Weird physics question - relativistic immobility

[KrazyHorse]

Is Ben saying retarded **** about relativity again?

[KrazyHorse]

In special relativity, acceleration is not relative. In general relativity it is

I know. That wasn't relevant to the specific point at hand, though.

It was, when you try to claim your own reference frame as evidence that you are always at rest relative to some reference frame despite the fact that you're accelerating. In fact, that's true, but you're at rest relative to a bunch of DIFFERENT reference frames, each for one instant in time.

[KrazyHorse]

It's no surprise that he flunked out of undergrad physics or astro or whatever he claims to have done.

[Ben Kenobi]

Dude, I know more physics than you do. Stop embarrassing yourself. Your response doesn't even make sense in context.

KH already said it before, and in not so many words. c is invarient to all reference frames.

[Ben Kenobi]

It's no surprise that he flunked out of undergrad physics or astro or whatever he claims to have done.

Yeah, I suppose you recall how to do conversions from Spherical to Cartesian coords.

What are the formulas for it? I'd like to convert RA and dec to XYZ.

I failed, yes it's true, but because of the maths, not the astro. Always did fine, 90 percent in my astro courses.

[KrazyHorse]

June 24, 2009, 16:57
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[Ben Kenobi]

Why is the behaviour of isolatitude tesselations nice in the context of spherical harmonics?

Because they conform to platonic solids?

[Kuciwalker]

Yeah, I suppose you recall how to do conversions from Spherical to Cartesian coords.

What are the formulas for it? I'd like to convert RA and dec to XYZ.

I failed, yes it's true, but because of the maths, not the astro. Always did fine, 90 percent in my astro courses.

You flunked because you didn't have an HS-level understanding of trigonometry?

[KrazyHorse]

Did he just ask me if I remember how to write down the most common coordinate transform in the world?

What a ****ing ****.

[Kuciwalker]

Isolatitude means what it sounds like: that lines of latitude run through large numbers of (the centers) of pixels.

In other words, I'll have a bunch of pixels centered at latitude a, then a bunch at latitude b, etc. etc.

If I just tesselated the sphere without thinking too hard then there's no reason that I'd end up with this arrangement. I'd end up with one at latitude a, one at latitude b etc.

Why is the behaviour of isolatitude tesselations nice in the context of spherical harmonics?

[rephrasing what I was trying to say before] don't the latitudes correspond to lines of constant amplitude of your wave function?

[Ben Kenobi]

You flunked because you didn't have an HS-level understanding of trigonometry?

I'm asking a question, oh wise and noble wizards.

[Ben Kenobi]

Did he just ask me if I remember how to write down the most common coordinate transform in the world?

I could look it up, but I was curious if you could recall it off hand, oh wise and noble wizards.

[KrazyHorse]

[rephrasing what I was trying to say before] don't the latitudes correspond to lines of constant amplitude of your wave function?

That would only be true for the Yl0 (that's an L, not a 1) modes....but you're almost there....

[Kuciwalker]

x = r cos theta
y = r sin theta
z = r cos phi

Shall I add 2 and 2 and see what I get, while I'm at it?

[KrazyHorse]

The point is that the spherical harmonics are written as the product of an associated legendre polynomial of the cosine of latitude times trig functions of the longitude. The associated legendre polynomial is the computationally intense part. By arranging to have the number of lines of latitude grow as N^1/2 we can optimize the computer time to compute the spherical harmonic decomposition of the CMB (or anything else, for that matter). The key measurement from the CMB is its power spectrum, and to compute this you first decompose to spherical harmonics.