1
$\begingroup$

I'm having a hard time differing between the Christoffel symbols and the Riemann tensor.

I know their formulas and their uses but i'm searching for a physical interpretation on the general theory of relativity.

While i know that mathematically, Christoffel symbols have more to do with the connection and the Riemann tensor has more to do with the curvature of the space. Physically, within the theory, i can't find a difference.

I often read that the Christoffel symbols account for the change in basis vectors but isn't curvature defined as a change in the metric tensor (which in turn would be a change in the basis)?

I now that having a changing basis doesn't necessarily implies curvature (for example, in polar coordinates there's a changing basis but not necessarily curvature) and that while the Christoffel symbols account for changing basis vectors they don't detect curvature and that's why the Riemann tensor exists. Is this idea true?

Also, could someone explain why the formula for the Riemann tensor components is in terms of the Christoffel symbols like this

$ R^\sigma_{\mu \nu \rho} = \partial_\nu \Gamma^\sigma_{\rho \mu} - \partial_\rho \Gamma^\sigma_{\nu \mu} + \Gamma^\lambda_{\nu \mu} \Gamma^\sigma_{\nu \lambda} - \Gamma^\tau_{\nu \mu} \Gamma^\sigma_{\rho \tau} $

Thanks for answering in advance

$\endgroup$
1
  • $\begingroup$ a proper intro to differential geometry and differential forms might help. I recommend Vonk‘s a mini course in topological strings where he introduced differential forms in short section. Then you can read a bit of differential geometry to clear the dust. $\endgroup$
    – Rescy_
    Jan 18 at 3:54

1 Answer 1

1
$\begingroup$

Christoffel symbols allow you to do parallel transport in a given frame. The way to visualize this is as you move around, the tangent vector space "rotates around" as prescribed by the connection. The last index is the direction you move in while the first two prescribe the linear transformation measuring (infinitesimal) change from the identity in this direction. The Riemann tensor is a measure of the holonomy ie when you go around a loop and get back to where you started, since your tangent space rotated along the way, it may not look the same as it started. Specifically, it is the limit of the resulting linear transformation as you traverse in an infinitesimally small loop in a plane. The last two indices (which are antisymmetric) are those of a 2 form and specify the plane, while the first two indices specify the linear transformation. The formula results because if your loop is a rectangle, you can parallel transport in direction $a$ and then $b$ or direction $b$ and then $a$ and the curvature tensor measures the difference in answer of these two paths. In an equation, $R_{ab}=[\nabla_a,\nabla_b]$ where $\nabla_i$ is the covariant derivative in direction $i$ (I omitted the linear transformation indices to focus on the 1 and 2 form indices). You can check how this recovers the equation above. Morally, $\nabla$ is the infinitesimal generator for parallel transport, analogous to how $\exp(a\frac{d}{dx})f(x)=f(x+a)$.

Even more morally, $R$ has units of 1/area, and "integrating" it over an area, you get the parallel transport around the boundary loop. This is analogous to Stoke's theorem with $R=d\nabla$, but I believe it gets technical since $O(n)$ is nonabelian, so this doesn't strictly work. The integral over the loop can be interpreted using the "path ordered exponential", but I'm not sure what the integral of $R$ over the area should be. One case in which this works literally is in the case of surfaces: the group is $O(2)=S^1$ which is abelian, and you have that the integral of curvature over an area corresponds to holonomy over a loop. It's a fun thing to check that the holonomy over a (piecewise linear) loop corresponds to the angular excess (better to look at external angles than internal ones). This is the classical local Gauss-Bonnet formula! It also gives intuition for things like a cone: it's clearly flat away from the point, but there is nontrivial holonomy for loops around the point, and moreso if the point is sharper. The curvature here is a "dirac delta" at the point and hence holonomy is picked up only for loops whose interior contains the vertex.

$\endgroup$
1
  • $\begingroup$ Although the most intuitive way to understand why the Riemann tensor represents curvature is its appearance in the geodesic deviation (Jacobi) equation. I still don't quite understand the relationship of this to holonomy, but I suspect it's something like the angle counting argument I mentioned above $\endgroup$ Jan 18 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.