The h Bar

@IceLord I use it quite a bit. It's on my desk, but not good enough to be on the shelf.

user218912

I'm using it every day.

user218912

I take it to QFT.

I think my dad randomly sent it to me

Sent an Amazon link, that is

Then he got it for me

My copy is not in good shape lol

user218912

in like grade 11?

12:11 AM

Yes

user218912

my copy is in brand new condition because I was too dumb to use it properly in grade 11.

user218912

now I realized how useful it is.

user218912

If I could go back in time and tell my gr11 self something it would be "don't be retarded by complaining and wasting time and just read the damn books @0celo7 recommended"

You were complaining that the books I told you read weren't "graduate level"

user218912

I was dumb af that time, didn't even know what I was doing.

user218912

12:15 AM

you laid out the path for me and I just kept going on tangents lol

user218912

sigh...

@IceLord I could have sworn you wanted to be an aeronautical engineer

user218912

@0celo7 at some point yeah, but that's irrelevant because I wanted to be many things in the past year and a half.

user218912

but I always ended up back in physics... idk why.

user218912

fuck now that I think of it I should have just listened to you and did physics all the way.

Alfred Centauri

12:28 AM

@JohnRennie I regularly take my copy off the bookshelf and read a section or three again.

@IceLord now that you're convinced listening to me is a good idea...

read Chow, Lu & Ni and do every exercise.

user218912

@0celo7 :o

@IceLord after you do that I will have no more to teach you

amazon.com/Hamiltons-Ricci-Graduate-Studies-Mathematics/dp/…

@IceLord Try not to read any other geometry books.

They will only dilute the purity of CLN.

@ACuriousMind Fun fact: $f(x)=x^2\exp(-x^8\sin^2x)$ is $L^2$, but not bounded!

@ACuriousMind Oh god I'm reading a paper on the pitfalls of physicist math in QM. I'm having a crisis...

zed111

1:03 AM

Hi how do I get the value of magnetic field strength H when it is given in Gauss. Gauss is the unit of magnetic field. (I need only CGS units)

1:17 AM

@0celo7 What's wrong with $f(x) = 1/\sqrt[4]{x}$, on $(0, 1)$? That's $L^2$, but not bounded.

@BalarkaSen $L^2(\Bbb R)$.

Not convinced that's any harder after one has done it on $(0, 1)$. One should be able to compose with a diffeomorphism $(0, 1) \to \Bbb R$ and use change of variables theorem.

Seems reasonable.

But in QM we integrate by parts and drop boundary terms. That's not allowed for general $L^2(\Bbb R)$ functions.

Not sure though. I'll try to plot your $f(x)$.

arxiv.org/abs/quant-ph/9907069

check that paper for it

page 5

1:22 AM

If I have $f,g \in \mathcal{C^{\infty}}([0,1])$. Isn't $|f(x)-g(x)|=0$ has to hold for all $x \in [0,1]$ thus $f=g$?

The plot is very interesting!

@0celo7 Aright.

@Secret What? Why should $|f-g|=0$?

Are you asking if $|f-g|=0\implies f=g$?

That works for any two functions $\Bbb R\to\Bbb R$.

$|x|=0$ iff $x=0$.

Yes, since you should expect f=g for all x in $[0,1]$ as $f,g \in \mathcal{C}^{\infty}([0,1])$

@Secret Smoothness has nothing to do with this.

@Secret Huh? $x^2$ and $x$ are $C^\infty([0, 1])$ but $x^2 \neq x$ for all $x \in [0, 1]$.

1:25 AM

@BalarkaSen What? They're equal at the endpoints.

FOR ALL

Aha.

Confusing wording.

Well the actual question is I don't understand the argument I saw in this [topology notes] (math.colostate.edu/~renzo/teaching/Topology10/Notes.pdf)

Example 1.1.4. Suppose f and g are functions in a space X = {f : [0, 1] → R}. Does
d(f, g) =max|f − g| define a metric?
Again, in order to check that d(f, g) is a metric, we must check that this function satisfies the
above criteria. But in this case property number 2 does not hold, as can be shown by considering
two arbitrary functions at any point within the interval [0, 1]. If |f(x) − g(x)| = 0, this does
not imply that f = g because f and g could intersect at one, and only one, point. Therefore,

You misread.

Or the question you're asking is not the one you want to ask.

@0celo7 It wouldn't do you any good if I wrote $\lnot$($x^2 = x$ for all $x \in [0, 1]$).

1:28 AM

I don't really understand the note's example
But if you want |f(x)-g(x)|=0, then don't you need f to agree with g over the entire interval [0,1], as any one disagreement in the values of f and g will cause the expression |f(x)-g(x)| to become nonzero?

Yes, @Secret.

|f - g| = 0 over [0, 1] means f = g over [0, 1].

But the note's claiming |f(x) - g(x)| = 0 at just one x does not mean f = g.

So they are implicitly assuming their ' alleged metric' is taken at only some particular x of the given f and g, and not the interval [0,1] as arguments (thus only f and g can vary), which is why the d(f,g) will break down because it is only taken at some x?

@BalarkaSen I think that's very confusing wording, still.

Or put it in another way, is their $d(f,g)$ of the form $d(f(x),g(x))$ or $d(f([0,1]),g([0,1]))$ (the later case cannot have $d(f,g)=0$ for $f\neq g$)?

@Secret Consider the following example, which might be more helpful

for a metric, what does $d(f,g)=0$ imply?

1:35 AM

they are the same points, which means the two elements are equal

yes

:)

Hmm, maybe I agree with you. If $d(f,g)=0$, that does mean $|f(x)-g(x)|=0$ for all $x$.

Because if $|f(x)-g(x)|=\epsilon>0$ for some $x$, then we'd have $d(f,g)\ge \epsilon\neq0$.

yup

And the absolute value is never negative, obviously

So the only way I can make sense of that note's example is that their d(f,g) is defined for individual x instead of the whole interval of x where the function domain is defined to be...

I sometimes find function space very confusing, since you can either have single points x in the interval as arguments, or have the entire interval as an argument...

and it seems it is not explicitly shown when is which case is used

Perhaps there's a typo there, they probably mean d(f, g) = min|f - g|

Also, d(f, g) for an individual x makes no sense at all, sorry.

d(f, g) = max |f - g| is a perfectly fine metric for continuous real valued functions on [0, 1]. It's known as the uniform norm.

norm/metric whatever.

1:50 AM

Hmm, in that case I guess min|f-g| breaks subadditivity, despite the other 3 rules holds, thus it is not a metric?

No, I mean, property 2 does not hold, like they said.

If f, g agree at an x, |f - g| = 0 at x, hence d(f, g) = 0. But clearly f, g need not be the same.

ah I see, that is different from the max case, where nonzero |f-g| does matter

I have not been paying attention to the discussions here though, and only replying based on the paragraph you quoted.

That is accurate enough, because that's where the original question stem from. The one that 0celo7 and I discussed basically explained why max|f-g| obeys property 2 as expected

which will be very weird had the option of a possible typo not came to mind

@BalarkaSen Is it not a metric for all functions?

oh you need continuity for the max to exist

2:01 AM

also $\mathcal{C}(\mathbb{R})$ is not compact because it is unbounded, and that prevent defining a supremum

3:36 AM

"It took some of the greatest minds our species has ever produced thousands of years to come up with our modern understanding of physics,"
~my astronomy teacher

Really mindblowing when you think about it

Not really.

It took 10,000 years for someone to define "1"

3:51 AM

@0celo7 Okay, how do you define it?

@SirCumference $1:=\{\emptyset\}$.

@0celo7 Uh, I was just gonna say the lowest ordinal

what's an ordinal?

*First

Goddammit man

Don't play that

You know what an ordinal is

I don't think it makes sense to know about PhD level set theory before knowing basic PDE, Riemannian geometry, etc.

@SirCumference I actually don't.

3:54 AM

It's not PhD set theory

It's basic set theory

Which can be a bit healthy to know

"basic" set theory is way too abstract for me right now.

Ordinals describe ordered numbers ("1st, 2nd, 3rd"). They basically have to be integers.

That's not a definition.

Cardinals describe numerical quantities

Also not a definition.

3:56 AM

What?

It's a basic idea. Go research PhD level set theory if you want a better definition.

You've not defined either.

Too advanced, but I'm glad you understand it.

Can you help me with Lie theory?

It's very interesting that you can take the set of all real numbers, and have numbers come after that

When we're talking about ordinals

@0celo7 Uh, why ask me?

Because you're studying advanced mathematics?

That involves higher level calculus

@0celo7 Knowing some basic set theory ≠ studying advanced mathematics

set theory is quite advanced

3:59 AM

I haven't studied it. I just happen to know the difference between ordinals and cardinals, and some basic consequences of their differences

such as?

Such as counting past aleph null

(the set of all real numbers)

what does that mean

user116211

@0celo7 Aren't you following Jech?

Or sets that, by definition, can't be counted past (theta)

@0celo7 What's confusing? You know what the set of all real numbers refers to

4:01 AM

@MAFIA36790 No, I gave it up.

user116211

:(

Page 5 was too hard, I was taking multiple hours per line

@MAFIA36790 When did I say I was?

user116211

@SirCumference Didn't ping you.

Crud

user116211

4:03 AM

@0celo7 You mean the logical formulas and definition of class?

user116211

He briefs all the symbols we use in First-order logic in that very page.

no

everything

set theory is beyond me

user116211

hmm.

I simply do not understand it

user116211

@0celo7 Then start with Helmos...

4:05 AM

@MAFIA36790 getting two AMS Chelsea books soon

user116211

@0celo7 Wow!!

best looking books

user116211

great ;)

user116211

@0celo7 And it took over 350 pages to define $1+1= 2$ by Russell in 1910.

user116211

Thanks to Bourbaki that they didn't pay too much attention to logic.

user116211

4:32 AM

Hey @0celo, do you know the proof of Minimal Element in Totally ordered set is unique?

user116211

I'm checking $\mathsf{Pr}\infty\mathsf{fWiki}$ for that...

It follows from trichotomy, no?

Yes, it does.

user116211

I'm seeing they first took an totally ordered set $(S, \preceq)$.

user116211

Then they took a minimal $m\,.$

user116211

Then they started the proof writing this:

4:36 AM

what's that symbol

user116211

> By definition of minimal element: $$\forall y\in S: y\preceq m\implies m= y$$

the TeX, I mean

user116211

@0celo7 \preceq

the hell does that mean

user116211

@0celo7 It's a symbol of ordered relation.

4:37 AM

no, what does preceq mean

user116211

@0celo7 hmm.

@MAFIA36790 Does $y\in S\implies m\preceq y$ not work as a minimal element definition?

user116211

@0celo7 See, I saw the definition as this:

user116211

Let $(S, \preceq)$ be an ordered set.

user116211

Let $T \subseteq S$.

4:41 AM

@MAFIA36790 I got a Cambridge "Studies in Advanced Mathematics" book, and I must say the print quality is not up to snuff with the outrageous price they wanted.

I got a good deal, but still.

user116211

An element $x$ of $T\subseteq S$ is a minimal of $T$ iff $$\forall y\in T: y\preceq x\implies x= y\,.$$

is $x$ the minimal element?

user116211

@0celo7 yup.

Hmm, that definition seems to imply uniqueness immediately.

user116211

Notice that the minimal element was defined for a subset of $S$ and not for the $S\,.$

4:43 AM

@MAFIA36790 I don't see how that's important.

user116211

@0celo7 Why?

But maybe there needs to be some work, it does not follow immediately.

@MAFIA36790 You have to tell me why it's important first.

user116211

@0celo7 Because it is defined so. You can't skip that part while deriving something else.

Orders behave nicely wrt. restrictions, like metrics and other things

So, we let $z\in T$ be another minimal element.

user116211

okay.

4:45 AM

Then $\forall y\in T:y\preceq z\implies y=z.$

Just writing that down to collect my thoughts.

user116211

sure.

Wlog, $z\prec x$. Are you ok with that?

We are assuming they are nonequal.

user116211

Wait... can we have more than one minimal for a subset $T\,?$

No, aren't we trying to prove that?

user116211

@0celo7 sure, go on; got it.

4:51 AM

@MAFIA36790 Assuming $x\neq z$, we have $z\prec x$ by definition of order. In particular, $z\preceq x$. Thus $z=x$ by definition of $x$ (see above).

Wait, not completely.

We could have $x\prec z$.

But then $x=z$.

Either way, we get uniqueness.

@BernardMeurer Cool :-)

Crap. John's up, time for me to leave.

It has nothing to do with his smell, I have to sleep.

NeuroFuzzy

5:11 AM

@0celo7 Just had to use Arnold's classical mechanics as a reference for drawing the phase flows of cats

user228700

@0celo7 I saw ur question about provability.

NeuroFuzzy

it must be a good book after all

user228700

First question: Are u allowed to use calculators? Second question: Seriously, in which year of university are u?

user116211

@0celo7 Okay, i see the very definition implies the uniqueness.

user116211

@0celo7 Wait, at what price did you grab it?

user116211

5:15 AM

Is it this:

user116211

amazon.com/Algebraic-Cambridge-Studies-Advanced-Mathematics/dp/…

user116211

?

user116211

I didn't know you are interested in algebraic number theory.

user228700

5:29 AM

Hello everyone else :-)

DanielSank

@0celo7 did you figure out your probability issue?

@KaumudiHarikumar Hi.

user116211

0celo went to sleep early ;/

user116211

@0celo7: I've taken $T= S\,.$ That clears the proof.

user116211

@DanielSank o/

DanielSank

@MAFIA36790 \o

I have a photo that I think is really cool from a wedding yesterday:

It's not every day you see physicists in formal wear.

user116211

5:44 AM

@DanielSank Now this is what I call a badass groupie ;P

DanielSank

For the curious: Hartmut Neven, Eddie Farhi, John Martinis, Daniel Sank, Alireza Shabani, Ted White, Dvir Kafri, Ryan Babbush, Julian Kelly, Sergio Boixo, Rami Barends, Austin Fowler, Damian Steiger

user116211

@DanielSank Aha!

user228700

@DanielSank He got an answer of 0.0058 something, I think and I think it's correct.

user228700

@MAFIA36790 Agreed.

knzhou

6:28 AM

@DanielSank @EmilioPisanty Hey, I'm pinging you because you do AMO, and I have a sort-of related question about how decoherence works. Does the question make sense? Is there a nice answer to it?

user116211

@knzhou You should also ping @yuggib.

user228700

6:40 AM

Quick question: Angular displacement-vector or not? I've read that infinitesimally small $d\theta$ s are vectors because they obey the commutative law but that finite angular displacements do not, and hence, are not vectors...

user228700

I'm not able to make much sense of this. How do we measure the direction of this vector? Clockwise vs. Anticlockwise..?

user228700

Please help if anyone is free :-)

user116211

@KaumudiHarikumar They are not vectors.

user116211

For an informal explanation, open your Resnik, Halliday; it is explicitly written in Angular Displacement subsection.

user116211

@KaumudiHarikumar Since you are not new here, let me tell you this; you don't have to write this every time to get help; anyone interested would respond to your query.

peterh

6:48 AM

@ACuriousMind I am still very curious to your vision about the site, and I still hold my old statement that I could support your stance if you would have a clear one. Unfortunately, you meta posts aren't really descriptive in this sense, although I've learned a lot from them.

@ACuriousMind Of course our communication would be much easier if you wouldn't cling to this childish ignore thing. Despite our obvious disagreement about the optimal review mechanism, I still think your posts are irreplaceable and I honor your work here. Kindly regards, peterh

@KaumudiHarikumar There is indeed a way to formalize that notion, though I don't really have enough time to go into it in depth.

In short, what you refer to as "angular displacements" is really just a rotation.

The set of all rotations is a group, not a vector space (and a non-commutative group, at that)

user116211

@EmilioPisanty There are posts already on it in Phys.SE.

user116211

@EmilioPisanty yup.

@MAFIA36790 In which case you might be kind enough to provide those references to @Kaumudi instead of having a go at me.

user116211

@EmilioPisanty sure.

Q: If angular velocity & angular acceleration are vectors, why not angular displacement?

6:55 AM

@KaumudiHarikumar This set of all rotations is smooth (in a suitably defined way) so it can locally be approximated by a flat space.

That flat space (the "Lie algebra" of the group) is where the vector properties show up.

The "direction" is the axis of rotation, and the magnitude is the angle.

user116211

3

Are angular quantities vector? ... It is not easy to get used to representing angular quantities as vectors. We instinctively expect that something should be moving along the direction of a vector. That is not the case here. Instead, something(the rigid body) is rotating around the direction...

rotational-kinematics

user116211

12

Q: How is it that angular velocities are vectors, while rotations aren't?

Does anyone have an intuitive explanation of why this is the case?

vectors group-theory rotation rotational-kinematics angular-velocity

user116211

3

Q: Can an angle be defined as a vector?

In Classical Mechanics angular velocity, angular acceleration, torque and angular momentum can be defined as vectors with clear advantages such as the possibility to use vector product to simplify expressions. As someone who appreciates the symmetry between translational and rotational dynamics ...

vectors rotation geometry conventions angular-velocity

user116211

Related: Euler's Axis-Rotation Theorem:

user116211

1

A: Understanding Euler's rotation theorem

Let us say you have a sequence of rotations about two axes, $\hat{x}$ and $\hat{y}$ by the angles $\varphi$ and $\theta$. You use rotation matrices to find the final orientation $$ \mathtt{E} = {\rm Rot}(\hat{x},\varphi)\,{\rm Rot}(\hat{y},\theta) $$ Now lets add some motion, and give the angle...

user116211

7:01 AM

3

Q: What's a pseudo-rotation?

I'm sorry for this lexical, probably extremely elementary, question. But what is a pseudo-rotation? I just read this term for the first time, in the beginning of the 4th chapter book of CFT by Di Francesco & al. I would say it may be an hyperbolic rotation or a rotation followed by a parity opera...

terminology geometry definition group-theory rotation

user116211

The last one is not fully relevant to your discussion, though.

user116211

Anyways, as I said, there are many posts on or related to this in Phys.SE.

user228700

7:16 AM

@EmilioPisanty OK, erm, I didn't fully understand all of that tbh.

user228700

But thank you. I will look it up some more :-)

user228700

@MAFIA36790 Thanks for these links. I will go through them :-)

user228700

@MAFIA36790 Yes, I know but the answerers are people, not machines-I just want to be friendly; after all, nobody has any obligation to help me or anything.

user228700

I can see how this is redundant when posting actual questions but not here, in the chat.

user116211

@KaumudiHarikumar That's why I said interested.

user116211

7:19 AM

But surely that's your call.

user228700

@MAFIA36790 OK :-) I'ma keep doing it.

Shing

7:47 AM

I am thinking the physics interpretation of $sin(x^2 - t^2)$ (if there is one). Is it a wave function (classical) with one source? (like keep hitting the water surface with a pen)

um... please pretend the dimension is correct...

Shing

8:18 AM

never mind, I asked a dump question. it is obvious a bad thing to describe "hitting the water surface with a pen" (unless the freq decreases with time)

9:33 AM

@KaumudiHarikumar Have a look at Mafia's links and their Related and Linked questions. If you're still confused, ask a separate question and feel free to ping me here about it. I'm travelling so my access will be spotty but I'll try and find time.

A: Cosmic Microwave Background seen from a hypothetical foreign Galaxy?

This is a deliberate piss take shirley? No-one could seriously believe this.

-1

The CMBR at present emanates from the remnant of the Primordial Universe which is a spherical Iron anulus at the centre of the Universe. The current temperature of the Primordial Universe remnant is 3K and is cooling towards Absolute Zero when the CMBR will cease. The CMBR is produced when free ...

user116211

@JohnRennie I never got why people leave a signature at the end of their posts....

https://www.facebook.com/peopleareawesome/videos/1096388957076948/
The joy of (basically) comoving frames

9:50 AM

@MAFIA36790 you're looking for rational behaviour from the person who posted that answer???

user228700

@EmilioPisanty I decided to abandon my textbook for a bit and read Resnick Halliday and Walker instead. After I finish, I will get to those links and then, if I am still confused, I will ping you. Thank you :-)

Swapnil Das

Hello.

Hi Swapnil.

Why do you think the lowest energy state would be one in which mass is zero?

user228700

10:06 AM

Hm. Interesting. Looking forward to this discussion :-)

user116211

@JohnRennie Hmm, I'm not talking specifically with this troll in the mind; I've encountered many first posts, irrespective of the quality of the post, have the signature of the poster. I mean what mentality would allow one to do this? This seems to be quite silly to me.

user116211

@KaumudiHarikumar As I mentioned earlier, there is a pretty good informal brief explanation there.

user228700

10:36 AM

I've gone through lo those many links and I definitely understand the answer to my question much better but my brain has a new question now (:P)

user228700

I don't quite understand why we define vectors for angular velocity and acceleration in this absurd manner; clockwise vs. anticlockwise ie.

user116211

@KaumudiHarikumar So, what do you they should be defined to be?

user228700

@MAFIA36790 I don't understand the need to define vectors for rotational properties of bodies, such as velocity and acceleration...

user116211

O.o

user228700

@JohnRennie: Erm, thoughts..?

10:41 AM

Because infintesimally speaking, rotation is a linear transformation. In simpler terms, it can be treated as linear motion in the infintesimal level.

this is why notions such as angular velocity and angular acceleration make sense

You also need these to be vectors, because the rotating plane itself has an orientation in space, and that clockwse anticlockwise convention and the vector notion allow you to denote that orientation

user228700

@Secret "Orientation of the rotating plane"..?

Consider the following diagram

You have different objects rotating anticlockwise ad clockwise. But note how they are pointing in space

user228700

@Secret Hm, yes, I see ur point.

Thus rotation has both a direction and an orientation. One way to specify all of this is to take a vector normal to it, and then using the anticlockwise and clockwise covention to specify whether that vector shoudl be pointing up or down the plane

user228700

@Secret Hm, OK. So vectors in rotational motion are not used in the traditional sense(as relating to linear motion I mean) yeah..?

10:50 AM

Well the above example is one way to denote some rotating quantity (such as angular momentum). Using this one can use the usual vector sum rules to add up angular momenta together

As for the type concern about linear motion, the best place to start is to consider circular motion

user228700

@Secret Yeah, that's what I'm learning now...

Consider the scenario where a ball attached to a rope is being flung around in a circle. You get something like this

Now, it seems nothing looks very linear and easy to deal with at all. Physicists likes to consider infintesimal quantities, in order to see what happens. Now consider

s.harp

Note that it is only in $\mathbb R^3$ that a vector can define a rotation "along its axis", and even more of a coincidence that every rotation in $\mathbb R^3$ can be given by such a vector :)
"basic rotations" are determined in an oriented plane, one of the coincidences is that in $\mathbb R^3$ planes have "codimension" $3-2=1$, and every oriented plane is determined by a unit vector

Sujith Sizon

can anyone here please help me with a norton circuit problem

user228700

@Secret OK...

10:55 AM

@s.harp Thanks for noting, however for the high school level, we will stick to $\mathbb{R}^3$ and postpone the mentioning about more general things later

user228700

@s.harp Didn't understand anything, sorry :/

Now suppose I take an infintesimal slice of this circle. Since it is so small, the sector is basically a triangle. Now from your high school maths, one can easily find the area of this thin piece of triangle

as $\frac{1}{2}r^2d\theta$

and the arc length is given by $rd\theta$

user228700

@Secret Yeah, OK...

Denote this arc length by $ds$, then you basically have $ds=rd\theta$

Now (I will show you the more mathematically rigorous version shortly), since this $d\theta$ is swepted out at an infintesimal interval of time, we can formally "divide both sides by $dt$ to obtain our first formula: $v=\frac{ds}{dt}=r\frac{d\theta}{dt}$

Therefore the velocity of the ball is determined by the infintesimal angle it traced out at infintesimal time

Now more rigorously, one can start with the known equation for arc length $s=r\theta$

ACuriousMind

nevermind me, I should wake up before writing anything :P

11:08 AM

Now using the diagram, it is easily seen that the ball remains at a fixed distance away from the centre. Hence $r$ is independent of time. Therefore differentiating both sides of the equation by $t$, only $s$ and $\theta$ are time dependent quantities. Hence you arrive at the same equation above

And now, the angular velocity, which is the rate of change of the angle being swept out, can be defined as $\omega=\frac{d\theta}{dt}$

user228700

@Secret It seems that ur messages aren't being displayed on time for me :/ So, sorry for the lack of response but please go on...

user228700

@Secret Right, OK...

Now if you differentiate this equation by t again, you get $\frac{dv}{dt}=r\frac{d^2 \theta}{dt^2}$. We can then call the $\frac{d^2 \theta}{dt^2}$ as angular acceleration $\alpha$. Now since we knew that $\frac{dv}{dt}=a$ is the acceleration, this give us the relation between the magnitude of acceleration and angular acceleration in circular motion $a=r\alpha$

user228700

@Secret OK...

As for the derivation for centripetal acceleration... unfortunately my memory slipped thus someone else have to fill this in for you as I went to revise it

user228700

11:16 AM

Before u proceed, I'd like to ask; these linear quantities, they're vectors too? Like, the acceleration $a$ that u've related as $a=r\alpha$?

In the full treatment, yes they are vectors. In particular, an important result is that the centripetal acceleration and the tangential velocity are always perpendicular to each ohher in circular motion

user228700

@Secret Yeah, OK...

The a above, is tangential acceleration, which controls how much v is changing at a unit itmie

For the simplest circular motion, this a is set to zero so that it is a uniform circular motion

(i.e. the magnitude ofthe tangential velocity is not changed, only its direction does

user228700

@Secret I see. I'm slightly confused about the direction of these vectors and how they are relating to the angular vectors. I mean, since the notion of vector itself is different for rotational motion and linear motion...

user116211

@yuggib o/

yuggib

11:22 AM

\o

user228700

11:49 AM

BTW, any "The Mountain Goats" fans here?

Swapnil Das

11:59 AM

@johnre

@johnRennie that's because mass is a form of energy

I am damn confused now, to be honest. Why is nature so partial?

ACuriousMind

@SwapnilDas I think what you're confused about is what "Systems tend to occupy states of lowest energy" actually means. By conservation of energy, the total energy is always the same. In classical mechanics, this principle should more properly read "Systems tend to occupy states of lowest potential energy" - there's no principle that systems move toward a particular amount of kinetic energy, for instance.

When you say "mass is a form of energy", that's in some sense ($E=mc^2$) correct, but it's completely irrelevant for the idea that systems move towards states of lowest potential.

Caption: Electromagnetism is interesting to me whenever things started to go around in circles...

This problem is a lot more symmetric for a circular ring than a sphere owing to the hairy ball theorem which said I cannot comb a sphere but can comb a circle

(Might have to ask Engg SE later about this. I wonder what it feels like to put my hand on such object...?)

12:23 PM

@NeuroFuzzy I've seen that cat in two places, don't remember where the other place was though.

12:49 PM

Hmm, I might be able to generate it easily. I only need to subject a polystyrene ball in the middle of a circular electric field produced by an AC current passing through a coil. Let's see...

John Duffield

@MAFIA36790 : As you are to joel savory, so am I to you.

@Secret : however you can comb a torus.

Aaron Abraham

Thoughts on this......(something bothers me about it....but I can't seem to put a finger on it....)

John Duffield

@JohnDuffield Hmm, in that case, all I need now is a stable source of a circular electric field. The issue is that circular electric fields can only be produced by a changing magnetic field thus I need to change it in a certain way and avoid all the radial components from the coil itself

ACuriousMind

@Secret What are you talking about? If you want to make a charge go round in circles, we've known how to do that for a long time, it's called a synchrotron.

John Duffield

12:59 PM

@Secret : the field is the electromagnetic field. Electromagnetic field interactions result in linear and/or rotational motion. When we only see the former because rotational forces cancel, we talk about an electric field. When we only see the latter because linear forces cancel, we talk about a magnetic field. But in both cases we are dealing with electromagnetic fields interacting.