What does it mean intuitively for a metric and connection to be compatible?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

Perhaps a 'noob' interpretation but this is what I understand.

We write the condition as:

$$ \nabla_X g = 0$$

For all vector field $X$ on the surface. Now let us recall what the idea behind $\nabla_X$ is, we say it is the generalization of the directional derivative of scalars. What new information does $\nabla_X$ have than the directional derivative? Well, it has the information of the Christoffel symbols which tell us about the geodesics (recall parallel transport).

So, the idea here is that the rule for computing length and angle of vectors doesn't change if one moves along the geodesics.

I guess in some sense, this is like a weaker isotropy of space condition. If we are in a Euclidean space, then we can understand this equation as saying that wherever you are in the $\mathbb{R^3}$ universe, the lengths and angles between two rulers you can carry around would be same. In the case here, isotropy condition is not true for no matter how you move, only if you move in Geodesic if that makes sense.