I apologise in advance that my knowledge of differential geometry and GR is very limited. In general relativity the equation of motion for a particle moving only under the influence of gravity is given by the geodesic equation:

$$ \ddot{x}^\lambda + \Gamma^\lambda_{\mu\nu}\dot{x}^\mu\dot{x}^\nu =0. $$

I am looking for a conceptual description of the role of the affine connection, $\Gamma^\lambda_{\mu\nu}$ in this equation. I understand that it is something to do with notion of a straight line in curved space.

Comparing it to the equation for a free particle according to Newtonian gravity:

$$ \ddot{x}_i = -\nabla\Phi, $$

Then it kind of looks like the affine connection is our equivalent of how to differentiate, except that our scalar field is now some kind of velocity?


1 Answer 1


You're on the right track, but there's more that can be said. For an introduction to this topic, I highly recommend Sean Carroll's Spacetime and Geometry, which I'll follow below for the purpose of illustrating where that $\Gamma$ comes from. The book grew out of lecture notes, the relevant chapter of which can be found online here.

Algebraic background

Note: This is long, but only because each step has been broken down into very simple parts.

You want a directional derivative - something that tells you how a tensor changes as you move along some path in your manifold. Just like in standard multivariable calculus, you define this as the sum of the derivatives of your object in each direction, weighted by how much your coordinate is changing in that direction: $$ \frac{\mathrm{D}}{\mathrm{D}\lambda} := \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \nabla_\mu. $$ Here $\lambda$ parameterizes your path, and the appropriate derivative to use is the covariant derivative. The scalar function $x^\mu$ is being differentiated with the standard directional derivative: $$ \frac{\mathrm{d}}{\mathrm{d}\lambda} := \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \partial_\mu. $$ Note that this is consistent with the fully covariant directional derivative in the case of scalar functions, and, despite appearances, this is not a circular definition (you should be able to differentiate known functions of $\lambda$ with respect to $\lambda$).

The affine connection is simply defined as the set of coefficients needed to augment the partial derivative in order to make a covariant derivative: $$ \nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\lambda} V^\lambda $$ for any vector $\vec{V}$. You don't even need to know how to generalize this to covariant derivatives of arbitrary tensors.

The geodesic equation arises from imposing the requirement that your curve parallel-transport its own tangent vector.1 That is, the directional derivative of the tangent vector along the direction it is pointing vanishes. A path will be a set of smooth functions \begin{align} x^\mu : \mathbb{R} & \to M \\ \lambda & \to x^\mu(\lambda), \end{align} one for each coordinate $\mu$. If you denote the tangent vector at $\vec{x}(\lambda)$ by $\vec{T}(\lambda)$, then $$ T^\mu = \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}. $$

Putting all this together, we have \begin{align} 0 & = \frac{\mathrm{D}}{\mathrm{D}\lambda} \left(\frac{\mathrm{d}x^\sigma}{\mathrm{d}\lambda}\right) \\ & = \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \nabla_\mu \left(\frac{\mathrm{d}x^\sigma}{\mathrm{d}\lambda}\right) \\ & = \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \left(\partial_\mu \left(\frac{\mathrm{d}x^\sigma}{\mathrm{d}\lambda}\right) + \Gamma^\sigma_{\mu\nu} \frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}\right) \\ & = \frac{\mathrm{d}^2x^\sigma}{\mathrm{d}\lambda^2} + \Gamma^\sigma_{\mu\nu} \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda} \end{align}


After all this (possibly too much) algebra, the take-home message is that the connection arose from the difference between partial and covariant differentiation along a curve. It's not that we want our curve's tangent to never point in a different coordinate-dependent direction ($\ddot{x}^\lambda = 0$, which would be sufficient for "straight line" in Euclidean space). Rather we want the change in the coordinate-dependent direction we are pointing to be compensated by the fact that our tangent space is rotating with respect to our coordinates as we move along the curve.

1 Some people define the geodesic in a more global sense, as the path that extremizes the arc length between two points. These definitions agree if and only if the affine connection you are using is indeed the metric compatible, torsion free Christoffel connection.

  • $\begingroup$ Thankyou very much for your answer, those notes look really helpful as well, so thankyou! $\endgroup$
    – rgvcorley
    Commented Apr 20, 2013 at 8:17

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