59 minutes ago, Markus Hanke said:

  49 minutes ago, KJW said:

This btw is why there is no absolute time in relativity.

Great insight! Never thought about it from this particular angle, though in retrospect it seems obvious +1

People intuitively accept that the distance between two points in three-dimensional space depends on the path between them. But for whatever reason, this intuition doesn't seem to extend to four-dimensional spacetime, although it is perhaps the notion of spacetime itself that is where the intuition fails.

Actually, the failure of ds in the metric to be an exact differential is easier to prove than usual. Usually, one would be considering integrability conditions involving the commutativity of mixed second-order partial derivatives. But in the case of the metric, the proof is algebraic (an invertible matrix cannot be the tensor product of two vectors).

Nice explanation @KJW +1

10 hours ago, KJW said:

The intrinsic properties of a manifold depend entirely on the manifold itself without any reference to a higher-dimensional embedding manifold

Thanks for the explanation KJW; excellent as always.
I still have a few questions ...

Consider a line segment 20 units long with point A at x=5 and B at x=15.
We now curl that segment such that the two ends identify with each other into a circle.
The distance between the points A and B doesn't change; it is still 10 units.
This is what I would term intrinsic curvature, and separation.

Now draw that same line segment on a sheet of graph paper.
When the segment is straight the distance between the points A and B is similarly 10 units.
But upon performing the equivalent curling operation as before ( into a circle ), the distance between the points A and B is no longer 10 units, but approx. 20/3.14 units, as it depends on both, x and y co-ordinates.
I would term this extrinsic curvature, and separation.

What am I missing ?

In the first example when you set the lines on a graph paper prior to bending this is intrinsically flat ( it is independant ) Once you curl the paper your curve is extrinsic as you need an extra dimension in order to curl the plane.

Im not sure you missed anything tbh.

Cylinder can simply be described as Eucludean flat is the internal geometry with extrinsic curvature. A sphere for example however has an intrinsic positive gaussian curvature ie circumference of the sphere. Intrinsic curvature K=1/r^2. With extrinsic curvature you need a higher dimension embedding. the 2 principle curvatures being \(k_1=K_2=1/R\) with mean curvature being \(H=1/2 (k_1+k_2)=1/R\).

with \(K_{a,b}\) being the second fundamental form

\[K_{\theta\theta}=R\]

\[K_{\phi\phi}=Rsin^2\Theta\]

\[k_{\theta\phi}=0\]

under GR the extrinsic curvature tensor is the projection of the gradient of the hypersurface.

\[K_{a,b}=-\nabla_\mu^\nu\]

\[K_{\theta\theta}=\frac{r}{\sqrt{g(r)}}\]

\[K_{\phi\phi}=\frac{r\sin^2\theta}{\sqrt{g(r)}}\]

mean curvature bieng \[k=h^{a,b}k_{a,b}=\frac{2}{r\sqrt{g(r)}}\]

K being a surface of a hypersphere where all affine normals intersect at the center

above ties into n sphere aka hypersphere

https://en.wikipedia.org/wiki/3-sphere

edit: I was at work earlier decided when I got home to go into greater detail

further detail in same format as above

https://en.wikipedia.org/wiki/Gaussian_curvature

https://faculty.sites.iastate.edu/jia/files/inline-files/gaussian-curvature.pdf

https://arxiv.org/pdf/1209.3845

Edited Wednesday at 02:50 AM2 days by Mordred

OK.
So, in the case of the x co-ordinate line segment, drawn on x-y co-ordinate graph paper, where the intrinsic separation between points doesn't change when curled to a circle, but the extrinsic separation does, which is the manifold that ...

22 hours ago, KJW said:

The intrinsic properties of a manifold depend entirely on the manifold itself without any reference to a higher-dimensional embedding manifold

Is it the line segment or the graph paper ?

Its the set that can be continuously parameterized where each parameter is a coordinate. Line segment is one example. The association of points/coordinates with their measured values can be thought as the mappings of the manifold.

However you may not be able to parameretize the entire manifold with the same parameters. Some manifolds are degenerate. Simple case a finite set of R^n in Euclidean space is non degenerate. However in Cartesian coordinates involving angle the origin or center is degenerate as at zero the angle is indeterminate. This is where the use of coordinate patches get involved. A manifold can have different coordinate systems as per above on the same manifold. With no preference to any coordinate system.

The set of coordinate patches that covers the entire manifold is called an atlas.

The saddle shape for negative curvature would be a good example.

Edit scratch that last example it can be continously parameterised under the same coordinate set. The Cartesian coordinate requires 2 sets.For reasons provided above.

Hyperbolic paraboloid

\[z=x^2-y^2\] can be parameterized by one coordinate set. Though multiple sets can optionally be used it isn't required.

Edited Wednesday at 04:05 AM2 days by Mordred

It should be noted that a simple curve (1D manifold) has no intrinsic curvature - the Riemann tensor vanishes identically in 1D. But it can of course have extrinsic curvature when embedded in a higher dimensional space.

There's another key detail when it comes to the intrinsic curvature it is independent of any higher dimensional embedding. Very useful for invariant functions. Particularly when it comes to applying the tangent vector to the line element ds^s. Of key note is the basis vectors. Taking the infinisimal distance between P and Q (local) this can be shown independent on coordinate transformations. So the basis vectors are independent. Subsequently this equates to the covariant and contravarient vectors. As well as the Christoffel connections.

7 hours ago, Markus Hanke said:

It should be noted that a simple curve (1D manifold) has no intrinsic curvature - the Riemann tensor vanishes identically in 1D. But it can of course have extrinsic curvature when embedded in a higher dimensional space.

The Riemann curvature tensor in one dimension does not vanish but rather is undefined. It has no components in 1D.

16 hours ago, Genady said:

The Riemann curvature tensor in one dimension does not vanish but rather is undefined. It has no components in 1D.

+1

You are completely right, this is actually an important distinction - thanks for correcting me on this 👍