Circle.

[ydoaPs]

i thought we have already determined a rotating circle is not bound by euclidean geometry. what do you want a reason for?

[osram]

The radius is just a distance between the center and the circumference. If the circumference is shortened, your measurement will show that the distance between the center and the circumference have shrinked (the radius).

 

Again: The radius is not an object and is thereby not affected by relativity.

[ydoaPs]

again, that is euclidean geometry, which we have already said doesn't apply in this case.

 

edit: the radius is more of an object than the circumference, for it is in the equation.

[Aeschylus]

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.

[Sayonara]

I'm sure that if you know your physics it does

[ydoaPs]

i haven't read that yet, but i am sorry for overlooking it

[ydoaPs]

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' date=' 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.[/quote']

 

could you explain it again is english? not english, american. :Þ

[Aeschylus]

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.

[ydoaPs]

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

[Aeschylus]

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).

[osram]

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

 

Maybe it still applies?

[ydoaPs]

so pi is NOT a constant?

[Aeschylus]

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.

[Aeschylus]

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).

[osram]

A disc is bounded by a circle.

Yes, 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).

[ydoaPs]

but if the circumference shrinks and radius stays the same, pi no longer equals 3.14......

[Aeschylus]

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.

[Sayonara]

I think everyone should declare themselves right.

 

My brain hurts and I want bed now.

[ydoaPs]

yay. I am the winner! :Þ

 

edit: sayo's way of saying i win. he doesn't like to admit it when i do

[osram]

I don't get it... but...

[ydoaPs]

but wat?

[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".

[osram]

but wat?

But I guess I'm wrong, even if I don't understand why.

[ydoaPs]

read the thread and you will find out why you are wrong.

[Cap'n Refsmmat]

I do not understand a word of this thread from the start of page 6.