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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:

  1. 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.

  2. 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.

  3. 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 $p$.

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  • $\begingroup$ A metric is a tensor defining intervals between events. If spacetime is static, you could devise experiments measuring specific intervals, but they wouldn't give you the full tensor. Physics is concerned with physical effects that depend on metric derivatives. This question seems to be about mathematics just using physics as an excuse. $\endgroup$ – safesphere Jun 28 at 21:40
  • $\begingroup$ What would you consider physics directly involving the metric? The Lagrangian of a free particle is $L=1/2 g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu$ but you're right that when you go to look at the equations of motion you'll mostly have derivatives of $g$ (mixed up with some $g$'s as well). $\endgroup$ – jacob1729 Jun 30 at 18:45
  • $\begingroup$ @jacob1729 I'm looking for basically anywhere where two different metrics on a spacetime manifold would result in different particle trajectories. Finding the geodesic of a free particle comes close, but it gives you the same results if you change the metric up to a derivative (or am I wrong on that?). I strongly suspect that Maxwell's equations (with the lorentz force) give you different particle trajectories if you change any part of the metric, but I'm having a difficult time proving this. $\endgroup$ – SpiralRain Jun 30 at 19:52
  • $\begingroup$ @SpiralRain but the trajectories are by definition extended so depend on $g$ not only at a point but also nearby. (Also btw I think you mean 'change the metric by a constant' not 'up to a derivative'). Particle trajectories certainly depend on the field $g$ even if much of the dependance is through $\partial g$. $\endgroup$ – jacob1729 Jul 1 at 18:25
  • $\begingroup$ The squared line element directly appears in e.g. the work energy theorem for a nonrelativistic particle on a surface, right? $\endgroup$ – alexarvanitakis Aug 3 at 22:37
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In general relativity the spacetime interval $ds$ is stated as $ds^2 = g_{\mu \nu} dx^\mu dx^\nu$. The components of the metric tensor are defined as the scalar product of the basis vectors in that coordinate system, that is $g_{\mu \nu} = \partial_\mu \cdot \partial_\nu$.

However the single components of the metric do not have a physical meaning, as you may change the coordinate system, e.g. from cartesian coordinates to spherical coordinates. What is physically meaningful is the spacetime interval $ds$.

If you are a stationary observer you use the metric to measure the proper time. At constant time you use the metric to measure a path length. To measure those physical entities you use specific components of the metric.

A way to measure the curvature is the geodesic deviation, which expresses the relative acceleration between two neighboring geodesics (e.g. two free-falling test particles) as proportional to the curvature described by the Riemann tensor $R^\rho_{\sigma \mu \nu}$, which depends however on the first derivatives and second derivatives of the metric.

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The components of the metric are not directly observable at a point, because in any sensible local coordinate system, the metric is $\operatorname{diag}(1,-1,-1,-1)$. This should also be clear in the Newtonian limit, where the metric becomes the gravitational potential, and the gravitational potential is clearly not measurable at a point, since you can add an arbitrary constant to it.

By analogy with Newtonian gravity, you might then imagine that the first derivatives of the metric would be observable, because in Newtonian gravity, the gradient of the potential is the gravitational field. But in GR, the equivalence principle tells us that the gravitational field is never measurable either. In GR, the derivatives of the metric are how you get the Christoffel symbols, which are not observable.

So what's measurable is second derivatives of the metric. These are what are used in calculating measures of curvature, which are observable. In the Newtonian analogy, they give the strength of tidal forces.

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  • $\begingroup$ The comparison with the gravitational potential is a very helpful analogy I haven't thought of. However, I have two quibbles here. First, isn't our notion of perpendicularity (which no one would ever doubt can be measured) derived from the metric? So it seems like some aspects of the metric are observable. Second, couldn't you construct a coordinate system for which the metric is Minkowski at a specific point? That would count as measuring it to me at least. I apologize for being contrarian, because this is a helpful answer, but I'm still mulling over this. $\endgroup$ – SpiralRain Aug 3 at 19:36
  • $\begingroup$ "But in GR, the equivalence principle tells us that the gravitational field is never measurable either. In GR, the derivatives of the metric are how you get the Christoffel symbols, which are not observable." I agree with this if you're only confined to one body/particle. However, if you're allowed to invoke more than one body to compare different points in spacetime, then it seems you can measure the gravitational field and Christoffel symbols. (If I place two satellites, one close to Earth and one away form Earth, and they communicated with each, they could tell which one is falling, no?) $\endgroup$ – SpiralRain Aug 3 at 20:04

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