I am wondering where in physics does the metric have direct physical significance.
The standard answer is that the metric gives us the "lengths" and "angles" of four-vectors. This is a good answer on one level, but it doesn't seem satisfying, because the next question one can ask is, where does the notion of perpendicularity (and lengths and so on) come from. This quickly turns into circular reasoning, as you can say angles and lengths come from the metric.
I'm wondering if there is a much more bottoms-up approach to thinking about this. Is there any piece of physics that directly makes use of the metric?
As I say in the thoughts section, many parts of physics make use of the connection $\nabla$, but from what I know that is not enough to determine the metric. Posts like this, this, and this seem to say the connection may correspond to metrics under some circumstances, but it won't pick out a unique metric.
Note that I'll assume all connections $\nabla$ here will be symmetric. This way, you can use geodesics and the geodesic equation to determine the Christoffel symbols (and hence the connection) uniquely.
Main Questions
These are basically the same question restated in a couple of different ways:
- Is there any piece of physics that directly makes use of the metric?
- What experiments with particles can you do to uniquely determine the metric at any point in spacetime?
- Is there an operational definition of the metric?
Thoughts
So far, I have the following thoughts:
Newton's Laws. In a subtle way, Newton's laws are just as much about geometry as they are about motion. For example, by postulating that an inertial frame exists by Newton's first law, we obtain the notion of straight lines in space, so we say the Christoffel symbols are zero in an inertial frame. Moreover, Newton's second law talks about acceleration, but in order to even begin talking about acceleration, one must define a connection on space. This tells me that Newton's laws define the connection, but it doesn't seem to me that they specify a unique metric.
General Relativity. From this answer, the metric is not uniquely specified by Einstein's equation. Also, the geodesic equation relies on us computing the Christoffel symbols, and that is written in terms of derivatives of the metric, not the metric itself. Again, the exact metric doesn't seem to have direct physical relevance.
Lorentz Force. Electrodynamics is the only place where I suspect I might have my answer. However, there are a lot of complications. First, what are we allowed to know and what are we not allowed to know?
- Let's assume we're given coordinates $(x^{\mu})$ and we know the positions/times of all particles. Also assume we know the proper time of all particles, so we may define the four-velocity of particles as $$ u^{\mu} = \frac{d x^{\mu}}{d\tau}. $$
- The four-acceleration of particles is given as $a = \nabla_{u(\tau)}(u(\tau))$, which we're allowed to use because we know what the connection is from points #1 and #2 above. Assume the rest masses are always known, so the four-force is $f^{\mu} = ma^{\mu}$.
- In order to simplify things for the time bring, let's assume the electromagnetic field $F^{\mu\nu}$ is known. Physically, one has to use test charges to find the field (and this changes our analysis). However, things get so overwhelmingly complicated that I'm willing to grant us knowledge of $F^{\mu\nu}$.
- The Lorentz force is given as $f^{\mu} = qu_{\nu}F^{\mu\nu}$, but we must pay close attention to the indices, since we're trying to avoid presuppose that we know the metric. Instead we write $$ f^{\mu} = qu^{\lambda}F^{\mu\nu}\, g_{\lambda\nu}. $$ From here, we see $\det F^{\mu\nu} = (\vec{B}\cdot\vec{E})^{2}$, so in the case where $F^{\mu\nu}$ is invertible, we can take the inverse $G_{\mu\nu}$, and obtain $$ f^{\mu}G_{\mu\kappa} = q u^{\lambda} \underbrace{F^{\mu\nu}G_{\kappa\mu}}_{\delta_{\kappa}^{\nu}} g_{\lambda\nu} = qu^{\lambda}g_{\lambda\kappa}. $$ The left hand-side is known, and for the right-hand side, we may assume we have an infinite supply of test charges with any velocity. If we just consider 4 linearly independent four-velocities, we can solve for the metric directly.
Is my reasoning so far sound? Can one do the same thing with Maxwell's equations as in #3 to solve for the metric? What if you're not allowed to assume that the field $F^{\mu\nu}$ is known (you're only allowed to use source charges and test charges)?
Where else in physics can you uniquely determine the metric?
Edit: I think I would be satisfied with the following possible answer. We know in GR and (pseudo-)Riemannian geometry that at any point $p\in M$ we can choose coordinates such that the metric at $p$ is Minkowski and the Christoffel symbols of the metric at $p$ are zero.
Assuming the metric is static or varies slowly, can we use electrodynamics to physically construct such a coordinate system at a point $p$?
If we constructed our coordinate system $(x^{i})$, and if we knew $g_{\mu\nu}(p) = \eta_{\mu\nu}$ and $\Gamma_{\mu\nu}^{\lambda}(p) = 0$ at that specific point, then I think we would have full knowledge about length scales and perpendicularity at that point $p$.