The h Bar

DanielSank

00:02

Hey guys, you know how e.g. LIGO is listed as an entity for paper author lists?

Same thing with CERN etc. etc.

Are those registered somewhere official?

My group basically needs to do this.

Emilio Pisanty

00:18

@DanielSank ooooh, good question, no idea

the APS submission form has an entry for when you want to submit as a collaboration

maybe try and see what happens?

DanielSank

00:45

@EmilioPisanty yeah

rob

01:31

@DanielSank For conferences I think it's more or less free-form ... I've submitted several APS papers as "rob for the [whatever] collaboration" and there was never any automated effort to confirm that the collaboration existed.

@DanielSank No mention of a collaborations database in the author guidelines.

user228700

02:38

Hey @SirCumference, watch this video and more importantly read the comments:

user228700

My Post-Election Thoughts

►

Obliv

03:12

anyone know how our eyes are able to process 'images'? Specifically, how do they know to treat converging light waves as images and non-converging light rays as useless? Please let me know if my understanding is misguided/wrong.

rob

@Obliv A "focusing" lens takes all the light that enters it from a particular direction and sends that light through a particular point in space.

Your eye has a network of light-sensitive nerve cells one focal length behind the lens.

So each nerve cell on your retina only gets excited by light coming from a particular direction.

Obliv

@rob once the nerve cell becomes excited, I imagine it sends some sort of signal to the brain at that point?

rob

@Obliv Yep. It's the same topology as "something is touching my left ring finger", but instead it's "blue light is coming from the right edge of my field of view"

Obliv

@rob you know so much...

the topic of sensory input & processing is really interesting to me for some reason

rob

@Obliv That particular example is part of an astronomy lab that I do. The students project images of the world outside the window onto a wall to measure the focal lengths of some lenses, then build a little telescope with them.

@Obliv I know just what you mean ... I get very strong feelings from sensory input

Obliv

03:23

these puns are overwhelming

GPhys

03:34

okay

because I hate myself I wrote that n body simualtor

Obliv

03:58

@GPhys what is it like

GPhys

slow

Obliv

lol

Why did you make it?

Also, do they collide with each other?

GPhys

pffft particles can't collide with each other

that's ridiculous

Obliv

is that sarcasm I'm too ill-informed to tell :p

GPhys

I feel like hoping these will magically pick a rotation direction in the timeframe I'm able to simulate is wishful thinking

user228700

04:10

I've a quick question for anybody here who might know some organic chemistry. Whilst drawing resonance structures, why is that structures that have smaller separation of opposite charges are more stable than those structures in which opposite charges are delocalized on atoms that are farther away from each other?
My textbook explains this by saying that it requires energy to separate opposite charges...but why?

user228700

Googling has not yielded any answers :-|

David Z

04:32

@Sanya no worries, I and other people will look at whatever you post whenever you decide to post it. You're not on a deadline or anything.

user228700

04:50

I've some questions about functions...Firstly, my textbook says:

user228700

> "If a set $A$ contains $n$ distinct elements, then the number of different functions defined from $A→A$ is $n^n$, out of which $n!$ are one-one"

user228700

Can anybody pls help me to prove these results?

dmckee

05:14

@Kaumudi In the first case there $n$ possible choices for each of the $n$ elements in the range. In the second you are doing permutations (because each element in the range must appear exactly once); so $n$ choices for the first $n-1$ for the second and so on.

Wrapping it formal language is (at least for me) harder than figuring our what is going on.

user228700

Ah, OK...

user228700

I will proceed from here, thanks very much :-)

user228700

I have another question.

user228700

My textbook says:

user228700

> "The composite of two bijections is a bijection off $f$ and $g$ are two bijections such that $gof$ is defined, then $gof$ is also a bijection only when codomain of $f$ is equal to the domain of $g$"

user228700

05:23

The wording is a bit confusing, but why is it that the codomain of $f$ must be equal to the domain of $g$?

user228700

I'm unable to figure out what the condition should be. For any one-one function $f(x)$, isn't it bijective if its range=its codomain?

user116211

05:43

@Kaumudi Are $f$ and $g$ bijective?

user228700

@MAFIA36790 :Yes, I have mentioned this...

user116211

Can you show if $g\circ f$ is onto, $g$ is onto?

user228700

@MAFIA36790 I don't think I can :-|

user116211

Use the definition.

user116211

You are half-way done if you can show this.

user228700

05:55

Ermf.

user116211

06:09

@DanielSank o/

DanielSank

@MAFIA36790 \o

Sir Cumference

Trump lost the popular vote

Still won the electoral vote

user116211

@SirCumference Popular vote is meaningless.

Sir Cumference

@MAFIA36790 That's the fucked up part

user116211

@DanielSank, Could you help me comprehend some linear algebra notation?

user116211

06:19

@SirCumference Accept it, Trump is The One.

Sir Cumference

@MAFIA36790 He's fitting for a country that doesn't take the people into account

user116211

@SirCumference Those people chose him.

Sir Cumference

@MAFIA36790 Not the majority of voters

DanielSank

@MAFIA36790 Probably

user116211

Whatever; he is not the president of US.

Sir Cumference

06:20

@MAFIA36790 He will be

user116211

@DanielSank :)

user116211

@SirCumference okay.

DanielSank

@SirCumference Can we have a rule: no talking about the election in hbar for a month?

Just make another room.

Otherwise this is going to get very repetitive very quickly.

Sir Cumference

@DanielSank I'm against that, we can be open to plenty of different discussions

Election talk might get repetitive but it's no less relevant than some of the other nonsense brought up here

DanielSank

@SirCumference Dude, I've made the same suggestion when people were taking up huge amounts of the chat talking about video games.

Sir Cumference

06:22

@DanielSank I don't play games, but is that a problem?

DanielSank

It's annoying.

That's the problem.

Sir Cumference

To you?

user116211

Anyways, I was reading about how the set of column matrices of an $n\times n$ invertible matrix over field $F$ spans all the column matrices $F^{n\times 1}\,.$

DanielSank

I'm not playing this game with you @SirCumference.

I'll just use the ignore feature, temporarily.

Sir Cumference

@DanielSank I'm not being pedantic

DanielSank

06:23

No?

Sir Cumference

I just think it's going to die out soon enough on its own

DanielSank

@MAFIA36790 Yeah ok.

user116211

Let $P$ be the concerned invertible $n\times n$ matrix.

DanielSank

@MAFIA36790 ok

@MAFIA36790 Do you understand geometrically why that's true?

It's actually quite intuitive!

user116211

Then $P_1, P_2,\ldots,P_n$ forms the basis of the concerned vector space.

user116211

06:24

$P_i$ is the $i$th column of $P\,.$

user116211

@DanielSank Well, the author proceeded to prove it but mentioned no geometrical significance, I fear.

user116211

Anyways, then he wrote the product $PX$ where $X$ is a column matrix as $$PX = x_1P_1 + \ldots + x_nP_n\,.$$

user116211

This is what I'm not getting.

user116211

Won't the product be $P_1x_1 + \ldots+ P_nx_n\,?$

Dvij

Hello. Does anyone know why people (even physicists) refer to Weak interaction as the fourth type of interaction along with Gravitational, Electromagnetic, and Strong? I mean isn't the electro-weak unification a standard established piece of today;s Physics?

user116211

06:32

What is going is this:

user116211

$$\begin{bmatrix}P_1 & \ldots& P_n\end{bmatrix}\cdot \begin{bmatrix}x_1 \\ \vdots \\ x_n\end{bmatrix}\,.$$

user116211

@DanielSank, any insight?

DanielSank

@MAFIA36790 How does that differ from the author's expression?

user116211

hmm.

user116211

I don't know but why he would write in that way.

DanielSank

06:33

@Dvij History. Lots of terminology used in science is very stupid if you think about it, but people use it because of history.

@MAFIA36790 How is $x_1 P_1$ different from $P_1 x_1$?

Scalar multiplication is commutative.

user116211

Okay!

user116211

I forgot $x$ is scalar.

DanielSank

@MAFIA36790 let me know if you want to talk about the geometrical way of thinking about matrices.

user116211

@DanielSank, Isn't every element of a matrix an element of a field? I mean field of complex numbers; they are scalars; aren't they?

DanielSank

@MAFIA36790 Yes.

user116211

06:41

Suppose $AX = Y\,.$ Then $y_i = A_{i1}x_1 + \ldots+ A_{in}x_n\,.$

user116211

You can't write $y_i = x_1A_{i1} +\ldots+ x_n A_{in},$ can you?

DanielSank

@MAFIA36790 Of course you can.

$$A = \left( \begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array} \right)$$

$$X = \left( \begin{array}{cc} 2 & 4 \\ 8 & 16 \end{array} \right)$$

$$y_1 = 3 \cdot 2 + 5 \cdot 8 = 2 \cdot 3 + 8 \cdot 5$$

@MAFIA36790 Wait a second.

What does $y_i$ mean?

Your notation is inconsistent.

user116211

$i$th row of $Y\,.$

user116211

@DanielSank Sorry, I should have mentioned that.

DanielSank

What is x_1?

user116211

06:49

$i$th row of $X\,.$

user116211

@JohnRennie morning.

John Rennie

Morning

DanielSank

The statement to remember is $$Y_{nm} = \sum_i A_{ni} X_{im}$$

user116211

@DanielSank Correct.

DanielSank

That defines matrix multiplication.

user116211

06:50

yes.

DanielSank

soooo

I don't understand your question.

user116211

In a nutshell, I'm surprised as the author switched the position like saying $Y_{nm} =\displaystyle \sum_i X_{im}A_{ni}\,.$

user116211

Matrix multiplication is not commutative, as everyone knows.

DanielSank

@MAFIA36790 If that's all the author did, it's just silly, but not wrong.

$$X_{im} A_{ni} = A_{ni} X_{im}$$

user228700

Man, I do not get springs >.<

Secret

07:05

[Example of circular logic] I want the feeling of hopelessness to feel the pain of the feeling of hopelessness itself, so it will stop bothering me by making me to feel hopeless, which is painful.

user116211

@DanielSank okay.

DanielSank

@Kaumudi Hi.

Bonjour, Monsieur LeRennie.

That "R" sound is just delicious.

user116211

@DanielSank yes, I would surely love to; but let me complete what the author has to say.

user228700

@DanielSank Ello :-)

user116211

Got it; that's easy.

user116211

07:16

If you have time @DanielSank, you can go.

user116211

They first proved, the set is linearly independent; then they showed why it spans $F^{n\times 1}\,.$

John Rennie

@Kaumudi Hookes law? You've done much harder stuff than that ...

DanielSank

@MAFIA36790 Imagine you live in a 2D plane.

user116211

done.

DanielSank

On that plane there is a set of coordinate axes.

user116211

07:22

okay.

user228700

@JohnRennie Hooke's law isn't so bad. It's when I've to deal with springs attached to blocks, performing S.H.M >.<

DanielSank

Those axes could be thought of as a set of vectors $$\left( \begin{array}{c} 1 \\ 0 \end{array} \right) \qquad \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$

user116211

okay.

John Rennie

@Kaumudi SHM is easy, and the spring is just an excuse to introduce the type of force required for SHM.

DanielSank

Now, suppose you have a matrix $A$ with components $A_{ij}$.

What do you get if you act $A$ on the first coordinate (basis) vector?

user228700

07:24

@JohnRennie Yeah, but I'm finding it difficult to internalize everything that's happening when there are multiple springs and blocks and what not.

DanielSank

@MAFIA36790 you dead?

user228700

It's very interesting and all, but it's a little difficult :-|

user228700

@JohnRennie: Are u especially busy at the moment? (Or are u just drinking coffee and chilling? :-P)

user116211

@DanielSank sorry, connectivity interruption.

user116211

@DanielSank Order of $A\,?$

John Rennie

07:32

@Kaumudi no. Are you still perplexed by springs? :-)

DanielSank

@MAFIA36790 huh?

user116211

I mean $m\times n\,.$

user228700

Yep yep yep. I've an actual problem (and a crappy solution :-|) Maybe it will help to see how it should be approached..? (By superhuman geniuses like urself :-P)

DanielSank

@MAFIA36790 $2 \times 2$.

user116211

okay.

John Rennie

07:34

Ok let's see the problem ...

Secret

0

Q: Is it possible for $\nabla f=y\mathbf{i}-x\mathbf{j}$

is it possible for a function $f(x,y)$ satisfying $\nabla f=y\mathbf{i}-x\mathbf{j}$ ? Since $\frac{\partial^2 f}{\partial x\partial y}=-1$ and $\frac{\partial^2 f}{\partial y\partial x}=1$, thus $f$ should be singular. I find a similar answer in wikipedia "Symmetry of second derivatives", $f(...

Off to MSE with you

DanielSank

@MAFIA36790 I'm going to sleep, unless you want to go through this now.

user116211

$(A\textrm{Basis}_1)_{11} = A_{11}\cdot 1 + A_{12} \cdot 0 $

user228700

John Rennie

That looks straightforward ...

user116211

07:38

$\begin{bmatrix}A_{11}\\ A_{21}\end{bmatrix}$

user228700

Of course :-P

John Rennie

The system is symmetrical about the mid point, so you need only consider the motion of one block.

user228700

( ^ Sarcasm)

Anonymous

@JohnRennie did you mean centre of mass ?

Anonymous

or literally mid point ?

John Rennie

07:40

The mid point is the centre of mass

user116211

Am I right @DanielSank?

user228700

But $m_1≠m_2$

Anonymous

@JohnRennie But the masses are different

John Rennie

@S007 Oops, well spotted, I assumed the masses were identical :-)

Anonymous

@JohnRennie :-)

John Rennie

07:42

@Kaumudi as S007 says, find the centre of mass and put your origin at that point.

DanielSank

@MAFIA36790 Yeah

Right. What about for the second basis vector?

John Rennie

@Kaumudi I have something to do for the next ten minutes or so, but if you don't mind waiting I can go through the detail.

user228700

@JohnRennie I'll try and wrap my head around it while I wait...

user116211

$\begin{bmatrix}A_{12}\\ A_{22}\end{bmatrix}$

DanielSank

Very good.

user228700

07:44

Perhaps we should continue later...I need to do something else in 15 mins :-|

DanielSank

Notice this simple fact: The $i^\text{th}$ column of a matrix is simply the vector you get by acting that matrix on the $i^\text{th}$ basis vector.

user116211

yes!

Anonymous

@Kaumudi The question is simple....see this youtube.com/… will get it

DanielSank

Ok, so, think about what happens if we act $A$ on our pair of basis vectors.

We get two new vectors.

user116211

The sum of which is ....

DanielSank

07:45

Don't worry about the sum.

user116211

okay.

DanielSank

The point is that we start with two orthogonal vectors, and we wind up with two new vectors.

user116211

yes.

DanielSank

Now, what does it mean for $A$ to be invertible?

user116211

$A$ is row-equivalent to $2\times 2$ identity matrix?

DanielSank

07:46

$A$ is arbitrary.

If $A$ is invertible, then given any vector $v$ in the 2D plane, there is some vector $u$ such that $Au = v$.

@MAFIA36790 No, $A$ is not the identity.

user116211

yes.

DanielSank

$A$ is something arbitrary.

user116211

gotcha.

DanielSank

Ok, now...

Let's call the two basis vectors $e_1$ and $e_2$.

user116211

okay.

DanielSank

07:49

Suppose $v$ can be written as a sum of $Ae_1$ and $Ae_2$.

So, e.g. $v = \alpha A e_1 + \beta A e_2$.

user116211

yeh.

DanielSank

In that case, can you tell me what the inverse of $v$ is?

Uh, not the inverse... I mean the thing that $A$ turns into $v$.

"Inverse" is the wrong word.

user116211

$\alpha e_1 + \beta e_2\,?$

DanielSank

Exactly

So, it seems like we can always find some $u$ such that $Au = v$. All we do is write $v$ in terms of $Ae_1$ and $Ae_2$, and then the value of $u$ is obvious.

Right?

user116211

yeh, sure.

DanielSank

07:52

Ok! Now if that were true that we could always do this, then every matrix $A$ would be invertible.

However, what happens if, for some $v$ we cannot write $v$ as a sum of $Ae_1$ and $Ae_2$?

user116211

yes, there is always a $u$ for which the relation, above, is true.

DanielSank

For example, what if $Ae_1$ and $Ae_2$ are parallel?

Then there is a direction in which we could point $v$ such that $v$ cannot be expressed as a sum of $Ae_1$ and $Ae_2$.

...and so our strategy fails.

user116211

Then our assumption should be wrong.

user116211

@DanielSank yes.

DanielSank

So we see the following: we can't invert $A$ if $A$ takes the two basis vectors and makes them parallel.

Think of $e_1$ and $e_2$ as cutting out a square in the 2D plane.

user116211

07:56

yes.

DanielSank

Another way to say what I said above just now is "If the shape cut out by $Ae_1$ and $Ae_2$ has nonzero area, then $A$ is invertible".

Mew

@heather, what is it?

@S007, let me know when you're going to behave sensibly and we can chat

DanielSank

@Mew Yeah, that's a useful tone which is likely to encourage everyone to get along and make progress.

Mew

@DanielSank, he spammed the new site with foul language and offensive posts

most of it has been hidden by moderators

DanielSank

@Mew ok.

Mew

07:59

I don't think I need to be polite to him when he's showing this behaviour

DanielSank

Still not seeing how sarcasm is going to help.

@Mew I strongly disagree. This site has a "be nice" policy.

Mew

@DanielSank, even to those who are trying to destroy it?

DanielSank

This site's rules don't care about users' activities on other sites.

user116211

@DanielSank cutting a square? I'm failing to visualise :(

DanielSank

@Mew Yes.

Mew

08:00

well we aren't on the site are we and i'm not being mean

DanielSank

Mew

I'm just saying when he starts being sensibly, we can chat

behaving*

DanielSank

@MAFIA36790 see picture

Mew

and clearly he isn't behaving sensibly

user116211

@DanielSank got it.

DanielSank

08:01

Ok, so we've established that if the shape cut out by the transformed basis vectors has zero area, then the matrix is not invertible.

user116211

@DanielSank Yes, when they are parallel.

DanielSank

Another way to see why is like this: If the transformation enacted by the matrix squashes the basis vectors together, then there's no way to invert because a point on the squashed vectors could have come from either $e_1$ or $e_2$.

Let me know if that makes sense. It's a critical thing to understand.

user116211

Okay, I'm seeing what you wanted to mean by the geometrical significance.

DanielSank

Ok, @MAFIA36790 now there's one last point to make that's very interesting.

user116211

tell tell.

DanielSank

08:03

This whole reasoning works in any dimension. If we take a 3x3 matrix, then we could act it on the x, y, and z basis vectors.

user116211

yes.

user116211

That's very true.

DanielSank

If the action of the matrix turns the xyz cube into something with zero volume, then the matrix is not invertible.

Ok so far?

user116211

@DanielSank okay.

DanielSank

Now to the final point: do you know what a determinant is?

user116211

08:05

yes from high school; but haven't read it rigorously now.

user116211

It is in chapter 5 in Hoffman, Kunze.

user116211

And I'm in chapter 2: Vector Spaces.

DanielSank

ok ok

There is a particular number associated to each matrix called its determinant. Never mind how you compute it for now.

Here's the important thing to remember:

user116211

okay.

DanielSank

The determinant of a matrix is the volume/area/whatever of the geometrical object produced by acting that matrix on the set of basis vectors.

user116211

08:07

okay!

user116211

This is new to me!

DanielSank

SO! If the determinant of a matrix is zero, the matrix is not invertible!

If the determinant of the matrix is not zero, the matrix is invertible.

user116211

Yes, I know that.

DanielSank

But now you know why. It's because the determinant is the volume of the transformation.

user116211

Singular matrices has no inverse matrix.

DanielSank

08:08

If the transformation squashes the volume, then you can't find an inverse for any point in the squashed area.

user116211

@DanielSank yes, I didn't know about the geometrical motivation behind it; now I got the whole point.

DanielSank

cool

Now I need to sleep.

ciao

user116211

Thanks @DanielSank! I'm now getting the picture of what your linear algebra book would have..... Lots of geometrical intuitions and motivations :)

user116211

I'm going to bookmark that.

user116211

@DanielSank o/

John Rennie

08:47

@BernardMeurer: see, it's not just you :-)

Anonymous

@JohnRennie Could you help me with an electromagnetism problem ? :-D

Mark Mitchison

@DanielSank I had a look and the answer makes sense to me, +1. Sorry it took me so long to get round to it, I've been kind of busy finishing off my PhD :)

Anonymous

@JohnRennie I mean I could'nt understand how in the absence of any external magnetic field how potential difference appears

John Rennie

Electrodynamics isn't my no. 1 favourite subject, but I'd guess it's because the rod is effectively in a gravitational field with the acceleration $a = (F_1 - F_2)/M$.

That gravitational field means that the energy of an electron at the top end of the rod is increased by $maL$.

Where that potential energy difference is just the usual expression for gravitational potential energy ina gravitational field with acceleration $a$. Does this make sense so far?

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