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".

Yes' date=' but if it's a circle only the circumference is affected by relativity. If it's a disk the whole plane within the circumference is affected.

That means it'll be misformed (if it's a disk).[/quote']

 

You still have a non-Euclidean circle on your hands.

so pi is NOT a constant?

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).

But that's only for a disk and not a circle' date=' or am I wrong?

 

Maybe it still applies?[/quote']

 

A disc is bounded by a circle.

Well since u say that i might just be mistaken by the exact year altough it was written somewhere along 50-60s might just have been late 60:s to then.. haven't really opened that book for long so my mistake if so then

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.

ok, so it means what? it means that space isn't euclidean?

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.

i don't know the "answer." it went from being a flat circle to being on a globe. how did that happen?

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.

I'm guessing that taking the shortest radius of the circle described by the circumference of the great circle is cheating?

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 it go from being flat, to being a circle drawn on a sphere?

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

A much better explanation.

 

My question now is: can we consider a circle to be equal to a great circle?

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

Do you know of any representations of that geometry I can take a look at?

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.

if it started before us and is faster than us, how hasn't it passed seconds after the big bang. actually, it should have passed before matter was even created.

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

So the big question we all want answering is "is it still a circle?"

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.

then how the bloody hell did we get in front of it???????????????

 

radiation was created before matter. radiation travels faster than matter. we should not be able to see radiation from the big bang.

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.

There's a difference between an outside observer sees and what actually happens, you know. That's the effect of time dilation.

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.

Ya I think so too. Otherwise' date=' the answer is over several hundred cents.

And so... What's the question, can anyone tell me what it is asking?[/quote']

 

The answer is 8 cents which I proved above.

What you said is:

So when I said "I'm not sure if yourdad meant to say that the circumference was actually affected' date=' or if it just appears that way to someone in a different relatavistic frame"[/i'] way back in post #20, you ignored it because...?

 

The changing view of the circle to an outside observer is not the same as the circle's geometry actually changing.

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.

You can buy so many 3 stamps and so many 5 stamps. Obviously you can't pay 1 or 2 cents for any combination. You can pay three cents (1 3cent stamp). You can't pay 4 cents, but you can pay 5 cents. After what point will every price become available through some combination of the two stamps. That is what the question is asking.

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.