Most topics in general relativity speak about the gravity, equivalence principle and the relation of energy mass tensor. How about the simple equation of the metric in 4D space-time of an accelerating frame, or a rotating frame? And as an application, will the clock tick the same rate for two observers moving with difference accelerations when they compare their clocks?


My first edit was done rather fast, so it was probably not so clear. Now in the second edit I improved the text and added more explanations.

Let's take as an example a rotating (non-initial) frame with the rotation axis $z$ and as a second system a initial system. In polar coordinates $(t,z,r,\phi)$ the invariant 4-distance in the initial system is:

$ds^2 = c^2 dt^2-dz^2-dr^2 -r^2 d\phi^2$.

In the initial system a clock at rest will measure the time $\frac{ds}{c}=dt$.

Now you change the reference system to the rotating frame. For this you have to apply a coordinate transformation $t=t'$, $z=z'$, $r=r'$ and $\phi=\phi' +\omega t'$. In particular $ r d\phi = r' d\phi'+ \omega r' dt'$ Then you get: $ds^2 = (1-\frac{\omega^2 r'^2}{c^2})c^2 dt'^2 -2\omega r'^2 d\phi'\, dt' -dz'^2-dr'^2 -r'^2 d\phi'^2$

Considering a clock at rest in the system $(t',r',z',\phi')$ means $dr'=0$, $d\phi'=0$ and $dz'=0$ (this clock does not change the space coordinates) we get for the invariant 4-distance $ds^2 = (1-\frac{\omega^2 r^2}{c^2})c^2dt'^2$. That means the clock in $(t',r',z',\phi')$ will measure the proper time $\frac{ds}{c}$:

$\frac{ds}{c} =\sqrt{1-\frac{\omega^2 r^2}{c^2}}dt'$

So in analogy to the initial system the time measured by a clock at rest in the rotating system is again $\frac{ds}{c}$ of the line element the clock is moving on in time (and not $dt'$ which is just a coordinate differential). There can be many different coordinate times $t'$, but $ds$ (for the same line element) is always the same.

The key point is here that after a change to curvilinear (in particular curvilinear in time) coordinates to measure distances the coordinates alone don't suffice, the right scaling factor have to be applied. Example: $d\sigma^2 = dx^2 + dy^2$ changed to $d\sigma^2 = dr^2 + r^2 d\phi^2$ the new coordinate $\phi$ does not suffice to measure the distance (it has to be $r d\phi$).

Of course the assumption $\frac{\omega r}{c}$<<1$ has to be made, i.a. the transformation has to be only applied locally.

  • $\begingroup$ But why $t=t`$?. Also, does the term; 2ωr′2dϕ′dt′ mean there is a correlation between the coordinates ϕ′ and t′ which implies those coordinates can not be the orthogonal dimensions of the rotating frame? $\endgroup$
    – Isaacadel
    Jul 15 '17 at 11:19
  • $\begingroup$ I improved my edit, it should be clearer now. To your question: Of course the transformation to get a coordinate system in which the second clock rests does not require necessarily an orthogonal result. There are coordinate changes which maintain orthogonality, but then the second clock will move on a different path as in this example. $\endgroup$ Jul 15 '17 at 18:35
  • $\begingroup$ This is interesting. But the clock which is fixed in position in the rotating frame, what dt` means? For it, how dt` is physically different from $d\tau$? $\endgroup$
    – Isaacadel
    Jul 16 '17 at 13:27
  • $\begingroup$ $ds$ measures the distance the clock travels on the worldline whereas $dt'$ is just a coordinate. Let's take the example of polar coordinates. If you are traveling on a circle line, the distance you travel is $ds= r d \phi$, not $\phi$ which is just a coordinate. For rectiliniear coordinates this problem does not exist as there are no coefficients different from 1. $ds^2=dx^2 +dy^2$ $\endgroup$ Jul 17 '17 at 17:42
  • $\begingroup$ I understand the polar coordinate example because of rdϕ is a length while dϕ is an angle. For our stationary observer in the accelerating frame, dτ can be measured by the time between two ticks, how can dt′ be measured? Aren`t they label the time between the same two events which are just 2 ticks in the clock? $\endgroup$
    – Isaacadel
    Jul 17 '17 at 18:55

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