Sunset Geometry (2016) | Hacker News Reader

Sunset Geometry (2016)

shapeoperator.com

153 pointsjohndcook30 comments

> I have never seen someone try to use Pauli matrices to solve a trigonometry problem, but it can certainly be done.

The Pauli matrices in this context are isomorphic to quaternions, which certainly have been used to solve geometric problems in 3 dimensions (although not necessarily by physicists), as has been discussed many many times here on HN. The property of describing spin-1/2 particles (i.e. generating SU(2)) is precisely the same property that makes the quaternions amenable for use in reasoning about 3D rotations!

The relationship between 3D rotations, spin-1/2 particles, and unit quaternions appears on one of the best-named Wikipedia pages: Exceptional Isomorphisms! [0]

[0] - https://en.wikipedia.org/wiki/Exceptional_isomorphism#Spin_g...

> describing spin-1/2 particles (i.e. generating SU(2))

This has the relationship backwards. Particles have this spin group because they can rotate in 3-dimensional space, not because they have some mysterious association with the complex plane.

That we can represent rotations using 2x2 unitary matrices of unit determinant is just a coincidence. There are a bunch of such coincidental isomorphism between groups. The explanation is comparable to the “strong law of small numbers” https://en.wikipedia.org/wiki/Strong_law_of_small_numbers

(The unitary matrices come about when we take points on the 2-sphere and stereographically project them onto the plane, representing points in the plane by complex numbers. Then concentric rotations of 3-space of correspond to particular Möbius transformations of the complex plane, which can be represented as the special unitary matrices. This works out cleanly when projecting onto a 2-dimensional plane, but representing rotations gets trickier in higher dimensions.)

* * *

But you are missing Jason’s point. There are clearly many possible ways of representing the same relationships. The underlying relationships don’t change if you e.g. use classical spherical trigonometry. The question he is trying to ask instead is: which representation is most conceptually clear and intuitive to work with (after some experience), has the nicest and easiest to manipulate notation, etc.

There are valid reasons to use a stereographic representation of points on the sphere. (For instance, it involves 2 coordinates instead of 3.) And from there, representing spherical rotations as Möbius transformations is convenient and effective. But conceptually, if trying to solve arbitrary problems with pen and paper, representing points as vectors and rotations as scalar+bivector "quaternions" is a lot more natural. Especially if you have a problem where some parts are not confined to the sphere.

For more on representing spherical geometry stereographically, cf. https://observablehq.com/@jrus/planisphere – there are some tools even here where we get leverage out of treating stereographically projected points as 2D vectors rather than as complex numbers, and separating the concepts of scalar, vector, and bivecor.

Somewhat related: How many of you have ever seen the green flash [0] ? I never have unfortunately even though I've looked for it many times.

[0] https://en.wikipedia.org/wiki/Green_flash

I believe I saw it once, at the beach looking out on the ocean. It was very brief and a long time ago.

> This is 20% larger than the true value, 3960 mi

Whoa, that is not insignificant error. What contributes to an error of this size?

My bet would be refraction of the sunlight through the atmosphere. At sunrise and sunset light that reaches you on the ground is taking just about the longest path through the atmosphere that it can, and gradient of the atmosphere's density is definitely going to add some bending to the light coming in from space.

I was going to say something similar

"we can see the Sun even when it is geometrically just below the horizon, at both sunrise and sunset. This is because of the refraction of the light from the Sun by the Earth's atmosphere--the Earth's atmosphere bends the path of the light so that we see the Sun in a position slightly different from where it really is."

see http://curious.astro.cornell.edu/our-solar-system/52-our-sol...

Yeah, it's significant enough that my astro mount has a "King tracking rate" to account for this effect. Named for Edward King.

https://en.wikipedia.org/wiki/Edward_Skinner_King

I don't think refraction would impact the error as much as a focal length of the camera. Probably both however.

What’s the effect of the focal length?

> What’s the effect of the focal length?

Not sure what the GP intended, but one possible effect is that increasing the focal length ('zooming in') would change the relative size of two objects (one closer, one further away) depending on their relative distance to the camera (I think the closer object might appear relatively larger).

But if the sun and the horizon are the intended objects, they are both effectively at infinity, from the camera's perspective, so I wouldn't expect focal length to change anything in this case.

Changing focal length of a lens ('zooming in') does not change relative size of two objects, it is a common misconception. What changes relative size of objects are your legs when you move from one spot to another to get roughly the same framing with different focal lengths. If you stand in one spot and zoom or change prime lenses relative object sizes stay the same.

Fair point - I should have tried this with an actual camera. But I suspect my line of reasoning still stands, that the focal length doesn't affect the phenomena being discussed here

https://fstoppers.com/architecture/how-lens-compression-and-...

That article specifically clarifies that moving the camera is what causes changes to relative sizes, not the focal length. But in the experiment described, the camera doesn’t move a significant difference with respect to the sun, the camera can’t really get anywhere close enough to the sun or the reflection on the horizon to make any difference, certainly not 20%.

Vanderbei’s linked articles have some discussion of sources of error. The main one seems to be waves making the reflection appear longer.

The moon.

I expect the curvature of the earth has much less effect than refraction and mirage effects in causing this phenomenon.

> If the earth was flat, photographs of the sun setting over water would look like this:

I am curious: has this argument, historally, ever been used against the idea that the world is flat?

The idea that the world is flat is not something that exists due to lack of arguments.

Too much math..

'Flat-earth', i wrote this to give you a smile, a couple of years ago i'd seen a wristwatch -i thought that is must have been made vor classic painters, not using a photoshot or better their own mind, but a sketch drawn outside (in the wild), its display showed a model of the time-dependent shaddow cast, using two overlaying layers which build the shaddow. It took a while and reading till i fiddling out how the time-setting to get a correct shaddow-cast will be done (hypothetical) by using a list on the internet. Hint: 'What may be possible done with building a watch' ^^

Great question. Skimming through the Spherical Earth page on Wikipedia I don't see it, though it could conceivably be among the phenomena listed by e.g. Strabo and not enumerated there.

Sebastian Lague made a fantastic video about approximating/simulating Rayleigh scattering in real-time (using Unity) to simulate sky color and sun sets. Really interesting stuff

https://www.youtube.com/watch?v=DxfEbulyFcY

Does geometric algebra provide an alternative to pseudovectors for representing things like angular velocity?

The fact that you have to flip the sign of pseudovectors sometimes feels like a hint that they aren't the right representation, somehow.

I think angular velocities are precisely the bivectors (which I take to be the grade-2 elements of an exterior algebra or Geometric Algebra). The change of basis is indeed different than for ordinary vectors. The exponential map (the obvious generalisation of e^x) then takes bivectors to the "rotors", which represent rotation operators.

In terms of Lie theory, the rotors are a Lie group and the bivectors are the corresponding Lie algebra.

Yes (to repeat the other reply here more confidently). Pseudovectors are (n-1)-graded vectors in exterior algebra, ie bivectors in 3d and trivectors in 4d. No sign flip is required required if you write them this way. Algebraically, P(x^y) = P(x)^P(y) (basically by definition), where P is a parity transformation.

Yes, a Gibbs/Heaviside “pseudovector” (a.k.a. “axial vector”) is actually a bivector – a magnitude with a planar orientation, rather than linear orientation. Quantities representing a rotation are naturally oriented with a plane of rotation; there is only a unique perpendicular linear “axis” of rotation in 3d space (e.g. in 4D, the subspace fixed under a simple plane-oriented rotation is a perpendicular plane).

When you properly treat bivectors as bivectors instead of cramming both bivectors and vectors down into one type of object, all of the trickiness about sign flips under reflection etc. go away.

Almost every time you see a cross product in a formula, you should really have a wedge product, and the result should be a bivector.

In GA, vector identities all generalize cleanly to any dimension (the trick of representing bivectors as vectors only works in 3D).

See http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf

> When you properly treat bivectors as bivectors instead of cramming both bivectors and vectors down into one type of object, all of the trickiness about sign flips under reflection etc. go away.

Okay, so next question: if you formulate the laws of physics in terms of geometric algebra, does CP violation still exist?

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