Explain the Riemann-Christoffel curvature tensor
February 14, 2013 9:04 AM Subscribe
Greetings Ask Metafilter:
Can anyone suggest a book or article that provides a simple, yet comprehensive explanation of the Riemann-Christoffel curvature tensor? I've studied the explanation provided by Lillian Lieber in her book The Einstein Theory of Relativity, but for several reasons I am suspicious of the validity of her derivation, and her claim that the tensor can be used to distinguish between Euclidean and non-Euclidean space.
Lieber gives the Rieman-Christoffel curvature as follows:
{σρ,ε} {ετ,α}- ∂/(∂x_ρ ) {στ,α}+∂/(∂x_τ ) {σρ,α}-{στ,ε} {ερ,α}
Lieber derives the curvature tensor from a covariant tensor of rank one that has undergone two different sequences of covariant differentiation. This supposedly produces two different versions of the second, partial, covariant derivative of a covariant tensor of rank one.
The curvature tensor is produced when the first version of the covariant derivative is subtracted from the second version of the covariant derivative.
The first version of the covariant derivative is produced when a covariant tensor of rank one is covariantly differentiated with respect to x_τ and then that quantity is covariantly differentiated with respect to x_ρ.
The second version of the covariant derivative is produced when the same covariant tensor of rank one is covariantly differentiated with respect to x_ρ, and then that quantity is covariantly differentiated with respect to x_τ.
Interestingly, x_τ and x_ρ seem to be equivalent terms so the first version of the tensor seems to be equivalent to the second version of the tensor. This indicates that when the first version of the covariant derivative is subtracted from the second version of the covariant derivative the difference will be zero.
This undercuts the claim that the curvature tensor can be used to distinguish between Euclidean and non-Euclidean space, since the claim is that in Euclidean space the curvature tensor will be equal to zero and in non-Euclidean space the curvature tensor will not be equal to zero.
The curvature tensor is produced when the first version of the covariant derivative is subtracted from the second version of the covariant derivative.
The first version of the covariant derivative is produced when a covariant tensor of rank one is covariantly differentiated with respect to x_τ and then that quantity is covariantly differentiated with respect to x_ρ.
The second version of the covariant derivative is produced when the same covariant tensor of rank one is covariantly differentiated with respect to x_ρ, and then that quantity is covariantly differentiated with respect to x_τ.
Interestingly, x_τ and x_ρ seem to be equivalent terms so the first version of the tensor seems to be equivalent to the second version of the tensor. This indicates that when the first version of the covariant derivative is subtracted from the second version of the covariant derivative the difference will be zero.
This undercuts the claim that the curvature tensor can be used to distinguish between Euclidean and non-Euclidean space, since the claim is that in Euclidean space the curvature tensor will be equal to zero and in non-Euclidean space the curvature tensor will not be equal to zero.
Johnny has it. Wald is good, so is Carroll's "Spacetime and Geometry."
As for an explicit example of detecting the non-Euclidean nature of a surface via parallel transport (which is what the Christoffel tensor is dealing with), look no further than the South-Pointing Chariot, which intrinsically drifts from south-pointing due to parallel transport on an non-Euclidean surface (the Earth).
posted by physicsmatt at 10:37 AM on February 14, 2013
As for an explicit example of detecting the non-Euclidean nature of a surface via parallel transport (which is what the Christoffel tensor is dealing with), look no further than the South-Pointing Chariot, which intrinsically drifts from south-pointing due to parallel transport on an non-Euclidean surface (the Earth).
posted by physicsmatt at 10:37 AM on February 14, 2013
I think the most succinct way of expressing the curvature tensor is this (despite slight abuse of notation):
R(U,V) = [∇ᵤ, ∇ᵥ] - ∇₍ᵤ, ᵥ₎
Where the [ , ] here is the commutator of the covariant derivative operators, and the subscript ( , ) is the Lie bracket, or the commutator of vector fields (actually I couldn't find unicode subscript [ , ] or I would have used those). This I got from the excellent book "The Geometry of Physics" by Theodore Frankel.
posted by Astragalus at 5:24 PM on February 14, 2013
R(U,V) = [∇ᵤ, ∇ᵥ] - ∇₍ᵤ, ᵥ₎
Where the [ , ] here is the commutator of the covariant derivative operators, and the subscript ( , ) is the Lie bracket, or the commutator of vector fields (actually I couldn't find unicode subscript [ , ] or I would have used those). This I got from the excellent book "The Geometry of Physics" by Theodore Frankel.
posted by Astragalus at 5:24 PM on February 14, 2013
You really want to ask this on math.stackexchange.com, which is much better set up than MetaFilter to ask and answer math questions.
posted by number9dream at 6:03 PM on February 14, 2013 [1 favorite]
posted by number9dream at 6:03 PM on February 14, 2013 [1 favorite]
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This is your problem: it's not true that ∇τ∇ρ, when applied to a tensor field, will give you the same as ∇ρ∇τ. (It's true for scalar fields, so long as you're not allowing torsion, but not for an arbitrary tensor field.) It's best to think of covariant differentiation as an operator that takes rank-n tensor fields to rank-(n+1) tensor fields, and operators aren't in general commutative with each other.
I'd be happy to expand on this later, but I have to run off to supervise lab right now. For the record, my favorite text on this subject is probably Wald's General Relativity, which does everything in as carefully and rigorously a way as you're likely to find in a physics textbook. (Mathematicians tend to use rather different language in talking about these notions, but you seem to be coming at this from a physics perspective.)
posted by Johnny Assay at 10:01 AM on February 14, 2013 [2 favorites]