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A circle embedded in a plane has extrinsic curvature, but it has no intrinsic curvature, and it is the latter type of curvature that relativity uses. Generally, you seem to be thinking of extrinsic curvature, the idea that a space is curved relative to a space in which it is embedded. We have no evidence that spacetime is embedded in anything. The curvature it has is intrinsic curvature which can be defined and measured without reference to an external space.Curvature can be informally defined as the deviation from a straight line in the context of curves, a circle in R^2 has curvature

No. A geodesic is the path with extremal distance between two events. And it is not a curve, it is by definition a straight line. The projection of the geodesic on to a subspace (such as the 3d spatial path of an object following a geodesic in 4d spacetime) may be curved in that subspace. The length-squared of a geodesic may be negative in Lorentzian manifolds, but I don't think it makes sense to talk of a negative length. That's just a positive length with a needless sign convention.Another topic is about geodesics and their length, a geodesic informally could be defined as the shortest distance between two points in a geometric space like a surface,it could be a curve.

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This is the definition in the context of general relativity. I am talking here about differentiable riemannian manifolds.The negative length of a geodesic should have applications to be more interested in it or not?A circle embedded in a plane has extrinsic curvature, but it has no intrinsic curvature, and it is the latter type of curvature that relativity uses. Generally, you seem to be thinking of extrinsic curvature, the idea that a space is curved relative to a space in which it is embedded. We have no evidence that spacetime is embedded in anything. The curvature it has is intrinsic curvature which can be defined and measured without reference to an external space.

No. A geodesic is the path with extremal distance between two events. And it is not a curve, it is by definition a straight line. The projection of the geodesic on to a subspace (such as the 3d spatial path of an object following a geodesic in 4d spacetime) may be curved in that subspace. The length-squared of a geodesic may be negative in Lorentzian manifolds, but I don't think it makes sense to talk of a negative length. That's just a positive length with a needless sign convention.

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The length of a spacetime interval may be positive and negative, corresponsing to spacelike and timelike separation.This is the definition in the context of general relativity. I am talking here about differentiable riemannian manifolds.The negative length of a geodesic should have applications to be more interested in it or not?

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ChinleShale

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https://www.preposterousuniverse.com/grnotes/

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ChinleShale

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Yes but things get a little complicated. At a point where the curvature of the curve is not zero, the covariant derivative of the unit normal vector ##∇_{T}N## has a non-zero component parallel to the curve. The length of this component is just the curvature back again.

If one ignores this component and just takes the length of the component that is perpendicular to the plane spanned by ##T## and ##N## then one gets the tendency of the curve to move out of that plane. This is called the torsion of the curve.

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ChinleShale

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Here is the calculation of the covariant derivative of the unit normal to the curve. Try doing the calculation for the unit torsion vector.

##0=T⋅<T,N>=<∇_{T}T,N> +<T,∇_{T}N> = ##κ + ##<T,∇_{T}N>## so the component of the covariant derivative of ##N## parallel to ##T## is negative the curvature times ##T##. For the length of the component perpendicular to the plane spanned by ##T## and ##N## one just calls it torsion. So ##∇_{T}N = ##-κ##T + τB##.

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