If 2 objects connected by a massless rod or wire, rotate around the center of mass, do they experience time dilation ? I'm thinking that the smaller one will move faster so time will pass slowly, but the bigger one being bigger time will also pass slowly. Do the 2 effects cancel out and they both experience time at the same rate ?
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3 Answers
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Let's say for illustration, the connecting rod is 3 light seconds long and the following parameters apply:
| M1 | M2 | |
|---|---|---|
| Mass | 100kg | 200Kg |
| Distance to COM | 2 lightsecond | 1 lightsecond |
| Tangential Velocity | 0.8c | 0.4c |
The COM observer will see a clock :
on the faster rotating M1, ticking at rate of $\sqrt{1-0.8^2} = 0.6$ ticks/second
on the slower rotating M2, ticking at rate of $\sqrt{1-0.4^2} \approx 0.9165$ ticks/second due to time dilation.
M1 does not see her own clock as time dilated and as far as she is concerned, her clock is ticking at a rate of 1 tick/second. Instead she sees signals coming from M2 as coming at a rate of 0.9165/0.6 $\approx$ 1.5275 ticks/second.
By similar reasoning, M2 sees his clock ticking at a rate of 1 tick per second and sees signals coming from M1 at a rate of 0.8/0.9165 $\approx$ 0.6546 ticks/second.
While all 3 observers measure different rates, they all agree that the clock on M1 is ticking at $\approx 0.6546$ of the rate of the clock on M2.
M1 sees M2's clock ticking faster than his own. M2 sees M1's clock as ticking slower than his own. The central COM observer also sees M1 ticking slower than M2.
It might seem counter intuitive to have such a clear result that all observer's agree on, because in relativity sometimes it is difficult to even define which object is faster and usually everything is symmetrical, but in this case rotation (and proper acceleration) is involved and measurements are absolute.
It might also seem slightly odd that M1 sees M2's clock ticking at a rate that is greater than unity, which seems counter to the idea of time dilation.
The reason is that M1 and M2 are not inertial observers but are experiencing proper acceleration and by the equivalence principle they see things as if they are in a gravitational field. M1 has the greater proper acceleration due to traveling the fastest on a longer radius, so M1 feels like she is lower in a gravitational field and sees signals coming from M2 higher up as blue shifted and sped up. Similarly, M2 with the lesser proper acceleration feels like he is higher up in a gravitational field and sees signals coming from M1 lower down as "gravitationally" time dilated and red shifted.
In the above diagram, M1 is lower down in the effective equivalent gravitational field, M2 is midway and the COM is equivalent to an observer that is infinitely far from the "gravitational source" and does not experience any time dilation.
Here is another simple example of non reciprocal time dilation. Consider the Einstein carousel thought experiment. The inertial clock at the centre ticks at a rate of 1 tick per second. The clock on the perimeter ticks at a slower rate due to velocity time dilation. The observer on the perimeter of the carousel that experiences proper acceleration is non inertial and considers the clock at the centre to be ticking faster than his own clock.
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Note: this is an answer about gravitational versus motion time dilation, because I misunderstood the original question.
No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).
In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer, or $M/r=3c^2/(2G)$ (that is, never, because that is smaller than the Schwarzschild radius).
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To calculate the effect of time dilation due to rotation for a rotating object with mass m, you need to consider the rotational speed and the distance from the axis of rotation.
The formula for the time dilation due to rotation, is given by ?
$$ \Delta t'=\Delta t\,\sqrt{1-\frac{4\,G\,J}{c^2\,R^3}}\tag 1$$
where
- $\Delta t~$ is the time interval measured by a distant observer (proper time).
- $\Delta t'~$ is the time interval measured by an observer close to the rotating object.
- G is the gravitational constant.
- J is the angular momentum (spin) of the rotating object.
- R is the distance from the axis of rotation.
- c is the speed of light in vacuum.
applying equation (1) with $~J_m=m\,\omega\,R_m^2~$ you obtain for the mass m
$$\Delta t'_m=\Delta t\,\sqrt{1-\frac{4\,G\,m\,\omega}{R_m\,c^2}}$$
and for the mass M
$$\Delta t'_M=\Delta t\,\sqrt{1-\frac{4\,G\,M\omega}{R_M\,c^2}}$$
now, the distance from the axis of rotation $~R_m~$ to the mass m is bigger then the distance from the axis of rotation $~R_M~$ to the mass M
thus the time dilation $$~\Delta t'_m > \Delta t'_M$$
