In General relativity, the metric tensor that satisfies Einstein's equations induces the Levi-Civita connection of that metric.

It is said that this connection is somehow "compatible" with the metric.

Technically Im told this means that straight lines (according to the connection) coincide with geodesics (according to the metric). However, this seems like an arbitrarily restrictive assumption from a mathematical point of view. Shouldn't it be possible for a single specific metric manifold with connection to have straight lines that are not necessarily geodesics? Why not?

So my main question is: what does this notion of "compatibility" of metric and connection really mean intuitively? Does it mean there cannot exist a metric manifold with connection whose metric and connection are incompatible? (I.e. is it a necessary condition?). Why is the intuitiv enotion of "compatibilty" captured formally by the "straight lines = geodesics" criterium?

  • $\begingroup$ It can be understood as saying parallel transport is an isometry of tangent spaces. $\endgroup$ – Pedro Feb 25 '17 at 17:28

We can use a pseudo-Riemannian metric $g$ to identify one-forms and vector fields, by raising and lowering indexes. For example, if $\omega$, the vector field $\omega^\sharp$ is the only one satisfying $\omega(Y) = g(\omega^\sharp,Y)$ for all $Y$, and if $X$ is a vector field, we have a one-form $X_\flat$ defined by $X_\flat(Y) = g(X,Y)$, for all $Y$. The non-degeneracy of the metric ensures that $\sharp$ and $\flat$ are isomorphisms.

Every connection induces total covariant derivatives, and so it makes sense to look at $\nabla_V\omega$ and $\nabla_VX$, for a given vector field $V$.

The condition that the connection is compatible with the metric or, in other words, that the metric is a parallel tensor ($\nabla g = 0$) tell us that $$\left(\nabla_V\omega\right)^\sharp = \nabla_V \omega^\sharp \quad\mbox{and}\quad \left(\nabla_VX\right)_\flat = \nabla_VX_\flat. $$Meaning that not only $\nabla$ commutes with contractions, $\nabla$ will also commute with metric contractions.

A priori this doesn't have any relation with geodesics, since you can define what is a geodesic with respect to an arbitrary connection (not necessarily the Levi-Civita connection), and you don't need a metric to define a connection. Similarly, you can define a metric in a manifold which has no connection. The motivation of the Levi-Civita connection is to have a good enough connection that relates these two concepts.

  • $\begingroup$ So "straight lines = geodesics" is merely a result of the two criteria you described, rather than a fundamental criterium? $\endgroup$ – user56834 Feb 25 '17 at 16:32
  • $\begingroup$ Geodesics in general need not be straight lines. For example, in Poincaré's half-plane the geodesics are vertical lines and semi-circles orthogonal to the half-plane's boundary. What happens is that if you write a geodesic in coordinates, and if you have any reparametrization of it which is still a geodesic, then the change of parameters is an affine function. $\endgroup$ – Ivo Terek Feb 25 '17 at 16:34
  • $\begingroup$ If in suitable coordinates, the Christoffel symbols all vanish, then the geodesic equations reduce to $$\frac{{\rm d}^2x^\mu}{{\rm d}t^2} = 0,$$so they correspond to the image of straight lines under the parametrization. $\endgroup$ – Ivo Terek Feb 25 '17 at 16:36
  • $\begingroup$ And fixed a geodesic, it is always possible to find such a coordinate system. See this. $\endgroup$ – Ivo Terek Feb 25 '17 at 16:38
  • $\begingroup$ First of all, note that by straight lines I mean "straight" according to a connection, not "straight" in the natural, Euclidean sense. Does that change your answers? If not, then what did my professor mean when he said that the Levi-Civita connection is obtained by assuming the general relativity metric and requiring that straight lines according to the connection coincide with geodesics according to the metric? $\endgroup$ – user56834 Feb 25 '17 at 16:46

Here is the physicist's reason for choosing the Levi-Civita connection.

In general relativity, freely falling particles travel along geodesics. A physicist would consider a geodesic to be a path $x^\mu(t)$ minimising proper time, i.e. minimising the integral $$\Delta \tau = \int_{t_0}^{t_1} \sqrt{g_{\mu\nu} \frac{d x^\mu}{d t} \frac{dx^\nu}{d t}} dt$$

Using the calculus of variations, you can show that this criterion is equivalent to the equation of motion, $$ \frac{D^2 x^\mu}{dt^2} + \Gamma^{\mu}_{\nu\rho} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt} = 0, $$ where $\Gamma^{\mu}_{\nu\rho}$ is the Levi-Civita connection.

Edit: While the Levi-Civita connection works here, levap points out that it is not the unique connection with this property.

Here is another comment that directly addresses the fact that the connection is metric-compatible: In general relativity, it is important that $dx^\mu / d \tau$ has length $1$ at all times, i.e. $g_{\mu \nu} dx^\mu / d\tau dx^\nu / d\tau = 1$, and in particular, it is constant. The fact that the connection is metric-compatible implies that, for geodesics, $g_{\mu \nu} dx^\mu / d\tau dx^\nu / d\tau$ is constant over time.

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    $\begingroup$ This can't be true. What you call the proper time (and is called the length in the Riemannian setting) is invariant under reparametrizations and there are many connections that have the same geodesics as the Levi-Civita connection up to reparametrization. Even if you minimize the energy (which is not invariant under arbitrary parametrizations) and not the length, you can add any anti-symmetric tensor to the connection and get the same geodesics. $\endgroup$ – levap Feb 26 '17 at 18:04
  • $\begingroup$ I see! I didn't notice that. So torsion doesn't affect the geodesics. $\endgroup$ – Kenny Wong Feb 26 '17 at 18:32
  • $\begingroup$ For future readers, it is NOT true that torsion doesn't affect geodesics. Instead, a completely antisymmetric torsion tensor (after "lowering" one of the indices with the metric to consider it a (0,3)-tensor) doesn't affect geodesics. In general, the antisymmetry built into the definition of the torsion tensor is in the wrong two indices to give the above conclusion immediately. $\endgroup$ – jawheele Jun 8 '18 at 16:57

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