Proving that the sphere cannot be expressed in flat coordinates

[demonelite123]

proving that the sphere cannot be expressed in "flat coordinates"

for the sphere in R^3 i have that ds^2 = dϕ^2 + sin^2(ϕ)dθ^2. using the definition of a flat space as one where given a set of curvilinear coordinates, one can find a metric such that ds^2 = dx^2 + dy^2, how would one prove that the sphere is not a flat space which means it cannot have "flat coordinates"?

i tried setting dϕ^2 + sin^2(ϕ)dθ^2 = dx^2 + dy^2 to find a contradiction. i was thinking of taking two points and finding the difference between them to approximate the differentials. however i have no idea what to do with the dx and dy on the right side as the coordinate system i am in do not involve them at all. any help is greatly appreciated!

 

[tiny-tim]

hi demonelite123!

in flat coordinates, the sum of the angles of a triangle is 180° (this is one definition of "flat")

on a sphere, it's always more than 180° …

a metric doesn't change distances or angles, so a sphere can't have a flat metric

 

[demonelite123]

hi demonelite123!

in flat coordinates, the sum of the angles of a triangle is 180° (this is one definition of "flat")

on a sphere, it's always more than 180° …

a metric doesn't change distances or angles, so a sphere can't have a flat metric

do you mean a flat metric doesn't change distances or angles, or do all metrics not change distances and angles.

also, is there a way to show that in a triangle on the sphere the angles add up to more than 180 degrees?

 

[tiny-tim]

hi demonelite123!

do you mean a flat metric doesn't change distances or angles, or do all metrics not change distances and angles.

if you want to keep spherical geometry, the metric has to preserve distances and angles

also, is there a way to show that in a triangle on the sphere the angles add up to more than 180 degrees?

extend sides AB and AC to meet at A', the opposite pole of A … we call that shape a lune

make two similar lunes by extending sides BC and BA to meet at B', and extending sides CA and CB to meet at C'

that should cover the whole surface in six pieces, with some overlaps

then count areas and angles ​

 

[Dale]

Another thing you can do is to parallel transport around a line of lattitude and show that the vector does not generally return to itself.

 

[HallsofIvy]

Or- probably the hard way- use the metric tensor in spherical coordinates to calculate the Riemann curvature tensor and observe that it is not the null tensor.

 

[_PJ_]

I was thinking about simil,ar things the other day, how the sum of the angles of a triangle are distorted according to the curvature, and I reasoned that PI must have a direct relationship to this curvature.
Is this right? If so, if only there was a means to identify the mean curvature of the universe according to our value of PI!

 

[Dale]

Pi is a purely mathematical constant. There is no "our value of pi".

 

[_PJ_]

Pi is a purely mathematical constant. There is no "our value of pi".

Of course.

BUT if you understood what I was referring to:
"IF" spacetime was curved any differently, PI would be different. It's only a constant so long as spacetime curvature is (which I'm not so sure of, if it's expanding - though admittedly, the difference would be tiny)

In a theoretically different-curved spacetime, the ratio between the rasdius and circumference of a circle would be different from PI - It was just easier to distinguish the difference by referring to each as PI, as in "our PI" and the theoretical universe's ratio.

 

[Dale]

"IF" spacetime was curved any differently, PI would be different.

No, pi is pi regardless of spacetime curvature. You could be just outside the singularity of a black hole and pi would still be pi. It is a purely mathematical constant meaning that it is calculated, not measured. Because it is not measured it cannot depend on anything physical.

Now, you could construct a physical circle and physically measure the circumference and physically measure the diameter. If you did that you might get a number different from pi. That would not change pi, but could give you a measure of the spacetime curvature inside the circle.

 

[_PJ_]

If you did that you might get a number different from pi.

Yeah, I can accept that, and understand the difference. It's getting this different value that I meant, sorry for misunderstanding and thanks for clarifying.

Incidentally, is it even possible to calculate such a different number, (Consider drawing a circle oin the surface of a sphere, where the radius would then be an arc), If only a ratio beteen the circumference of circle and curvature of the sphere was known?

 

[WannabeNewton]

I think the most intuitive check would be to show that vectors being parallel transported along curves other than a great circle will not necessarily return as the same vector indicating a non - flat geometry.