Aeschylus

IIRC the disc is dealt with (though more as a discussion) in the Feynman lectures and the part that deals with the disc also appears in "Six Not-So-Easy Pieces".

I'd say pi is a constant, because it's defined by plane geometry (but it's a matter of defintio). For example pi appears in Einstein's fild equations wich don't necessarily describe spacetime with Euclidean spatially slices, but it's still the same number (i.e. the numebr defined by plane geomerty and the trigonmetric functions).

Perhaps some parts of the model had already been propsed by then, I really don't know. The actual word 'quark' comes from a poem by James Joyce.

It's complicatd, but space in general is not Euclidean (it's exact geometry depnds on gravity and the motion of the observer).

Okay I'll simplfy it to a few points becasue the explanation on the site really isn' that complicated but the some of the terminolgy might throw some people not overly famlair with the terminolgy of relativty: 1) the length of the circumfenrce changes but not the radius 2) therfore the circle is non-Euclidean 3) but space in special relatitvity is alwaus Euclidean 4) Therefore a rigid disc of this type is not allowed in spcial relativty 5) but Einstein uses the disc as an argument for the non-Euclidean spaces of GR.

I'll post the article again as it pretty much explains what's going on: http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html to summarize the circumfenrce changes but not the radius, so the spatial slice cannot be Eucldean, but the spatial slice of the Minkpowski metric is Eucldean so there exists no such rigid disc in specil relativity. However Einstein uses this as an argument for the non-Minkowskian (except in the limiting case obviously) Lorentzian metrics of GR which have non-Euclidean (again except in the limiting case) spatial slices.

Well, quarks weren't proposed until 1964 (though the excat history of QCD is a little sketchy to me) and the first experimental evidnce for them came in 1975.

If your talking about the relativstic disc, I posted a link a few pages back which explains the scenatio in detail. I'm just using spherical geometry as an example of non-Eucldean geometry.

Well on sphere of radius R (though rember that R can just be parameter describing the geomery it needn't be the radius of any actual sphere) the circumeference of the great cricle is: 2pi*R, but the radius is pi*R/2

how did what go from flat......?

We can consider a great circle to be a special case of circles in spherical geometry.

If you have a globe in your home that is problem one of the best ways to look at it, because it has the great circles marked in it (the equator and the lines of longitude or 'meridians'). You should see that any circle drawn om the globe has a radius larger than a Eucldean circle's radius of the same circumfenrce (though the ratio tends to pi as r tends to zero) simalirly you can see that the angles of triangles always add up to more than 180 degrees (again though the sum tends to 180 as the area of the triangle temds to zero). http://mathworld.wolfram.com/SphericalGeometry.html

Okay: During the recombination era the unievrse was very homogenous and isotropic, so the CMBR was emitted by all areas of the unievrse during pretty much the same tim period. Now you know that, due to the fact that light has a finite speed, the further we look, the furtehr back in time we look. If we look in any direction far enough we we 'see' the unievsre 300,000 yrs after the BB (infact this is the furtherest back we can see as before this time the unievrse was opaque to EM radiation) and so we will see the CMBR as it was emitted 300,000yrs after the BB.

Rubbish, buy a decent book on cosmology and you'll understand why that is not the case.

Yes, for example the geodesics on the surface of a sphere run along what we call still call 'great circles'. It also meets the defintion as a set of points equidistant from a given point.

For a global event such as recombination (which is where the radiation comes from 300,000yrs after the BB), we should always be able to see the radiation. The radiation is far too homogenous to come from the anihilation of matter and antimatter and it is alos of the wrong wavelngh.

What happens is that thegeomtry changes, the geometry (as demonstrated by a link earlier) can no longer be Euclidean so the ratio is no longer pi.

No, there is no difference as what 'actually' happens is frame dependent. What you are doing is suggesting that there is a prefered frame in relativty, this goes against the fundamentals of the theory of relativty.

You can extend the solution by saying: if z = nx + my is given by the smallest z that satisfies: z +1 = n(x + p) + m(y - q) = n(x - r) + m(y + s) Tho' I'm not 100% on this oen and I'm sure ther's a neater solution.

What you said is: This is incorrect, it is exactly the same as the geometry of that circle changing in that frame of reference and I'll say it again :you should not equate the proper dimensions (the dimensionsn of the disc in it's own frame) with the actual dimensions of the disc as they are frame dependent.

No, what does x and y equal for 7 cents then? see my post above.

Yourdad. is actually correct, the dimensions of the disc have changed, you can't say that the proper diemnsions of the disc are it's actual dimensions, it is not an illusion.