Consider the theory only posseses GR.

If an observer lives in a curved spacetime, how can he measure the metric of the spacetime without knowing the matter distribution? Suppose he can establish a coordinate system, say $( t,x,y,z)$(don't have to be flat), then he can determine the metric by measure the length within this coordinate system, by watching a free falling objects and give the geodesic equation, or by looking at the geodesic deviation to give the curvature, etc. But how can an observer establish such a coordinate system?

Note that here we can only use GR, which means we cannot use an atomic clock or such things to give a time measurement, because the atomic clock rely on the dynamics given by other theory(time evolution of quantum mechanics...). If we can use other theory to give a time measurement, then we can use the observer's proper time as time coordinate, then follow the normal procedure to give the space coordinate(for example define the length as the light travels unit time).

In curved spacetime, normal measurement becomes vague, can you define the unit length as the length of the "straight" rod you have? We can use a straight rod as the length unit in flat spacetime, but what do this mean in curved spacetime?

Here's my opinion. If the "spatial part" of the metric is not a function of time, which means: \begin{equation*} \mathrm{d} s^{2} =f( t,\boldsymbol{x})\mathrm{d} t^{2} +g(\boldsymbol{x})_{ij}\mathrm{d} x^{i}\mathrm{d} x^{j} , \end{equation*} then we can observe a free falling object, which gives a bunch of geodesics, and choose three (linearly) independent directions as a basis. Then label the geodesics with length labels to get a spatial coordinate and measure the metric to give $g(\boldsymbol{x})$. (I this there is still some vagueness in this process) Then define the unit length which light travels as unit times, and finally give the metric.

Then what about a general metric? I don't know how to establish a coordinate system in this case: \begin{equation*} \mathrm{d} s^{2} =g( t,\boldsymbol{x})\mathrm{_{\mu \nu } d} x^{\mu }\mathrm{d} x^{\nu } . \end{equation*}

This question is different from what we have in reality. In the real world, we examine GR by knowing the energy-momentum tensor $T_{\mu \nu }$ and use the field equation to give the metric, then observe the motion of an object(e.g. gravitational lens effect) to testify the metric. The coordinate system, especally the time coordinate is given by other theory independent of GR, such as the atomic clock. In this question, we do not know $T_{\mu \nu }$ and only have GR.

So my question:

  1. How can an observer establish a coordinate system in a metric whose spatial part is independent of time? Take Schwartzchild metric as example: \begin{equation*} \mathrm{d} s^{2} =-\left( 1-\frac{2GM}{r}\right)\mathrm{d} t^{2} +\frac{1}{( 1-2GM/r)}\mathrm{d} r^{2} +r^{2}\left(\mathrm{d} \theta ^{2} +\sin^{2} \theta \mathrm{d} \varphi ^{2}\right) , \end{equation*} how can an observer in this spacetime know this metric?

  2. How can an observer establish a coordinate system in a general metric $g( t,\boldsymbol{x})$?

  • $\begingroup$ If you cannot use another theory to establish a measurement of time or length, and all you have left for measurements are say angles, you cannot measure the metric of spacetime, but only its affine structure, which is roughly the metric up to a conformal transformation. You can use for instance the EPS construction to recover the spacetime metric in radar coordinates : link.springer.com/article/10.1007/s10714-012-1353-4 $\endgroup$
    – Slereah
    Nov 26 '21 at 8:55
  • $\begingroup$ @Slereah I see. I wonder if we can use other theory, how can we establish the length coordinates? For time coordinate we can use atomic clock, but what about length? $\endgroup$
    – Photon-gjq
    Nov 26 '21 at 9:19
  • $\begingroup$ @Slereah, If you only have GR, you can still have black holes, and you can use one to set a reference length/time scale. Using very light black holes as test particles, and gravitational waves, you can set up pretty much any experiment that you could with EM. $\endgroup$
    – mmeent
    Nov 26 '21 at 9:29

Your Answer

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

Browse other questions tagged or ask your own question.