Conversation started Sep 20, 2016 at 6:17.
John Rennie
John Rennie
Sep 20, 2016 06:17
@MAFIA36790 Suppose you have a particle moving in 3D Euclidean space. The metric is obviously flat.
user116211
yep.
John Rennie
John Rennie
Now add a constraint that the particle must remaina fixed distance froma specified point, so the particle now moves on the surface of a sphere.
user116211
okay...
John Rennie
John Rennie
The metric of a sphere is not flat
user116211
sure.
John Rennie
John Rennie
Sep 20, 2016 06:18
So there's an example of how imposing a constraint selects a submanifold that is not Euclidean.
user116211
okay!
user116211
@JohnRennie, As I know, for the line-element to be Euclidean, the terms like $\mathrm dx_i\mathrm dx_j$ must vanish. Is it true for all coordinate-system? As for the first case, where there was no constraint the line-element was in Riemannian form with those terms; but the geometry was still Euclidean. So, does that mean even with the presence of those terms, the geometry can be Euclidean?
John Rennie
John Rennie
Remember that you can choose any coordinate system you want. The presence of cross terms doesn't necessarily mean the space isn't flat, it could just mean you've chosen screwy coordinates.
user116211
@JohnRennie You made the things quite clear, thanks!!
John Rennie
John Rennie
The only way to tell for certain is to calculate the Riemann curvature tensor. If space is flat this is zero in all coordinate systems.
user116211
Sep 20, 2016 06:24
@JohnRennie yeh, yeh, sure.
 
Conversation ended Sep 20, 2016 at 6:24.