Feynman Problem

[ninguen]

I'm reading "6 not-so-easy pieces" by Feynman.

In an early chapter, they're discussing symmetry and rotation. And they present the following image.



Everything here seems to make sense. I understand most of this, but I have no idea where they got the YsinΘ from.

Maybe I'm overlooking some basic geometric property. But I just don't get it. Or maybe I'm just not good at visualizing problems.

 

[mmm4444bot]

The measure of angle BAP is theta. (Try to figure out why.)

sin(theta) = opposite/hypotenuse

In triangle BAP, the hypotenuse is y.

PS: Here is a rhetorical question. Would your opinion be that, when Einstein was puzzled about something, he should have wondered whether he was lousy at visualizations? :cool: (My opinion would be that those types of expressions probably serve no useful purpose.)

 

[ninguen]

The measure of angle BAP is theta. (Try to figure out why.)

sin(theta) = opposite/hypotenuse

In triangle BAP, the hypotenuse is y.

Thanks. Maybe I need a refresher.

According to the book,

X' = XcosΘ + YsinΘ
Y' = YcosΘ - XsinΘ

Can you point me to the geometric theorems that prove these lengths to be correct? I've got a fuzzy understanding, but I'd like to have a more solid foundation before I continue.

 

[ninguen]

I found the answer, and it's not intuitive at all. It's something called a rotation matrix.

[video=youtube;h11ljFJeaLo]http://www.youtube.com/watch?v=h11ljFJeaLo[/video]

Hopefully this will be helpful to someone else who runs into this problem.

I'm going to practice this one a few times until I've got the logic down cold.

 

[mmm4444bot]

The rotation matrix is not needed to answer your original question; basic geometry and trigonometry are good enough.

The rotation matrix is a linear-algebra tool for quickly converting coordinates in one system to another (rotated) system.

Cheers ~ Mark

 

[ninguen]

The rotation matrix is not needed to answer your original question; basic geometry and trigonometry are good enough.

The rotation matrix is a linear-algebra tool for quickly converting coordinates in one system to another (rotated) system.

Cheers ~ Mark

Now I'm all caught-up and ready to tackle the next few chapters of the book. Thanks for your help and patience.

 

[Deleted member 4993]

That matrix is valid where the rotation is about z-axis (i.e. z' = z)

or


X' = |cosΘ.......sinΘ.......0|
Y' = |cosΘ... ...-sinΘ.......0|
Z' = | 0............0..........1 |

Now find the rotational matrix where the system rotates Θ about z, Φ about y and µ about x.

 

[ninguen]

That matrix is valid where the rotation is about z-axis (i.e. z' = z)

or


X' = |cosΘ.......sinΘ.......0|
Y' = |cosΘ... ...sinΘ.......0|
Z' = | 0............0..........1 |

Now find the rotational matrix where the system rotates Θ about z, Φ about y and µ about x.

Hmm... identity matrix stuff. I'll get to work on that and see what I come up with. I'll bet it has something to do with Cos^2 + sin^2 = 1.

But I think the x' and y' would be linearly dependent then. So I'd wind up with zeroes across the first 2 rows. I'll get to work on this and let you know. [edit: because the sign is wrong on the second row sin.]

Also, I've figured out how to derrive the Lorentz transformation and apply it to this rotation. (I think)

 

[ninguen]

So I played around and got

(X' + Y')/(2cosΘ) = ..|1.......0.......0|
(Y' - X')/(-sinΘ)= ....|0... ...1.......0|
Z' = ......................| 0......0.......1 |

Does this mean anything? Is it useful for something? Or did I go off in the wrong direction?

 

[ninguen]

That matrix is valid where the rotation is about z-axis (i.e. z' = z)

or


X' = |cosΘ.......sinΘ.......0|
Y' = |cosΘ... ...-sinΘ.......0|
Z' = | 0............0..........1 |

Now find the rotational matrix where the system rotates Θ about z, Φ about y and µ about x.

I've been over-thinking things.

x' = xcosΘ + ysinΘ
y' = ycosΘ - xsinΘ

x' = xcosΦ + zsinΦ
z' = zcosΦ - xsinΦ

y' = ycosµ + zsinµ
z' = zcosµ - ysinµ


Is this it? Did I nail it? (Sorry I didn't put this in linear algebra form)