Mathematics

12:00 AM

today's intellectual high-tide mark for me: realizing complex exponentiation makes more sense on the universal cover of the punctured plane

a close second: proving two point groups defined on an elliptical curve are isomorphic regardless of choice of "base point"

Link

That's great, but I have no idea what that means, since I'm just in calc II :p

David Wheeler

i bet you know what polar coordinates are, right?

Link

yes

David Wheeler

well, the usual way to define exponentiation is $a^b = e^{b\log(a)}$

but that doesn't work so well on the complex plane, because the complex exponential function is not 1-1

so there's not just "one" log function...there's "lots"

Link

okay

David Wheeler

12:09 AM

when we write a point in the plane as (r cos(t), r sin(t))....the problem is with t

if you add an extra dimension for t, this problem goes away, the circles becomes helices

this turns the complex plane into something like a giant spiral ramp, and on this 3-d structure, log is now a single-valued function again

Link

sweet

that's pretty cool

David Wheeler

if you pick a "base range" of interval length 2pi for "level one", the transformation to get from one level to any other (a deck transformation) is just a vertical shift of k*2pi for some integer k, the "winding number" of our circular circuit on a circle in the plane

the elliptic curve thing is harder to explain: but here's a "baby example": say you have a circle, pick a point on it, call it O.

pick two more points, A and B, and draw a line between them.

then define A+B to be the point on the circle where the line parallel to AB through O intersects.

voila! now we can "add points" on a circle!

Link

12:25 AM

That is prettu damn cool

know any physics?

David Wheeler

only a very tiny amount.

Link

well more of a trig question

docs.google.com/file/d/0B9uJ-WmpI8m1SzkySk42UXFhQTg/edit

I need to get the top bar, in terms of the slanted bar

using theta

It ends up with slant/tan theta = top bar

but why is it a tangent?

David Wheeler

hmm....so the weight is suspended from the slanted bar, and attached by something horizontally as well?

Link

well

the weight is halfay up

yea

just like the picture

David Wheeler

do we have any dimensions at all?

Link

12:32 AM

well

here's the problem

it's like a torque problem

so the slanted bar is wiehgtless

so T_cw = T_ccw

then the T_cw = 100*.5 length of slant

and T_ccW = the top bar (T)*trig function(theta)

then T = 50/trig(theta)

but why is the trig function tan?

David Wheeler

ok, torque at what point?

Link

the pin

at little circle?

at bottom left?

thats the torque point

David Wheeler

so where the top bar joins the vertical bar is a rigid connection?

Link

12:49 AM

it's a wire

but essentially rigid

David Wheeler

well i'm confused. suppose the weight is w, and the length of the slant is L. why isn't the T_cw = (wL/2)cos(θ)?

Link

1:04 AM

then the wieght is going down, so it is 1/2 the length away, so it's t_cw=(W*(l/2))

wait

that might be it

you said it's cos(theta)?

why did you say that?

David Wheeler

because i thought only the perpendicular component contributed torque

anorton

Back from dinner... :\

Yes, only the perpendicular component contributes to the torque.

Link

hmm, wait, what?

what perpendicular component?

oh, wait

nvm, waht?

David Wheeler

perpendicular to the slant bar

Link

what's perpeinduclar to that?

David Wheeler

1:17 AM

the angle between "straight down" and perpendicular to the slant bar is also θ, so i get (wL/2)cos(θ) for the force perpendicular to the slant bar from the weight

Link

So, wait

If the angle is like that |\

straight down and perpedicular

the angle in between them?

David Wheeler

the rest of the weight gets transferred ALONG the slanted bar, which doesn't produce any torque (but might push the vertical support sideways, if it's not stiff enough...this produces torque at the base of the upright....but we're not there)

yes, like you drew it

anorton

@Link Not the angle in between. When we say perpendicular, we mean perpendicular to the lever arm.

Link

So, one sec

I'll draw you it

David Wheeler

when he drew |\, i assumed he meant for the "\" to be perpendicular to the slant

Link

1:25 AM

docs.google.com/file/d/0B9uJ-WmpI8m1SzkySk42UXFhQTg/edit

what angle do I use

a b, or c?

anorton

@Link b is equal to theta

Link

okay

why?

David Wheeler

angle b = θ

Link

and why are we using cos theta?

anorton

Hang on... drawing something up real quick

David Wheeler

1:31 AM

draw a right triangle with "base" the perpendicular to the slant piece, "origin" at the point the weight is attached (or the wire its hanging by, whatever), so the horizontal leg is perpendicular to to the slant piece

Link

wait, what?

David Wheeler

the components of the force due to the weight are (wL/2)cos(θ) perpendicular, and (wL/2)sin(θ) in the direction of the slanted bar.

Link

So, think of it as a inclind ramp?

David Wheeler

the angle b is θ because we have similar triangles

anorton

Here:
https://docs.google.com/drawings/d/1BQmQBYX0QZ2GVBoMBkLd7eRRiNGjTfpwOkHauna96Mk/edit

Red is the force of gravity.
Green is the perpendicular component
Blue is a connecting line (just to help visualize).

Why is b = theta? We know that a+theta+90 must equal 180. Note that we have a+b+90 = 180 as well in the upper part of the triange.

David Wheeler

1:37 AM

@anorton nice drawing

anorton

@DavidWheeler thanks! :)

Link

what therom is that though

i know certain angles are the same as each other

David Wheeler

what theorem is what?

Link

with intersecting lines

David Wheeler

if you have a force F at an angle ψ to a reference line....the resolvant components parallel and perpendicular to your reference line are Fcos(ψ) and Fsin(ψ), respectively. in this case, ψ is the complementary angle to θ, so we can swap the sine and cosine when we swap the angles

anorton

1:41 AM

I've added something to the diagram to show why b=$\theta$

Link

thanks

I get it now

anorton

Glad to help.

ok. bye for now--I've got my own physics and engineering hw to do. :)

Link

wait

it's cos theta?