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?


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 $p$.

  • $\begingroup$ 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, 2019 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, 2019 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. $\endgroup$ Jun 30, 2019 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, 2019 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? $\endgroup$
    – user21299
    Aug 3, 2019 at 22:37

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$ Aug 3, 2019 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?) $\endgroup$ Aug 3, 2019 at 20:04

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.


I am sorry in advance that my answer will be long. First, I want to clarify one thing:

Difference between laws and definitions of physical quantities: I will explain what I mean on example about Newton's laws using dialogue between physicist and relativist.

-Relativist- What is the definition of mass of a body?

-Physicist- Mass is the measure of inertia of a body (wiki definition).

-Relativist- What does "inertia" of a body means?

-Physicist- Inertia is the tendency of a body to resist any change in its motion.

-Relativist- But what do you mean by "resist" and "change of motion"?

-Physicist- A body resists more if you have to push it harder to change its motion and I think the meaning of the "change of motion" is pretty clear. We are not philosophers, we are physicists.

-Relativist- Precisely due to us being physicists I am asking these questions. I am trying to make you realize that you used FORCE (pushing harder means applying greater force) and ACCELERATION (changing motion means accelerating) in your definition of mass. What would then be your definition of force?

-Physicist- I think I am starting to see your point. Force is the measure of how hard we push bodies, but that is hardly having any meaning if by "hard pushing" we don't attach extra sense. Then I would add that harder pushing means that the bodies of the same mass will accelerate more.

-Relativist- Exactly! So, mass is the ratio of force and acceleration and force is the product of mass and acceleration. It appears then that Newton's laws are basically the definition of quantities mass and force. If so, then I can no more give them a LAW status. They are merely definitions!

In order that our laws have appropriate status they have to be experimentally confirmable. Physics is experimental science! What that means is that we have to define physical quantities by measurement procedure. They do not have more meaning in the beginning than just being a number obtained by definite series of steps using some (pretty arbitrarily) standard units. Only after we know how to measure mass, force, acceleration etc. independently, we can start searching for mathematical relations between those numbers. Then we will observe in Nature that mass of a body is equal to the ratio of the force applied to the body and acceleration of that body. Then we can state we discovered some law (Newton's law). Only now after we have seen some relations between our quantities we can start attaching meaning to them. Now and only now we can say that mass is measure of inertia...

I will talk now about GENERAL RELATVITY without law of gravity (Eienstein's equation). We first have to DEFINE metric using measurements and then we can talk about relations between metric and mass, energy... (law of gravity etc.). It will turn out that this is more difficult task as we discover some underlaying laws along the way and people usually mix those up with definitions.

What is only observable in our world are EVENTS. Events are (no more and no less) point-coincidences of OBJECTS (a bullet meeting a wall, end of our ruler meeting end of some object etc.). This is the "existence of absolutes" law of relativity - absolutes exists and they are merely point-coincidences.

We want to attach coordinates (ordered four numbers) to every event (at least in some neighborhood) such that those coordinates are unique and close events (at this point this have loose meaning) have coordinates differing little - mathematically, all of this means that we want to describe our world by 4-dimensional smooth manifold. Coordinates are simply numbers, they do not have meaning of lengths and time (hence the GENERAL in general relativity). Already we have assumptions (laws) here:

  1. Why 4 coordinates instead of 3? It turns out that 3 is just not enough to describe events (satisfying above criteria) and that 5 is too much numbers. This is experimentally verifiable hence we have the dimensionality law of relativity - our world of events is FOUR dimensional smooth manifold.

  2. If it turns out that we can describe all of events this way (not just locally) than our manifold would be topologically simple (this is impossible to confirm experimentally as we cannot describe events much far from Earth so we do not want to assume that we can do that.

  3. It is possible to attach to same events more than one system of coordinates and if we explicitly do so, we can immediately see relations between those systems i.e. we know how to translate coordinates of one observer to coordinates of another.

Next what we want is not only to talk about coordinates of events, but to talk about "distance" (interval) between them. We want to somehow attach a number for any pair of events, a number that is unique and independent of coordinate systems, a number representing something absolute. This is where metric comes into picture and where light plays the main role. We assume that interval is quadratic form - ds^2=gμνdxμdxν.

Now here is the setup: We place a light bulb somewhere, surround closely that light bulb with circular wall and at one point turn on the light. Light will then go from the bulb and hit the wall at different points. Let the turning on of the light bulb be Event0 and events describing light hitting the wall be Event1,Event2,Event3... All observers (with their system of coordinates describing Event0, Event1 etc.) do the following procedure: First they set quadratic interval between Event0 and Event1 to be ZERO, then they set interval between Event0 and Event2 to be ZERO etc. From those they CALCULATE metric components (in their coordinate system). This can always be done because we have enough number of equations. For example, they have done that using only events from 1 do 16 (as metric has 16 components). Now, USING OBTAINED metric components, they check if the intervals between Event0 and Event17, between Event0 and Event18 etc. is again zero. All of them will get positive answer meaning the light law of relativity - there is something absolute about light and events connected by light. Light is truly manifested the same way for ALL of the observers and light is that one absolute thing allowing us to define metric and claim we all talk about same things just having different perspectives.

They found one more thing along the way. They all found out that gμν is symmetric having only 10 independent components. Next they compare their components of metric and they also find out that components transform exactly like a tensor (remember, they see each others coordinates and know how to transform from one system to another). Their next task is completing the story about our spacetime by observing events not connected by light signals. They all pick two same, but otherwise arbitrary, events, say events A i B. They calculate (using their metric components) interval between A i B and check if they all get the same number (number will now not be zero as in light case, but it will be the same for all observers). After getting again the positive result and after considering all of the above, they conclude the metric postulate of relativity - our smooth 4-dimensional manifold have absolute metric structure defined by the light allowing us to talk about distance between events.

You can see now why this story was complicated - we have laws mixed up with definitions. To be more precise, all of these laws (which are experimentally checkable) allows us to define in a definite way quantities used in general relativity. Now and only now, since we know how to measure metric and interval, we can start searching for laws such as laws of gravitation (dependence of metric on other measurable quantities) or checking if particles do follow geodesics (geodesics is defined using Christoffel connection which is defined using metric) etc.

I hope I managed to clarify meaning of general relativity without gravity. I would gladly give you the literature, but what I wrote here you can't find anywhere written the way I wrote it (all in one piece of text). Related to: Are Newton's "laws" of motion laws or definitions of force and mass?


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