Spacetime interval in one co-ordinate system is given by : $$g_{\mu \nu} dx^{\mu} dx^{\nu} \tag{1}$$ $dx$ is some infinitesimal displacement vector between two events.

Spacetime interval after a change of co-ordinate system is give by the algorithm : Change the basis of the matrix $g_{\mu \nu}$, Change the basis of the vector $dx^{\mu}$, and then calculate the same quantity as in $(1)$.

So of course a change of basis leaves the spacetime interval invariant. This is a purely mathematical fact.

What's the physical content here? I understand the physical content of spacetime interval in $SR$, because there the four components of $dx^{\mu}$ refer to actual space and time measurements using clocks and sticks.

In GR however, $dx^{u}$ is more abstract as the four indices don't refer to space and time measurements but to generalised co-ordinates....

In fact, even if we assume that the four indices of $dx^{\mu}$ in GR refer to actual spacetime-measurements by an observer, then a "change to another generalised co-ordinate system" need not mean a "change to another physical situation". Let me explain.

Suppose a GR observer measures $dx^{\mu}={dt, dx, dy, dz}$ using sticks and clocks, and calculates $(1)$. Then we switch to another co-ordinate system : $(dt, r, \theta, \phi)$, and then we calculate $(1)$ again and find it to be invariant. But this is no surprise as the change of co-ordinates was purely mathematical. The "new co-ordinates" refer to the same observer using different variables to parametrise spacetime.

In SR, the invariance of $(1)$ relates spacetime measurements made by observers in two different physical situations. In GR, this isn't the case either.

  • $\begingroup$ I'm not 100% sure what the actual question here is, but see this question and its answers for discussion of what the special symmetry in GR actually is. $\endgroup$
    – ACuriousMind
    Apr 17 at 0:50
  • $\begingroup$ @ACuriousMind Idk anything about Gauge transformations. Can't understand your answer $\endgroup$
    – Rain Deer
    Apr 17 at 1:39
  • $\begingroup$ The physical content is an underlying assumption that there is an 'objective truth' in the universe. That all observers, however they may measure things, are measuring the same underlying mathematical object. Symmetries refer to those changes of coordinates where the physics 'look' the same to you, and you can't actually tell that you're using a weird coordinate system. E.g. in classical physics, the Lagrangian $\mathcal L$ reperesents an 'objective' scalar field over the manifold. If that one Lagrangian $\mathcal L$ 'looks the same' written in two coordinate systems, that's a symmetry. $\endgroup$
    – Myridium
    Apr 17 at 13:34

2 Answers 2


At any spacetime point $x$, you can always go to locally inertial coordinates where $g_{\mu\nu}$ at $x$ is simply the Minkowski metric, $\eta_{\mu\nu}$, and the Christoffel symbols vanish. Hopefully it is clear that in these coordinates, in a small neighborhood around $x$, the interpretation of the invariant spacetime interval in a neighborhood of $x$ is the same as the interpretation of the invariant spacetime interval in special relativity. The fact that the interval is invariant under general choices of coordinate means that we can calculate the interval in any coordinate system, without having to explicitly go to locally inertial coordinates.

In particular, whether the separation between two nearby points is spacelike, timelike, or null can be calculated in any coordinates. Since the value of the interval is invariant, we know without having to do any calculations that we always could go to a locally inertial frame where we can use our intuition about spacelike, timelike, and null separations from special relativity.

  • $\begingroup$ but this invariance is just a mathematical truth. It just says that you can calculate the same numbers while working in different vector bases. The interval invariance in SR is a fact about spacetime measurements made by different observers in relative motion. That's not just a mathematical truth. $\endgroup$
    – Rain Deer
    Apr 17 at 7:41
  • 1
    $\begingroup$ @RainDeer . You should distinguish between the coordinates $t,x,y,z$ or $t,r,\theta,\phi$ that one observer is free to choose, and the primed coordinates $t',x',y',z'$ or $t',r',\theta',\phi'$ that another observer is free to choose. The physical content of space time interval invariance is that it is the same in $S$ and $S'$. The transition from SR to GR is just that you integrate infinitesimal space time intervals to end up with, say, proper time $\tau$. That proper time is the same for all observers. $\endgroup$
    – Kurt G.
    Apr 17 at 7:51
  • $\begingroup$ @RainDeer You are hitting on an important point. In SR, you can interpret the coordinates in a physical way in terms of what you would measure with a clock and ruler. In GR, the coordinates are not physical. We switch between coordinate systems depending on what is useful. However, if you go to the locally inertial coordinates at a point $x$ in GR, then (near $x$), you can think of the coordinates as what an observer near $x$ would see, and so you can use your SR intuition in those coordinates. $\endgroup$
    – Andrew
    Apr 17 at 14:02
  • $\begingroup$ However, while the locally inertial coordinates are conceptually very nice, in practice, it is a pain to transform to those coordinates every time you want to ask a question about what an observer would see at point $x$. So the usual trick is to find coordinate invariant quantities which reduce to something with a known meaning in a coordinate system that is easy to interpret. Then you can calculate the coordinate invariant quantity in any coordinate system. $\endgroup$
    – Andrew
    Apr 17 at 14:04
  • $\begingroup$ Would you say that Lorentz invariance is a discovery about the behavior of spacetime, while diffeomorphism invariance is more of just a mathematical truth? $\endgroup$
    – Rain Deer
    Apr 29 at 13:46

What's the physical content here?

The physical content is that measurements of distance and time do not depend on the choice of coordinates.

even if we assume that the four indices of 𝑑𝑥𝜇 in GR refer to actual spacetime-measurements by an observer,

No, that is not a valid assumption. The coordinates are not measurements, they are just labels. The actual spacetime-measurements are invariants like $g_{\mu\nu}dx^\mu dx^\nu$

  • $\begingroup$ Either a nit or I might be wrong: $g_{\mu\nu}dx^\mu dx^\nu$ is an actual measurement only if the interval is timelike/lightlike, right? For the spacelike interval, any actual measurement of length would be a "two-way" measurement and it won't be simply $g_{\mu\nu}dx^\mu dx^\nu$. For example, Zee, Chapter V.3, Eq (9). I realize that one can define a proper length along a spacelike curve and it will be the integral of $\sqrt{g_{\mu\nu} dx^\mu dx^\nu}$ but I'm not sure if it's something one can measure. $\endgroup$
    – ACat
    Apr 17 at 6:50
  • $\begingroup$ Measurements of distance and time do depend on the choice of coordinates; that's what the choice is! We have so-called inertial frames as those special subset of coordinate choices which reproduce the same physics in a 'form-invariant' way (e.g. a choice of Lagrangian, or path integral in QFT, 'look' the same when written in two different coordinate systems-- this is a symmetry). More generally, other choices of coordinates may be symmetries of some theories, e.g. so-called 'scale invariance' in QED with a massless fermion. $\endgroup$
    – Myridium
    Apr 17 at 13:29
  • $\begingroup$ @Myridium I disagree. A clock measures the same amount of time between two events on its worldline regardless of the coordinates used. In fact, you can even use coordinates that have no timelike coordinates or multiple timelike coordinates. Regardless of the coordinates, the clock measures the same $\endgroup$
    – Dale
    Apr 17 at 13:43
  • $\begingroup$ So the measurements of distance and time (the coordinates, tangent vectors) change, but the metric changes covariantly, so that the spacetime interval is invariant. Or, you may say that the physics change, so that all process proceed at half the speed (accounting for a scaling of how time is measured). $\endgroup$
    – Myridium
    Apr 17 at 13:47
  • $\begingroup$ @DivjDC it is a little tricky for spacelike intervals. Suppose that you have lengths marked on an ideal string which you have pulled taut. The string forms a worldsheet which measures $g_{\mu\nu}$ along certain geodesics lying in the worldsheet. Those geodesics are found by taking any event on the worldsheet and going “straight” in the direction perpendicular to the string’s congruence and parallel to the string. It is definitely not trivial which length is being measured, but it is indeed a measurement of length. $\endgroup$
    – Dale
    Apr 17 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.