The h Bar

DanielSank

15:00

@heather, remember, the ith column of a matrix is what that matrix spits out when acting on the ith basis vector...

heather

right

$\begin{bmatrix}-1&0\\0&1\end{bmatrix}$

would be a 90 degree rotation to the left, I think

but then you said as a function of $\theta$

DanielSank

Well, that matrix reverses the sign of the first basis vector, and does nothing to the second.

Kenshin

$\begin{bmatrix}cos \theta&-sin\theta\\sin\theta&cos\theta\end{bmatrix}$

DanielSank

Seems like a reflection about the vertical axis to me.

heather

i know i'm doing this wrong somewhere...

DanielSank

15:01

Hey!

No cheating!

Kenshin

I believe in direct instruction

not cheating, just efficient learning

AccidentalFourierTransform

here, have this \

DanielSank

I believe in Socratic method.

heather

@Kenshin, who is teaching here, you or @DanielSank?

DanielSank

I'm putting you on ignore, @Kenshin. heather should too.

Kenshin

15:02

ok then, why do you believe what you believe @DanielSank?

heather

no offense, but i'd have liked to figure it out myself.

@DanielSank, you can ignore people?

DanielSank

Yah.

Click their avatar thingy.

heather

@Kenshin, this is not a philosophy discussion, this is a request to let us have a discussion without interference.

Kenshin

@heather this is a public chat room

heather

whoa

vanished...

DanielSank

15:03

Anyway heather, can you figure out the matrix?

heather

let me try again

DanielSank

Focus on one column at a time.

heather

okay, i want to take the first basis vector

DanielSank

If we rotate $$\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$$ by some angle $\theta$, what do we get?

Initially, it points along the x axis, but then we rotate it...

AccidentalFourierTransform

(you should have a piece of paper with you, and draw the vectors)

heather

15:04

@AccidentalFourierTransform i do

DanielSank

Oh yeah, I always assume everyone here has paper at hand.

heather

well it depends on how big theta is

it could be rotated to where it is on the y-axis

DanielSank

ok well suppose it's kinda small to get started.

Yes, it could.

But what happens if it's just some angle that's not too big?

heather

well, the vector's just kind of pointing up and to the right

but it's still pretty close to the x-axis.

well, i guess since the original vector's on the x axis you could drop a line down from the tip of the new vector to the tip of the old vector

DanielSank

Yes!

heather

15:07

and you'd have a right triangle

DanielSank

yes!

heather

and then trig and all that

DanielSank

Yes yes yes.

go on

heather

so, hmm

we're trying to find the tip of the new vector

DanielSank

You want to know the x and y components.

heather

15:08

right.

so sine

that'd be opposite hypotenuse

Slereah

Rarita Schwinger stuff is annoying because it's back to basics differential geometry :V

I forgot all that stuff

Normal vectors of characteristic surfaces

DanielSank

@heather Remember that $\sin(0) = 0$.

heather

well, okay, the x-position is the same, right? of the tip of the new vector? because it's a right angle so perpendicular and all that?

so we really just need to find the y component?

DanielSank

If you rotate the x-basis-vector a bit, the resulting x component must decrease a little. You can see that because if you rotate, say, 45 degrees, you only have half of the original x-component.

And obviously the y component increases at first as you start rotating. But then, as you rotate past 90 degrees, the y component starts decreasing again. But anyway, think about the small angle case and you'll get it right.

heather

okay, for the y component, would it be tan(theta)?

because, tan is opposite over adjacent

and the adjacent is the original vector, and we know it has a length of one (?)

so it's really opposite over 1

and the y component is the height of the triangle?

DanielSank

15:15

heather

yeah

DanielSank

Think about the rotated vector. The hypotenuse is length 1.

Adjacent is not length 1 after you rotate it some!

heather

i'm confused, i'm sorry

DanielSank

That's ok.

heather

the adjacent is the original vector

the basis vector

right?

DanielSank

15:16

No it's not.

That's where you're confused.

heather

the adjacent is the x-axis?

DanielSank

It's the x-component of the rotated vector.

heather

what happened to the original vector?

DanielSank

I'm looking for an illustration.

The original vector is the horizontal bit of dotted line between the center of the circle and the circle itself.

It's not moving.

The red thing is the result of rotating the original vector.

Ooooh I found a good picture!

Look at the blue thingy. That's the rotated vector. The original vector is still sitting there on the x axis.

heather

the blue thingy

okay

but wouldn't the adjacent line up with the original vector then?

DanielSank

15:23

Line up, yes. Have the same length, no.

AccidentalFourierTransform

the norm of the vector is always $1$. If upon a rotation its $y$ component increases, its $x$ component must decrease

DanielSank

^ That

I have a feeling @heather has a piece of sand in one of her mental gears that will come out soon...

Probably just confuzzled over the diagram.

AccidentalFourierTransform

heather

oh

AccidentalFourierTransform

maybe?

heather

15:27

i think i see the light

the adjacent is on the x-axis, but it is not the same as the original vector

DanielSank

Correct.

heather

the new vector does have a norm of one

it has to because it's a basis vector

DanielSank

Well, it has to because it's a rotated version of the original basis vector.

Rotation doesn't change length.

heather

that too =)

i just kind of need to forget about the original vector.

and think about the components of the new vector.

DanielSank

yep

heather

15:29

and the hypotenuse does have a length of one

so

DanielSank

yep

heather

you want to have sine and cosine

so it'd be

$\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}$

right?

AccidentalFourierTransform

zacly

DanielSank

Yes.

AccidentalFourierTransform

now plug $\theta=0$ to make sure that the result makes sense

DanielSank

15:31

Now do the other column.

AccidentalFourierTransform

and $\theta=\pi/2$

heather

okay

DanielSank

Yes, as @AccidentalFourierTransform always check a few cases!

heather

well, sin(0) = 0

and cos(0) = 1

so you'd have [1,0]

DanielSank

(@AccidentalFourierTransform I like the way you operate)

heather

15:32

which makes sense

and then pi/2

so sin(pi/2) = 1

AccidentalFourierTransform

(@DanielSank I know, I'm a beautiful human being)

heather

and cos(pi/2) = 0

so that also makes sense!

=D

DanielSank

yay!

Now do the other column.

heather

okay

wait

would it be the same thing?

AccidentalFourierTransform

Analogous, yes. Same, no

heather

15:34

oh...huh. okay

AccidentalFourierTransform

its the same philosophy as before

just a bit of trig

DanielSank

@heather It's not the same because you're starting from $$\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \, . $$

heather

hmm, okay

that makes sense

so then...i don't suppose it'd be the same thing just flipped?

DanielSank

Check some cases and see if it works.

heather

so it works for the original vector

Physics Meta

15:36

0

Q: Secret Hats - what is going on

So what's the big secret? Why would I wear, a hat that, presumably, tells people something about me or my actions, without getting to know what it is that is being said. I'm not prone to paranoia, but how is the secrecy intended to be contributing to the site?

discussion

DanielSank

Well that died quickly.

heather

yup

DanielSank

@heather Ok put in $\pi/2$ or something.

heather

right

so you'd have [1, 0] right?

DanielSank

Yeah.

Is that what you want?

AccidentalFourierTransform

15:39

@heather i.imgur.com/KJyYmaG.png you were faster for a couple of seconds =P

heather

@AccidentalFourierTransform lol =P

@DanielSank i don't think so

DanielSank

Right. What do you want?

What happens when you rotate the y axis $\pi/2$ counterclockwise?

heather

(-1, 0)

AccidentalFourierTransform

If you have some actual understanding of CMB physics, then I would appreciate an answer. If you don't, then please assume that you just don't know enough about the subject matter to comment on either the original question or this one. — Mike Doonsebury 2 mins ago

nice people everywhere!

DanielSank

@heather Right, so...

heather

15:43

so the flipped thing isn't right

DanielSank

It's easiest in my mind to think about a small angle and use the triangles to get everything right.

heather

yeah

i'm gonna redo it

DanielSank

In fact, the green triangle in the picture above will help.

heather

oh!

i was thinking about it in the wrong direction! would it be

$\begin{bmatrix}-\sin(\theta)\\\cos(\theta)\end{bmatrix}$

AccidentalFourierTransform

yep

heather

15:45

=D

i'll plug in a few to double check

so putting in 0, you get (0,1)

AccidentalFourierTransform

therefore, the rotation matrix is...?

heather

which makes sense

putting in pi/2 you get (-1, 0) which is also right

DanielSank

right

Yeah, so your matrix is $$\left[ \begin{array}{cc} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array} \right] \, . $$

heather

yeah

that makes tons of sense

DanielSank

(I typed it because I could do it quick, no disrespect, @heather)

Ok now here's the reason we did this.

AccidentalFourierTransform

15:46

Now show that its a Lie Group =P

heather

@DanielSank, none taken

DanielSank

@AccidentalFourierTransform lol

heather

@AccidentalFourierTransform a what now?

DanielSank

Ignore him.

heather

i'm going to assume that has nothing to do with honesty

=P

DanielSank

15:47

For a minute. Actually what he said was sort of relevant to what I wanted to show you.

@heather Hahahaha it's pronounced "Lee".

AccidentalFourierTransform

actually, showing that rotations compose is fun and interesting

DanielSank

Yeah that's exactly where we're going with this because you can prove trig identities that way!!!

Now @heather, suppose we rotate by some angle $\theta$, and then after that we rotate by a different angle $\phi$.

heather

@DanielSank ::tucks that away in the "don't embarrass myself" file::

@DanielSank would that be composing matrices, by any chance?

or...matrix multiplication?

DanielSank

Indeed it would.

heather

=D

DanielSank

15:48

Call the rotation by $\theta$ $R(\theta)$.

If we start with some vector $|v \rangle$, then rotating by theta is $R(\theta) | v \rangle$.

AccidentalFourierTransform

oh no youre one of those

DanielSank

Then rotating by phi is $R(\phi) R(\theta) |v \rangle$, but matrix multiplication is associative so it is also $$(R(\phi) R(\theta)) | v \rangle$$

In other words, you can multiply the matrices together first and then act the result on the vector, i.e. you multiply the matrices to get the rotation by angle $\phi + \theta$.

heather

okay

DanielSank

Now, dearest @heather, could you please do the matrix multiplication?

heather

sure, one moment

DanielSank

15:50

Why? Well, note that the result of that multiplication will be $R(\theta + \phi)$.

AccidentalFourierTransform

should be

heather

wait, do i need to do any math on paper at all? or is there some trick to this?

AccidentalFourierTransform

lets find out!

heather

sounds good

DanielSank

So, the result must be equal to $$\left[ \begin{array}{cc} \cos(\theta + \phi) & - \sin(\theta + \phi) \\ \sin(\theta + \phi) & \cos(\theta + \phi) \end{array} \right] \, . $$

heather

15:51

seriously? that's a good trick to know.

cool.

DanielSank

But you will get something different, and by equating what you get and what I just wrote, we'll see something very neat.

heather

oh, okay.

DanielSank

@AccidentalFourierTransform were you going to instead use the trig identities to show that the rotations form a group?

(lie group)

I guess I'm coming at it from the point of view that the matrix I just wrote has to be correct, and you can prove the trig identities from that.

ACuriousMind

@DanielSank But why "has" it to be correct?

AccidentalFourierTransform

@DanielSank "assuming that the result is correct, let us prove the hypotheses" :P

ACuriousMind

15:54

There are some people who think rotations "should" be commutative, but they aren't!

AccidentalFourierTransform

@ACuriousMind De Moivre says so

DanielSank

@ACuriousMind Basic logic.

@AccidentalFourierTransform Not really.

@ACuriousMind In 2D they are.

ACuriousMind

@DanielSank I think you mean intuition, not logic. I agree that it is intuitively clear that $R(\theta)R(\phi) = R(\theta + \phi)$, but you have not actually shown, logically, from some axioms, that this is the case.

DanielSank

Indeed not.

That is not my intent.

But it will be rather obvious that it's true once @heather does the matrix multiplication.

So I'm not worried about offending the math gods.

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