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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?

Qmechanic
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Isaacadel
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1 Answers1

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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.

Frederic Thomas
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  • 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? – Isaacadel Jul 15 '17 at 11:19
  • 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. – Frederic Thomas Jul 15 '17 at 18:35
  • This is interesting. But the clock which is fixed in position in the rotating frame, what dtmeans? For it, how dt is physically different from $d\tau$? – Isaacadel Jul 16 '17 at 13:27
  • $ds$ 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. $ds^2=dx^2 +dy^2$ – Frederic Thomas Jul 17 '17 at 17:42
  • 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? – Isaacadel Jul 17 '17 at 18:55