vec r = \ddot r ?
 
12:53 AM
c1cos(wt)+c2sin(wt) = Acos(wt+delta) for clever combinations of A, delta and c1 and c2
it's a trig identity
I'm just giving a different way of expressing the answer
oh they used polar coordinates
 
oh c1 and c2 are vectors too
yeah they assumed the coordinate system to be like
 
then they need to specify an equation for theta too
oh wait, your r is a vector
 
yes
the derivation is actually here: http://www.madscitech.org/tm/slns/l5e2.pdf

bottom of the first page it gets all confusing
 
in cartesian coordinates, the solution is x=Acos(wt+delta), y=B(wt+epsilon)
in polar coordinates the solution is r=Acos(wt+delta), theta=B*t+C
choose one, and stick with it for understanding :)
 
which could be written as \vec r = \vec b cos(wt+delta) right?
in cartesian coordinates
 
12:58 AM
oh, the pdf is solving it in a vectorized way
lol
 
yes, whats with the substitution of b1 and b2 vectors and adding phi?
all of a sudden xd
 
Yeah, the cartesian solution is pretty much there
You can see it directly from that equation really, without having to do so much fancy stuff.
 
why the dot product and the replacing r_0 and v_0 by b1 and b2?
how?
 
You just have to be comfortable plotting a parametric curve
 
@ACuriousMind I feel like I know nuclear physics after understanding this homework
 
1:07 AM
it is not apparent what that pdf is trying to do lol
 
its hard to understand this xD idk what'd happen when i come to principle of least action
 
full EoM was already given
the rest is just them trying to find the shapes of the orbits, but to find the "shapes" you have to make certain assumptions about the initial conditions
different initial conditions give rise to different orbits
 
yeah the \vec r(t) equation with r_0 and v_0/w was the answer i stopped at
dont understand what's with all the dot products and stuff
 
that stuff is just trying to find the shapes of orbits given some certain initial conditions
 
yeah but cant seem to see how its doing that :D
 
1:15 AM
@JohnRennie This meme of "being a mod sucks" doesn't make sense to me. Am I to believe that @ACuriousMind, @Qmechanic, @DavidZ, @rob, and whoever else enjoy pain? There must be some benefit.
 
yeah it's not very clear in that explaination
 
@DanielSank Someone's got to do it. I'm a mod because I like this site and want it to run smoothly, not because there's something intrinsically enjoyable about being a mod
Most of it isn't "pain" either, though, just routine stuff
 
what is ur avatar @ACuriousMind
A dota character?
 
I got a new avatar
 
is yours an onion?
 
1:29 AM
@enumaris Tunon the Adjudicator, Archon of Law from Tyranny.
 
wat
oh, video game character heh
 
It hasn’t updated in chat yet
It was inspired by @ACuriousMind and @BernardoMeurer
 
is it an onion tho
 
No
 
:(
 
1:41 AM
j
 
vzn
@DanielSank the statement was semi-ironic by JR who has quasi mod duties. but do think if he finds some major part of it objectionable, maybe its time to "unvolunteer" (& same with any others).
 
@0celo7 Hehehehe
Nice one
 
in Mathematics, Nov 20 at 12:59, by ACuriousMind
Masochism and curiosity.
Proof
 
@0celo7 god meme right there
literally
 
2:04 AM
@BalarkaSen what?
 
your avatar
it's Very Nice 审 美 观
 
@BalarkaSen I went on that website for that Santa in the lagoon video
and there was a hyperlink called "Copyright Notice"
So I click
Dude
This is the motherfucker's copyright notice
 
lmao
 
I can't
I'm not worthy
 
vzn
@BernardoMeurer hey whats new playin any video games these days?
 
2:10 AM
@vzn Heya, not really
Not doing much
 
vzn
@BernardoMeurer ah thought that was mandatory for geeks. are you in school?
 
You mean right now or in general?
 
vzn
@BernardoMeurer lol either
 
Right now, no. In general, yes
I'm doing two years in purgatory
 
vzn
do you wanna go to new school? did you try one in santa barbara? how was it? "purgatory"?
 
2:14 AM
I'm going to community college right now
Yeah, purgatory
 
vzn
hey better than nothing commend you :)
 
Cause I was in Lisbon, and now I want to go to a real school but I gotta pit stop here for two years and transfer
So it's kind of like purgatory :P
 
vzn
hey santa barbara is far from purgatory in a lot of ways eh? how far are you from the beach? :)
 
5 minutes by bike?
 
vzn
awesome, do you like it? tried surfing? :)
 
2:16 AM
The beach is cool, yeah
And I got Daniel around which is nice
I can barely stand on my feet, let alone on a surfboard
 
vzn
yep DS can be nice :) :P
 
But yeah, CC is not very challenging
Which is good in one hand because I get to do projects
But bad in the other because I get really sloppy
 
vzn
CC is probably a bargain...? compared to cost of most uni's these days... student debt, o_O
 
Yeah, it's pretty cheap
I mean, I got into UCSD and VATech before I went to Lisbon, I just couldn't afford it at all
 
vzn
dont think the debt is always worth it. (although ofc the semipropaganda marketing/ cultural imperatives etc say differently.)
 
2:30 AM
Chris Ray Gun is pretty good
 
@DanielSank To be fair, most memes don't make sense to most people.
But I think it's true that there are plenty of people who would hate the job. It goes better with some types of personalities than others.
 
@DavidZ I sense an unintentional Rick and Morty meme
 
Could be
I haven't seen enough of that show that I would get it. I need to fix that....
 
uhhhh
 
2:48 AM
@BalarkaSen you like it?
 
@BalarkaSen to be fair you need a really high iq to identify an unintentional rick and morty meme
 
@BernardoMeurer you could have gone to VT and hung out with bob
 
@0celo7 No, I didn't have money for that and I don't have a car
 
3:06 AM
Some thoughts about space and time perceptions:
Given some motion. It can look frozen by two different ways:
1. It moves slowly i.e. $\frac{dx}{dt}$ is small
 
2. Relative motion of others objects is the same xD
 
3. Imaging the motion at very short time scales e.g. a fast camera
We can then use a similar idea, to comprehend special relativity phenomenon:
 
Are you two twins or something?
 
He/she just happened to respond, and thus I change my 2 to 3
 
He* :P
And not twins
jbtw
 
3:15 AM
(This is the underlying reason why my messages look like blog messages, because they only don't look like a blog post if there exists spontaneous responses like the above)
 
@Secret Have you ever met a woman called Mohammad?
 
and no I have not
(and my messages are basically capture those sponteneous responses, of which I have no way to filter away uninterested audience)
 
Ive never met someone called secret too
 
but given we can have Kelly as a male name, it would not surprise me there's a female mohammed somewhere in the world
 
@Secret I'd be pretty surprised if there is a female mohammad with their head on still tbh
@Secret Are you chinese?
 
3:17 AM
I am from HK, which is next to China
 
holy shit how did u guess
 
@MohammadAreebSiddiqui I'm a skillfull ninja tracker
skillllllllllfull
skiing
 
oh so u are a russian
 
Nope, I'm Brazilian
 
@MohammadAreebSiddiqui Maybe you did meet someone, but they didn't tell you they were called that because...it's a secret.
 
3:20 AM
@ACuriousMind Is this some sort of rebound from leaving academia?
:P
 
@BernardoMeurer What?
 
@MohammadAreebSiddiqui @ACuriousMind Just left his position as a researcher to become CEO of SAP
 
so everyone i have met is equally likely to be 'the secret'
 
@ACuriousMind I don't know, I'm bored
 
@BernardoMeurer Don't confuse the newcomers, please :P
 
3:21 AM
except for secret himself
 
@ACuriousMind I speak nothing but the truth (assuming multiverse theory is correct)
 
hi, someone nows how to define the disocioaton energy of the H2+ ion?
 
@ACuriousMind my avatar is in honor of you
 
for the H2 molecule I know it is E[molecule] - 2E[isolated-atom]
but in the case of the ion, I am not sure anymore
 
@0celo7 I'm afraid I'm not sure what it's supposed to depict
 
3:28 AM
@MiguelCarvajal It's easy, go on random.org and click "Generate" until the correct answer shows up
 
@ACuriousMind it's the GDP formula!
 
@ACuriousMind You inspired him by choosing a career that makes more GDP
 
whatever, this is not a serious chat
 
@MiguelCarvajal It is, @ACuriousMind help him
 
@0celo7 Ahhh
 
3:50 AM
Attention!
Grothendieck spotted in Brazil
 
4:48 AM
(cont.) We can then use a similar idea, to comprehend special relativity phenomenon:
Take muon lifetime as an example
In the frame of the muon, time as measured by the observer in the muon's frame ticks normally because it is proper time. However, space get contracted in the direction of travel (and time elsewhere is dilated as well)
In Earth's frame, the surrounding space is normal (because that's the proper distance). The time of the muon becomes dilated (and if a muon does have some notion of size, should look squished in the direction of travel)
In usual explanations, the muon's ability to survive to the earth's surface despite having a lifetime too short to do so can be explained in these two frames as follows:
1. In the muon's frame, the distance required to reach the ground is shortened
2. In Earth's frame, the muon's time get dilated, thus the onset of the decay effectively occured later
However, we can say something more than that (and perhaps also a bit more intuitive(?))
If we use our intuition that the muon can survive reaching the surface of the earth because time slows down at the muon in Earth's frame, and wrongly conclude that the time dilation is just an apparent effect explained by at the muon's frame, time flows normally but only a much shorter distance is needed to travel through
(thus in a relative sense, time seemed to slow down while it is ticking normally in the muon's frame, is because for each tick, a larger distance is being covered as seen from the Earth's frame)
then this will not work. This is because:
We would have expected the time elapsed at earth to be much faster when we switched to the muon's frame from the earth's frame, but what is observed is that time is also dilated in Earth's frame. Likewise, length get shorten but not expanded at the muon as seen from Earth's frame
This relationship where for A and B travelling relative to each other, A looks shorten and time dilated to B and B looks shorten and time dilated to A, suggests the transformation that relate these two frames cannot be galelian and more generally, cannot be expressed as a (I have no idea what is the terminology) mapping, since in such maps, if A > B as seen from space A, then B < A as seen from space B
Therefore, it must be a (I have no idea what that terminology is) mapping, which in the case of special relativity, is a hyperbolic rotation, such that if A > B as seen from space A, then B > A as seen from space B
typo: is that time is also dilated at Earth as seen from the muon's frame
 
5:06 AM
holy jeez wall of text
 
probably much shorter and rarer than those in the past after so many users told me to be careful with the flooding issue
It would have been some conversation, but the Mohammad discussion plus the fact I am typing this in a symposium means the conversation flow is cut off
 
I find most of the confusion in SR arises from going back and forth between frames wily nily
just pick a frame and do calculations in that frame, or simply just compute frame invariant quantities like proper time.
 
yeah, but I think understanding how quantities are related between two or more frames is key to understand why the transformation is Lorentian
All the stuff in SR took place in a 3D+T spacetime, but at every moment in proper time in our own frames, we can only ever see the spatial information of events one tick at a time
so all that length contraction and time dilation, are in some sense projections of the hyperbolic rotations that happened in the 4 dimensional spacetime
The hyperbolic (instead of spherical or euclidean) transformation also further adds to the confusion since in hyperbolic space things don't scale proportionally in the sense that A can looks smaller to B and B looks smaller to A (Unlike our intuition that if A looks smaller than B, then B must looks larger than A)
 
5:59 AM
@SirCumference how dy
 
6:34 AM
hmmm
 
hlo
 
 
2 hours later…
8:20 AM
@BernardoMeurer Very interesting folk impression music you have there, including the use of water as a modulation
 
@Secret Somehow it's unsurprising you'd enjoy it :P
Here, have more
 
some kind of beatboxing
4:57 and beyond sounds really out of alignment with the rest of the rhythm, though
 
1st 'music' lack a clear melody to be called music, although it is also not noise because it is bland. 2nd is music because I can hear the melody clearly, 3rd is a bit strange
 
Wait wait
you think the first one isn't music but the second one is?
 
8:30 AM
What's a good model for an evaporating black hole, is it just the Vaidya metric
and is there a specific function $M(t)$ for Hawking radiation
 
@BernardoMeurer I tend to easily pick up something as musical if there's some melody (significant but not sharp variation in pitch), and/or if it is rhythmic like drumbeats and other percussions
If both of these are not clear, then it is just ordinary sound scape to me
Of course, there are exceptions: If it makes me felt some emotions and is not noise, then it will be music regardless of the above criteria. But such is rare
 
8:45 AM
@Slereah I've got one for a Schwarzchild BH
 
well obviously
For Schwarzschild it's just $M(t) = M$!
 
@Slereah Mass of BH as a function of time?
 
Schwarzschild is specifically a static black hole :p
hence it will always be constant
(was the joke)
 
@Slereah OK... I'll blame a mixture of: it's the morning and I haven't had breakfast yet and 2: American humour :P
 
I'm not American
Damn rosbeef!
 
8:54 AM
@Slereah ... Bah! Fine! I haven't had breakfast yet :P
$$M(t) = \left(\frac{3c^4\hbar}{15360\pi G^2}\left(t_0 - t\right)\right)^\frac{1}{3}$$
 
Some dude on stack exchange says the Vaidya outgoing metric corresponds to a white hole solution
But from looking at papers on the internest
It doesn't seem true
 
(the constant's often just written as $3\alpha\hbar$)
 
any mod around?
 
Also I think that function probably needs a $\theta()$ somewhere
Lest the mass becomes negative
 
@Slereah yeah, this is only valid until $t_0$, when it evaporates completely (in theory)
 
8:59 AM
@Federico We killed them all
 
@BernardoMeurer should I call my bodyguards?
 
@BernardoMeurer shhh... They're a mod... Don't tell them that
 
8
Q: Nothing falls into an evaporating black hole?

DvijThe Vaidya Metric is the metric that can be used to describe the spacetime geometry of a varying mass black hole. This metric reads $$d\tau^2=\bigg(1-\dfrac{2M(\nu)}{r}\bigg)d\nu^2+2d\nu dr - r^2 d\Omega_2^2$$ For simplicity, I will assume $d\Omega_2=0$. Now, the condition for any interval to be...

Wondering due to this question
 
@Federico They won't help you, you're already dead
 
@Slereah I don't see a problem... Nothing falls into an evaporating BH... Because that's what it means to be an evaporating BH, surely...
 
9:05 AM
The answer guy seems to imply it is a problem
which made me suspiscious
 
I do disagree with that answer...
 
9:16 AM
"Meanwhile, the Hawking radiation is a quantum phenomenon that introduces radiation even from a black hole" yeah, fine "and cannot be accurately described by the outgoing Vadiya metric" That's not what the paper says - it looks like someone has implication statements mixed up again - they didn't even mention the outgoing metric, so that's a massive assumption
 
9:49 AM
Am I being a pedant when I dislike that people say, implicitly without defining their number system first, that $(-1)^{0.5} = i$?
I feel like I am but on the other hand if you have multiple complex numbers it no longer applies right?
 
@Phase I'd say that it depends on context
 
Just normies on facebook
 
So, for many 'everyday' uses, to say that $\sqrt{-1} = i$ is fine
(from a physics perspective at least)
 
It does feel strange though, that while $i^2 = -1$ I have no problem with, I feel like writing $i = \sqrt {-1}$ is inaccurate
 
@Phase You're definitely a mathematician :P
 
9:56 AM
@Mithrandir24601 is there a cure :(
 
Yeah, the other 'non-everyday' (from a physics perspective) uses do require some sort of 'my starting point is at $i$ and everything else will be referenced from there' kind of thing
@Phase Learn complex analysis :P
 
I haven't even learnt Real analysis yet
I'm a plebian
 
That kind of thing never bothered me, so ¯_(ツ)_/¯
 

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