2:57 AM
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I wish to add a tikz diagram to one of my questions, but the website doesn't seem to compile. Some help please?

3:20 AM
Well, today - we celebrate Pongal :D

Kite flying? My Indian coworkers had the day off.

What?!!! Nah, not kite-flying :D

Also, I thought you celebrated it yesterday? My coworkers had Monday off. Does it land on tuesday

It's a Tamil-ian thing
@BrandonEnright Today for us, tmrw we celebrate for the cows and buffalos (when we paint their horns with different colors, so & so) :D
Thai Pongal (தைப்பொங்கல்) or Pongal (பொங்கல்) is a harvest festival celebrated by Tamil people at the end of the harvest season. Pongal is a four day festival which usually takes place from January 13 to 16 in the Gregorian calendar i.e., the last day of the Tamil month Maargazhi to the third day of Tamil month Thai. The second of the four days i.e., first day of the Tamil month Thai is the second and main day of the festival and this day is known as Thai Pongal. This day coincides with Makara Sankranthi which is a winter harvest festival celebrated throughout India. The day marks th...
@BrandonEnright We usually celebrate on 14th Jan. Farmers celebrate 14 & 15 :)

Regarding bovines, I've always wondered what the religious significance is. Is it a Hindu thing? I haven't had the balls to ask any of my coworkers (many of whom I suspect are atheist) for the details / history.

3:27 AM
Yest., not much of us actually celebrate, because yest. is a welcoming day for Pongal (called Bhogi), when we burn all the old things and pollute the environment (which is why, I compel my family not to) :D
@BrandonEnright Nah, it's an encouraging festival for the farmers
It's celebrated only in our state :)
Not necessarily "Hindus" - we don't think of any Gods today ;-)
So (unfortunately), I had to bath early in the morning...

I was asking where the significance of cows came from. Sorry I can't respond to specific lines, I'm on the mobile chat site.

I dunno much about the significance. But, I do know that it's for the farmers. Let's say, cows are somewhat favorite for the farmers (you know, it's a harvesting festival - and, buffalos & their family are responsible for growing the crops, thereby leading to the harvest) :D
Maybe sentimental :P

6 hours later…
9:06 AM
Hey guys :D
I just noticed something about the work-energy theorem
Change in Kinetic Energy is a vector while work is a scalar
Even if you tell me that the W-E theorem is only true when Force and Displacement are in the same direction, it still doesn't explain how you're equating a vector to a scalar.

@Nick no it isn't
kinetic energy is a scalar, so is the change in it
K.E is $\frac12 m \vec{v}\cdot\vec{v}$, which is a dot product and thus a scalar

O_O So, you're telling me that I've been making a mistake by putting an arrow on top of all those kinetic motion formulas
Dang
@Manish: Is it just me or are people liberal with these vector and scalar product things? I just get so frustrated when I make these sort of mistakes.

@Nick Don't know what you mean. After a point in writing one may stop specifying if a quantity is a vector because it usually is obvious. Books usually continue to use boldface of arrows.
@Nick why would there be an arrow?
Think about it, why does energy need a direction

... mhh, you're right, energy is best packed up.
@Manish: Arrows $\implies \vec{v^2} - \vec{u^2} = 2\vec{a} \vec{s}$

hello @ManishEarth, can I ask if there is anything wrong with my answer here physics.stackexchange.com/questions/72711/…?

9:20 AM
@Nick That's a $\vec{v}\cdot\vec{v}$

Dang!

You can't square a vector and get another
@Amaterasu looks fine

good good... was just a bit worried that's all

@Manish: Really? never?

9:21 AM

It just means that people didn't notice, or people weren't interested in the question (can happen if the title is confusing)
@Nick as in?

Null vector x Null vector = (Null vector)^2
Never say never

or they don't like, thus ignore the user, perhaps?

@Nick wut?
That makes no sence
There's no point of vector-cross-product-squaring because you get a zero
Dot products spit out scalars
Tensor product squaring is used often, but that spits out a tensor

But wasn't what I said right? 0*0 = 0^2 = 0

9:24 AM
@Nick Multiplying a vector by itself using the cross product is not squaring

Ah zero, everything's possible with you.... except division

For that matter $\hat{i}\times\hat{i}=\vec{0}$
but all of these are useless

Yeah sin(0), I get it

So the operation "squaring a vector" always refers to the dot product

uh-huh

9:25 AM
You will never see a cross-product vector square in formulae. In proofs, yes, because it can be cancelled out, but never in formulae because zero

Well, that was the most I've learnt about vectors this whole year.
@Manish: hugs

heh

But really, I get very annoyed when I learn a formula and I have to relearn it because it's a vector product
L = mvr
I had to relearn it as L = r x P
never to be confused as P x r

I tend not to care much about which vector comes first. It's fine to use L= p x r as long as you flip all the other cross products
Of course, it's best to remember the formula as is for school purposes :)

You're right... it's confusing as it is

9:32 AM
Once I forgot the vector order for magnetic fields, I just quickly ran a check and ended up using the opposite convention -- which still worked for calculating final
. Of course, that doesn't happen to me now :p

@Manish: I tried finding torque by differentiating L = r x P, I have this weird $\frac{d\vec{r}}{dt} * \vec{P}$ which has to be zero... I have no clue how it's got to be zero.
@Manish: Is $\vec{r}$ constant? Is that it?

@Nick Usually the force is applied on a single point, so yes
Unless the force is moving around, in which case $\vec \tau \neq \vec r\times \vec F$

O_o the force is moving around? what do you mean?

@Nick If the force being applied was not applied at a point at a fixed distance away from the center of rotation

mhh, my physics teacher said that $\frac{d\vec{r}}{dt}$ is $\vec{v}$ ... Ok, sure it's kind of like velocity (the derivative of distance with respect to time)

9:41 AM
@Nick Not always
What is r?
If r is the position vector of a particle, its derivative is the velocity of that particle

and by differentiating v = omega * r with respect to time, We got a = alpha*r + omega*v

If r is the relative position of a point on a spinning disk, then its derivative is the velocity of that point

Yes it is the postion vector

@Nick Actually, r is not constant here
However, usually dr/dt will be perpendicular to the momentum

O_O but it was in L = r * P, right?

9:43 AM
@Nick So that is the formula for the angular momentum of a particle

yes :D

In this case, its derivative is the velocity of the particle

O_O

(I was just saying that dr/dt is not always the velocity, sorry if that confused you)
However, what is dr/dt x p?
It's dr/dt x m v
=m dr/dt x v
= m.0
@Nick I initially confused your forumula with something else for rigid bodies

I'm sorry for no stating what it was. XD I thought all you physicists had these notations so you wouldn't get things confused

9:46 AM
Nah, just doing other things in parallel
Got a qmech assignment that I'm texifying, plus some code

1 hour later…
11:00 AM
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For every question, there is a number of views. What does it mean ? How many times a question has been viewed/opened ? How many times it has been opened by a unique member of community ? Or something else ? How is it determined ? Is a log of IP addresses visiting the question kept ? If it is, w...

3 hours later…
1:34 PM
@Manish: Dang, fell into a time portal and traveled into the future. (actually, internet failure but whatevs)
I've been playing around with some equations and I have something that's really weird.
From the definition of a radian,
theta = x/r
x = theta * r

@Nick okay... (???) That doesn't seem weird to me. I'm used to that already :P

@Nick okay...

Differentiate that, and you get v = omega*r + (something weird)

Are you going to ask how omega is a vector?

@Nick Only in the small angle approximation.

1:38 PM
Not really...

Wow... Manish predicts the question nowadays :D

d(theta) is a vector but theta is not

@Nick nah

It's an approximation, I guess

This is just some fudging of definitions, basically
d(theta) is a scalar, as is omega when written as a scalar (magnitude)
however, omega vector is defined as omega scalar with a direction

1:40 PM
.... who the heck gave omega direction?

Someone
@Nick For that matter, you can give theta direction, too

@Nick Who cares? But, it has. Doesn't it?... clock & anticlock!!???

Theta has a sense... that sense is all curvy..
I thought vectors are liney

Quite a few vectors are cobbled together with the direction and magnitude coming from different sources

O_O sounds like a linear approximation of a curve.

1:43 PM
The "area vector" is a similar vector where the direction is tacked on to the area scalar to make a vector
@Nick ?

Linear approximation is not cobbled together. Without that math would be hard

It's like the directions are arbitrary

Not really
The directions are convenient
Once you start using them you'll see why they were added on

It's like somebody wanted their scouters to read an energy level and made that energy level up just so it could be measured.
@Manish: Are these guys really vectors? As in really physically there or did we just make em up so that we could measure something.

1:48 PM
@Nick What do you mean with a quantity "really being a vector"
Is position "really a vector"?

... It's got direction... as in I change position with respect to time, the position is changed in a certain direction

After all, you've taken "distance from the origin" and joined it with "direction from origin"
@Nick So does angular velocity. I can speed up a disk, I can also reorient it
So does area. It points somewhere

Ok, but do they really exist?

What does it mean for a vector to "really exist"?
Vectors are mathematical constructs

So, they're just abstract stuff for measuring other stuff?

1:50 PM
We take a scalar quantity and a related direction and associate the two
@Nick that and calculating things and making manipulation in general easier

mh.. i guess so.

In mathematics you may prove things like the Cauchy inequality by just using an N-dimensional vector. A vector without physical meaning

woah that's wierd. XD
speaking of weird...Hey, nobody's addressed my "weird thing"
dx/dt = d(theta)/dt * r + theta*dr/dt
theta*dr/dt has to be zero
Idk how

@Nick theta*dr/dt is the radial velocity
If your particle is sliding up and down a rotating wire, then it has radial velocity too
v=omega x r only works when the particle is at a fixed radius
Otherwise it is v_t = omega x r where t stands for tangential

@Manish: ...what do you mean sliding up and down a rotation wire? Like the solar system around the center of the galaxy?

1:57 PM
@Nick nah
Take a wire or a rod, whirl it around
Put a robotic bead on the wire.
Have the robot move up and down the wire
It has both a radial and tangential velocity

Up and down as in like how the torque vector goes up and down? Or to and fro the length of the wire.

@Nick to and fro

@Manish: Can you sell me that robot?

heh nope

@Manish: Is there any particular difference between angular velocity and angular frequency?

2:03 PM
2pi

XD

Angular frequency is 1 if it goes one rotation per second

It's same concept isn't it?

Angular velocity for 1 rps is 2pi
Yep
Just a different scale
Angular velocity is radians per second, angular frequency is rotations per second

Um remember last we talked, I said about a = alpha x r + omega x v
Which is the radial and which is the tangential?
Is there a rule of thumb?
@Manish: Is that too dumb a question?
maybe...

2:11 PM
@Nick radial is the one that points towards/away from the center
tangential is the one that is perpendicular to that
Here, alpha x r is tangential

Oh!

That is so cool, I got it!

@ManishEarth There's also a $2 \omega v$ which comes out as a coriolis component (which overlaps the tangential comp. most of the time) :D

That's only if you carefully differentiate it and keep track of the frames

2:13 PM
I say this because, that's what we're doing this sem. Analyze the rotational/translational kinematics of links & chains :P
@ManishEarth yeah :)

@Waffle'sCrazyPeanut: How the heck did you get that? (tell me only if it's simple enough for my comprehension)

9

I was wondering how easily these two pseudo-forces can be derived mathematically in order to exhibit a clear physical meaning. How would you proceed?

@Nick ^^^ It's simple :P

@Nick ...you don't want to know yet :p

hehe, yeah :D

@Manish: I'm going to take your word for it.

2:17 PM
@Nick: You can think of those this way, "radial is the thing that we use to pull the object, thereby make it go along the circle, while tangential is the thing that comes out as a result of the inertia of the object - that the object tries to break the connection and fly off... ;-)

@Manish: Why is the centrifugal force a psuedo force? It's in that question that Waffles linked.

I still wonder how far that analogy holds...

@Waffle'sCrazyPeanut: I'd love to see the limitation of that analogy as well but I think I'm gonna just think of it as a planar tangent in the direction of velocity.

@Nick Because, that's just equal and opposite to centripetal force (say, the reaction in rotating frame)
@Nick yeah, that's good :)

@Waffle'sCrazyPeanut: So, all reaction forces are called pseudo forces?

2:22 PM
@Nick Dunno for sure... Because, the pseudo forces are defined only for non-inertial frames -_-
A fictitious force, also called a phantom force, pseudo force, d'Alembert force or inertial force, is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame. The force F does not arise from any physical interaction between two objects, but rather from the acceleration a of the non-inertial reference frame itself. As stated by Iro: According to Newton's second law in the form F = ma, fictitious forces always are proportional to the acceleration a. A fictitious force on an object arises when the...

If I'm given a question, where a body is in uniform circular motion, where the mass and velocity of the body and the radius of the circular path is given and I'm asked to find the centrifugal force acting on the body, should I just calculate the negative of centripetal force?

Yeah, just add a minus sign to that of centripetal... :P

um, that brings me to a greater question
Clearly since the two forces belong in different frames, they do not cancel out each other in your frame
1.From the viewpoint of an inertial reference frame stationary with respect to the road, the car is accelerating toward the center of the circle. This acceleration is necessary because the direction of the velocity is changing, despite a constant speed. This inward acceleration is called centripetal acceleration and requires a centripetal force to maintain the circular motion. This force is exerted by the ground upon the wheels, in this case thanks to the friction between the wheels and the road
The car is accelerating, due to the unbalanced force, which causes it to move in a circle.
2.From the viewpoint of a rotating frame, moving with the car, there is a fictitious centrifugal force that tends to push the car toward the outside of the road (and to push the occupants toward the outside of the car). The centrifugal force balances the friction between wheels and road, making the car stationary in this non-inertial frame.
@Waffle'sCrazyPeanut: I'm assuming these were the frames he was talking about.

Yeah, disagreement between frames :D

@Waffle'sCrazyPeanut: Yeah, I don't get the first frame. What is the inertial reference frame stationary with respect to the road?

2:36 PM
@Nick that's the rest frame... (relative to the road)

O_O oh, disagreement with the frames of rest and motion. That makes so much sense!

Whoooo!!! FINALLY!!! :D
@Nick a correction -- actually non-inertial motion :)

It's like the angels are singing hallelujah!

Inertial frames always agree... (that's ) :D

dang
Isn't that what stops the kid running in the train moving 99.9% the speed of light from not surpassing the speed of light?
Yeah, I never got that

2:40 PM
@Nick I guess it's enough for you today :D

XD
Indeed!

coz I'm off for a while... will be back for chat session :D

i got to sleep sometime too.
Well thanks for the greatest most informative day ever in my life.
@Manish: A special thanks to you
@Waffle'sCrazyPeanut: and a special relativity thanks to you!
In honor of Kyle Kanos,
Goodnight everybody! :D

3:02 PM
@ManishEarth: pssst, about the radial component of angular acceleration. $a_{\text{radial}} = \vec{ \omega } \times \vec{v}$
I know $\vec{v} = \vec{\omega} \times \vec{r}$
So, $a_{\text{radial}} = \vec{ \omega } \times (\vec{\omega} \times \vec{r})$
Um, won't that become zero?
Why do I have a formula in my book that says $a = {\omega}^2 r$?
With my knowledge of vectors, $a_{\text{radial}} = \vec{ \omega } \times (\vec{\omega} \times \vec{r}) = \vec{ \omega } \times (\omega r \sin \theta \hat{n}) = {\omega}^2 r \sin^2 \theta \hat{n}$
Even if I only consider only the magnitude, I still have that that $\sin^2(\theta)$
Unless... angular velocity and position from origin are always perpendicular to eachother
@Manish, @Waffle'sCrazyPeanut : Is that true?

3:27 PM
@Nick Not necessarily, no
But usually yes, we consider rotation in a plane
@Nick Because it's not a real force. It's a force that your mind thinks you experience while rotating

3:41 PM
@ManishEarth, and other 10k'ers, care to show this some love? meta.stackoverflow.com/questions/212375/…

@EmilioPisanty have a bounty

4:07 PM
Welcome to our chat session!!! :)

user54412
so, anything people want to discuss?

I'm here
and not here

Me too :)

4:29 PM
Hello goodbye, ending in 30 seconds??
On the other hand I was the fourth of four entries, so I gather I did not miss much :)

Start speaking @TerryBollinger & @ChrisWhite -- BANG BANG!!! I suppose it'd be interesting :D

user54412
I'm afraid I'm out of things to say

user54412
also, i just realized my 534 consecutive visit streak may soon be broken

@ChrisWhite Why so? Going somewhere?
@ChrisWhite And, we differ only by 5 days (529 cons. <-- me) :D

user54412
yeah, and I may not have internet access for a couple days, apparently for the first time in over a year

4:42 PM
@ManishEarth Oh, wow! Thanks!

@ChrisWhite where? out of town? or out of country? :D

Hi Waffle, actually I'm working and my 10 minute lunch break is up, so I'm out of here (multiple "very soon" deadlines). Er, bang bang? Hmm, deciphering your meaning can at times be an excellent and edifying exercise in complex cognitive perception using very loose binding of the semantic input variables... :) In any case, my parting wisdom today is: Eschew Obfuscation (from an old bumper sticker) or its positive form, Perseverate Pellucidty (I just now made that up).
6

user54412
@Waffle'sCrazyPeanut out of town - I haven't left the country in many years, and I don't really have any reason or money to travel internationally

@TerryBollinger Ugh..!!! I'm unaware of that language (Lotsssss of new words) >_<

user54412
@Waffle'sCrazyPeanut don't worry, I don't think I've ever seen the word "pellucidity" before in my life

user54412
4:47 PM
though apparently it's a real English word o.O

@ChrisWhite "haven't left in many years" - so you've left before?
I've never left my state :D

user54412
I guess I traveled to east Asia 3 years ago, and Spain 10? years ago, and Canada ~15 years ago

Pellucidity: a single-word contradiction. Perhaps a new record?

@ChrisWhite Oh, that's nice :)

user54412
I suppose I could look for international conferences, but the thing about being in the US is that most of the astronomy is here anyway

user54412
4:54 PM
Observers get to travel, but theorists don't

Man, maybe I should have been an astronomer
:-P

Hey there @DavidZ

The research group I was in traveled constantly. The security side of computer science happens everywhere.

Looks like you're punctual

5:16 PM
okay... I'm off to bed (unfortunately, unexpectedly, unlikely early) -_-
c'yall later :)

user54412
5:40 PM
In other news, looks like Congress has given the astro/space community about as much money as they hoped to get

user54412
particularly good news for LSST, WFIRST, and JWST

user54412
seems these are the days of Big Science, Big Data, and Big Money
4

6:17 PM
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I want to add something like FA (the force at point A.), I've been doing it this way: $F$<sub>$A$</sub> but inside an equation starting with 2 dollar signs(I want it to be written on a separate line Like: $$F&lt;sub>A&lt;/sub>$$ this is causing some troubles, and it seems that it couldn't be don...

7:00 PM
@PhysicsMeta That's faintly horrifying.

7:12 PM
@EmilioPisanty because it shows a lack of even the tiniest effort to answer the question themselves?
Or because of their use of <sub> and the mess that must be?

1 hour later…
8:24 PM
I got all excited there were new questions. And then I read them and was promptly disappointed by the questions, comments and answers :(

2 hours later…
user54412
10:05 PM
so they're trying to have a beer stackexchange, in addition to the will-never-leave-beta homebrewing site?

10:21 PM
Are you into homebrewing?

10:54 PM
I can't really speak for Chris but I think his point is that there is way too much overlap and extremely niche topics with proposals / betas.

11:39 PM