# Curvature and geodesics

Hello there.Curvature can be informally defined as the deviation from a straight line in the context of curves, a circle in R^2 has curvature, then if we get higher dimensions than three we can't see the manifolds because it is their nature and the nature of our eyes that it is bounded by the three dimensions, but we can study them with math or physics by using theorems,proofs etc. So curves can be in R^2 but this is RxR what if we could find another set like R and try to see it as a geometric space?Perhaps it could be a generalisation of R, but in the same dimensions, could we then generalise curvature as we know it in R^2 for curves? Perhaps it could be about numbers also this set I do not know.I think curvature could be generalised in this way.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.What about its length being negative?could it be?Could we define it and have some results then with theorems and proofs?Thank you.

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PeroK
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universe function
Ibix
Curvature can be informally defined as the deviation from a straight line in the context of curves, a circle in R^2 has curvature
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.
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.
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.

geordief
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.
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?

PeroK
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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?
The length of a spacetime interval may be positive and negative, corresponsing to spacelike and timelike separation.

universe function
How about the generalisation of the curvature in two dimensions?We talked about extrinsic and intrinsic curvatures and it is about embedding and if it depends on it or not, but regarding the curvature of curves?Could it be generalised but staying in the same two dimensions?

On a Riemannian manifold the curvature of a curve is defined in the same way as in the plane. It is the length of the "acceleration vector" of the curve when parameterized by arc length. Specifically, it is the length of the covariant derivative ##∇_{T}T## where ##T## is the unit tangent to the curve. This is called the "geodesic curvature". It is a measure of the deviation of the curve from being a geodesic because the condition for being a geodesic is ##∇_{T}T=0##.

universe function
If we take the length of the derivative of the acceleration vector do we get something related to curvature?How did they define curvature this way?What helped the mathematicians do it?Does anyone know?

PeroK
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If we take the length of the derivative of the acceleration vector do we get something related to curvature?How did they define curvature this way?What helped the mathematicians do it?Does anyone know?
https://www.preposterousuniverse.com/grnotes/

universe function
If we take the length of the derivative of the acceleration vector do we get something related to curvature?How did they define curvature this way?What helped the mathematicians do it?Does anyone know?
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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@universe function

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##.