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123
4:24 AM
GooD MoRning @Azmuth
 
Good Morning :)
 
123
what about corona situation in your region.
 
123
4:52 AM
What is the difference between projectile and circular motion is a sense that.
In circular motion velocity is constant it changes direction only. $\vec{a}_c = \frac{\vec{v}^2}{\vec{r}}$
In Projectile we also see direction is changing but acceleration does velocity change. Why?
 
 
1 hour later…
6:00 AM
You divided by a vector...
 
123
No $v^2$ is not vector
$|\vec{a}| = \frac{v^2}{r}$
 
You wrote $\frac{1}{\vec{r}}$
You can compare projectile and circular motion by asking when does projectile motion become circular motion i.e. when does a thrown object start orbiting the earth
 
123
my question was that in circular motion there is constant velocity. mean acceleration does two things change in velocity and change in direction.
In circular motion we can see that acceleration does only change in direction.
But in projectile motion acceleration does only change in velocity but we see that there is also changing direction. why this change in direction not in calculation.
All the acceleration we use in change in velocity but there is also direction changes continuously.
 
6:17 AM
The acceleration always points down in projectile motion, it always points towards the center of a circle in uniform circular motion
In circular motion, as the particle moves, the acceleration changes direction, this does not happen in projectile motion
 
123
@bolbteppa I know this frnd. I am confused there is also direction changes in projectile why we don't encounter this situation?
 
What does that mean
The acceleration does not change direction
 
123
In curvilinear motion we take acceleration as $a = a_{parallel} + a_{perp}$
$a_{parallel}$ changes velocity and $a_{perp}$ changes direction in terms of polar components.
 
In projectile motion, if you expand those vectors in terms of their x and y components, $\vec{a} = \vec{a}_{parallel} + \vec{a}_{perp} = (0,a_y)_{parallel} + (0,a_y)_{perp}$ you see they only have a y component that is non-zero
 
123
acceleration means Force here it is just multiplication of scalar mass. Consider it as 1
Means in projectile it can only changes direction? right
But physically when i see the projectile trajectory it is parabolic it changes direction. How we explain this? It conflict with our direction problem.
 
6:26 AM
In projectile motion, the acceleration always points down, but the initial velocity can point in any direction, so the y coordinate can go up and down because there is a velocity in the y direction also the acceleration has to make it change direction, while the particles e.g. moves right so you get parabolic shapes
 
123
Hi, @Azmuth
 
Hey :)
 
123
You also take a glance on my question. As @bolbteppa also helping me
 
$| \vec a | = v^2/r$
 
123
6:29 AM
@Azmuth Pls see my question above.
 
Velocity changes in both Projectile and circular motion.
@123 check these notes
 
123
@bolbteppa Hmm.. It means it is not necessary what we see in motion changes direction due to acceleration?
 
123
@Azmuth I know all the calculation of projectile and circular motion. I was asking about what we see.
What i think now after discussion with @bolbteppa pls correct me if i am wrong
If motion is curved like in projectile it is not necessary which is due to acceleration. right or wrong
 
Motion is curved due to Force not acting in the direction of motion and that causes acceleration not being in the direction of motion and that causes curved path.
 
123
6:36 AM
This is my question. Whenever force is not in the direction of motion is always curved. right
 
yes
 
123
and we divide acceleration in two parts $a = a_{cetripetal} + a_{tangential}$
 
@123 wait, 1 sec... Someone else is asking me doubts in other room.
@123 yes, in case of circular motion
 
123
$a_c$ responsible of changes direction and $a_t$ responsible changes velocity
This could also be the case in projectile.
 
@123 check these books, they are explained very well here: archive.org/details/FundamentalsOfPhysicsHalliday9thEdition
 
123
6:41 AM
I have read @Azmuth Halliday, Resnick
 
@123 I didn't say motion is curved due to 2 different acceleration components.
 
123
I know calculations are correct. I am asking question because of confusion.
 
It's just because direction of accelertation being not aligned to velocity vector
there is only centripetal acceleration and tangential accleration is = 0 (in uniform circular motion)
 
123
@Azmuth Yes. Also explain in projectile
I was thought whenever we see direction changes there is always two components of acceleration radial and tangential.
 
@123 In projectie there is only one component of acceleration, that's downwards, the sideways has no acceleration, only velocity
 
123
6:43 AM
In projectile this is not the case.
 
@123 No, not necessarily, the tangential components aren't necessarily required to be non zero to have change in direction.
 
123
@Azmuth yes. i didn't mean changes direction mean tangential is zero
If direction changes there is always a radial component of acceleration or not? which is responsible of changing direction.
 
@123 always radial component would be non zero.... (if the motion is on a plane)
 
123
OK let me clear one what in my mind.
In circular motion always radial component of acceleration which called centripetal acceleration $a_c = \frac{v^2}{r}$ at which velocity is constant. Take mass = 1 to take force and acceleration same.
In this case velocity is not changing acceleration due to force serves only changes direction.
 
@123 No, that's true only for *uniform circular motion)
 
123
6:49 AM
Is this correct or wrong
Yes yes in uniform circular motion we are taking ideal cases for both circular and projectile
 
@123 this i right
correct :)
 
123
If there is any tangential component of acceleration in circular motion it also changes tangential velocity and angular velocity.
With this idea i have in my mind whenever i see direction is changing (not necessarily circular) there is always radial component and tangential component. radial at which velocity is constant and tangential at which velocity changes.
Is this correct or wrong?
 
does anyone know when ACM comes online?
 
Uniform circular motion is defined as a a special case of circular motion where tangential direction is $0$. But it doesn't mean that it'd always remain $0$. There can be tangential components too, but then in such case, $a = v^2/r$ won't be applicable. You need velocity at an instant to make this true
@satan29 usually after 3 IST (my best guess, although, he doesn't has a regular time) @ACuriousMind
 
123
@Azmuth My confusion in projectile motion.
I see the direction changes in projectile. Why there is no radial component at which velocity constant which is responsible for changing direction of projectile. Using above analogy
 
6:56 AM
@123 $v^2/r$ will also work for projectile motion to find the instantaneous radius, but that's something radically different from the radius you see in circular motion
also, instantaneous radius is true for all kind of curved motion.
check this:
3
Q: Calculating force through ICOR (instantaneous centre of rotation)

Vasu Goyal A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure). If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ re...

 
123
@Azmuth Yes this is what i wanted to know what happened in projectile it is actually changing direction
 
This is explained in the picture I sent above, I don't know what the problem is, the acceleration pointing down is making the velocity vector go from pointing up at an angle to pointing down at an angle until it hits the ground
 
123
@Azmuth Yes i agree with instantaneous. Because in curved we always need a support of calculus.
 
@123 ICOR is a cheap and dirty method to approximate a very small/tiny section of any curved motion as a curved motion to use rotational dyanamics there.
@123 Even straight line paths require calculus (such as non uniform).
 
123
Yes you are right. I know the difference. Thanks
@Azmuth Yes it depend what parameters we are plotting whether it is behaving linear or not
 
7:00 AM
Unfortunately ICOR won't be any of any help in projectile path/motion calculus, you should look into the picture @bolbteppa sent above. The strategies to decode them into Newton's Laws are well pictured there.
@123 There are certain non linear cases which doesn't requires caculus either. (But as a physicist, a theoretical physicist, one should use calculus), god knows what complications a question has to offer,.
 
123
If we use radial and tangential component of acceleration due to gravity which is always pointing downward along y-axis. Can we see both components separately radial changes direction and tangential changes velocity?
 
@Azmuth hey!
Qm is gping great
 
123
@Azmuth Yes
 
@JackRod Hey, sup buddy?
 
Good
Wbu?
 
7:04 AM
@123 Hell no! That's why I told, ICOR is useless here, separating them as tangential and radial components, you are hitting yourself. Better separate them as 1. Direction of gravity and 2. Tangent to direction of gravity and not as radial and tangential ones.
@JackRod doing fine. Reading Algorithms, they are pretty hard. @PM2Ring suggested me a PhD level book and I'm still struck in first few pages.
Donald Knuth...! (God of Algorithm programming!)
 
123
@Azmuth OoKay ... That's why i am confusing. The idea of radial and tangential component did not work in projectile. right. It means it is not universal method for calculating the curves?
What is ICOR stand for?
 
@123 There's no single universal method for curves. Although there are some approximation ones (like I mentioned), areal velocity, coordinate rotation, etc, etc... but all are useless....
@123 Instantaneous Center of Rotation)
3
Q: Calculating force through ICOR (instantaneous centre of rotation)

Vasu Goyal A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure). If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ re...

@JackRod Which branch you are in?
also called instantaneous radius method
 
123
Can we explain limitation of ICOR?
So, we can say what method is good for what type of motion.
 
@Azmuth did not I mentioned earlier?
 
@123 There are only limitations of ICOR.... only one advantage that it helps to visualize any curved motion as circular motion and helps us with circular dyanamics..
@JackRod I guess I forgot :P
@123 There's no single good method. Each situation can be resolved into multiple good ones.
@JackRod uhhhhhh!? Computer Science?!!!!
really?
 
123
7:12 AM
Is it possible to write most used method for calcuation. like cartesian, polar, ICOR etc.. what else
 
sorry, I'm involved in 2 rooms simultaneously....
@123 no
 
123
When you have time. list few of them. Or give me a link
 
@JackRod Have you all completed algorithms?
 
@Azmuth busy then we can talk later?
Mostly
 
Okay :)
@123 Rectangular coordinates, polar and circular motion method are the most common ones...
 
123
7:14 AM
Ookay. Thanks
 
I just realized that the series $\sum \limits_{\text{p primes}} \frac 1 p$ diverges... :O
DAMN! Editing
 
 
2 hours later…
123
8:59 AM
Is there book which explains energy with history. Like I have kimbell which explains electrostatic with history
 
9:43 AM
0
Q: Why are there close votes on this question?

FoundABetterNameI recently asked this question and noticed a close vote. So I simply commented on the post that maybe someone would share the reasoning for the close vote however no one did and today I noticed another close vote. I have the close vote privilege so I could see the reason is lack of clarity but I ...

 
 
2 hours later…
11:22 AM
did I write anything objectionable here
 
I will read that after I have finished waking up :P
 
That question basically seems to be asking why is $e^{ipx}$ different to $u_p e^{ipx}$
 
@bolbteppa well they are isomorphic and all
you can do it
it's just not a particularly wise decision
 
@Slereah I think you're correct and likely wrote a lot of what Charlie needed to hear but I think you missed the core of the question
oh, no wait, you wrote the right thing
 
I probably should have written more on irreps I guess
 
11:29 AM
Are you saying the space of plane wave solutions of Klein-Gordon is isomorphic to the space of plane wave solutions of the Dirac equation
 
@bolbteppa Well the Hilbert spaces are, so... maybe?
The Hilbert space theorem is for the orthonormal basis, I don't know if it extends to the rigged Hilbert space
I'm not 100% sure what would be the definition of the "good Hilbert space"
 
The problem is that "isomorphic" has two different meaning in the QFT context: You can talk about the Hilbert spaces being isomorphic as Hilbert spaces, or you can talk about the physical state spaces being isomorphic as representations of the respective algebra of observables
 
something something faithful irrep of the Poincaré group
 
"The Fundamental Theorem of Infinite Dimensional Vector Spaces states that all Hilbert spaces with a countable infinite orthonormal basis are isomorphic"
 
when people say that two state spaces are "not the same", they are using the latter meaning, not the former
 
11:32 AM
but then again, you can just translate the symmetry operators to that weird Hilbert space, so
I don't know
 
well hello
 
you kind of danced around that when talking about spin and the mapping of basic vectors being nonsense
 
So even though a KG state with momentum $p$ is attached to two spinor states with momentum $p$ (different polarizations) they are isomorphic?
 
what makes $L^2 \otimes V_n$ a better group than $L^2$, as far as the representation goes
@bolbteppa In the sense that they can fundamentally be described as a big list of numbers (wrt to the eigenvalues), yes
they are big lists with the same number of eigenvalues
it's just not a very compelling space to put them in
 
As in a big infinite list
so everything is isomorphic as they are all countable lists
 
11:35 AM
Pretty much, yes
unless it's finite dimensional or not separable
But then that's getting into "µ-recursive functions, lambda calculus and set theory are the same" territory
Technically true but perhaps not very helpful
 
123
Is there any book which gives history of energy and work. Like i read kimbell which has history of electrostatic.
 
@Charlie I would really advise chapter 1 of Walecka Fetter's Many Particles QM book on second quantization
 
Are you just looking for the names of scientists that invented certain concepts @123?
 
123
I know few the scientist and history. but need to read what problem occurred to develop the idea energy KE , PE and work done.
 
Energy was mostly investigated by Mr. Hamilton, IIRC?
 
11:39 AM
I see @bolbteppa, thanks I'll have a read of that today
 
Like the modern notion of energy was invented by him
 
123
I learnt many things yesterday from you guys.
 
Although there were plenty of precursors
 
123
Like K&K give one idea as physicist need force in term of position. i need more history
 
IIRC Hall's book has a whole section on why that Hilbert space is a good Hilbert space
I should read it in more details
 
11:41 AM
When they introduce creation and annihilation operators out of the blue, skip ahead to where it links up with the previous section they reanalyze what they got and express it in terms of creation operators that's basically the way to even motivate bringing them in as some operators in the occupation number formalism which do what's written there
 
@Slereah what does $\aleph$ and $\aleph_0$ represent in the answer you gave, I meant to ask. Does it represent a countable infinity?
 
123
why i dont see upload button
 
Perhaps you need over a certain reputation on the main site @123
to post images in the chat
 
@Charlie aleph null is countable infinity
 
I see, ty
 
123
11:43 AM
But it was there 2 days ago.
 
aleph 1 is the next highest infinity, aleph 2 is the next highest, etc
 
wait there more than two kinds of infinity?
 
wait nvm, i'm pretty sure it's aleph 1 that's the size of $\mathbb{R}$
 
@Charlie oh boy, are you in for a treat :-)
 
D:
 
11:44 AM
@Charlie yeah, some infinities are bigger than others. the smallest infinity has the size of the set of all integers
it's called countable infinity
 
I'm aware of only countable and uncountable infinities
 
but there are more numbers in $[0, 1]$ than there are integers in the whole number line. e.g. if i had infinite time, i could count all the integers, but i couldn't even begin to count all the tiny numbers in $[0, 1]$
 
@SirCumference yes, what is unknown (and undecidable) is if there is an infinity in between $\aleph_0$ and $\aleph_1$.
 
My assumption was that there was no way to interchange them by arithmetic operations. But you could for instance double the cardinality of the natural numbers and obtain a set with the same cardinality, countable infinity
 
@Slereah I've written another answer that focuses on that
 
11:46 AM
@Charlie you could consider the set of all subsets of $\mathbb{Z}$
for finite sets $S$, the set of all subsets has cardinality $2^{|S|}$. for infinite sets, the set of all subsets has a strictly bigger infinity
so you can always generate a higher one by looking at its set of subsets
 
But, $2^{|S|}$ is necessarily finite for finite $|S|$, right?
 
if an infinite set has cardinality aleph $n$, then its power set (i.e. set of all subsets)'s cardinality is named aleph $n+1$
 
So there is a "bigger" infinity than uncountable?
 
@Charlie yeah
 
:O
 
11:49 AM
@Charlie uncountable just refers to any infinity that's bigger than aleph 0 (i.e. the smallest infinity)
 
Do they exist for all $n$?
 
so there's only one countable infinity, but the rest are uncountable infinities
 
Sure
 
@Charlie yeah
 
For any set, the power set is larger
 
11:50 AM
Well that is definitely some new information
 
iirc though, this infinity is bigger than any other:
In set theory, an uncountable cardinal is inaccessible if it cannot be obtained from smaller cardinals by the usual operations of cardinal arithmetic. More precisely, a cardinal κ {\displaystyle \kappa } is strongly inaccessible if it is uncountable, it is not a sum of fewer than κ {\displaystyle \kappa } cardinals that are less than κ {\displaystyle \kappa } , and α < κ {\displaystyle \alpha <\kappa } implies...
 
It is, but it's not very often useful
 
@123 You need 100 reputation to be able to upload images to chat, cf. meta.stackexchange.com/q/317853/263383
 
Usually it stops being useful beyond $\aleph_2$
$\aleph_2$ is the cardinality of all the real valued functions
 
what about the cardinality of all functionals on real-valued functions? :P
 
11:51 AM
(Assuming the continuum hypothesis, don't at me)
 
123
@ACuriousMind Thanks, It means my reputation is not GooD :D
 
@ACuriousMind I mean $\aleph_2$ is already too big for most purpose, since that's counting every discontinuous function
I'm sure the cardinality of distributions isn't that large
 
@ACM ty for posting a second answer, if I've understood what you and Slereah are basically saying, it's that the Hilbert spaces of the free qfts are isomorphic at the Hilbert space level, but that this doesn't imply they are isomorphic in every sense?
 
@Charlie this also proves that a power set can't have a cardinality of aleph null, btw. bc any finite set's power set is finite, and any infinite set's power set is a bigger infinity (and therefore wouldn't be the smallest infinity)
 
@Charlie yes
 
11:53 AM
I could live with that, it would make sense
ok nice, ty for taking the time write those
 
one of the big issue of math is that there's a lot of meaning of "are the same"
 
123
What is the benefit of stackexcange. You guy earn money from to answer or any other?
 
hahahahaha...no, we're all volunteers here
we're exclusively paid in fake internet points
 
Yeah I guess my wording could have been more precise
 
You can send money, though
 
123
11:55 AM
@ACuriousMind It is no doubt you all guys gives lot of lot of help to me and others.
 
There should be a government scheme to trade in your internet points for money, like a casino
 
@Slereah i guess that's what equivalence relations are for tho
 
123
@Slereah I am also looking for money :D
 
@SirCumference there are many
Is $U(1)$ the same as $S$???
 
123
What is the charm and harm of being physicist?
 
11:59 AM
$U(1) = S^1 \neq S$
 
The Hilbert spaces in regular qm, once a basis is chosen, can the entries in their vectors (albeit infinite ones) be written countably?
 
@Charlie Yes, that is the basic idea of it being separable
 
123
How do we calculate the radial and tangential components of acceleration due to gravity in projectile motion?
 
The usual basis people use (the momentum basis) isn't technically a basis, but there are actual orthonormal basis that are indeed countable
for instance, Hermite polynomials
 
Ok that's kind of what I suspected, in that case how is it possible to establish an isomorphism between that and the uncountable real line(s)?
 
12:03 PM
Laguerre polynomials
etc etc
@Charlie There is no such isomorphism
 
this question has probably been asked before, I've seen things resembling it slightly on se before
Yeah actually let me rephrase that
When we take $\langle x|\Psi\rangle$, this assigns to each point on (at least some subset of) the real line a value $\psi(x)$, which seems to require uncountable infiniteness
this has actually been bugging me for a while
 
@123 you can project the constant gravitational vertical acceleration vector onto each component but it's a silly thing to do this is not the way to think about it at all
It's very simple, see the picture I sent above, that's all there is to it
 
Jul 22 at 16:30, by ACuriousMind
@Charlie the game is rigged
or, if you just need to be reminded: $\lvert x\rangle$ does not actually lie inside the Hilbert space
 
Ah I see
 
12:15 PM
So the whole $\int F_{\mu \nu} \tilde{F}^{\mu \nu}$ thing being a total derivative vanishing at the endpoints for Maxwell, in the non-abelian case $\int F_{\mu \nu}^a \tilde{F}^{\mu \nu}_a$ can be expressed as a total derivative which does not have to vanish at the endpoints, and this actually leads to something that can be measured (but hasn't?)...
 
Once you have messed around with rigged Hilbert spaces for long enough, non-relativistic QM is mathematically rigorous in its entirety, right?
last question, before I go and get a sandwich :P
 
I doubt it, I'd guess it can't be because it's all equivalent to path integrals which are a bunch of nonsense rigorously
You can work that integral out explicitly for $\text{su}(2)$ and it relates/motivates instantons somehow... this is pretty cool
 
Oh I didn't realise the path integral wasn't well defined
 
@Charlie The rigorous versions usually avoid talking about rigged spaces entirely, but yes, QM with finitely many degrees of freedom (i.e. non-QFT) is entirely rigorous
@bolbteppa the path integral is only ill-defined for generic field theories, for non-field QM it is perfectly rigorous
 
ah
Ok, thanks, I can eat in peace now :P
 
12:19 PM
I should do that, too, actually
 
@JohnRennie in the example of a pendulum in an accelerating car as seen from an inertial frame , does the pendulum stay at its position due to inertia ?
 
@Ankit you mean if the pendulum is just hanging stationary i.e. not oscillating?
 
12:34 PM
@Ankit I have to go now. I'll be back tomorrow as usual.
 
I think the rigorous qft analogue is one of those 'lost causes' in that book
 
I believe in it
and nlab will do it
 
"the quantum propagator – in FQFT the value of the functor $U:\text{Cob} \to \text{Vect}$ on a certain cobordism"
Clearly if you're not doing functorial qft you're not doing qft
 
well you don't have to use path integrals
 
12:50 PM
I love the framing "A simple form of the path integral is realized in quantum mechanics, where it was originally dreamed up by Richard Feynman and then made precise using the Feynman-Kac formula" i.e. you're an idiot if you just get the 'simple' one and haven't done the “holonomy integrated against the Wiener measure” one "in the slice (infinity,1)-topos of smooth infinity-groupoids over the developing groupoid $\textbf{B } \text{U}(1)$"
 
tough but fair
 
It's so complicated it didn't even copy properly...
 
Can someone help me here .. the pendulum is not oscillating as John Rennie asked for ...
 
nlab uses a lot of mathjax definitions, I think
 
Even the word developing didn't go right the first time :p
 
 
1 hour later…
2:00 PM
"I think it was Belinfante who first showed that the most general possibility, again, assuming that no orbital angular momentum is transferred in the fundamental decay process, was a mixture of five types of current: scalar, vector, tensor, axial vector, and pseudoscalar. This may be unfamiliar to today’s graduate students in physics, because the work of Marshak and Sudarshan made this kind of theoretical classification obsolete.
We now know what the right answer is, so who studies these other possibilities anymore? But, I find history fascinating, so allow me here to look back on this bit of history. It was clear that you had to have scalar or vector, because the other currents would not produce transitions between nuclear states of 0 spin, and there are transitions like the decay of Oxygen-14 into Nitrogen-14 where both nuclei have spin 0.
That is allowed by the scalar and vector current interactions, but would not be allowed by the other interactions. Then there are transitions like the one I mentioned when Boron-12 and Nitrogen-12 decay into the ground state of Carbon-12, which require the presence of tensor or axial vector currents.
So you have scalar and/or vector plus tensor and/or axial vector currents - in some mixture that no one knew. (The pseudoscalar interaction doesn’t produce allowed beta decay, so it is a somewhat tangential possibility.)"
That is very cool
 
2:42 PM
I remember that it was the big thing Feynman was known for for a while
Finding out that the weak interaction was a vector-pseudovector interaction
or whatever it was
 
Tell mah your halloween ideas!
now!
 
Stay at home
 
and?
 
that's about it
I mean I will do a variety of things, ie breathe, eat, metabolize
 
2:51 PM
I'm didn't step out of home for last 4 months.
 
I will perform a lot of Krebs cycles
 
I'd probably go to roof and capture #BlueMoon
 
The citric acid cycle (CAC) – also known as the TCA cycle (tricarboxylic acid cycle) or the Krebs cycle – is a series of chemical reactions used by all aerobic organisms to release stored energy through the oxidation of acetyl-CoA derived from carbohydrates, fats, and proteins. In addition, the cycle provides precursors of certain amino acids, as well as the reducing agent NADH, that are used in numerous other reactions. Its central importance to many biochemical pathways suggests that it was one of the earliest components of metabolism and may have originated abiogenically. Even though it is branded...
I do that shit all day long
 
Cool!
You won't be capturing #BlueMoon?
Dun worry, If I did, I'll post my pictures here.
It happens in every 2.7 years
 
mb I will watch a spooky movie
 
2:52 PM
Me too!
 
If I am feeling particularly in the holiday mood
 
conjuring 2?
How's it?
 
Although Halloween isn't really a holiday here
 
Me too, holiday mood
@Slereah where?
 
The France
 
2:53 PM
@ACuriousMind you give ideas!
 
I concur with Slereah's plan of staying at home :P
 
@Slereah cool place, I'd dress as Iffel Towe
or somethign same name
@ACuriousMind no movies?
 
I've got an RPG session later actually, but we do that about every two weeks, so it's not special.
 
Role Playin Game?
Let's to a virtual party here
using Mozilla Hubs!
 
@bolbteppa : see Charles Galton Darwin's 1927 Nature paper "The Electron as a Vector Wave": nature.com/articles/119282a0
 
03:00 - 15:0015:00 - 22:00

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