special relativity - A simple pendulum moving at a relativistic speed - how does the period change?

I've been pondering the precise mechanism of time dilation for the example of a simple pendulum in two different situations:

1. 
   
   

   The observer and ground are at rest in one frame of reference; the pendulum is moving at high speed with respect to that frame.

   
   


2. 
   

   The observer is at rest in one frame of reference; the pendulum and the ground together move at high speed with respect to that frame.

   
   

user8260 has pointed out that in situation 1, in the pendulum's frame $g$ is greater by $\gamma^2$ compared to $g$ in the observer's frame. Thus in the pendulum's frame the period is less than it is in the observer's frame by a factor of $1/\gamma$, just as one would expect from time dilation.

But what about situation 2? Here, compared to the pendulum frame, the observer sees the pendulum with the same length in a stronger gravitational field, yet observes a longer period. Does the inertial mass of the pendulum change differently than its gravitational mass? Also, does the analysis depend on whether the plane of swing is perpendicular or parallel the direction of motion?

Answer

As in physics in general, a suitable choice of coordinates makes our life so much better. Time dilation in this problem is somewhat a more trivial effect, and the transformation of gravitational field is somewhat a more complicated phenomenon. With this in mind, let me reformulate slightly the two situations:

Case 2. Pendulum is at rest with respect to the Earth (and some observer moves with respect to them, observes time dilation etc etc)

Case 1. Pendulum is set above the Earth, which moves relativistically below it (and some observer moves with the Earth, observes time dilation etc)

So, let us settle the physics first, and the observer effects last.

Case 2: Classical physics problem, nothing to settle.

Case 1: From the pendulum's point of view, the gravitational field is generated by a moving body (=>the field is unknown). From the Earth frame, a relativistic body moves in a gravity field (=>the equations of motion are unknown).

One might transform the energy-momentum tensor of the Earth from the Earth rest frame to the pendulum frame, but special care should be taken about the fact that the Earth ceases to be spherical in the new frame (though its density does increase as $\gamma^2$). Additionally, it is not clear appriori that the motion of the Earth doesn't cause any additional forces.

I propose to use a straightforward yet more secure method of transforming the metric tensor from the Earth frame to the pendulum frame, and hence obtain the gravity, acting on the pendulum.

In the Earth rest frame the metric tensor is known to be $$g_{\mu\nu}=\left(\begin{array}{cccc} &1-2U & 0 & 0 & 0 &\\ &0 & 1-2U & 0 & 0 &\\ &0 & 0 & 1-2U & 0 &\\ &0 & 0 & 0 & -1-2U &\\ \end{array} \right),$$

where $U$ is the Newtonian potential of the Earth. This expression corresponds to the so called weak field limit, when the metric tensor is nearly flat. We use the standard notation of MTW ($c=1$, signature $(+++ -)$, Einstein's summation rule etc) and refer to this book for further details on linearized gravity.

Transformation of the field to the pendulum frame:

Lorentz tranformation matrix is given by:

$$ \Lambda_{\mu'}^{~\mu}=\left(\begin{array}{cccc} &\gamma & 0 & 0 & \beta \gamma &\\ &0 & 1 & 0 & 0 &\\ &0 & 0 & 1 & 0 &\\ &\beta\gamma & 0 & 0 & \gamma &\\ \end{array} \right), $$ with $\beta=\dfrac{v}{c}, \gamma=(1-\beta^2)^{-1/2}$ and $v$ being the relative velocity of the pendulum with respect to the Earth rest frame.

The transformed metric tensor is obtained by: $$g_{\mu'\nu'}=\Lambda_{\mu'}^{~\mu}\Lambda_{\nu'}^{~\nu} g_{\mu\nu}=\left(\begin{array}{cccc} &1-2U\dfrac{1+\beta^2}{1-\beta^2} & 0 & 0 & -\dfrac{4 U \beta}{1-\beta^2} &\\ &0 & 1-2U & 0 & 0 &\\ &0 & 0 & 1-2U & 0 &\\ &-\dfrac{4 U \beta}{1-\beta^2} & 0 & 0 & -1-2U\dfrac{1+\beta^2}{1-\beta^2} &\\ \end{array} \right)$$

In the pendulum frame (further primes in the indices are omitted!):

It is known that only the term $g_{44}$ determines the newtonian potential. One can see that by writing out the lagrangian for the pendulum:

$$ \mathcal{L}=\dfrac{1}{2}g_{\mu\nu} u^\mu u^\nu=\\ =\dfrac{1}{2}((u^1)^2+(u^2)^2+(u^3)^2-(u^4)^2)-\\ -U((u^2)^2+(u^3)^2+4 u^1 u^4 \beta \gamma^2+((u^1)^2+(u^4)^2)\dfrac{1+\beta^2}{1-\beta^2}) $$

Here $u^\mu$ is the 4-velocity of the pendulum. As the latter moves non-relativistically (in its own frame), we may consider $u^4\gg u^1,u^2,u^3$ and $u^4\approx \mathrm{const}$, which leaves:

$$ \mathcal{L}=\dfrac{1}{2}((u^1)^2+(u^2)^2+(u^3)^2)-U(u^4)^2\dfrac{1+\beta^2}{1-\beta^2} $$

If the pendulum as a whole didn't move with respect to the Earth, we would have $\beta = 0$ and

$$ \mathcal{L}_0=\dfrac{1}{2}((u^1)^2+(u^2)^2+(u^3)^2)-U(u^4)^2 $$

Effectively, therefore, the pendulum in its rest frame experiences the gravitational field magnified by the factor of $\dfrac{1+\beta^2}{1-\beta^2}$. The pendulum frequency is thus magnified by $\dfrac{(1+\beta^2)^{1/2}}{(1-\beta^2)^{1/2}}$.

Remarks: the neglected terms in the lagrangian are either $\dfrac{v}{c}$ or $(\dfrac{v}{c})^2$ smaller than the kept leading terms. Hence, up to $\dfrac{v}{c}$ accuracy the direction of motion doesn't affect the pendulum frequency.

Finally, lets add time dilations to get the final answers. Let the period of the pendulum in the case when observer, the Earth, and the pendulum do not move with respect to each other be $T_0$. Then:

Case 1: In the pendulum frame, as we have seen it has the period of $\dfrac{(1-\beta^2)^{1/2}}{(1+\beta^2)^{1/2}} T_0$. Then in the observer frame, due to time dilation, the period is $\dfrac{1}{(1+\beta^2)^{1/2}}T_0$.

Case 2: In the pendulum frame the period is $T_0$. In the observer frame the period is $\dfrac{T_0}{(1-\beta^2)^{1/2}}$.

To conclude, the two cases are quite different due to the different physics happening. In one case the observed period changes due to the change of the reference frame, whereas in the other there is an additional factor due to the fact that the gravity of a moving source is not the same as that of a still source.